Quarter Wit, Quarter Wisdom: Squares and Square Roots on the GMAT

Quarter Wit, Quarter WisdomIn today’s post, we will try to clear up your doubts regarding positive and negative solutions in the case of squares and square roots. We will explain the reasons behind each case, which will help you recall the fundamentals when you need to use them. While preparing for the GMAT, you have probably come across a discussion that says x^2 = 4 has two roots, 2 and -2, while √4 has only one value, 2.

Now, let’s try to understand why this is so:

1) x^2 = 4
Basic algebra tells us that quadratics have two roots. Here, x can be either 2 or -2; each, when squared, will give you 4.

x^2 – 4 = 0 and (x + 2)*(x – 2) = 0 when x equals -2 or 2.

2) √x is positive, only
Now this is odd, right? √4 must be 2. Why is that? Shouldn’t it be 2 or -2. After all, when we square both 2 and -2, we get 4 (as discussed above). So, √4 should be 2 or -2.

Here is the concept: √x denotes only the principal square root. x has two square roots – the positive square root (or principal square root) written as √x and the negative square root written as -√x. Therefore, when you take the square root of 4, you get two roots: √4 and -√4, which  is 2 and -2 respectively.

On a GMAT question, when you see √x, this is specifically referring to the positive square root of the number. So √4 is 2, only.

3) (√x)^2 = x
This is fairly straightforward – since x has a square root, it must be non-negative. When you square it, just the square root sign vanishes and you are left with x.

4) √(x^2) = |x|
Now this isn’t intuitive either. √(x^2) should simply be x – why do we have absolute value of x, then? Again, this has to do with the principal square root concept. First you will square x, and then when you write √, it is by default just the principal square root. The negative square root will be written as -√(x^2). So, irrespective of whether x was positive or negative initially, √(x^2) will definitely be positive x. Therefore, we will need to take the absolute value of x.

Here’s a quick recap with some examples:

  • √9 = 3
  • x^2 = 16 means x is either 4 or -4
  • √(5^2) = 5
  • √(-5^2) = 5
  • (√16)^2 = 16
  • √100 = 10

To see this concept in action, let’s take a look at a very simple official problem:

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

We know that √(x^2) is not simply x, but rather |x|. So, √(x^2)/x = |x|/x.

Depending on whether x is positive or negative, |x|/x will be 1 or -1 – we can’t say which one. Hence, there is no further simplification that we can do, and our answer must be E.

Now that you are all warmed up, let’s examine a higher-level question:

Is √[(x – 3)^2] = (3 – x)?

Statement 1: x is not 3
Statement 2: -x * |x| > 0

We know that √(x^2) = |x|, so √[(x – 3)^2] = |x – 3|.

This means that our question is basically:

Is |x – 3| = 3 – x?

Note that 3 – x can also be written as -(x – 3).

Is |x – 3| = -(x – 3)?

Recall the definition of absolute values: |a| = a if a is greater than or equal to 0, and -a if a < 0.

So, “Is |x – 3| = -(x – 3)?” depends on whether (x – 3) is positive or negative. If (x – 3) is negative (or 0), then |x – 3| is equal to -(x – 3).

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

This means we know that (x – 3) is not 0, but we still don’t know whether it is negative or positive. This statement is not sufficient.

Statement 2: -x * |x| > 0

|x| is always non-negative, so for the product to be positive, “-x” must also be positive. This means x must be negative. If x is negative, x – 3 must be negative, too.

If (x – 3) is negative, |x – 3| is equal to -(x – 3). Hence, this statement alone is sufficient, and our answer is B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Why Critical Reasoning Needs Your Complete Attention on the GMAT!

Quarter Wit, Quarter WisdomLet’s look at a tricky and time consuming official Critical Reasoning question today. We will learn how to focus on the important aspects of the question and quickly evaluate our answer choices:

Tiger beetles are such fast runners that they can capture virtually any nonflying insect. However, when running toward an insect, a tiger beetle will intermittently stop and then, a moment later, resume its attack. Perhaps the beetles cannot maintain their pace and must pause for a moment’s rest; but an alternative hypothesis is that while running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop. 

Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other? 

(A) When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately stops and runs away without its usual intermittent stopping. 
(B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.
(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insect’s direction, and it pauses equally frequently whether the chase is up or down an incline. 
(D) If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit. 
(E) The faster a beetle pursues an insect fleeing directly away from it, the more frequently the beetle stops.

First, take a look at the argument:

  • Tiger beetles are very fast runners.
  • When running toward an insect, a tiger beetle will intermittently stop and then, a moment later, resume its attack.

There are two hypotheses presented for this behavior:

  1. The beetles cannot maintain their pace and must pause for a moment’s rest.
  2. While running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.

We need to support one of the two hypotheses and undermine the other. We don’t know which one will be supported and which will be undermined. How will we support/undermine a hypothesis?

The beetles cannot maintain their pace and must pause for a moment’s rest.

Support: Something that tells us that they do get tired. e.g. going uphill they pause more.

Undermine: Something that says that fatigue plays no role e.g. the frequency of pauses do not increase as the chase continues.

While running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.

Support: Something that says that they are not able to process changing visual information e.g. as speed increases, frequency of pauses increases.

Undermine: Something that says that they are able to process changing visual information e.g. it doesn’t pause on turns.

Now, we need to look at each answer choice to see which one supports one hypothesis and undermines the other. Focus on the impact each option has on our two hypotheses:

(A) When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately stops and runs away without its usual intermittent stopping.

This undermines both hypotheses. If the beetle is able to run without stopping in some situations, it means that it is not a physical ailment that makes him take pauses. He is not trying to catch his breath – so to say – nor is he adjusting his field of vision.

(B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.

If the beetle alters its course while running, it is obviously processing changing visual information and changing its course accordingly while running. This undermines the hypothesis “it cannot process rapidly changing visual information”. However, if the beetle pauses more frequently as the chase progresses, it is tiring out more and more due to the long chase and, hence, is taking more frequent breaks. This supports the hypothesis, “it cannot maintain its speed and pauses for rest”.

Answer choice B strengthens one hypothesis and undermines the other. This must be the answer, but let’s check our other options, just to be sure:

(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insect’s direction, and it pauses equally frequently whether the chase is up or down an incline.

This answer choice undermines both hypotheses. If the beetle responds immediately to changes in direction, it is able to process changing visual information. In addition, if the beetle takes similar pauses going up or down, it is not the effort of running that is making it take the pauses (otherwise, going up, it would have taken more pauses since it takes more effort going up).

(D) If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit.

This answer choice might strengthen the hypothesis that the beetle is not able to respond to changing visual information since it decides whether it is giving up or not after pausing (in case there is a certain stance that tells us that it has paused), but it doesn’t actually undermine the hypothesis that the beetle pauses to rest. It is very possible that it pauses to rest, and at that time assesses the situation and decides whether it wants to continue the chase. Hence, this option doesn’t undermine either hypothesis and cannot be our answer.

(E) The faster a beetle pursues an insect fleeing directly away from it, the more frequently the beetle stops.

This answer choice strengthens both of the hypotheses. The faster the beetle runs, the more rest it would need, and the more rapidly visual information would change causing the beetle to pause. Because this option does not undermine either hypothesis, it also cannot be our answer.

Only answer choice B strengthens one hypothesis and undermines the other, therefore, our answer must be B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using Visual Symmetry to Solve GMAT Probability Problems

Quarter Wit, Quarter WisdomToday, let’s take a look at an official GMAT question involving visual skills. It takes a moment to understand the given diagram, but at close inspection, we’ll find that this question is just a simple probability question – the trick is in understanding the symmetry of the figure:

The figure shown represents a board with 4 rows of pegs, and at the bottom of the board are 4 cells numbered 1 to 4. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has the probability 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes through the first two pegs at the top it will end in cell 2?

qwqw pegs pic

 

 

 

 

(A) 1/16
(B) 1/8
(C) 1/4
(D) 3/8
(E) 1/2

First, understand the diagram. There are small pegs arranged in rows and columns. The ball falls between two adjacent pegs and hits the peg directly below. When it does, there are two ways it can go – either to the opening on the left or to the opening on the right. The probability of each move is equal, i.e. 1/2.

The arrow show the first path the ball takes. It is dropped between the top two pegs, hits the peg directly below it, and then either drops to the left side or to the right. The same process will be repeated until the ball falls into one of the four cells – 1, 2, 3 or 4.

Method 1: Using Symmetry
Now that we understand this process, let’s examine the symmetry in this diagram.

Say we flip the image along the vertical axis – what do we get? The figure is still exactly the same, but now the order of cells is reversed to be 4, 3, 2, 1. The pathways in which you could reach Cell 1 are now the pathways in which you can use to reach Cell 4.

OR think about it like this:

To reach Cell 1, the ball needs to turn left-left-left.

To reach Cell 4, the ball needs to turn right-right-right.

Since the probability of turning left or right is the same, the situations are symmetrical. This will be the same case for Cells 2 and 3. Therefore, by symmetry, we see that:

The probability of reaching Cell 1 = the probability of reaching Cell 4.

Similarly:

The probability of reaching Cell 2 = the probability of reaching Cell 3. (There will be multiple ways to reach Cell 2, but the ways of reaching Cell 3 will be similar, too.)

The total probability = the probability of reaching Cell 1 + the probability of reaching Cell 2 + the probability of reaching Cell 3 + the probability of reaching Cell 4 = 1

Because we know the probability of reaching Cells 1 and 4 are the same, and the probabilities of reaching Cells 2 and 3 are the same, this equation can be written as:

2*(the probability of reaching Cell 1) + 2*(the probability of reaching Cell 2) = 1

Let’s find the probability of reaching Cell 1:

After the first opening (not the peg, but the opening between pegs 1 and 2 in the first row), the ball moves left (between pegs 1 and 2 in second row) or right (between pegs 2 and 3 in second row). It must move left to reach Cell 1, and the probability of this = 1/2.

After that, the ball must move left again – the probability of this occurring is also 1/2, since probability of moving left or right is equal. Finally, the ball must turn left again to reach Cell 1 – the probability of this occurring is, again, 1/2. This means that the total probability of the ball reaching Cell 1 = (1/2)*(1/2)*(1/2) = 1/8

Plugging this value into the equation above:

2*(1/8) + 2 * probability of reaching Cell 2 = 1

Therefore, the probability of reaching Cell 2 = 3/8

Method 2: Enumerating the Cases
You can also answer this question by simply enumerating the cases.

At every step after the first drop between pegs 1 and 2 in the first row, there are two different paths available to the ball – either it can go left or it can go right. This happens three times and, hence, the total number of ways in which the ball can travel is 2*2*2 = 8

The ways in which the ball can reach Cell 2 are:

Left-Left-Right

Left-Right-Left

Right-Left-Left

So, the probability of the ball reaching Cell 2 is 3/8.

Note that here there is a chance that we might miss some case(s), especially in problems that involve many different probability options. Hence, enumerating should be the last option you use when tackling these types of questions on the GMAT.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using the Standard Deviation Formula on the GMAT

Quarter Wit, Quarter WisdomWe have discussed standard deviation (SD) in detail before. We know what the formula is for finding the standard deviation of a set of numbers, but we also know that GMAT will not ask us to actually calculate the standard deviation because the calculations involved would be way too cumbersome. It is still a good idea to know this formula, though, as it will help us compare standard deviations across various sets – a concept we should know well.

Today, we will look at some GMAT questions that involve sets with similar standard deviations such that it is hard to tell which will have a higher SD without properly understanding the way it is calculated. Take a look at the following question:

Which of the following distribution of numbers has the greatest standard deviation? 

(A) {-3, 1, 2} 
(B) {-2, -1, 1, 2} 
(C) {3, 5, 7} 
(D) {-1, 2, 3, 4} 
(E) {0, 2, 4}

At first glance, these sets all look very similar. If we try to plot them on a number line, we will see that they also have similar distributions, so it is hard to say which will have a higher SD than the others. Let’s quickly review their deviations from the arithmetic means:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2
For answer choice C, the mean = 5 and the deviations are 2, 0, 2
For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2
For answer choice E, the mean = 2 and the deviations are 2, 0, 2

We don’t need to worry about the arithmetic means (they just help us calculate the deviation of each element from the mean); our focus should be on the deviations. The SD formula squares the individual deviations and then adds them, then the sum is divided by the number of elements and finally, we find the square root of the whole term. So if a deviation is greater, its square will be even greater and that will increase the SD.

If the deviation increases and the number of elements increases, too, then we cannot be sure what the final effect will be – an increased deviation increases the SD but an increase in the number of elements increases the denominator and hence, actually decreases the SD. The overall effect as to whether the SD increases or decreases will vary from case to case.

First, we should note that answers C and E have identical deviations and numbers of elements, hence, their SDs will be identical. This means the answer is certainly not C or E, since Problem Solving questions have a single correct answer.

Let’s move on to the other three options:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2
For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2

Comparing answer choices A and D, we see that they both have the same deviations, but D has more elements. This means its denominator will be greater, and therefore, the SD of answer D is smaller than the SD of answer A. This leaves us with options A and B:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2

Now notice that although two deviations of answers A and B are the same, answer choice A has a higher deviation of 3 but fewer elements than answer choice B. This means the SD of A will be higher than the SD of B, so the SD of A will be the highest. Hence, our answer must be A.

Let’s try another one:

Which of the following data sets has the third largest standard deviation?

(A) {1, 2, 3, 4, 5} 
(B) {2, 3, 3, 3, 4} 
(C) {2, 2, 2, 4, 5} 
(D) {0, 2, 3, 4, 6} 
(E) {-1, 1, 3, 5, 7}

How would you answer this question without calculating the SDs? We need to arrange the sets in increasing SD order. Upon careful examination, you will see that the number of elements in each set is the same, and the mean of each set is 3.

Deviations of answer choice A: 2, 1, 0, 1, 2
Deviations of answer choice B: 1, 0, 0, 0, 1 (lowest SD)
Deviations of answer choice C: 1, 1, 1, 1, 2
Deviations of answer choice D: 3, 1, 0, 1, 3
Deviations of answer choice E: 4, 2, 0, 2, 4 (highest SD)

Obviously, option B has the lowest SD (the deviations are the smallest) and option E has the highest SD (the deviations are the greatest). This means we can automatically rule these answers out, as they cannot have the third largest SD.

Deviations of answer choice A: 2, 1, 0, 1, 2
Deviations of answer choice C: 1, 1, 1, 1, 2
Deviations of answer choice D: 3, 1, 0, 1, 3

Out of these options, answer choice D has a higher SD than answer choice A, since it has higher deviations of two 3s (whereas A has deviations of two 2s). Also, C is more tightly packed than A, with four deviations of 1. If you are not sure why, consider this:

The square of deviations for C will be 1 + 1+ 1 + 1  + 4 = 8
The square of deviations for A will be 4 + 1 + 0 + 1 + 4 = 10

So, A will have a higher SD than C but a lower SD than D. Arranging from lowest to highest SD’s, we get: B, C, A, D, E. Answer choice A has the third highest SD, and therefore, A is our answer

Although we didn’t need to calculate the actual SD, we used the concepts of the standard deviation formula to answer these questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: An Innovative Use of the Slope of a Line on the GMAT

Quarter Wit, Quarter WisdomLet’s continue our discussion on coordinate geometry today.

The concept of slope is extremely important on the GMAT – it is not sufficient to just know how to calculate it using (y2 – y1)/(x2 – x1).

In simple terms, the slope of a line specifies the units by which the y-coordinate changes and the direction in which it changes with each 1 unit increase in the x-coordinate. If the slope (m) is positive, the y-coordinate changes in the same direction as the x-coordinate. If m is negative, however, the y-coordinate changes in the opposite direction.

For example, if the slope of a line is 2, it means that every time the x-coordinate increases by 1 unit, the y-coordinate increases by 2 units. So if the point (3, 5) lies on a line with a slope of 2, the point (4, 7) will also lie on it. Here, when the x-coordinate increases from 3 to 4, the y-coordinate increases from 5 to 7 (by an increase of 2 units). Similarly,  the point (2, 3) will also lie on this same line – if the x-coordinate decreases by 1 unit (from 3 to 2), the y-coordinate will decrease by 2 units (from 5 to 3). Since the slope is positive, the direction of change of the x-coordinate will be the same as the direction of change of the y-coordinate.

Now, if we have a line where the slope is -2 and the point (3, 5) lies on it, when the x-coordinate increases by 1 unit, the y-coordinate DECREASES by 2 units – the point (4, 3) will also lie on this line. Similarly, if the x-coordinate decreases by 1 unit, the y-coordinate will increase by 2 units. So, for example, the point (2, 7) will also lie on this line.

This understanding of the concept of slope can be very helpful, as we will see in this GMAT question:

Line L and line K have slopes -2 and 1/2 respectively. If line L and line K intersect at (6,8), what is the distance between the x-intercept of line L and the y-intercept of line K? 

(A) 5
(B) 10
(C) 5√(5)
(D) 15
(E) 10√(5)

Method 1: The Traditional Approach
Traditionally, one would solve this question like this:

The equation of a line with slope m and constant c is given as y = mx + c. Therefore, the equations of lines L and K would be:

Line L: y = (-2)x + a
and
Line K: y = (1/2)x + b

As both these lines pass through (6,8), we would substitute x=6 and y=8 to get the values of a and b.

Line L: 8 = (-2)*6 + a
a = 20

Line K: 8 = (1/2)*6 + b
b = 5

Thus, the equations of the 2 lines become:

Line L: y = (-2)x + 20
and
Line K: y = (1/2)x + 5

The x-intercept of a line is given by the point where y = 0. So, the x-intercept of line L is given by:

0 = (-2)x + 20
x = 10

This means line L intersects the x-axis at the point (10, 0).

Similarly, the y-intercept of a line is given by the point where x = 0. So, y-intercept of line K is given by:

y = (1/2)*0 + 5
y = 5

This means that line K intersects the y-axis at the point (0, 5).

Looking back at our original question, the distance between these two points is given by √((10 – 0)^2 + (0 – 5)^2) = 5√(5). Therefore, our answer is C.

Method 2: Using the Slope Concept
Although the using the traditional method is effective, we can answer this question much quicker using the concept we discussed above.

Line L has a slope of -2, which means that for every 1 unit the x-coordinate increases, the y-coordinate decreases by 2. Line L also passes through the point (6, 8). We know the line must intersect the x-axis at y = 0, which is a decrease of 8 y-coordinates from the given point (6,8). If y increases by 8, according to our slope concept, x will increase by 4 to give 6 + 4 = 10. So the x-intercept of line L is at (10, 0).

Line K has slope of 1/2 and also passes through (6, 8). We know the this line must intersect the y-axis at x = 0, which is a decrease of 6 x-coordinates from the given point (6,8). This means y will decrease by 1/2 of that (6*1/2 = 3) and will become 8 – 3 = 5. So the y-intercept of line K is at (0, 5).

The distance between the two points can now be found using the Pythagorean Theorem – √(10^2 + 5^2) = 5√(5), therefore our answer is, again, C.

Using the slope concept makes solving this question much less tedious and saves us a lot of precious time. That is the advantage of using holistic approaches over the more traditional approaches in tackling GMAT questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Coordinate Geometry: Solving GMAT Problems With Lines Crossing Either the X-Axis or the Y-Axis

Quarter Wit, Quarter WisdomToday let’s learn about the cases in which lines on the XY plane cross, or do not cross, the x- or y-axis. Students often struggle with questions such as this:

Does the line with equation ax+by = c, where a,b and c are real constants, cross the x-axis?

What concepts will you use here? How will you find whether or not a line crosses the x-axis? What conditions should it meet? Think about this a little before you move ahead.

We know that most lines on the XY plane cross the x-axis as well as the y-axis. Even if it looks like a given line doesn’t cross either of these axes, eventually, it will if it has a slope other than 0 or infinity.

QWQW pic 1

 

 

 

 

 

Note that by definition, a line extends infinitely in both directions – it has no end points (otherwise it would be a line “segment”). We cannot depict a line extending infinitely, which is why we will only show a small section of it. Ideally, a line on the XY plane should be shown with arrowheads to depict that it extends infinitely on both sides, but we often omit them for our convenience. For instance, if we try to extend the example line above, we see that it does, in fact, cross the x-axis:

QWQW pic 2

 

 

 

 

 

So what kind of lines do not cross either the x-axis or the y-xis? We know that the equation of a line on the XY plane is given by ax + by  + c = 0. We also know that if we want to find the slope of a line, we can use the equation y = (-a/b)x – c/b, where the slope of the line is -a/b.

A line with a slope of 0 is parallel to the x-axis. For the slope (i.e. -a/b) to be 0, a must equal 0. So if a = 0, the line will not cross the x-axis – it is parallel to the x-axis. The equation of the line, in this case, will become y = k. In all other cases, a line will cross the x-axis at some point.

Similarly, it might appear that a line doesn’t cross the y-axis but it does at some point if its slope is anything other than infinity. A line with a slope of infinity is parallel to the y-axis. For -a/b to be infinity, b must equal 0. So if b = 0, the line will not cross the y-axis. The equation of the line in this case will become x = k. In all other cases, a line will cross the y-axis at some point.

Now, we can easily solve this official question:

Does the line with equation ax+by = c, where a, b and c are real constants, cross the x-axis?

Statement 1: b not equal to 0

Statement 2: ab > 0

As we discussed earlier, all lines cross the x-axis except lines which have a slope of 0, i.e. a = 0.

Statement 1: b not equal to 0

This tells statement us that b is not 0 – which means the line is not parallel to y-axis – but it doesn’t tell us whether or not a is 0, so we don’t know whether the line is parallel to the x-axis or crosses it. Therefore, this statement alone is not sufficient.

Statement 2: ab>0

If ab > 0, it means that neither a nor b is 0 (since any number times 0 will equal 0). This means the line is parallel to neither x-axis nor the y-xis, and therefore must cross the x-axis. This statement alone is sufficient and our answer is B.

Hopefully this has helped clear up some coordinate geometry concepts today.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Be Tolerant Towards Pronoun Ambiguity on the GMAT

Quarter Wit, Quarter WisdomWe encounter many different types of pronoun errors on the GMAT Verbal Section. Some of the most common errors include:

Using a pronoun without an antecedent. For example, the sentence, “Although Jack is very rich, he makes poor use of it,” is incorrect because “it” has no antecedent. The antecedent should instead be “money” or “wealth.”

Error in matching the pronoun to its antecedent in number and gender. For example, the sentence, “Pack away the unused packets, and save it for the next game,” is incorrect because the antecedent of “it” is referring to “unused packets,” which is plural.

Using a nominative/objective case pronoun when the antecedent is possessive. For example, the sentence, “The client called the lawyer’s office, but he did not answer,” is incorrect because the antecedent of “he” should be referring to “lawyer,” but it appears only in the possessive case. Official GMAT questions will not give you this rule as the only decision point between two options.

But note that the rules governing pronoun ambiguity are not as strict as other rules! Pronoun ambiguity should be the last decision point for eliminating an option after we have taken care of SV agreements, tenses, modifiers, parallelism etc.

Every sentence that has two nouns before a pronoun does not fall under the “pronoun ambiguity error” category. If the pronoun agrees with two nouns in number and gender, and both nouns could be the antecedent of the pronoun, then there is a possibility of pronoun ambiguity. But in other cases, logic can dictate that only one of the nouns can really perform (or receive) an action, and so it is logically clear to which noun the pronoun refers.

For example, “Take the bag out of the car and get it fixed.”

What needs to get fixed? The bag or the car? Either is possible. Here we have a pronoun ambiguity, but it is highly unlikely you will see something like this on the GMAT.

A special mention should be made here about the role nouns play in the sentence. Often, a pronoun which acts as the subject of a clause refers to the noun which acts as a subject of the previous clause. In such sentences, you will often find that the antecedent is unambiguous. Similarly, if the pronoun acts as the direct object of a clause, it could refer to the direct object of the  previous clause. If the pronoun and its antecedent play parallel roles, a lot of clarity is added to the sentence. But it is not necessary that the pronoun and its antecedent will play parallel roles.

Let’s look at a different example, “The car needs to be taken out of the driveway and its brakes need to get fixed.”

Here, obviously the antecedent of “its” must be the car since only it has brakes, not the driveway. Besides, the car is the subject of the previous clause and “its” refers to the subject. Hence, this sentence would be acceptable.

A good rule of thumb would be to look at the options. If no options sort out the pronoun issue by replacing it with the relevant noun, just forget about pronoun ambiguity. If there are options that clarify the pronoun issue by replacing it with the relevant noun, consider all other grammatical issues first and then finally zero in on pronoun ambiguity.

Let’s take a quick look at some official GMAT questions involving pronouns now:

Congress is debating a bill requiring certain employers provide workers with unpaid leave so as to care for sick or newborn children. 

(A) provide workers with unpaid leave so as to 
(B) to provide workers with unpaid leave so as to 
(C) provide workers with unpaid leave in order that they 
(D) to provide workers with unpaid leave so that they can 
(E) provide workers with unpaid leave and 

The answer is (D). Why? The correct sentence would use “to provide” (not “provide”) and “so that” (not “so as to”), and should read, “Congress is debating a bill requiring certain employers to provide workers with unpaid leave so that they can care for sick or newborn children.” In this sentence, “they” logically refers to “workers.” Even though “they” could refer to employers, too, after you sort out the rest of the errors, you are left with (D) only, hence answer must be (D).

Let’s look at another question:

While depressed property values can hurt some large investors, they are potentially devastating for homeowners, whose equity – in many cases representing a life’s savings – can plunge or even disappear.

(A) they are potentially devastating for homeowners, whose
(B) they can potentially devastate homeowners in that their
(C) for homeowners they are potentially devastating, because their
(D) for homeowners, it is potentially devastating in that their
(E) it can potentially devastate homeowners, whose

The correct answer is (A). The correct sentence should read, “While depressed property values can hurt some large investors, they are potentially devastating for homeowners, whose equity – in many cases representing a life’s savings – can plunge or even disappear.” The pronoun “they” logically refers to “depressed property values.” Both the pronoun and its antecedent serve as subjects in their respective clauses, so the pronoun antecedent is quite clear.

One more question:

Although Napoleon’s army entered Russia with far more supplies than they had in their previous campaigns, it had provisions for only twenty-four days. 

(A) they had in their previous campaigns 
(B) their previous campaigns had had 
(C) they had for any previous campaign 
(D) in their previous campaigns 
(E) for any previous campaign

The correct answer is (E). The correct sentence should read, “Although Napoleon’s army entered Russia with far more supplies than for any previous campaign, it had provisions for only twenty-four days.”

The pronoun “it” logically refers to “Napolean’s army” and not Russia. Both the pronoun and its antecedent serve as subjects in their respective clauses, so the pronoun antecedent is quite clear. Note that the pronoun and its antecedent are a part of the non-underlined portion of the sentence so we don’t need to worry about the usage here but it strengthens our understanding of pronoun ambiguity.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using a Venn Diagram vs. a Double Set Matrix on the GMAT

Quarter Wit, Quarter WisdomCritics may have given a rotten rating to the recently released “Batman v. Superman” movie, but we sure can use it to learn a valuable GMAT lesson. A difficult decision point for GMAT test takers is picking the probable winner between Venn diagrams and Double Set Matrices for complicated sets questions. If that is true for you too, then the onscreen rivalry between Batman and Superman will help you remember this trick:

Venn diagrams are like Superman – all powerful. They can help you solve almost all questions involving either 2 or 3 overlapping sets. But then, there are some situations in which double set matrix method (aka Batman with his amazing weaponry) might be easier to use. It is possible to solve these questions using Venn diagrams, too, but it is more convenient to solve them using a Double Set Matrix.

We have discussed solving three overlapping sets using Venn diagrams here.

Today, we will look at the case in which using a Double Set Matrix is easier than using a Venn diagram – in instances where we have two sets of variables, such as English/Math and Middle School/High School, or Cake/Ice cream and Boys/Girls, etc.

Eventually, we will solve our question again using a Venn diagram, for those who like to use a single method for all similar questions. First, take a look at our question:

A business school event invites all of its graduate and undergraduate students to attend. Of the students who attend, male graduate students outnumber male undergraduates by a ratio of 7 to 2, and females constitute 70% of the group. If undergraduate students make up 1/6 of the group, which of the following CANNOT represent the number of female graduate students at the event?

(A) 18
(B) 27
(C) 36
(D) 72
(E) 180

To solve this problem using a Double Set Matrix, first jot down one set of variables as the row headings and the other as the column headings, as well as a row and column for “totals.” Now all you need to do is add in the information line by line as you read through the question.

“…male graduate students outnumber male undergraduates by a ratio of 7 to 2…
QWQW graph 1

 

 

“…females constitute 70% of the group.

Female students make up 70% of the group, which implies that male students (total of 9x) make up 30% of the group.

9x = (30/100)*Total Students

Total Students = 30x

Since 9x is the total number of male students while 30x is the total number of all students, the total number of female students must be 30x – 9x = 21x.

QWQW graph 2

 

 

If undergraduate students make up 1/6 of the group…

Undergrad students make 1/6 of the group, i.e. (1/6)*30x = 5x

If the total number of undergrad students is 5x and the number of male undergrad students is 2x, the number of female undergrad students must be 5x – 2x = 3x.

This implies that the number of graduate females must be 18x, since the total number of females is 21x.

QWQW graph 3

 

 

Therefore, the number of graduate females must be a multiple of 18. 27 is the only answer choice that is not a multiple of 18, so it cannot be the number of graduate females – therefore, our answer must be B.

Now, here is how Superman can rescue us in this question. An analysis similar to the one above will give us a Venn diagram which looks like this:

qwqw pic

 

 

 

 

 

 

Of course, we will get the same answer: the number of graduate females must be a multiple of 18. We know 27 is not a multiple of 18, so it cannot be the number of graduate females and therefore, our answer is still B.

Hopefully, next time you come across an overlapping sets question, you will know exactly who your superhero is!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

What Makes GMAT Quant Questions So Hard?

Quarter Wit, Quarter WisdomWe know that the essentials of the GMAT Quant section are pretty simple: advanced topics such as derivatives, complex numbers, matrices and trigonometry are not included, while fundamentals we all learned from our high school math books are included. So it would be natural to think that the GMAT Quant section should not pose much of a problem for most test-takers (especially for engineering students, who have actually covered far more advanced math during their past studies).

Hence, it often comes as a shock when many test-takers, including engineering students, receive a dismal Quant score on the first practice test they take. Of course, with practice, they usually wise up to the treachery of the GMAT, but until then, the Quant section is responsible for many a nightmare!

Today, let’s see what kind of treachery we are talking about – problems like this make some people laugh out loud and others pull at their hair!

Is the product pqr divisible by 12?
Statement 1: p is a multiple of 3
Statement 2: q is a multiple of 4

This seems like an easy C (Statements 1 and 2 together are sufficient, but alone are not sufficient), doesn’t it? P is a multiple of 3 and q is a multiple of 4, so together, p*q would be a multiple of 3*4 = 12. If p * q is already a multiple of 12, then obviously it would seem that p*q*r would be a multiple of 12, too.

But here is the catch – where is it mentioned that r must be an integer? Just because p and q are integers (multiples of 3 and 4 respectively), it does not imply that r must also be an integer.

If r is an integer, then sure, p*q*r will be divisible by 12. Imagine, however, that p = 3, q = 4 and r = 1/12. Now the product p*q*r = 3*4*(1/12) = 1. 1 is not divisible by 12, so in this case, pqr is not divisible by 12. Hence, both statements together are not sufficient to answer the question, and our answer is in fact E!

This question is very basic, but it still tricks us because we want to assume that p, q and r are clean integer values.

Along these same lines, let’s try the another one:

If 10^a * 3^b * 5^c = 450^n, what is the value of c?
Statement 1: a is 1.
Statement 2:  b is 2.

The first thing most of us will do here is split 450 into its prime factors:

450 = 2 * 3^2 * 5^2

450^n = 2^n * 3^2n * 5^2n

And do the same thing with the left side of the equation:

10^a * 3^b * 5^c = 2^a * 3^b * 5^(a+c)

Bringing the given equation back, we get:

2^a * 3^b * 5^(a+c) = 2^n * 3^2n * 5^2n

Statement 1: a is 1.

Equating the power of 2 on both sides, we see that a = n = 1.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

Statement 2:  b is 2.

Equating the power of 3 on both sides, we see that b = 2n = 2, so n = 1.

If n = 1, a = 1 by equating the powers of 2 on both sides.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

So it seems that both statements are separately sufficient. But hold on – again, the variables here don’t need to be cleanly fitting integers. The variables could pan out the way discussed in our first problem, or very differently.

Say, n = 1. When Statement 1 gives you that a = 1, you get 10^1 * 3^b * 5^c = 450^1.

3^b * 5^c = 45

Now note that value of c depends on the value of b, which needn’t be 2.

If b  = 3, then 3^3 * 5^c = 45.

5^c = 45/27

C will take a non-integer value here.

c = .3174

The question does not mention that all variables are integers, therefore there are infinite values that c can take depending on the values of b. Similarly, we can see that Statement 2 alone is also not sufficient. Using both statements together, you will get:

2^a * 3^b * 5^(a+c) = 450^n

2^1 * 3^2 * 5^(1 + c) = 450^n

5^(1 + c) = 450^n/18

By now, you’ve probably realized that depending on the value of n, c can take infinite different values. If n = 1, c = 1. If n = 2, c = 4.8. And so on… We don’t need to actually find these values – it is enough to know that different values of n will give different values of c.

With this in mind, we can see that both statements together are not sufficient, and therefore our answer must be E.

Hopefully, in future, this sneaky trick will not get you!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Dealing with Tangents on the GMAT

Quarter Wit, Quarter WisdomConsidering a two dimensional figure, a tangent is a line that touches a curve at a single point.  Here are some examples of tangents:
QWQW 1

 


 

In each of these cases, the line touches the curve at a single point. In the case of a circle, when you draw the radius of the circle from the center to the point of contact with the tangent, the radius is perpendicular to the tangent (as demonstrated in the figure on the right, above). A question discussing this concept is given in our post here.

Today, we will look at a question involving a tangent to a parabola:

If f(x) = 3x^2 – tx + 5 is tangent to the x-axis, what is the value of the positive number t?

(A) 2√15
(B) 4√15
(C) 3√13
(D) 4√13
(E) 6√15

Let’s first try to understand what the question is saying.

f(x) is a tangent to the x-axis. We know that the x-axis is a straight line, so f(x) must be a curve. A quadratic equation, such as our given equation of f(x) = 3x^2 -tx +5, gives a parabola. Since the x^2 term in the equation is positive, the parabola would be facing upwards and touching the x-axis at a single point, such as:

QWQW 2

 

 

 

 

 

Since the parabola touches the x-axis in only one point, it means the quadratic has only one root, or in other words, the quadratic must be a perfect square.

Therefore, f(x) = 3x^2 – tx + 5 = √3(x)^2 – tx + (√5)^2

To get f(x) in the form a^2 – 2ab + b^2 = (a – b)^2,

tx = 2ab = (2√3)x * √5

t = 2√15

Note that if t takes this value, the quadratic will have only one root.

Plugging this value of t back into our equation, we will get: f(x) = √3(x)^2 – 2(√15)(x) + (√5)^2

f(x) = (√3)x – (√5)^2

We know that the root of f(x) is the point where the value of the y coordinate is 0. Therefore:

(√3)x – (√5)^2  = 0

x = (√5)/(√3)

At this x co-ordinate, the parabola will touch the x axis.

[This calculation was shown only to help you completely understand the question. We could have easily stopped at t = 2(√15).]

Therefore, our answer is A.

The question can be solved in various other ways – think of how, and write your thoughts in the comments below!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Understanding Absolute Values with Two Variables

Quarter Wit, Quarter WisdomWe have looked at quite a few absolute value and inequality concepts. (Check out our discussion on the basics of absolute values and inequalities, here, and our discussion on how to handle inequalities with multiple absolute value terms in a single variable, here.) Today let’s look at an absolute value concept involving two variables. It is unlikely that you will see such a question on the actual GMAT, since it involves multiple steps, but it will help you understand absolute values better.

Recall the definition of absolute value:

|x| = x if x ≥ 0

|x| = -x if x < 0

So, to remove the absolute value sign, you will need to consider two cases – one when x is positive or 0, and another when it is negative.

Say, you are given an inequality, such as |x – y| < |x|. Here, you have two absolute value expressions: |x – y| and |x|. You need to get rid of the absolute value signs, but how will you do that?

You know that to remove the absolute value sign, you need to consider the two cases. Therefore:

|x – y| = (x – y) if (x – y) ≥ 0

|x – y| = – (x – y) if (x – y) < 0

But don’t forget, we also need to remove the absolute value sign that |x| has. Therefore:

|x| = x if x ≥ 0

|x| = -x if x < 0

In all we will get four cases to consider:

Case 1: (x – y) ≥ 0 and x ≥ 0

Case 2: (x – y) < 0 and x ≥ 0

Case 3: (x – y) ≥ 0 and x < 0

Case 4: (x – y) < 0 and x < 0

Let’s look at each case separately:

Case 1: (x – y) ≥ 0 (which implies x ≥ y) and x ≥ 0

|x – y| < |x|

(x – y) < x

-y < 0

Multiply by -1 to get:

y > 0

In this case, we will get 0 < y ≤ x.

Case 2: (x – y) < 0 (which implies x < y) and x ≥ 0

|x – y| < |x|

-(x – y) < x

2x > y

x > y/2

In this case, we will get 0 < y/2 < x < y.

Case 3: (x – y) ≥ 0 (which implies x ≥ y) and x < 0

|x – y| < |x|

(x – y) < -x

2x < y

x < y/2

In this case, we will get y ≤ x < y/2 < 0.

Case 4: (x – y) < 0 (which implies x < y) and x < 0

|x – y| < |x|

-(x – y) < -x

-x + y < -x

y < 0

In this case, we will get x < y < 0.

Considering all four cases, we get that both x and y are either positive or both are negative. Case 1 and Case 2 imply that if both x and y are positive, then x > y/2, and Case 3 and Case 4 imply that if both x and y are negative, then x < y/2. With these in mind, there is a range of values in which the inequality will hold. Both x and y should have the same sign – if they are both positive, x > y/2, and if they are both negative, x < y/2.

Here are some examples of values for which the inequality will hold:

x = 4, y = 5

x = 8, y = 2

x = -2, y = -1

x = -5, y = -6

etc.

Here are some examples of values for which the inequality will not hold:

x = 4, y = -5 (x and y have opposite signs)

x = 5, y = 15 (x is not greater than y/2)

x = -5, y = 9 (x and y have opposite signs)

x = -6, y = -14 (x is not less than y/2)

etc.

As said before, don’t worry about going through this method during the actual GMAT exam – if you do get a similar question, some strategies such as plugging in values and/or using answer choices to your advantage will work. Overall, this example hopefully helped you understand absolute values a little better.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: The Case of a Correct Answer Despite Incorrect Logic!

Quarter Wit, Quarter WisdomIt is common for GMAT test-takers to think in the right direction, understand what a question gives and what it is asking to be found out, but still get the wrong answer. Mistakes made during the execution of a problem are common on the GMAT, but what is rather rare is going with incorrect logic and still getting the correct answer! If only life was this rosy so often!

Today, we will look at a question in which exactly this phenomenon occurs – we will find the flaw in the logic that test-takers often come up with and then learn how to correct that flaw:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did had he driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100

(B) 120

(C) 140

(D) 150

(E) 160

This little gem (and it’s detailed algebra solution) is from our Advanced Word Problems book. We will post its solution here, too, for the sake of a comprehensive discussion:

Method 1: Algebra
Let’s start with the basic “Distance = Rate * Time” formula:

D = R*T ……….(I)

From here, the first theoretical trip can be represented as D + 70 = (R + 5)(T + 1), (the motorist travels for 1 extra hour at a rate of 5 mph faster), which can be expanded to D + 70 = RT + R + 5T +5.

We can then eliminate “D” by plugging in the value of “D” from our equation (I):

RT + 70 = RT + R + 5T + 5, which simplifies to 70 = R + 5T + 5 and then to 65 = R + 5T ……….. (II)

The second theoretical trip can be represented as (R+10)(T+2), which expands to RT + 2R + 10T + 20 (not that we only have an expression since we don’t know what the distance is).

The two middle terms (2R + 10T) can be factored to 2(R+5T), which allows us to use equation (II) here:

RT + 2(R+5T) + 20 = RT + 2(65) + 20 = RT + 150.

Since the original distance was RT, the additional distance is 150 more miles, or answer choice D.

We totally understand that this solution is a bit convoluted – algebra often is. So, understandably, students often look for a more direct logical solution.

Here is one they sometimes employ:

Method 2: Logic (Incorrect)
If the motorist had driven 1 hour longer at a rate 5 mph faster, then his original speed would be 70 miles subtracted by the extra 5 miles he drove in that hour to get 70 – 5 = 65 mph. If he drives at a rate 10 mph faster (i.e. at 65 + 10 = 75) * 2 for the extra hours, he/she would have driven 150 miles extra.

But here is the catch in this logic:

The motorist drove for an average rate of 5 mph extra. So the 70 includes not only the extra distance covered in the last hour, but also the extra 5 miles covered every hour for which he drove. Hence, his original speed is not 65. Now, let’s see the correct logical method of solving this:

Method 3: Logic (Correct)
Let’s review the original problem first. Say, speed is “S” mph – we don’t know the number of hours for which this speed was maintained.

STEP 1:

S + S + S + … + S + S = TOTAL DISTANCE COVERED

In the first hypothetical case, the motorist drove for an extra hour at a speed of 5 mph faster. This means he covered 5 extra miles every hour and then covered another S + 5 miles in the last hour. The underlined distances are the extra ones which all add up to 70.

STEP 2:

S + S + S + … + S + S = TOTAL DISTANCE COVERED

+5 +5 +5 + … + 5 + 5 = +70

In the second hypothetical case, in which the motorist drove for two hours longer at a speed of 10 mph faster,  he adds another 5 mph to his hourly speed and covers yet another distance of “S” in the second extra hour. In addition to S, he also covers another 10 miles in the second extra hour. The additional distances are shown in red  in the third case – every hour, the speed is 10 mph faster and he drove for two extra hours in this case (compared with Step 1).

STEP 3:

S + S + S + … + S + S + S + S = TOTAL DISTANCE COVERED

+5  +5  +5 + …  +5  +5  +5 = +70

+5  +5  +5 + …  +5  +5  +5 + 10 = +70 + 10

Note that the +5s and the S all add up to 70 (as seen in Step 2). We also separately add the extra 10 from the last hour. This is the logic of getting the additional distance of 70 + 70 + 10 = 150. It involves no calculations, but does require you to understand the logic. Therefore, our answer is still D.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: How to Find Composite Numbers on the GMAT

Quarter Wit, Quarter WisdomWe love to talk about prime numbers and their various properties for GMAT preparation, but composite numbers usually aren’t mentioned. Composite numbers are often viewed as whatever is leftover after prime numbers are removed from a set of positive integers (except 1 because 1 is neither prime, nor composite), but it is important to understand how these numbers are made, what makes them special and what should come to mind when we read “composite numbers.”

Principle: Every composite number is made up of 2 or more prime numbers. The prime numbers could be the same or they could be distinct.

For example:

2*2 = 4 (Composite number)

2*3*11 = 66 (Composite number)

5*23 = 115 (Composite number)

and so on…

Look at any composite number. You will always be able to split it into 2 or more prime numbers (not necessarily distinct). For example:

72 = 2*2*2*3*3

140 = 2*2*5*7

166 = 2*83

and so on…

This principle does look quite simple and intuitive at first, but when tested, we could face problems because we don’t think much about it. Let’s look at it with the help of one of our 700+ level GMAT questions:

x is the smallest integer greater than 1000 that is not prime and that has only one factor in common with 30!. What is x?

(A) 1009

(B) 1021

(C) 1147

(D) 1273

(E) 50! + 1

If we start with the answer choices, the way we often do when dealing with prime/composite numbers, we will get stuck. If we were looking for a prime number, we would use the method of elimination – we would find factors of all other numbers and the number that was left over would be the prime number.

But in this question, we are instead looking for a composite number – a specific composite number – and some of the answer choices are probably prime. Try as we might, we will not find a factor for them, and by the time we realize that it is prime, we will have wasted a lot of precious time. Let’s start from the question stem, instead.

We need a composite number that has only one factor in common with 30!. Every positive integer will have 1 as a factor, as will 30!, hence the only factor our answer and 30! will have in common is 1.

30! = 1*2*3*…*28*29*30

30! is the product of all integers from 1 to 30, so all prime numbers less than 30 are factors of 30!.

To make a composite number which has no prime factor in common with 30!, we must use prime numbers greater than 30. The first prime number greater than 30 is 31.

(As an aside, note that if we were looking for the smallest number with no factor other than 1 in common with 31!, we would skip to 37. All integers between 31 and 37 are composite and hence, would have factors lying between 1 and 31. Similarly, if we were looking for the smallest number with no factor other than 1 in common with 50!, 53 would be the answer.)

Let’s get back to our question. If we want to make a composite number without using any primes until 30, we must use two or more prime numbers greater than 30, and the smallest prime greater than 30 is 31. If we use two 31’s to get the smallest composite number, we get 31*31 = 961 But 961 is not greater than 1000, so it cannot be our answer.

So, let’s find the next prime number after 31 – it is 37. Multiplying 31 and 37, we get 31*37 = 1147. This is the smallest composite number greater than 1000 with no prime factors in common with 30! – the only factor it has in common with 30! is 1. Therefore, our answer is (C).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Ratios in GMAT Data Sufficiency

Quarter Wit, Quarter WisdomWe know that ratios are the building blocks for a lot of other concepts such as time/speed, work/rate and mixtures. As such, we spend a lot of time getting comfortable with understanding and manipulating ratios, so the GMAT questions that test ratios seem simple enough, but not always! Just like questions from all other test areas, questions on ratios can be tricky too, especially when they are formatted as Data Sufficiency questions.

Let’s look at two cases today: when a little bit of data is sufficient, and when a lot of data is insufficient.

When a little bit of data is sufficient!
Three brothers shared all the proceeds from the sale of their inherited property. If the eldest brother received exactly 5/8 of the total proceeds, how much money did the youngest brother (who received the smallest share) receive from the sale?

Statement 1: The youngest brother received exactly 1/5 the amount received by the middle brother.

Statement 2: The middle brother received exactly half of the two million dollars received by the eldest brother.

First impressions on reading this question? The question stem gives the fraction of money received by one brother. Statement 1 gives the fraction of money received by the youngest brother relative to the amount received by the middle brother. Statement 2 gives the fraction of money received by the middle brother relative to the eldest brother and an actual amount. It seems like the three of these together give us all the information we need. Let’s dig deeper now.

From the Question stem:

Eldest brother’s share = (5/8) of Total

Statement 1: Youngest Brother’s share = (1/5) * Middle brother’s share

We don’t have any actual number – all the information is in fraction/ratio form. Without an actual value, we cannot find the amount of money received by the youngest brother, therefore, Statement 1 alone is not sufficient.

Statement 2: Middle brother’s share = (1/2) * Eldest brother’s share, and the eldest brother’s share = 2 million dollars

Middle brother’s share = (1/2) * 2 million dollars = 1 million dollars

Now, we might be tempted to jump to Statement 1 where the relation between youngest brother’s share and middle brother’s share is given, but hold on: we don’t need that information. We know from the question stem that the eldest brother’s share is (5/8) of the total share.

So 2 million = (5/8) of the total share, therefore the total share = 3.2 million dollars.

We already know the share of the eldest and middle brothers, so we can subtract their shares out of the total and get the share of the youngest brother.

Youngest brother’s share = 3.2 million – 2 million – 1 million = 0.2 million dollars

Statement 2 alone is sufficient, therefore, the answer is B.

When a lot of data is insufficient!
A department manager distributed a number of books, calendars, and diaries among the staff in the department, with each staff member receiving x books, y calendars, and z diaries. How many staff members were in the department?

Statement 1: The numbers of books, calendars, and diaries that each staff member received were in the ratio 2:3:4, respectively.

Statement 2: The manager distributed a total of 18 books, 27 calendars, and 36 diaries.

First impressions on reading this question? The question stem tells us that each staff member received the same number of books, calendars, and diaries. Statement 1 gives us the ratio of books, calendars and diaries. Statement 2 gives us the actual numbers. It certainly seems that we should be able to obtain the answer. Let’s find out:

Looking at the question stem, Staff Member 1 recieved x books, y calendars, and z diaries, Staff Member 2 recieved x books, y calendars, and z diaries… and so on until Staff Member n (who also recieves x books, y calendars, and z diaries).

With this in mind, the total number of books = nx, the total number of calendars = ny, and the total number of diaries = nz.

Question: What is n?

Statement 1 tells us that x:y:z = 2:3:4. This means the values of x, y and z can be:

2, 3, and 4,

or 4, 6, and 8,

or 6, 9, and 12,

or any other values in the ratio 2:3:4.

They needn’t necessarily be 2, 3 and 4, they just need the required ratio of 2:3:4.

Obviously, n can be anything here, therefore, Statement 1 alone is not sufficient.

Statement 2 tell us that nx = 18, ny = 27, and nz = 36.

Now we know the actual values of nx, ny and nz, but we still don’t know the values of x, y, z and n.

They could be

2, 3, 4 and 9

or 6, 9, 12 and 3

Therefore, Statement 2 alone is also not sufficient.

Considering both statements together, note that Statement 2 tells us that nx:ny:nz = 18:27:36 = 2:3:4 (they had 9 as a common factor).

Since n is a common factor on left side, x:y:z = 2:3:4 (ratios are best expressed in the lowest form).

This is a case of what we call “we already knew that” – information given in Statement 1 is already a part of Statement 2, so it is not possible that Statement 2 alone is not sufficient but that together Statement 1 and 2 are. Hence, both statements together are not sufficient, and our answer must be E.

A question that arises often here is, “Why can’t we say that the number of staff members must be 9?”

This is because the ratio of 2:3:4 is same as the ratio of 6:9:12, which is same as 18:27:36 (when you multiply each number of a ratio by the same number, the ratio remains unchanged).

If 18 books, 27 calendars, and 36 diaries are distributed in the ratio 2:3:4, we could give them all to one person, or to 3 people (giving them each 6 books, 9 calendars and 12 diaries), or to 9 people (giving them each 2 books, 3 calendars and 4 diaries).

When we see 18, 27 and 36, what comes to mind is that the number of people could have been 9, which would mean that the department manager distributed 2 books, 3 calendars and 4 diaries to each person. But we know that 9 is divisible by 3, which should remind us that the number of people could also be 3, which would mean that the manager distributed 6 books, 9 calendars and 12 diaries to each person. As such, we still don’t know how many staff members there are, and our answer remians E.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Circular Reasoning in GMAT Critical Reasoning Questions

Quarter Wit, Quarter WisdomConsider this argument:

Anatomical bilateral symmetry is a common trait. It follows, therefore, that it confers survival advantages on organisms. After all, if bilateral symmetry did not confer such advantages, it would not be common.

What is the flaw here?

The argument restates rather than proves. The conclusion is a  premise, too – we start out by assuming that the conclusion is true and then state that the conclusion is true.

If A (bilateral symmetry) were not B (confer survival advantages), A (bilateral symmetry) would not be C (common).

A (bilateral symmetry) is C (common) so A (bilateral symmetry) is B (confer survival advantages).

Note that we did not try to prove that “A is C implies A is B”. We did not explain the connection between C and B. For our reasoning, all we said is that if A were not B, it would not be C, so we are starting out by taking the conclusion to be true.

This is called circular reasoning. It is a kind of logical fallacy – a flaw in the logic. You begin with what you are trying to prove, using your own conclusion as one of your premises.

Why is it good to understand circular reasoning for the GMAT? A critical reasoning question that asks you to mimic the reasoning argument could require you to identify such a flawed reasoning and find the argument that mimics it.

Continuing with the previous example:

Anatomical bilateral symmetry is a common trait. It follows, therefore, that it confers survival advantages on organisms. After all, if bilateral symmetry did not confer such advantages, it would not be common.

The pattern of reasoning in which one of the following arguments is most similar to that in the argument above?

(A) Since it is Sawyer who is negotiating for the city government, it must be true that the city takes the matter seriously. After all, if Sawyer had not been available, the city would have insisted that the negotiations be deferred.
(B) Clearly, no candidate is better qualified for the job than Trumbull. In fact, even to suggest that there might be a more highly qualified candidate seems absurd to those who have seen Trumbull at work.
(C) If Powell lacked superior negotiating skills, she would not have been appointed arbitrator in this case. As everyone knows, she is the appointed arbitrator, so her negotiating skills are, detractors notwithstanding, bound to be superior.
(D) Since Varga was away on vacation at the time, it must have been Rivers who conducted the secret negotiations. Any other scenario makes little sense, for Rivers never does the negotiating unless Varga is unavailable.
(E) If Wong is appointed arbitrator, a decision will be reached promptly. Since it would be absurd to appoint anyone other than Wong as arbitrator, a prompt decision can reasonably be expected.

We’ve established that the above pattern of reasoning has a circular reasoning flaw. Let’s consider each answer option to find the one which has similarly flawed reasoning.

(A) Since it is Sawyer who is negotiating for the city government, it must be true that the city takes the matter seriously. After all, if Sawyer had not been available, the city would have insisted that the negotiations be deferred.

Here is the structure of this argument:

If A (Sawyer) were not B (available), C (the city) would have D (insisted on deferring).

Since A (Sawyer) is B (available to the city), C (the city) does E (takes matter seriously).

Obviously, this argument structure is not the same as in the original argument.

(B) Clearly, no candidate is better qualified for the job than Trumbull. In fact, even to suggest that there might be a more highly qualified candidate seems absurd to those who have seen Trumbull at work.

Here is the structure of this argument:

A (people who have seen Trumbull at work) find B (Trumbull is not the best) absurd, therefore B (Trumbull is not the best) is false.

This is not circular reasoning. We have not assumed that B is false in our premises, we are simply saying that people think B is absurd. This is flawed logic too, but it is not circular reasoning.

(C) If Powell lacked superior negotiating skills, she would not have been appointed arbitrator in this case. As everyone knows, she is the appointed arbitrator, so her negotiating skills are, detractors notwithstanding, bound to be superior.

Here is the structure of this argument:

If A (Powell) were not B (had superior negotiating skills), A (Powell) would not have been C (appointed arbitrator).

A (Powell) is C (appointed arbitrator), therefore A (Powell) is B (had superior negotiating skills).

Note that the structure of the argument matches the structure of our original argument – this is circular reasoning, too. We are saying that if A were not B, A would not be C and concluding that since A is C, A is B. The conclusion is already taken to be true in the initial argument, so we can see it is is also an example of circular reasoning.

Hence (C) is the correct answer. Nevertheless, let’s look at the other two options and why they don’t work:

(D) Since Varga was away on vacation at the time, it must have been Rivers who conducted the secret negotiations. Any other scenario makes little sense, for Rivers never does the negotiating unless Varga is unavailable.

Here is the structure of this argument:

If A (Varga) is B (available), C (Rivers) does not do D (negotiate).

A (Varga) was not B (available), so C (Rivers) did D (negotiate).

This logic is flawed – the premise tells us what happens when A is B, however it does not tell us what happens when A is not B. We cannot conclude anything about what happens when A is not B. And because this is not circular reasoning, it cannot be the answer.

(E) If Wong is appointed arbitrator, a decision will be reached promptly. Since it would be absurd to appoint anyone other than Wong as arbitrator, a prompt decision can reasonably be expected.

Here is the structure of this argument:

If A (Wong) is B (appointed arbitrator), C (a decision) will be D (reached promptly).

A (Wong) not being B (appointed arbitrator) would be absurd, so C (a decision) will be D (reached promptly).

Again, this argument uses brute force, but it is not circular reasoning. “A not being B would be absurd” is not a convincing reason, so the argument is not strong as it is, but in any case, we don’t have to worry about it since it doesn’t use circular reasoning.

Take a look at this question for practice:

Dr. A: The new influenza vaccine is useless at best and possibly dangerous. I would never use it on a patient.
Dr. B: But three studies published in the Journal of Medical Associates have rated that vaccine as unusually effective.
Dr. A: The studies must have been faulty because the vaccine is worthless.

In which of the following is the reasoning most similar to that of Dr. A?

(A) Three of my patients have been harmed by that vaccine during the past three weeks, so the vaccine is unsafe.
(B) Jerrold Jersey recommends this milk, and I don’t trust Jerrold Jersey, so I won’t buy this milk.
(C) Wingz tennis balls perform best because they are far more effective than any other tennis balls.
(D) I’m buying Vim Vitamins. Doctors recommend them more often than they recommend any other vitamins, so Vim Vitamins must be good.
(E) Since University of Muldoon graduates score about 20 percent higher than average on the GMAT, Sheila Lee, a University of Muldoon graduate, will score about 20 percent higher than average when she takes the GMAT.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

A 750+ Level GMAT Geometry Question

Quarter Wit, Quarter WisdomToday we will discuss a pretty advanced GMAT question, because we can still use our basic GMAT concepts to find the answer. It may seem like we will need trigonometry to handle this question, but that is not so. In fact, the question will look familiar at first, but will present unforeseen problems later on.

While going through this exercise, we will learn a few tips and tricks which will be useful in our mainstream GMAT questions, hence, it will add value to our GMAT repertoire (especially in elimination techniques). Let’s go on to the question now:

In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = BC = X, what is the value of X?

(A) Sqrt(3) – 1

(B) Sqrt(3) + 2

(C) (Sqrt(3) – 1)/2

(D) (Sqrt(3) + 1)/2

(E) 2*(Sqrt(3) + 1)

What we see here is an isosceles triangle with one angle as 30 degrees and other two angles as (180 – 30)/2 = 75 degrees each.

The side opposite the 30 degrees angle is 2*sqrt(2). One simple observation is that X must be greater than 2*sqrt(2) because these sides are opposite the greater angles (75 degrees).

2*sqrt(2) is a bit less than 2*1.5 because Sqrt(2) = 1.414. So 2*sqrt(2) is a bit less than 3. Note that options (A), (C), and (D) are much smaller than 3, so these cannot be the value of X. We have already improved our chances of getting the correct answer by eliminating three options! Now we have to choose out of (B) and (E).
2Triangles

 

 

 

 

 

 

Here is what is given: Angle ABC = 30 degrees, and AC = 2*sqrt(2). We need to find the value of X. Now, our 30 degree angle reminds us of a 30-60-90 triangle in which we know the ratio of the sides – given one side, we can find the other two.

The problem is this: if we drop an altitude from angle B to AC, the angle 30 degrees will be split in half and we will actually get a 15-75-90 triangle, instead. We won’t have a 30-60-90 triangle anymore, so what do we do now? Let’s try to maintain the 30 degree angle as it is to get the 30-60-90 triangle, and drop an altitude from angle C to AB instead, calling it CE. Now we have a 30-60-90 triangle! Since BCE is a 30-60-90 triangle, its sides are in the ratio 1:sqrt(3):2. Side X corresponds to 2 on the ratio, so CE = x/2.

Area of triangle ABC = (1/2)*BD*AC = (1/2)*CE*AB

(1/2)*BD*2*sqrt(2) = (1/2)*(X/2)*X

BD = X^2/4*Sqrt(2)

Now DC = (1/2)AC = 2*sqrt(2)/2 = sqrt(2)

Let’s use the pythagorean theorem on triangle BDC:

BD^2 + DC^2 = BC^2

(X^2/4*Sqrt(2))^2 + (Sqrt(2))^2 = X^2

X^4/32 + 2 = X^2

X^4 – 32*X^2 + 64 = 0

X^4 – 16X^2 + 8^2 – 16X^2 = 0

(X^2 – 8)^2 – (4X)^2 = 0

(X^2 -8 + 4X) * (X^2 – 8 – 4X) = 0

Normally, this would require us to use the quadratic roots formula, but let’s not get that complicated. We can just plug in the the two shortlisted options and see if either of the factors is 0. If one of the factors becomes 0, the equation will be satisfied and we will have the root of the equation.

Since both options have both terms positive, it means the co-efficient corresponding to B in Ax^2 + Bx + C = 0 must be negative.

x = [-B +- Sqrt(B^2 – 4AC)]/2A

-B will give us a positive term if B is negative, so we will get the answer by plugging into (X^2 – 4X – 8):

Put X = Sqrt(3) + 2 in X^2 – 4X – 8 and you do not get 0.

Put X = 2*(Sqrt(3) + 1) in X^2 – 4X – 8 and you do get 0.

This means that X is 2*(Sqrt(3) + 1), so our answer must be (E).

To recap:

Tip 1: A greater side of a triangle is opposite a greater angle.

Tip 2: We can get the relation between sides and altitudes of a triangle by using the area of the triangle formula.

Tip 3: The quadratic formula can help identify the sign of the irrational roots.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving GMAT Critical Reasoning Questions Involving Rates

Quarter Wit, Quarter WisdomIn our “Quarter Wit, Quarter Wisdom” series, we have seen how to solve various rates questions – the basic ones as well as the complicated ones. But we haven’t considered critical reasoning questions involving rates, yet. In fact, the concept of rates makes these problems very difficult to both understand and explain. First, let’s look at what “rate” is.

Say my average driving speed is 60 miles/hr. Does it matter whether I drive for 2 hours or 4 hours? Will my average speed change if I drive more (theoretically speaking)? No, right? When I drive for more hours, the distance I cover is more. When I drive for fewer hours, the distance I cover is less. If I travel for a longer time, does it mean my average speed has decreased? No. For that, I need to know  what happened to the distance covered. If the distance covered is the same while time taken has increased, only then can I say that my speed was reduced.

Now we will look at an official question and hopefully convince you of the right answer:

The faster a car is traveling, the less time the driver has to avoid a potential accident, and if a car does crash, higher speeds increase the risk of a fatality. Between 1995 and 2000, average highway speeds increased significantly in the United States, yet, over that time, there was a drop in the number of car-crash fatalities per highway mile driven by cars.

Which of the following, if true about the United States between 1995 and 2000, most helps to explain why the fatality rate decreased in spite of the increase in average highway speeds?

(A) The average number of passengers per car on highways increased.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

(D) The average mileage driven on highways per car increased.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

Let’s break down the given argument:

  • The faster a car, the higher the risk of fatality.
  • In a span of 5 years, the average highway speed has increased.
  • In the same time, the number of car crash fatalities per highway mile driven by cars has reduced.

This is a paradox question. In last 5 years, the average highway speed has increased. This would have increased the risk of fatality, so we would expect the number of car crash fatalities per highway mile to go up. Instead, it actually goes down. We need to find an answer choice that explains why this happened.

(A) The average number of passengers per car on highways increased.

If there are more people in each car, the risk of fatality increases, if anything. More people are exposed to the possibility of a crash, and if a vehicle is in fact involved in an accident, more people are at risk. It certainly doesn’t explain why the rate of fatality actually decreases.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

This option tells us that the safety features in the cars have been enhanced. That certainly explains why the fatality rate has gone down. If the cars are safer now, the risk of fatality would have reduced, hence this option does help us in explaining the paradox. This is the answer, but let’s double-check by looking at the other options too.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

This option is irrelevant – why the average speed increased is not our concern at all. Our only concern is that average speed has, in fact, increased. This should logically increase the risk of fatality, and hence, our paradox still stands.

(D) The average mileage driven on highways per car increased.

This is the answer choice that troubles us the most. The rate we are concerned about is number of fatalities/highway mile driven, and this option tells us that mileage driven by cars has increased.

Now, let’s consider the parallel with our previous distance-rate-time example:

Rate = Distance/Time

We know that if I drive for more time, it doesn’t mean that my rate changes. Here, however:

Rate = Number of fatalities/highway mile driven

In this case, if more highway miles are driven, it doesn’t mean that the rate will change. It actually has no impact on the rate; we would need to know what happened to the number of fatalities to find out what happened to the rate. Hence this option does not explain the paradox.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

This answer choice tells us that on average, the trips were made more quickly, i.e. the speed increased. The given argument already tells us that, so this option does not help resolve the paradox.

Our answer is, therefore, (B).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Should You Use the Permutation or Combination Formula?

Quarter Wit, Quarter WisdomA recurring question from many students who are preparing for GMAT is this: When should one use the permutation formula and when should one use the combination formula?

People have tried to answer this question in various ways, but some students still remain unsure. So we will give you a rule of thumb to follow in all permutation/combination questions:

You never NEED to use the permutation formula! You can always use the combination formula quite conveniently. First let’s look at what these formulas do:

Permutation: nPr = n!/(n-r)!
Out of n items, select r and arrange them in r! ways.

Combination: nCr = n!/[(n-r)!*r!]
Out of n items, select r.

So the only difference between the two formulas is that nCr has an additional r! in the denominator (that is the number of ways in which you can arrange r elements in a row). So you can very well use the combinations formula in place of the permutation formula like this:

nPr = nCr * r!

The nCr formula is far more versatile than nPr, so if the two formulas confuse you, just forget about nPr.

Whenever you need to “select,” “pick,” or “choose” r things/people/letters… out of n, it’s straightaway nCr. What you do next depends on what the question asks of you. Do you need to arrange the r people in a row? Multiply by r!. Do you need to arrange them in a circle? Multiply by (r-1)!. Do you need to distribute them among m groups? Do that! You don’t need to think about whether it is a permutation problem or a combination problem at all. Let’s look at this concept more in depth with the use of a few examples.

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can the presenters be chosen?

In this question, you simply have to choose 3 of the 8 teachers, and you know that you can do that in 8C3 ways. That is all that is required.

8C3 = 8*7*6/3*2*1 = 56 ways

Not too bad, right? Let’s look at another question:

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can all three presentations be done?

This question is a little different. You need to find the ways in which the presentations can be done. Here the presentations will be different if the same three teachers give presentations in different order. Say Teacher 1 presents, then Teacher 2 and finally Teacher 3 — this will be different from Teacher 2 presenting first, then Teacher 3 and finally Teacher 1. So, not only do we need to select the three teachers, but we also need to arrange them in an order. Select 3 teachers out of 8 in 8C3 ways and then arrange them in 3! ways:

We get 8C3 * 3! = 56 * 6 = 336 ways

Let’s try another one:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook. How many groups of photographs are possible?

For this problem, out of 7 photographs, we just have to select 3 to make a group. This can be done in 7C3 ways:

7C3 = 7*6*5/3*2*1 = 35 ways

Here’s another variation:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook, 1 each on the walls of his three best friends. In how many ways can he do that?

Here, out of 7 photographs, we have to first select 3 photographs. This can be done in 7C3 ways. Thereafter, we need to put the photographs on the walls of his three chosen friends. In how many ways can he do that? Now there are three distinct spots in which he will put up the photographs, so basically, he needs to arrange the 3 photographs in 3 distinct spots, which that can be done in 3! ways:

Total number of ways = 7C3 * 3! = (7*6*5/3*2*1) * 6= 35 * 6 = 210 ways

Finally, our last problem:

12 athletes will run in a race. In how many ways can the gold, silver and bronze medals be awarded at the end of the race?

We will start with selecting 3 of the 12 athletes who will win some position in the race. This can be done in 12C3 ways. But just selecting 3 athletes is not enough — they will be awarded 3 distinct medals of gold, silver, and bronze. Athlete 1 getting gold, Athlete 2 getting silver, and Athlete 3 getting bronze is not the same as Athlete 1 getting silver, Athlete 2 getting gold and Athlete 3 getting bronze. So, the three athletes need to be arranged in 3 distinct spots (first, second and third) in 3! ways:

Total number of ways = 12C3 * 3! ways

Note that some of the questions above were permutation questions and some were combination questions, but remember, we don’t need to worry about which is which. All we need to think about is how to solve the question, which is usually by starting with nCr and then doing any other required steps. Break the question down — select people and then arrange if required. This will help you get rid of the “permutation or combination” puzzle once and for all.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Keeping an Open Mind in Critical Reasoning

Quarter Wit, Quarter WisdomToday we will discuss why it is important to keep an open mind while toiling away on your GMAT studying. Don’t go into test day with biases expecting that if a question tells us this, then it must ask that. GMAC testmakers are experts in surprising you and taking advantage of your preconceived notions, which is how they confuse you and convert a 600-level question to a 700-level one.

We have discussed necessary and sufficient conditions before; we have also discussed assumptions before. This question from our own curriculum is an innovative take on both of these concepts. Let’s take a look.

All of the athletes who will win a medal in competition have spent many hours training under an elite coach. Michael is coached by one of the world’s elite coaches; therefore it follows logically that Michael will win a medal in competition.

The argument above logically depends on which of the following assumptions?

(A) Michael has not suffered any major injuries in the past year.

(B) Michael’s competitors did not spend as much time in training as Michael did.

(C) Michael’s coach trained him for many hours.

(D) Most of the time Michael spent in training was productive.

(E) Michael performs as well in competition as he does in training.

First we must break down the argument into premises and conclusions:

Premises:

  • All of the athletes who will win a medal in competition have spent many hours training under an elite coach.
  • Michael is coached by one of the world’s elite coaches.

Conclusion: Michael will win a medal in competition.

Read the argument carefully:

All of the athletes who will win a medal in competition have spent many hours training under an elite coach.

Are you wondering, “How does one know that all athletes who will win (in the future) would have spent many hours training under an elite coach?”

The answer to this is that it doesn’t matter how one knows – it is a premise and it has to be taken as the truth. How the truth was established is none of our business and that is that. If we try to snoop around too much, we will waste precious time. Also, what may seem improbable may have a perfectly rational explanation. Perhaps all athletes who are competing have spent many hours under an elite coach – we don’t know.

Basically, what this statement tells us is that spending many hours under an elite coach is a NECESSARY condition for winning. What you need to take away from this statement is that “many hours training under an elite coach” is a necessary condition to win a medal. Don’t worry about the rest.

Michael is coached by one of the world’s elite coaches.

It seems that Michael satisfies one necessary condition: he is coached by an elite coach.

Conclusion: Michael will win a medal in competition.

Now this looks like our standard “gap in logic”. To get this conclusion, the necessary condition has been taken to be sufficient. So if we are asked for the flaw in the argument, we know what to say.

Anyway, let’s check out the question (this is usually our first step):

The argument above logically depends on which of the following assumptions?

Note the question carefully – it is asking for an assumption, or a necessary premise for the conclusion to hold.

We know that “many hours training under an elite coach” is a necessary condition to win a medal. We also know that Michael has been trained by an elite coach. Note that we don’t know whether he has spent “many hours” under his elite coach. The necessary condition requires “many hours” under an elite coach.

If Michael has spent many hours under the elite coach then he satisfies the necessary condition to win a medal. It is still not sufficient for him to win the medal, but our question only asks for an assumption – a necessary premise for the conclusion to hold. It does not ask for the flaw in the logic.

Focus on what you are asked and look at answer choice (C):

(C) Michael’s coach trained him for many hours.

This is a necessary condition for Michael to win a medal. Hence, it is an assumption and therefore, (C) is the correct answer.

Don’t worry that the argument is flawed. There could be another question on this argument which asks you to find the flaw in it, however this particular question asks you for the assumption and nothing more.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: An Interesting Property of Exponents

Quarter Wit, Quarter WisdomToday, let’s take a look at an interesting number property. Once we discuss it, you might think, “I always knew that!” and “Really, what’s new here?” So let me give you a question beforehand:

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

Think about it for a few seconds – could you come up with the answer in the blink of an eye? If yes, great! Close this window and wait for the next week’s post. If no, then read on. There is much to learn today and it is an eye-opener!

Let’s start by jotting down some powers of numbers:

Power of 2: 1, 2, 4, 8, 16, 32 …

Power of 3: 1, 3, 9, 27, 81, 243 …

Power of 4: 1, 4, 16, 64, 256, 1024 …

Power of 5: 1, 5, 25, 125, 625, 3125 …

and so on.

Obviously, for every power of 2, when you multiply the previous power by 2, you get the next power (4*2 = 8).

For every power of 3, when you multiply the previous power by 3, you get the next power (27*3 = 81), and so on.

Also, let’s recall that multiplication is basically repeated addition, so 4*2 is basically 4 + 4.

This leads us to the following conclusion using the power of 2:

4 * 2 = 8

4 + 4 = 8

2^2 + 2^2 = 2^3

(2 times 2^2 gives 2^3)

Similarly, for the power of 3:

27 * 3 = 81

27 + 27 + 27 = 81

3^3 + 3^3 + 3^3 = 3^4

(3 times 3^3 gives 3^4)

And for the power of 4:

4 * 4 = 16

4 + 4 + 4 + 4 = 16

4^1 + 4^1 + 4^1 + 4^1 = 4^2

(4 times 4^1 gives 4^2)

Finally, for the power of 5:

125 * 5 = 625

125 + 125 + 125 + 125 + 125 = 625

5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 5^4

(5 times 5^3 gives 5^4)

Quite natural and intuitive, isn’t it? Take a look at the previous question again now.

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

A) 18

(B) 32

(C) 35

(D) 64

(E) 70

Which two powers when added will give 2^(36)?

From our discussion above, we know they are 2^(35) and 2^(35).

2^(35) + 2^(35) = 2^(36)

So x = 35 and y = 35 will satisfy this equation.

x + y = 35 + 35 = 70

Therefore, our answer is E.

One question arises here: Is this the only possible sum of x and y? Can x and y take some other integer values such that the sum of 2^x and 2^y will be 2^(36)?

Well, we know that no matter which integer values x and y take, 2^x and 2^y will always be positive, which means both x and y must be less than 36. Now note that no matter which two powers of 2 you add, their sum will always be less than 2^(36). For example:

2^(35) + 2^(34) < 2^(35) + 2^(35)

2^(2) + 2^(35) < 2^(35) + 2^(35)

etc.

So if x and y are both integers, the only possible values that they can take are 35 and 35.

How about something like this: 2^x + 2^y + 2^z = 2^36? What integer values can x, y and z take here?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Calculating the Probability of Intersecting Events

Quarter Wit, Quarter WisdomWe know our basic probability formulas (for two events), which are very similar to the formulas for sets:

P(A or B) = P(A) + P(B) – P(A and B)

P(A) is the probability that event A will occur.

P(B) is the probability that event B will occur.

P(A or B) gives us the union; i.e. the probability that at least one of the two events will occur.

P(A and B) gives us the intersection; i.e. the probability that both events will occur.

Now, how do you find the value of P(A and B)? The value of P(A and B) depends on the relation between event A and event B. Let’s discuss three cases:

1) A and B are independent events

If A and B are independent events such as “the teacher will give math homework,” and “the temperature will exceed 30 degrees celsius,” the probability that both will occur is the product of their individual probabilities.

Say, P(A) = P(the teacher will give math homework) = 0.4

P(B) = P(the temperature will exceed 30 degrees celsius) = 0.3

P(A and B will occur) = 0.4 * 0.3 = 0.12

2) A and B are mutually exclusive events

If A and B are mutually exclusive events, this means they are events that cannot take place at the same time, such as “flipping a coin and getting heads” and “flipping a coin and getting tails.” You cannot get both heads and tails at the same time when you flip a coin. Similarly, “It will rain today” and “It will not rain today” are mutually exclusive events – only one of the two will happen.

In these cases, P(A and B will occur) = 0

3) A and B are related in some other way

Events A and B could be related but not in either of the two ways discussed above – “The stock market will rise by 100 points” and “Stock S will rise by 10 points” could be two related events, but are not independent or mutually exclusive. Here, the probability that both occur would need to be given to you. What we can find here is the range in which this probability must lie.

Maximum value of P(A and B):

The maximum value of P(A and B) is the lower of the two probabilities, P(A) and P(B).

Say P(A) = 0.4 and P(B) = 0.7

The maximum probability of intersection can be 0.4 because P(A) = 0.4. If probability of one event is 0.4, probability of both occurring can certainly not be more than 0.4.

Minimum value of P(A and B):

To find the minimum value of P(A and B), consider that any probability cannot exceed 1, so the maximum P(A or B) is 1.

Remember, P(A or B) = P(A) + P(B) – P(A and B)

1 = 0.4 + 0.7 – P(A and B)

P(A and B) = 0.1 (at least)

Therefore, the actual value of P(A and B) will lie somewhere between 0.1 and 0.4 (both inclusive).

Now let’s take a look at a GMAT question using these fundamentals:

There is a 10% chance that Tigers will not win at all during the whole season. There is a 20% chance that Federer will not play at all in the whole season. What is the greatest possible probability that the Tigers will win and Federer will play during the season?

(A) 55%

(B) 60%

(C) 70%

(D) 72%

(E) 80%

Let’s review what we are given.

P(Tigers will not win at all) = 0.1

P(Tigers will win) = 1 – 0.1 = 0.9

P(Federer will not play at all) = 0.2

P(Federer will play) = 1 – 0.2 = 0.8

Do we know the relation between the two events “Tigers will win” (A) and “Federer will play” (B)? No. They are not mutually exclusive and we do not know whether they are independent.

If they are independent, then the P(A and B) = 0.9 * 0.8 = 0.72

If the relation between the two events is unknown, then the maximum value of P(A and B) will be 0.8 because P(B), the lesser of the two given probabilities, is 0.8.

Since 0.8, or 80%, is the greater value, the greatest possibility that the Tigers will win and Federer will play during the season is 80%. Therefore, our answer is E.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Basic Operations for GMAT Inequalities

Quarter Wit, Quarter WisdomWe know that we can perform all basic operations of addition, subtraction, multiplication and division on two equations:

a = b

c = d

When these numbers are equal, we know that:

a + c = b + d (Valid)

a – c = b – d (Valid)

a * c = b * d (Valid)

a / c = b / d (Valid assuming c and d are not 0)

When can we add, subtract, multiply or divide two inequalities? There are rules that we need to follow for those. Today let’s discuss those rules and the concepts behind them.

Addition:

We can add two inequalities when they have the same inequality sign.

a < b

c < d

a + c < b + d (Valid)

Conceptually, it makes sense, right? If a is less than b and c is less than d, then the sum of a and c will be less than the sum of b and d.

On the same lines:

a > b

c > d

a + c > b + d (Valid)

Case 2: What happens when the inequalities have opposite signs?

a > b

c < d

We need to multiply one inequality by -1 to get the two to have the same inequality sign.

-c > -d

Now we can add them.

a – c > b – d

Subtraction:

We can subtract two inequalities when they have opposite signs:

a > b

c < d

a – c > b – d (The result will take the sign of the first inequality)

Conceptually, think about it like this: from a greater number (a is greater than b), if we subtract a smaller number (c is smaller than d), the result (a – c) will be greater than the result obtained when we subtract the greater number from the smaller number (b – d).

Note that this result is the same as that obtained when we added the two inequalities after changing the sign (see Case 2 above). We cannot subtract inequalities if they have the same sign, so it is better to always stick to addition. If the inequalities have the same sign, we simply add them. If the inequalities have opposite signs, we multiply one of them by -1 (to get the same sign) and then add them (in effect, we subtract them).

Why can we not subtract two inequalities when they have the same inequality sign, such as when a > b and c > d?

Say, we have 3 > 1 and 5 > 1.

If we subtract these two, we get 3 – 5 > 1 – 1, or -2 > 0 which is not valid.

If instead it were 3 > 1 and 2 > 1, we would get 1 > 0 which is valid.

We don’t know how much greater one term is from the other and hence we cannot subtract inequalities when their inequality signs are the same.

Multiplication:

Here, the constraint is the same as that in addition (the inequality signs should be the same) with an extra constraint: both sides of both inequalities should be non-negative. If we do not know whether both sides are non-negative or not, we cannot multiply the inequalities.

If a, b, c and d are all non negative,

a < b

c < d

a*c < b*d (Valid)

When two greater numbers are multiplied together, the result will be greater.

Take some examples to see what happens in Case 1, or more numbers are negative:

-2 < -1

10 < 30

Multiply to get: -20 < -30 (Not valid)

-2 < 7

-8 < 1

Multiply to get: 16 < 7 (Not valid)

Division:

Here, the constraint is the same as that in subtraction (the inequality signs should be opposite) with an extra constraint: both sides of both inequalities should be non-negative (obviously, 0 should not be in the denominator). If we do not know whether both sides are positive or not, we cannot divide the inequalities.

a < b

c > d

a/c < b/d (given all a, b, c and d are positive)

The final inequality takes the sign of the numerator.

Think of it conceptually: a smaller number is divided by a greater number, so the result will be a smaller number.

Take some examples to see what happens in Case 1, or more numbers are negative.

1 < 2

10 > -30

Divide to get 1/10 < -2/30 (Not valid)

Takeaways: 

Addition: We can add two inequalities when they have the same inequality signs.

Subtraction: We can subtract two inequalities when they have opposite inequality signs.

Multiplication: We can multiply two inequalities when they have the same inequality signs and both sides of both inequalities are non-negative.

Division: We can divide two inequalities when they have opposite inequality signs and both sides of both inequalities are non-negative (0 should not be in the denominator).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Grammatical Structure of Conditional Sentences on the GMAT

Quarter Wit, Quarter WisdomToday, we will take a look at the various “if/then” constructions in the GMAT Verbal section. Let us start out with some basic ideas on conditional sentences (though I know that most of you will be comfortable with these):

A conditional sentence (an if/then sentence) has two clauses – the “if clause” (conditional clause) and the “then clause” (main clause).  The “if clause” is the dependent clause, meaning the verbs we use in the clauses will depend on whether we are talking about a real or a hypothetical situation.

Often, conditional sentences are classified into first conditional, second conditional and third conditional (depending on the tense and possibility of the actions), but sometimes we have a separate zero conditional for facts. We will follow this classification and discuss four types of conditionals:

1) Zero Conditional

These sentences express facts; i.e. implications – “if this happens, then that happens.”

  • If the suns shines, the clothes dry quickly.
  • If he eats bananas, he gets a headache.
  • If it rains heavily, the temperature drops.

These conditionals establish universally known facts or something that happens habitually (every time he eats bananas, he gets a headache).

2) First Conditional

These sentences refer to predictive conditional sentences. They often use the present tense in the “if clause” and future tense (usually with the word “will”) in the main clause.

  • If you come to  my place, I will help you with your homework.
  • If I am able to save $10,000 by year end, I will go to France next year.

3) Second Conditional

These sentences refer to hypothetical or unlikely situations in the present or future. Here, the “if clause” often uses the past tense and the main clause uses conditional mood (usually with the word “would”).

  • If I were you, I would take her to the dance.
  • If I knew her phone number, I would tell you.
  • If I won the lottery, I would travel the whole world.

4) Third Conditional

These sentences refer to hypothetical situations in the past – what could have been different in the past. Here, the “if clause” uses the past perfect tense and the main clause uses the conditional perfect tense (often with the words “would have”).

  • If you had told me about the party, I would have attended it.
  • If I had not lied to my mother, I would not have hurt her.

Sometimes, mixed conditionals are used here, where the second and third conditionals are combined. The “if clause” then uses the past perfect and the main clause uses  the word “would”.

  • If you had helped me then, I would be in a much better spot today.

Now that you know which conditionals to use in which situation, let’s take a look at a GMAT question:

Botanists have proven that if plants extended laterally beyond the scope of their root system, they will grow slower than do those that are more vertically contained.

(A) extended laterally beyond the scope of their root system, they will grow slower than do

(B) extended laterally beyond the scope of their root system, they will grow slower than

(C) extend laterally beyond the scope of their root system, they grow more slowly than

(D) extend laterally beyond the scope of their root system, they would have grown more slowly than do

(E) extend laterally beyond the scope of their root system, they will grow more slowly than do

Now that we understand our conditionals, we should be able to answer this question quickly. Scientists have established something here; i.e. it is a fact. So we will use the zero conditional here – if this happens, then that happens.

…if plants extend laterally beyond the scope of their root system, they grow more slowly than do…

So the correct answer must be (C).

A note on slower vs. more slowly – we need to use an adverb here because “slow” describes “grow,” which is a verb. So we must use “grow slowly”. If we want to show comparison, we use “more slowly”, so the use of “slower” is incorrect here.

Let’s look at another question now:

If Dr. Wade was right, any apparent connection of the eating of highly processed foods and excelling at sports is purely coincidental.

(A) If Dr. Wade was right, any apparent connection of the eating of

(B) Should Dr. Wade be right, any apparent connection of eating

(C) If Dr. Wade is right, any connection that is apparent between eating of

(D) If Dr. Wade is right, any apparent connection between eating

(E) Should Dr. Wade have been right, any connection apparent between eating

Notice the non-underlined part “… is purely coincidental” in the main clause. This makes us think of the zero conditional.

Let’s see if it makes sense:

If Dr. Wade is right, any connection … is purely coincidental.

This is correct. It talks about a fact.

Also, “eating highly processed foods and excelling at sports” is correct.

Hence, our answer must be (D).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions (Part 2)

Quarter Wit, Quarter WisdomLast week, we reviewed the concepts of cyclicity and remainders and looked at some basic questions. Today, let’s jump right into some GMAT-relevant questions on these topics:

 

 

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2.

The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1

So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

Since the exponent a remainder of 3, the new cycle ends at 3, 9, 7. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9.

This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E.

Not so bad; let’s try a data sufficiency problem:

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Statement 1: k is divisible by 10

Statement 2: k is divisible by 4

With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k.

The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change:

If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6.

If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2.

If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4.

If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8.

If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case.

and so on…

So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4.

Statement 1: k is divisible by 10

With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient.

Statement 2: k is divisible by 4

This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B.

Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT:

What is the remainder of (3^7^11) divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}.

We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4.

We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4.

For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”):

The remainder of 7^{11} divided by 4

= The remainder of (4 + 3)^{11} divided by 4

= The remainder of 3^{11} divided by 4

= The remainder of 3*3^{10} divided by 4

= The remainder of 3*9^5 divided by 4

= The remainder of 3*(8+1)^5 divided by 4

= The remainder of 3*1^5 divided by 4

= The remainder of 3 divided by 4, which itself = 3

So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4.

Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions

Quarter Wit, Quarter WisdomUsually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Today we will look at the cases in which it can, and we will see why it helps us in these cases.

First let’s look at a pattern:

 

20/10 gives us a remainder of 0 (as 20 is exactly divisible by 10)

21/10 gives a remainder of 1

22/10 gives a remainder of 2

23/10 gives a remainder of 3

24/10 gives a remainder of 4

25/10 gives a remainder of 5

and so on…

In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on…

Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5).

With this in mind:

20/5 gives a remainder of 0 (as 20 is exactly divisible by 5)

21/5 gives a remainder of 1

22/5 gives a remainder of 2

23/5 gives a remainder of 3

24/5 gives a remainder of 4

25/5 gives a remainder of 0 (as 25 is exactly divisible by 5)

26/5 gives a remainder of 1

27/5 gives a remainder of 2

28/5 gives a remainder of 3

29/5 gives a remainder of 4

30/5 gives a remainder of 0 (as 30 is exactly divisible by 5)

and so on…

So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10.

Let’s take a few questions now:

What is the remainder when 86^(183) is divided by 10?

Here, we need to find the last digit of 86^(183) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is (previously discussed in this post).

So the units digit of 86^(183) will be 6. So when we divide this by 10, the remainder will also be 6.

Next question:

What is the remainder when 487^(191) is divided by 5?

Again, when considering division by 5, the units digit can help us.

The units digit of 487 is 7.

7 has a cyclicity of 7, 9, 3, 1.

Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term.

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3

So the units digit of 487^(191) is 3, and the number would look something like ……………..3

As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5.

Therefore, the remainder of 487^(191) divided by 5 is 3.

Last question:

If x is a positive integer, what is the remainder when 488^(6x) is divided by 2?

Take a minute to review the question first. If you start by analyzing the expression 488^(6x), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether 488^(6x) is odd or even.

488^(6x) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even.

So 488^(6x) is even and will give remainder 0 when it is divided by 2.

That is all for today. We will look at some GMAT remainders-cyclicity questions next week!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Cyclicity of Units Digits on the GMAT (Part 2)

Quarter Wit, Quarter WisdomAs discussed last week, all units digits have a cyclicity of 1 or 2 or 4. Digits 2, 3, 7 and 8 have a cyclicity of 4, i.e. the units digit repeats itself every 4 digit:

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

Digits 4 and 9 have a cyclicity of 2, i.e. the units digit repeats itself every 2 digits:

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1

Digits 0, 1, 5 and 6 have a cyclicity of 1, i.e. the units digit is 0, 1, 5, or 6 respectively.

Now let’s take a look at how to apply these fundamentals:

What is the units digit of 813^(27)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 3.

Remember, our cyclicity of 3 is 3, 9, 7, 1 (four numbers total).

We need the units digit of 3^(27). How many full cycles of 4 will be there in 27? There will be 6 full cycles because 27 divided by 4 gives 6 as quotient and 3 will be the remainder. So after 6 full cycles of 4 are complete, a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

… (6 full cycles)

3, 9, 7 (new cycle for remainder of 3)

7 will be the units digit of 3^(27), so 7 will be the units digit of 813^(27).

Let’s try another question:

What is the units digit of 24^(1098)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 4.

Remember, our cyclicity of 4 is 4 and 6 (this time, only 2 numbers).

We need the units digit of 24^(1098) – every odd power of 24 will end in 4 and every even power of 24 will end in 6.

Since 1098 is even, the units digit of 24^(1098) is 6.

Not too bad; let’s try something a little harder:

What is the units digit of 75^(25)^5

Note here that you have 75 raised to power 25 which is further raised to the power of 5.

25^5 is not the same as 25*5 – it is 25*25*25*25*25 which is far more complicated. However, the simplifying element of this question is that the last digit of the base 75 is 5, so it doesn’t matter what the positive integer exponent is, the last digit of the expression will always be 5.

Now let’s take a look at a Data Sufficiency question:

Given that x and y are positive integers, what is the units digit of (5*x*y)^(289)?

Statement 1: x is odd.

Statement 2: y is even.

Here there is a new complication – we don’t know what the base is exactly because the base depends on the value of x and y. As such, the real question should be can we figure out the units digit of the base? That is all we need to find the units digit of this expression.

When 5 is multiplied by an even integer, the product ends in 0.

When 5 is multiplied by an odd integer, the product ends in 5.

These are the only two possible cases: The units digit must be either 0 or 5.

With Statement 1, we do not know whether y is odd or even, we only know that x is odd. If y is odd, x*y will be odd. If y is even, x*y will be even. Since we don’t know whether x*y is odd or even, we don’t know whether 5*x*y will end in 5 or 0, so this statement alone is not sufficient.

With Statement 2, if y is even, x*y will certainly be even because an even * any integer will equal an even integer. Therefore, it doesn’t matter whether x is odd or even – regardless, 5*x*y will be even, hence, it will certainly end in 0.

As we know from our patterns of cyclicity, 0 has a cyclicity of 1, i.e. no matter what the positive integer exponent, the units digit will be 0. Therefore, this statement alone is sufficient and the answer is B (Statement 2 alone is sufficient but Statement 1 alone is not sufficient).

Finally, let’s take a question from our own book:

If n and a are positive integers, what is the units digit of n^(4a+2) – n^(8a)?

Statement 1: n = 3

Statement 2: a is odd.

We know that the cyclicity of every digit is either 1, 2 or 4. So to know the units digit of n^{4a+2} – n^{8a}, we need to know the units digit of n. This will tell us what the cyclicity of n is and what the units digit of each expression will be individually.

Statement 1: n = 3

As we know from our patterns of cyclicity, the cyclicity of 3 is 3, 9, 7, 1

Plugging 3 into “n”, n^{4a+2} = 3^{4a+2}

In the exponent, 4a accounts for “a” full cycles of 4, and then a new cycle begins to account for 2.

3, 9, 7, 1

3, 9, 7, 1

3, 9

The units digit here will be 9.

Again, plugging 3 into “n”, n^{8a} = 3^{8a}

8a is a multiple of 4, so there will be full cycles of 4 only. This means the units digit of 3^{8a} will be 1.

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

Plugging these answers back into our equation: n^{4a+2} – n^{8a} = 9 – 1

The units digit of the combined expression will be 9 – 1 = 8.

Therefore, this statement alone is sufficient.

In Statement 2, we are given what the exponents are but not what the value of n, the base, is. Therefore, this statement alone is not sufficient, and our answer is A (Statement 1 alone is sufficient but Statement 2 alone is not sufficient).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit Quarter Wisdom: Cyclicity of Units Digits on the GMAT

Quarter Wit, Quarter WisdomIn our algebra book, we have discussed finding and extrapolating patterns. In this post today, we will look at the patterns we get with various units digits.

The first thing you need to understand is that when we multiply two integers together, the last digit of the result depends only on the last digits of the two integers.

For example:

24 * 12 = 288

Note here: …4 * …2 = …8

So when we are looking at the units digit of the result of an integer raised to a certain exponent, all we need to worry about is the units digit of the integer.

Let’s look at the pattern when the units digit of a number is 2.

Units digit 2:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

2^9 = 512

2^10 = 1024

Note the units digits. Do you see a pattern? 2, 4, 8, 6, 2, 4, 8, 6, 2, 4 … and so on

So what will 2^11 end with? The pattern tells us that two full cycles of 2-4-8-6 will take us to 2^8, and then a new cycle starts at 2^9. 

2-4-8-6

2-4-8-6

2-4

The next digit in the pattern will be 8, which will belong to 2^11. 

In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base. 

So 652^(11) will end in 8,1896782^(11) will end in 8, and so on…

A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits. 

Units digit 3:

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on…

Units digit 4:

4^1 = 4

4^2 = 16

4^3 = 64

4^4 = 256

The pattern here is 4, 6, 4, 6, 4, 6, and so on… 

Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is:

1545^23 = ……..5

1650^19 = ……..0

161^28 = ………1

Hope you get the point.

Units digit 7:

7^1 = 7

7^2 = 49

7^3 = 343

7^4 = ….1 (Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit)

7^5 = ….7 (Now multiply 1 from above by 7 to get 7 as the units digit)

7^6 = ….9

The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on…

Units digit 8:

8^1 = 8

8^2 = 64

8^3 = …2

8^4 = …6

8^5 = …8

8^6 = …4

The pattern here is 8, 4, 2, 6, 8, 4, 2, 6, and so on…

Units digit 9: 

9^1 = 9

9^2 = 81

9^3 = 729

9^4 = …1

The pattern here is 9, 1, 9, 1, 9, 1, and so on…

Summing it all up:

1) Digits 2, 3, 7 and 8 have a cyclicity of 4; i.e. the units digit repeats itself every 4 digits.

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

2) Digits 4 and 9 have a cyclicity of 2; i.e. the units digit repeats itself every 2 digits.

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1 

3) Digits 0, 1, 5 and 6 have a cyclicity of 1.

Cyclicity of 0: 0

Cyclicity of 1: 1

Cyclicity of 5: 5

Cyclicity of 6: 6

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Permutation Involving Sum of Digits

Quarter Wit, Quarter WisdomWe have seen in previous posts how to deal with permutation and combination questions on the GMAT. There is a certain variety of questions that involve getting a bunch of numbers using permutation, and then doing some operations on the numbers we get. The questions can get a little overwhelming considering the sheer magnitude of the number of numbers involved! Let’s take a look at that concept today. We will explain it using an example and then take a question as an exercise:

 

What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 such that each digit is used exactly once in each integer?

First of all, we will use our basic counting principle to find the number of integers that are possible.

The first digit can be chosen in 4 ways. The next one in 3 ways since each digit can be used only once. The next one in 2 ways and there will be only one digit left for the last place.

This gives us a total of 4*3*2*1 = 24 ways of writing such a four digit number. This is what some of the numbers will look like:

1234

1243

1324

1342

2143

4321

Now we need to add these 24 integers to get their sum. Note that since each digit has an equal probability of occupying every place, out of the 24 integers, six integers will have 1 in the units place, six will have 2 in the units place, another six will have 3 in the units place and the rest of the six will have 4 in the units place. The same is true for all places – tens, hundreds and thousands.

Imagine every number written in expanded form such as:

1234 = 1000 + 200 + 30 + 4

2134 = 2000 + 100 + 30 + 4

…etc.

For the 24 numbers, we will get six 1000’s, six 2000’s, six 3000’s and six 4000’s.

In addition, we will get six 100’s, six 200’s, six 300’s and six 400’s.

For the tens place, will get six 10’s, six 20’s, six 30’s and six 40’s.

And finally, in the ones place we will get six 1’s, six 2’s, six 3’s and six 4’s.

Therefore, the total sum will be:

6*1000 + 6*2000 + 6*3000 + 6*4000 + 6*100 + 6*200 + … + 6*3 + 6*4

= 6*1000*(1 + 2 + 3 + 4) + 6*100*(1 + 2 + 3 + 4) + 6*10*(1 + 2 + 3 + 4) + 6*1*(1 + 2 + 3 + 4)

= 6*1000*10 + 6*100*10 + 6*10*10 + 6*10

= 6*10*(1000 + 100 + 10 + 1)

= 1111*6*10

= 66660

Note that finally, there aren’t too many actual calculations, but there is some manipulation involved. Let’s look at a GMAT question using this concept now:

What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A444440

B) 610000

C) 666640

D) 711040

E) 880000

Conceptually, this problem isn’t much different from the previous one.

Using the same basic counting principle to get the number of integers possible, the first digit can be chosen in 4 ways, the next one in 4 ways, the next one in again 4 ways and finally the last digit in 4 ways. This is what some of the numbers will look like:

1111

1112

1121

and so on till 4444.

As such, we will get a total of 4*4*4*4 = 256 different integers.

Now we need to add these 256 integers to get their sum. Since each digit has an equal probability of occupying every place, out of the 256 integers, 64 integers will have 1 in the units place, 64 will have 2 in the units place, another 64 integers will have 3 in the units place and the rest of the 64 integers will have 4 in the units place. The same is true for all places – tens, hundreds and thousands.

Therefore, the total sum will be:

64*1000 + 64*2000 + 64*3000 + 64*4000 + 64*100 + 64*200 + … + 64*3 + 64*4

= 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)

= (64*1 + 64*2 + 64*3 + 64*4) * (1000 + 100 + 10 + 1)

= 64*10*1111

= 711040

So our answer is D.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube and Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: The Tricky Critical Reasoning Conclusion

Quarter Wit, Quarter WisdomAs discussed previously, the most important aspect of a strengthen/weaken question on the GMAT is “identifying the conclusion,” but sometimes, that may not be enough. Even after you identify the conclusion, you must ensure that you have understood it well. Today, we will discuss the “tricky conclusions.”

First let’s take a look at some simple examples:

Conclusion 1: A Causes B.

We can strengthen the conclusion by saying that when A happens, B happens.

We can weaken the conclusion by saying that A happened but B did not happen.

How about a statement which suggests that “C causes B,” or, “B happened but A did not happen”?

Do these affect the conclusion? No, they don’t. The relationship here is that A causes B. Whether there are other factors that cause B too is not our concern, so whether B can happen without A is none of our business.

Conclusion 2: Only A Causes B.

This is an altogether different conclusion. It is apparent that A causes B but the point of contention is whether A is the only cause of B.

Now here, a statement suggesting, “C causes B,” or, “B happened but A did not happen,” does affect our conclusion. These weaken our conclusion – they suggest that A is not the only cause of B.

This distinction can be critical in solving the question. We will now illustrate this point with one of our own GMAT practice questions:

Two types of earthworm, one black and one red-brown, inhabit the woods near the town of Millerton. Because the red-brown worm’s coloring affords it better camouflage from predatory birds, its population in 1980 was approximately five times that of the black worm. In 1990, a factory was built in Millerton and emissions from the factory blackened much of the woods. The population of black earthworms is now almost equal to that of the red-brown earthworm, a result, say local ecologists, solely stemming from the blackening of the woods.

Which of the following, if true, would most strengthen the conclusion of the local ecologists?

(A) The number of red-brown earthworms in the Millerton woods has steadily dropped since the factory began operations.

(B) The birds that prey on earthworms prefer black worms to red-brown worms.

(C) Climate conditions since 1990 have been more favorable to the survival of the red-brown worm than to the black worm.

(D) The average life span of the earthworms has remained the same since the factory began operations.

(E) Since the factory took steps to reduce emissions six months ago, there has been a slight increase in the earthworm population.

Let’s look at the argument.

Premises:

  • There are two types of worms – Red and Black.
  • Red has better camouflage from predatory birds, hence its population was five times that of black.
  • The factory has blackened the woods and now the population of both worms is the same.

Conclusion:

From our premises, we can determine that the blackening of the woods is solely responsible for equalization of the population of the two earthworms.

We need to strengthen this conclusion. Note that there is no doubt that the blackening of the woods is responsible for equalization of populations; the question is whether it is solely responsible.

(A) The number of red-brown earthworms in the Millerton woods has steadily dropped since the factory began operations.

Our conclusion is that only the blackening of the woods caused the numbers to equalize (either black worms are able to hide better or red worms are not able to hide or both), therefore, we need to look for the option that strengthens that there is no other reason. Option A only tells us what the argument does anyway – the population of red worms is decreasing (or black worm population is increasing or both) due to the blackening of the woods. It doesn’t strengthen the claim that only blackening of the woods is responsible.

(B) The birds that prey on earthworms prefer black worms to red-brown worms.

The fact that birds prefer black worms doesn’t necessarily mean that they get to actually eat black worms. Even if we do assume that they do eat black worms over red worms when they can, this strengthens the idea that “the blackening of the woods is responsible for equalization of population,” but does not strengthen the idea that “the blackening of the woods is solely responsible for equalization,” hence, this is not our answer.

(C) Climate conditions since 1990 have been more favorable to the survival of the red-brown worm than to the black worm.

Option C tells us that another factor that could have had an effect on equalization (i.e. climate) is not responsible. This strengthens the conclusion that better camouflage is solely responsible – it doesn’t prove the conclusion beyond doubt, since there could be still another factor that could be responsible, but it does discard one of the other factors. Therefore, it does improve the probability that the conclusion is true.

(D) The average life span of the earthworms has remained the same since the factory began operations.

This option does not distinguish between the two types of earthworms. It just tells us that as a group, the average lifespan of the earthworms has remained the same. Hence, it doesn’t affect our conclusion, which is based on the population of two different earthworms.

(E) Since the factory took steps to reduce emissions six months ago, there has been a slight increase in the earthworm population.

Again, this option does not distinguish between the two types of earthworms. It just tells us that as a group, the earthworm population has increased, so it also does not affect our conclusion, which is based on the population of two different earthworms.

Therefore, our answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube and Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Critical Role of Reading in GMAT Critical Reasoning Questions!

Quarter Wit, Quarter WisdomMost non-native English users have one question: How do I improve my Verbal GMAT score?  There are lots of strategies and techniques we discuss in our books, in our class and on our blog. But one thing that we seriously encourage our students to do (that they need to do on their own) is read more – fiction, non fiction, magazines (mind you, good quality), national dailies, etc. Reading high quality material helps one develop an ear for correct English. It is also important to understand the idiomatic usage of English, which no one can teach in the class. At some time, most of us have thought how silly some things are in the English language, haven’t we?

For example:

Fat chance” and “slim chance’”mean the same thing – Really? Shouldn’t they mean opposite things?

But “wise man” and “wise guy” are opposites – Come on now!

A house burns up as it burns down and you fill in a form by filling it out?

And let’s not even get started on the multiple unrelated meanings many words have – The word on the top of the page, “critical,” could mean “serious” or “important” or “inclined to find fault” depending on the context!

Well, you really must read to understand these nuances or eccentricities, if you may, of the English language. Let’s look at an official question today which many people get wrong just because of the lack of familiarity with the common usage of phrases in English. But before we do that, some quick statistics on this question – 95% students find this question hard and more than half answer it incorrectly. And, on top of that, it is quite hard to convince test takers of the right answer.

Some species of Arctic birds are threatened by recent sharp increases in the population of snow geese, which breed in the Arctic and are displacing birds of less vigorous species. Although snow geese are a popular quarry for hunters in the southern regions where they winter, the hunting season ends if and when hunting has reduced the population by five percent, according to official estimates. Clearly, dropping this restriction would allow the other species to recover.

Which of the following, if true, most seriously undermines the argument?

(A) Hunting limits for snow geese were imposed many years ago in response to a sharp decline in the population of snow geese.

(B) It has been many years since the restriction led to the hunting season for snow geese being closed earlier than the scheduled date.

(C) The number of snow geese taken by hunters each year has grown every year for several years.

(D) As their population has increased, snow geese have recolonized wintering grounds that they had not used for several seasons.

(E) In the snow goose’s winter habitats, the goose faces no significant natural predation.

As usual, let’s start with the question stem – “… most seriously undermines the argument”

This is a weaken question. The golden rule is to focus on the conclusion and try to weaken it.

Let’s first understand the argument:

Snow geese breed in the Arctic and fly south for the winter. They are proliferating, and that is bad for other birds. Southern hunters reduce the number of geese when they fly south. There is a restriction in place that if the population of the geese that came in reduces by 5%, hunting will stop. So if 1000 birds flew south and 50 were hunted, hunting season will be stopped. The argument says that we should drop this restriction to help other Arctic birds flourish (conclusion), then hunters will hunt many more geese and reduce their numbers.

What is the conclusion here? It is: “Clearly, dropping this restriction would allow the other species to recover.”

You have to try to weaken it, i.e. give reasons why even after dropping this restriction, it is unlikely that other species will recover. Even if this restriction of “not hunting after 5%” is dropped and hunters are allowed to hunt as much as they want, the population of geese will still not reduce.

Now, first look at option (B);

(B) It has been many years since the restriction led to the hunting season for snow geese being closed earlier than the scheduled date.

What does this option really mean?

Does it mean the hunting season has been closing earlier than the scheduled date for many years? Or does it mean the exact opposite, that the restriction came into effect many years ago and since then, it has not come into effect.

It might be obvious to the native speakers and to the avid readers, but many non-native test takers actually fumble here and totally ignore option (B) – which, I am sure you have guessed by now, is the correct answer.

The correct meaning is the second one – the restriction has not come into effect for many years now. This means the restriction doesn’t really mean much. For many years, the restriction has not caused the hunting season to close down early because the population of geese hunted is less than 5% of the population flying in. So if the hunting season is from January to June, it has been closing in June, only, so even if hunters hunt for the entire hunting season, they still do not reach the 5% of the population limit (Southern hunters hunt less than 50 birds when 1000 birds fly down South).

Whether you have the restriction or not, the number of geese hunted is the same. So even if you drop the restriction and tell hunters that they can hunt as much as they want, it will not help as they will not want to hunt geese much anyway. This implies that even if the restriction is removed, it is likely that there will be no change in the situation. This definitely weakens our conclusion that dropping the restriction will help other species to recover.

So when people ignore (B), on which option do they zero in? Some fall for (C) but many fall for (D). Let’s look at all other options now:

(A) Hunting limits for snow geese were imposed many years ago in response to a sharp decline in the population of snow geese.

This is out of scope to our argument. It doesn’t really matter when and why the limits were imposed.

(C) The number of snow geese taken by hunters each year has grown every year for several years.

This doesn’t tell us how dropping the restriction would impact the population of geese, it just tells us what has happened in the past – the number of geese hunted has been increasing. If anything, it might strengthen our conclusion if the number of geese hunted is close to 5% of the population. When the population decreases by 5%, if the restriction is dropped, chances are that more geese will be hunted and other species will recover. We have to show that even after dropping the restriction, the other species may not recover.

(D) As their population has increased, snow geese have recolonised wintering grounds that they had not used for several seasons.

With this answer choice, “wintering grounds” implies the southern region (where they fly for winter). In the South, they have recolonised regions they had not occupied for a while now, which just tells you that the population has increased a lot and the geese are spreading. It doesn’t say that removing the restrictions and letting hunters hunt as much as they want will not help. In fact, if anything, it may make the argument a little stronger. If the geese are occupying more southern areas, hunting grounds may become easily accessible to more hunters and dropping hunting restrictions may actually help more!

(E) In the snow goose’s winter habitats, the goose faces no significant natural predation.

We are concerned about the effect of hunting, thus natural predation is out of scope.

Therefore, our answer is (B).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube and Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Complicated GMAT Work-Rate Questions Made Easy!

Quarter Wit, Quarter WisdomToday, we will take up a gem of a work-rate question from our own curriculum. Its basics lie in a post on joint variation that we discussed many weeks ago. Here is a quick recap of the actual methodology:

If 10 workers complete a work in 5 days working 8 hours a day, how much work will be done by 6 workers in 10 days working 2 hours a day?

Here is what it looks like:

 

10 workers……………..5 days …………….. 8 hours ……………1 work

6 workers………………10 days …………… 2 hours …………… ? work

We need to find the amount of work done, so we start with the work done in the first case and then multiply it by the respective ratios:

Work done = 1 * (6/10) * (10/5) * (2/8) = 3/10

We multiply by 6/10 because number of men decreases from 10 to 6. The work done will reduce, so we multiply by 6/10 (the fraction less than 1).

We also multiply by 10/5 because number of days increases from 5 to 10. Because of this, the work done will increase, so we multiply by 10/5 (the fraction more than 1).

We also multiply by 2/8 because number of hours decreases from 8 to 2. Because of this, the work done will decrease, hence, we multiply by 2/8 (the fraction less than 1).

So the process is super simple – start with what you need to find out, say x, and multiply it by the ratio of each thing that changes from A to B. Whether you multiply by A/B or B/A depends on whether with this change increases or reduces x. If x increases, you will multiply by the fraction that is greater than 1, however if x decreases, you will multiply by the fraction that is less than 1.

On this same concept, let’s look at the question:

16 horses can haul a load of lumber in 24 minutes. 12 horses started hauling a load and after 14 minutes, 12 mules joined the horses. Will it take less than a quarter-hour for all of them together to finish hauling the load?

Statement 1: Mules work more slowly than horses.

Statement 2: 48 mules can haul the same load of lumber in 16 minutes.

Let’s see what data we have in the question stem:

16 horses …….. 24 mins ………. 1 work

12 horses …….. 14 mins ………. ? work

Work done = 1*(14/24)*(12/16) = (7/16)th of the work

We multiply by 14/24 because if the time taken to do the work decreases, the work done will also decrease. 14/24 is less than 1 so it will decrease the work done.

We also multiply by 12/16 because if the number of horses decreases, the work done will also decrease. 12/16 is less than 1 so it will decrease the work done.

All in all, we now know that 12 horses complete 7/16th of the work in 14 mins. So there is still 1 – 7/16 = 9/16 of the work left to do.

Now let’s review the two statements.

Statement 1: Mules work more slowly than horses.

This statement doesn’t give us any figures, so how can we analyse it mathematically? What we can do is find the range in which the time taken by all the horses and mules together will lie according to this statement.

Case 1: When mules work at a rate that is infinitesimally smaller than the rate of horses.

In this case, 12 mules are equivalent to 12 horses. So we have a total of 12 + 12 = 24 horses working together to complete (9/16)th of the work.

16 horses …….. 24 mins ………. 1 work

24 horses ……… ? mins ………. 9/16 work

Time taken = 24*(16/24)*(9/16) = 9 mins

Since the mules are slower than the horses, the time taken to complete the work will be more than 9 minutes. How much more than 9 minutes, we do not know. Now look at the flip side:

Case 2: When the mules work at a rate close to 0.

If the mules work slower, time taken will be more till the point when mules work so slowly that they do almost no work.

16 horses …….. 24 mins ………. 1 work

12 horses ……… ? mins ………. 9/16 work

Time taken = 24*(16/12)*(9/16) = 18 minutes

Therefore, depending on how fast/slow the mules are, the time taken to do the rest of the work could be anywhere from 9 minutes to 18 minutes. Therefore the time taken could be either less or more than 15 minutes – this statement alone is not sufficient.

Statement 2: 48 mules can haul the same load of lumber in 16 minutes.

We now know exactly how fast the mules are, so this must be sufficient to say whether the time taken to do the rest of the work was less or more than 15 minutes – we don’t need to actually find the time taken here – therefore, the answer is B, Statement 2 alone is sufficient.

However, if you would like to find out for practice, just find the equivalence between the horses and the mules first.

To haul the load in 16 minutes, we need 48 mules

To haul the load in 24 minutes, we need 48 * (16/24) = 32 mules

So 32 mules are equivalent to 16 horses (because 16 horses haul the load in 24 minutes). This means that 2 mules are equivalent to 1 horse, and 12 mules are, therefore, equivalent to 6 horses.

So now, in effect we have a total of 12 + 6 = 18 horses, and the situation now becomes this:

16 horses …….. 24 mins ………. 1 work

18 horses ……… ? mins ………. 9/16 work

Time taken = 24*(16/18)*(9/16) = 12 minute – less than a quarter-hour to finish the work.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube and Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

7 Formulas for Tackling Three Overlapping Sets on the GMAT

Quarter Wit, Quarter WisdomIn a previous post, we saw how to solve three overlapping sets questions using venn diagrams. Today, we will look at all of the various formulas floating around on three overlapping sets. Most of these are self explanatory but we will look into the details of some of them.

 

 

 

There are two basic formulas that we already know:

1) Total = n(No Set) + n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets)

2) Total = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) + n(No Set)

From these two formulas, we can derive all others.

n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets) gives us n(At least one set). So we get:

3) Total = n(No Set) + n(At least one set)

From (3), we get n(At least one set) = Total – n(No Set)

Plugging this into (2), we then get:

4) n(At least one set) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)

Now let’s see how we can calculate the number of people in exactly two sets. There is a reason we jumped to n(Exactly two sets) instead of following the more logical next step of figuring out n(At least two sets) – it will be more intuitive to get n(At least two sets) after we find n(Exactly two sets).

n(A and B) includes people who are in both A and B and it also includes people who are in A, B and C. Because of this, we should remove n(A and B and C) from n(A and B) to get n(A and B only). Similarly, you get n(B and C only) and n(C and A only), so adding all these three will give us number of people in exactly 2 sets.

n(Exactly two sets) = n(A and B) – n(A and B and C) + n(B and C) – n(A and B and C) + n(C and A) – n(A and B and C). Therefore:

5) n(Exactly two sets) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)

Now we can easily get n(At least two sets):

6) n(At least two sets) = n(A and B) + n(B and C) + n(C and A) – 2*n(A and B and C)

This is just n(A and B and C) more than n(Exactly two sets). That makes sense, doesn’t it? Here, you include the people who are in all three sets once and n(Exactly two sets) converts to n(At least two sets)!

Now, we go on to find n(Exactly one set). From n(At least one set), let’s subtract n(At least two sets); i.e. we subtract (6) from (4)

n(Exactly one set) = n(At least one set) – n(At least two sets), therefore:

7) n(Exactly one set) = n(A) + n(B) + n(C) – 2*n(A and B) – 2*n(B and C) – 2*n(C and A) + 3*n(A and B and C)

You don’t need to learn all these formulas. Just focus on first two and know how you can arrive at the others if required. Let’s try this in an example problem:

Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185

(B) 180

(C) 175

(D) 190

(E) 195

You are given that:

n(At least one channel) = 250

n(Exactly two channels) = 50

So we know that n(At least one channel) = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels) = 250

250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let’s find the value of n(Exactly 3 channels) = x

We also know that n(At least one channel) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) = 250

Also, n(Exactly two channels) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)

So n(A and B) + n(B and C) + n(C and A) = n(Exactly two channels) + 3*n(A and B and C)

Plugging this into the equation above:

250 = n(A) + n(B) + n(C) – n(Exactly two channels) – 3*x + x

250 = 116 + 127 + 107 – 50 – 2x

x = 25

250 = n(Exactly 1 channel) + 50 + 25

n(Exactly 1 channel) = 175, so your answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube and Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Manipulating Standard Formulas on the GMAT

Quarter Wit, Quarter WisdomWe know the formula we need to use to find the sum of n consecutive positive integers starting from 1. The formula is given as n(n+1)/2.

So the sum of first four positive integers is 4 * (4 + 1)/2 = 10.

This might seem a bit cumbersome, since it is easy to see that 1 + 2 + 3 + 4 = 10, but we know that the formula comes in very handy when n is a large number. For example, the sum of first 50 positive integers = 50 * 51/2 = 1275. Obviously, this will be a lot harder when done the “1 + 2 + 3 + 4 … + 49 + 50” way.

Now the question is, how do we adjust the same formula to find the sum of consecutive integers which do not start from 1?

Say, how do we find the sum of all positive integers from 8 to 20? The formula assumes a starting point of 1, so then we insert only the last number, n. How do we manage the 8? Let’s try to figure it out

Say the sum of first 20 positive integers = 1 + 2 + 3 + 4 + …. + 19 + 20 = 20 * 21/2

(1 + 2 + 3 +… + 7) + (8 + 9 +10 + … + 19 + 20) = 20 * 21/2

We need the value of the part in red, let’s call it the required sum.

(1 + 2 + 3 +… + 7) + The Required Sum = 20 * 21/2

Note here that we know the sum of 1 + 2 + 3 + … + 7 = 7 * 8/2

So, 7*8/2 + The Required Sum = 20 * 21/2, therefore the Required Sum = 20*21/2 – 7*8/2

To get the sum of consecutive integers from 8 to 20, we find the sum of all integers from 1 to 20 (using the formula we know) and subtract the sum of integers from 1 to 7 out of it (using the same formula).

To generalize, the sum of all positive integers from m to n is given as:

n(n+1)/2 – (m-1)*m/2

Let’s look at a question based on this concept:

If the sum of the consecutive integers from –40 to n inclusive is 356, what is the value of n?

(A) 47

(B) 48

(C) 49

(D) 50

(E) 51

If you are thinking that we haven’t gone over how to adjust the formula for negative numbers, you are right, but what we have discussed is enough to solve this question.

Numbers around 0 are symmetrical. So 1 and -1 add up to equal 0. Similarly, 2 and -2 add up to equal 0, and so on…

-40, -39 … 0 … 39, 40, 41, 42, 43, 44, 45 …

The sum of all numbers from -40 to 40 will be 0. Or another way to look at it is that 0 is the mean of all numbers from -40 to 40. So the total sum of these numbers will also be 0.

The given sum is actually the sum of numbers from 41 to n only.

We know how to calculate that:

n(n+1)/2 – 40*41/2 = 356

n(n+1) = 2352

From the options, we see that n cannot be 49 or 50 because the product of 49*50 or 50*51 will end in 0, so plug in n = 48 to check whether 48*49 is equal to 2352. It is, therefore our answer is B

(Had we obtained a lower product than required, we could have said that n must be 51. Had we obtained a higher product than was required, we could have said that n is 47.)

Another method:

Use the concept of arithmetic mean and ballpark. The mean of numbers from 41 to 47 or 48 or 49… will be somewhere between 44 and 46.

Let’s estimate the number of integers we need to get the sum of about 356. Each additional integer is quite large (more than 45) therefore, a difference of about 10-15 in the sum due to the various possible values of the mean will be immaterial.

45*7 = 315

45*8 = 360

This brings us very close to the value of 356.

Assuming there are 8 integers, their values will be from 41 to 48. The average of these 8 numbers will be 44.5. The total sum will be 44.5 * 8 = 356. It matches, so our answer is still B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on FacebookYouTube and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How Understanding Sampling Can Help You Conquer the GMAT

Quarter Wit, Quarter WisdomToday, we will discuss the concept of sampling. People with a statistics background will be very comfortable with it, but if you have not studied statistics, a little bit of knowledge will be helpful. You are not required to know this for the GMAT, however there could be questions framed on the sampling premise, and you will be far more comfortable solving them with some understanding in place. A sample is a selection made from a larger group (the “population”) which helps you examine certain characteristics of the larger group using limited resources.

For example:

In a large population, say all the people in a state, it is difficult to find the number of people with a certain trait, such as red hair. So you pick up 100 people at random (from different families, different areas, different backgrounds) and find the number of people who have red hair in this selection of 100.

Let’s say 12 have red hair. You can then generalize that approximately 12% of the whole population has red hair. The more unbiased your sample, the better the approximation.

In this example, you found something about the entire population (12% has red hair) based on a small sample and hence, using few resources. To find the actual percentage of people who have red hair in the entire population, you would need far more effort, time and money. Usually the use of fewer resources justifies the use of sampling even though it comes with some error.

So that is a bit of background on sampling. It will help you make sense of the  official question given below:

In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate number of fish in the pond?

A) 400

B) 625

C) 1,250

D) 2,500

E) 10,000

This is what took place: From a pond, 50 fish were caught, tagged and returned to the pond. Then 50 were caught again and 2 of those were found to be tagged.

Why was this done?

The total number of fish in the pond is the population of the pond. It is unknown. Since counting the total number of fish in the pond was hard, they tagged 50 of them and let them disperse evenly in the population. This means they gave a certain trait to a known number of fish in the pond – they tagged 50 fish.

Then they caught 50 fish again and these fish became the sample. Out of these 50, 2 were found to be tagged. So 2 of the 50 fish caught were found to have the trait given (tagged) – 4% of our sample was tagged.

The question tells us that “… the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond …” that is, the question tells us that the sample is representative of the population. This implies that 50 (the number of fish we tagged) is 4% of the entire fish population of the pond.

50 = 4% of Total Fish Population, therefore, we can calculate that the Total Fish Population = 50 * 100/4 = 1250. Our answer is then C.

Using sampling, we were able to calculate the total population of the pond without actually counting each fish. For increased accuracy, often the exercise of taking samples is repeated many times and then some kind of average is used to get the best approximation.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on FacebookYouTube and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Exponent Properties for the GMAT

Quarter Wit, Quarter WisdomToday, let’s discuss the relative placements of exponents on the number line.

We know what the graph of 2^x looks like:

 

 

 

 

Graph of 2^x

 

 

 

 

 

 

 

It shows that when x is positive, with increasing value of x, 2^x increases very quickly (look at the first quadrant), but we don’t know exactly how it increases.

It also shows that when x is negative, 2^x stays very close to 0. As x decreases, the value of 2^x decreases by a very small amount.

Now note the spacing of the powers of 2 on the number line:

Number line jpg

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

and so on…

2^1 = 2 * 2^0 = 2^0 + 2^0

2^2 = 2 * 2^1 = 2^1 + 2^1

2^3 = 2 * 2^2 = 2^2 + 2^2

2^4 = 2 * 2^3 = 2^3 + 2^3

So every power of 2 is equidistant from 0 and the next power. This means that a power of 2 would be much closer to 0 than the next higher powers. For example, 2^2 is at the same distance from 0 as it is from 2^3.

But 2^2 is much closer to 0 than it is to 2^4, 2^5 etc.

Let’s look at a question based on this concept. Most people find it a bit tough if they do not understand this concept:

Given that x = 2^b – (8^30 + 16^5), which of the following values for b yields the lowest value for |x|?

A) 35

B) 90

C) 91

D) 95

E) 105

We need the lowest value of |x|. We know that the smallest value any absolute value function can take is 0. So 2^b should be as close as possible to (8^30 + 16^5) to get the lowest value of |x|.

Let’s try to simplify:

(8^30 + 16^5)

= (2^3)^30 + (2^4)^5

= 2^90 + 2^20

Which value should b take such that 2^b is as close as possible to 2^90 + 2^20?

2^90 + 2^20 is obviously larger than 2^90. But is it closer to 2^90 or 2^91 or higher powers of 2?

Let’s use the concept we have learned today – let’s compare 2^90 + 2^20 with 2^90 and 2^91.

2^90 = 2^90 + 0

2^91 = 2^90 + 2^90

So now if we compare these two with 2^90 + 2^20, we need to know whether 2^20 is closer to 0 or closer to 2^90.

We already know that 2^20 is equidistant from 0 and 2^21, so obviously it will be much closer to 0 than it will be to 2^90.

Hence, 2^90 + 2^20 is much closer to 2^90 than it is to 2^91 or any other higher powers.

We should take the value 90 to minimize |x|, therefore the answer is B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on FacebookYouTube and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Know the Concept of Cost Price for the GMAT

Quarter Wit, Quarter WisdomMost of us are quite comfortable with the concepts of percentages, cost price and sale price, but when we come across a toughie from these topics, we feel lost. Then we go back to the theory but there seems to be nothing new there – nothing new that could potentially help us tackle such questions with ease in the future. The point is, the basic theory of these topics is quite simple – there isn’t anything else to it – but it’s application to GMAT questions is an altogether different deal. There are small but critical things that you need to keep in mind, one of which we will discuss today: what is the cost price?

Let’s take a look at this with an official question:

A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer’s initial cost. What was the dealer’s approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

(A) 7% loss

(B) 13% loss

(C) 7% profit

(D) 13% profit

(E) 14% profit

Solution:

Here are the various data points:

  • 60 cameras bought at 20% markup.
  • Selling Price = $250
  • 6 not sold and 50% of initial cost refunded
  • Profit/Loss = ?

Now look at the solution:

The cost price per camera = 250/1.2 = 1250/6

The total cost price = (1250/6)*60 = $12,500

50% of the cost of 6 cameras was returned.

The cost price of 6 cameras = (1250/6)*6 = $1250

50% of this = 1250/2 = $625

This means the effective cost price = 12,500 – 625 = $11,875

If the selling price per camera = $250, the total selling price = 54 * 250 = $13500 (only 54 cameras were sold)

Hence, the profit % = [(13500 – 11875) / 11875] x 100 = (1625/11875) x 100 = 13.684%

This gives us approximately 14% as the answer (rounding up). But that is not correct. Before you move ahead, try to figure out the problem with this solution. If you are able to, it means you do understand this topic very well.

Here is the problem with the solution:

The cost price is the total initial cost price. You cannot subtract the refund out of it. The refund is effectively the price at which the 6 cameras were sold. You cannot cancel off your cost price with your sale price and have a smaller cost price. Your initial investment in the transaction is your cost price. When you reduce it by cancelling off some sale price (or refund), you are artificially increasing your profit percentage.

Say, we buy a few thing for $100. While selling them off, we get $50 for half of them. We reduce our cost price by $50 and get $50 as cost price. For the other half, we sell them for $60. We say that $50 is out cost price and $60 is our selling price. The profit we made is $10, which is fine. The issue is that our profit percentage is not (10/50) * 100 = 20%. Rather, our profit percentage will be (10/100) * 100 = 10% only, so $100 would be our actual cost price.

Keeping this in mind, here is the correct algebra solution:

The total cost price = (1250/6)*60 = $12,500

The total selling price = 54 * 250 + $625 = $13,500 + $625 = $14,125 (60 cameras were sold, 54 at $250 each and 6 at 50% of cost price)

The profit = 14,125 – 12,500 = $1625 (same as before)

The profit percentage = (1625/12,500) * 100 = 13%

Therefore, the answer is (D).

Obviously, we can always use our trusted weighted averages formula here for a quick and efficient solution:

Weighted Averages

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. The ratio of the cost price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Average Profit/Loss percentage = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook, YouTube and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Averages Concepts for the GMAT

Quarter Wit, Quarter WisdomLet’s discuss an advanced averages concept today.

Say, you have the following set of consecutive integers: 2, 3, 4, 5, 6, 7, 8

What is the average of this set? There are 7 consecutive integers here and the average is 5, the middle number.

Say the set is changed to: 2, 3, 4, 5, 6, 7, 8, 9 (another consecutive number is added to the extreme right). Now what is the average? It is the average of the two middle numbers (5+6)/2 = 5.5.

Let’s edit the set one more time: 1, 2, 3, 4, 5, 6, 7, 8, 9 (another consecutive number is added to the extreme left). The average now is 5 again.

Whenever you add a number on either side of a set of consecutive integers, the average changes by 0.5. This is obvious because odd number of consecutive integers have the middle number as the average and an even number of consecutive integers have the average of two middle numbers as the average. Since every time you add an integer, the number of integers changes from odd to even or from even to odd, the average changes by 0.5.

By the same logic, what happens when you remove an integer from either extreme?

Given a set 3, 4, 5, 6, 7, 8, 9, how will its average change if you remove 3?

The average of 3, 4, 5, 6, 7, 8, 9 is 6, and the average of 4, 5, 6, 7, 8, 9 is 6.5 — the average increases to 6.5 because you removed a small number.

Now how will the average change if you remove 9 instead of 3?

The average of 3, 4, 5, 6, 7, 8, 9 is 6, and the average of 3, 4, 5, 6, 7, 8 is 5.5 — here, the average decreases to 5.5 because you removed a large number.

So, every time you add or remove a number from one of the extremes, the average will move by 0.5.

What happens if you remove a number from somewhere in the middle?

The average changes but by how much? When you remove the greatest or the least number, the average changes by 0.5. So when you remove some other number, the average will change by something less than 0.5. For example, from the set 3, 4, 5, 6, 7, 8, 9, if you remove 8, the average changes from 6 to 5.667. If instead, you remove 7, the average changes to 5.833.

A few takeaways:

  1. When you remove an integer very close to the average, the average changes by very little. If you remove the average, the average doesn’t change (changes by 0). When you remove a number close to the extreme, the average changes by a larger number (up to a maximum of 0.5).
  2. When you remove a number less than the average, the average increases. When you remove a number more than the average, the average decreases.
  3. When you remove the smallest number, the average increases by 0.5. When you remove the greatest number, the average decreases by 0.5.

Now, a question based on this concept:

In a class, the teacher wrote a set of consecutive integers beginning with 1 on the blackboard. A student erased one number. The average of the remaining numbers was 29(14/19). What was the number that the student erased?

(A) 13

(B) 16

(C) 28

(D) 36

(E) 50

Solution:

The numbers on the board: 1, 2, 3, 4, …

The new average is 29(14/19). Since the average changes by not more than 0.5 when you remove an integer from a set of consecutive integers, the original average was either 29.5 or 30. So originally there were either 58 numbers (average 29.5) or 59 numbers (average 30).

When you remove a number, you are left with either 57 numbers or with 58 numbers. Now, the new average will tell you whether you are left with 57 numbers or 58 numbers. The denominator is 19 in the fraction, so when you divide the sum of all remaining integers by the number of integers, the number of integers (denominator) is 19 or a multiple of 19 — 57 is a multiple of 19, 58 is not. So you must have been left with 57 integers and the original number of integers must be 58. This means the original average must have been 29.5.

The original average of 29(1/2) increases to 29(14/19), i.e. an increase of 14/19 – 1/2 = 9/38.

When an integer was removed, the average increased by 9/38 so the integer must be less than the original average. Now use the concept of average that we have learned. One integer was bringing the rest of the numbers down by 9/38 each so the integer must have been (9/38)*57 = 13.5, which is less than the original average of 29.5.

This means the integer that was removed must have been (29.5 – 13.5) = 16, so the answer is B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook, YouTube and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

A Closer Look at Set and Ratio GMAT Quant Questions

Quarter Wit, Quarter WisdomWriting this post on Teacher’s Day made me dedicate this post to questions on teachers! Considering that all GMAT questions are written by teachers, oddly enough, I found very few questions actually involving them. Looks like we are a humble bunch! Today, we will discuss two GMAT Quant questions on two different topics of discussion – sets and ratios. Both questions are official and of higher difficulty.

Question 1: Of the 1400 college teachers surveyed, 42% said they considered engaging in research an essential goal. How many of the college teacher surveyed were women?

Statement I: In the survey 36% of men and 50% of women said that they consider engaging in research activity an essential goal.

Statement II: In the survey 288 men said that they consider engaging in research activity an essential goal.

Solution:

On reading the question stem we realise that this question involves two variables:

Research Essential – Not Essential

Men – Women

This should immediately make us think about a matrix. Not that we cannot solve the question without one, but you know that I am a huge proponent of visual approaches.

We are given that 42% of total teachers (1400) considered research essential. So this means that 58% did not consider it essential. No need to actually calculate the number right now, let’s wait and see what else we know (anyway, we love to procrastinate calculations in Data Sufficiency questions).

Statement I: In the survey 36% of men and 50% of women said that they consider engaging in research activity an essential goal.

Say the number of women is W. We need the value of W. The number of men must be ‘Total – W’ = 1400 – W. 36% of men and 50% of women consider research essential. Knowing this, we see that we get:

TeachersDay

36% * (1400 – W) + 50% * W = 42% * 1400

This is a linear equation in W so we can solve it to get the value of W. Therefore, this statement alone is sufficient.

Statement II: In the survey 288 men said that they consider engaging in research activity an essential goal.

This statement doesn’t tell us the number of women who consider research essential, so it is not sufficient alone, therefore the answer is A, Statement I alone is sufficient but Statement II is not.

Question 2: If the ratio of the number of teachers to the number of students is the same in School District A and School District B, what is the ratio of the number of students in School District A to the number of students in School District B?

Statement I: There are 10,000 more students in School District A than there are in School District B.

Statement II: The ratio of the number of teachers to the number of students in School District A is 1 to 20.

Solution:

In both schools, the ratio of the number of teachers : the number of students is the same.

Statement I: There are 10,000 more students in School District A than there are in School District B.

We don’t know the number of students in either school district, so it is not informative enough to know that School District A has 10,000 more students. Therefore, this statement alone is not sufficient.

Statement II: The ratio of the number of teachers to the number of students in School District A is 1 to 20.

With this statement, we know that the ratio of the number of teachers : the number of students in School District A = 1:20.

Say the number of teachers in A = a; the number of students in A = 20a. We also know the ratio of the number of teachers : the number of students in School District B = 1:20.

Say the number of teachers in B = b; the number of students in B = 20b. Mind you, we don’t know the value of a and b. All we know is that the teacher student ratio is 1:20 in both.

The ratio of the number of students in A: the number of students in B = 20a : 20b = a:b. With this ratio, we don’t know a:b (even using both statements, we just know that a – b = 10,000). Therefore, the answer is E, Statements 1 and 2 together are not sufficient.

Were you able to solve both questions effortlessly? No? Don’t worry, that’s what we are here for! (Ignore the preposition at the end. It sounds most natural this way.)

Not so humble anymore, eh? :)

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook, YouTube and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Compare Effectively During the GMAT

Quarter Wit, Quarter WisdomA lot of GMAT test takers complain about insufficient time. This is understandable as far as the Verbal section is concerned. We all have different reading speeds and that itself accounts for a lot of time issues in the Verbal section. Obviously then there are other factors – your comfort with the language, your comprehension skills, your conceptual understanding of the Verbal question types, etc.

However, timing issues should not arise in the Quant section. Your reading speed has very little effect on the overall timing scheme because most of the time during the Quant section is spent in solving the question. So if you are falling short on time, it means the methods you are using are not appropriate. We have said it before and will say it again – most GMAT Quant questions can be done in under one minute if you just look for the right thing.

For example, of the four listed numbers below, which number is the greatest and which is the least?

2/3

2^2/3^2

2^3/3^3

Sqrt(2)/Sqrt(3)

Now, how much time you take to solve this depends on how you approach this problem. If you get into ugly calculations, you will end up wasting a ton of time.

2/3 = .667

2^2/3^2 = 4/9 = .444

2^3/3^3 = 8/27 = .296

Sqrt(2)/Sqrt(3) = 1.414/1.732 = .816

So we know that the greatest is Sqrt(2)/Sqrt(3) and the least is 2^3/3^3. We got the answer but we wasted at least 2-3 mins in getting it.

We can do the same thing very quickly. We know that the squares/cubes/roots etc of numbers vary according to where the numbers lie on the number line.

2/3 lies in between 0 and 1, as does 1/4.

The Sqrt(1/4) = 1/2, which is greater than 1/4, so we know that the Sqrt(2/3) will be greater than 2/3 as well.

Also, the square and cube of 1/4 is less than 1/4, so the square and cube of 2/3 will also be less than 2/3. So the comparison will look like this:

(2/3)^3 < (2/3)^2 < 2/3 < Sqrt(2/3)

That is all you need to do! We arrived at the same answer using less than 30 secs.

Using this technique, let’s solve a question:

Which of the following represents the greatest value?

(A) Sqrt(3)/Sqrt(5) + Sqrt(5)/Sqrt(7) + Sqrt(7)/Sqrt(9)

(B) 3/5 + 5/7 + 7/9

(C) 3^2/5^2 + 5^2/7^2 + 7^2/9^2

(D) 3^3/5^3 + 5^3/7^3 + 7^3/9^3

(E) 3/5 + 1 – 5/7 + 7/9

Such a question can baffle someone who believes in calculating everything. We know better than that!

Note that the base values in all the options are 3/5, 5/7 and 7/9. This should hint that we need to compare term to term and not the entire expressions. Also, all values lie between 0 and 1 so they will behave the same way.

Sqrt(3)/Sqrt(5) is the same as Sqrt(3/5). The square root of a number between 0 and 1 is greater than the number itself.

3^2/5^2 is the same as (3/5)^2. The square (and cube) of a number between 0 and 1 is less than the number itself.

So, the comparison will look like this:

(3/5)^3 < (3/5)^2 < 3/5 < Sqrt(3/5)

(5/7)^3 < (5/7)^2 < 5/7 < Sqrt(5/7)

(7/9)^3 < (7/9)^2 < 7/9 < Sqrt(7/9)

This means that out of (A), (B), (C) and (D), the greatest one is (A).

Now we just need to analyse (E) and compare it with (B).

The first term is the same, 3/5.

The last term is the same, 7/9.

The only difference is that (B) has 5/7 in the middle and (E) has 1 – 5/7 = 2/7 in the middle. So (E) is certainly less than (B).

We already know that (A) is greater than (B), so we can say that (A) must be the greatest value.

A quick recap of important number properties:

Case 1: N > 1

N^2, N^3, etc. will be greater than N.

The Sqrt(N) and the CubeRoot(N) will be less than N.

The relation will look like this:

… CubeRoot(N) < Sqrt(N) < N < N^2 < N^3 …

Case II: 0 < N < 1

N^2, N^3 etc will be less than N.

The Sqrt(N) and the CubeRoot(N) will be greater than N.

The relation will look like this:

… N^3 < N^2 < N < Sqrt(N) < CubeRoot(N)  …

Case III: -1 < N < 0

Even powers will be greater than N and positive; Odd powers will be greater than N but negative.

The square root will not be defined, and the cube root of N will be less than N.

CubeRoot(N) < N < N^3 < 0 < N^2

Case IV: N < -1

Even powers will be greater than N and positive; Odd powers will be less than N.

The square root will not be defined, and the cube root of N will be greater than N.

N^3 < N < CubeRoot(N) < 0 < N^2

Note that you don’t need to actually remember these relations, just take a value in each range and you will know how all the numbers in that range behave.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Finding the Product of Factors on GMAT Questions

Quarter Wit, Quarter WisdomWe have discussed how to find the factors of a number and their properties in these two posts:

Writing Factors of an Ugly Number

Factors of Perfect Squares

Today let’s discuss the concept of ‘product of the factors of a number’.

From the two posts above, we know that the factors equidistant from the centre multiply to give the number. We also know that the behaviour is a little different for perfect squares. Let’s take two examples to understand this.

Example 1: Say N = 6

Factors of 6 are 1, 2, 3, 6

1*6 = 6 (first factor * last factor)

2*3 = 6 (second factor and second last factor)

Product of the four factors of 6 is given by 1*6 * 2*3 = 6*6 = 6^2 = [Sqrt(N)]^4

Example 2: Say N = 25 (a perfect square)

Factors of 25 are 1, 5, 25

1*25 = 25 (first factor * last factor)

5*5 = 25 (middle factor multiplied by itself)

Product of the three factors of 25 is given by 1*25 * 5 = 5^3 = [Sqrt(N)]^3

If a number, N, can be expressed as: 2^a * 3^b * 5^c *…

The total number of factors f = (a+1)*(b+1)*(c+1)…

The product of all factors of N is given by [Sqrt(N)]^f i.e. N^(f/2)

Let’s look at a couple of questions based on this principle:

Question 1: If the product of all the factors of a positive integer, N, is

2^(18) * 3^(12), how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(18) * 3^(12)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(18) * 3^(12)

a*(a+1)*(b+1)/2 = 18

b*(a+1)*(b+1)/2 = 12

Dividing the two equations, we get a/b = 3/2

Smallest values: a = 3, b = 2. It satisfies our two equations.

Can we have more values for a and b? Can a = 6 and b = 4? No. Then the product a*(a+1)*(b+1)/2 would be much larger than 18.

So N = 2^3 * 3^2

There is only one such value of N.

Answer (B)

Question 2: If the product of all the factors of a positive integer, N, is 2^9 * 3^9, how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(9) * 3^(9)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(9) * 3^(9)

a*(a+1)*(b+1)/2 = 9

b*(a+1)*(b+1)/2 = 9

Dividing the two equations, we get a/b = 1/1

Smallest values: a = 1, b = 1 – Does not satisfy our equation

Next set of values: a = 2, b = 2 – Satisfies our equations

All larger values will not satisfy our equations.

Answer (B)

Note that we can easily use hit and trial in these questions without actually working through the equations.

This is how we will do it:

N^(f/2) = 2^(18) * 3^(12)

Case 1: Assume values of f/2 from common factors of 18 and 12 – say 2

[2^9 * 3^6]^2

Can f/2 = 2 i.e. can f = 4?

If N = 2^9 * 3^6, total number of factors f = (9+1)*(6+1) = 70

This doesn’t work.

Case 2: Assume f/2 is 6

[2^3 * 3^2]^6

Can f/2 = 6 i.e. can f = 12?

If N = 2^3 * 3^2, total number of factors f = (3+1)*(2+1) = 12

This works.

The reason hit and trial isn’t a bad idea is that there will be only one such set of values. If we can quickly find it, we are done.

Why should we then bother to find it at all. Shouldn’t we just answer with option ‘B’ in both cases? Think of a case in which the product of all factors is given as 2^(16) * 3^(14). Will there be any value of N in such a case?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure tofind us on Facebook and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!