The post Quarter Wit, Quarter Wisdom: Squares and Square Roots on the GMAT appeared first on Veritas Prep Blog.

]]>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 Facebook, YouTube, 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!*

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]]>The post Quarter Wit, Quarter Wisdom: Why Critical Reasoning Needs Your Complete Attention on the GMAT! appeared first on Veritas Prep Blog.

]]>*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:

- The beetles cannot maintain their pace and must pause for a moment’s rest.
- 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 Facebook, YouTube, 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 post Quarter Wit, Quarter Wisdom: Why Critical Reasoning Needs Your Complete Attention on the GMAT! appeared first on Veritas Prep Blog.

]]>The post Quarter Wit, Quarter Wisdom: Using Visual Symmetry to Solve GMAT Probability Problems appeared first on Veritas Prep Blog.

]]>*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?*

*(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 Facebook, YouTube, 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!*

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]]>The post Quarter Wit, Quarter Wisdom: Using the Standard Deviation Formula on the GMAT appeared first on Veritas Prep Blog.

]]>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.

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]]>The post Quarter Wit, Quarter Wisdom: An Innovative Use of the Slope of a Line on the GMAT appeared first on Veritas Prep Blog.

]]>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.

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]]>The post Coordinate Geometry: Solving GMAT Problems With Lines Crossing Either the X-Axis or the Y-Axis appeared first on Veritas Prep Blog.

]]>*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.

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:

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.

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]]>The post Quarter Wit, Quarter Wisdom: Be Tolerant Towards Pronoun Ambiguity on the GMAT appeared first on Veritas Prep Blog.

]]>**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.

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The post Quarter Wit, Quarter Wisdom: Be Tolerant Towards Pronoun Ambiguity on the GMAT appeared first on Veritas Prep Blog.

]]>The post Quarter Wit, Quarter Wisdom: Using a Venn Diagram vs. a Double Set Matrix on the GMAT appeared first on Veritas Prep Blog.

]]>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…*”

“…*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.

“*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.

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:

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!

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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]]>The post What Makes GMAT Quant Questions So Hard? appeared first on Veritas Prep Blog.

]]>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!

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]]>The post Quarter Wit, Quarter Wisdom: Dealing with Tangents on the GMAT appeared first on Veritas Prep Blog.

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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:

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!

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The post Quarter Wit, Quarter Wisdom: Dealing with Tangents on the GMAT appeared first on Veritas Prep Blog.

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