# The Holistic Approach to Absolute Values – Part V

We will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)

Let’s look at the following GMAT question:

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

In this question, we are given the inequality |x – 6| > 3*|x + 2|

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

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!

# The Holistic Approach to Absolute Values – Part IV

Last week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

• |x – 3| will be the distance covered by the friend at 3 to reach x.
• |x + 1| will be the distance covered by the friend at -1 to reach x.
• |x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.

Today, let’s look at how to solve a more advanced GMAT question using the same logic:

For how many integer values of x, is |x – 5| + |x + 1| + |x|  + |x – 7| < 15?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x|  + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

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!

# The Holistic Approach to Absolute Values – Part III

A while back, we discussed some holistic approaches to answering absolute value questions. Today, we will enhance our understanding of absolute values with some variations that you might see on the GMAT.

Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.

A quick review:

• |x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.
• |x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or  -3.
• |x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.

Let’s move ahead now and see how we can use these concepts to solve inequalities:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:

What will be the total distance at any value of x between these two points?

Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4

Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7

In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).

Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.

Along the same lines, consider a slight variation of this question:

For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.

We hope this logic is clear. We will look at some other variations of this concept 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 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!

# Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient

Today we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?

Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.

A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?

(1) The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday

We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.

There are four variables to consider here:

1. Number of compact cars rented (this is what we need to find)
2. Number of luxury cars rented
3. Daily rental rate of compact cars
4. Daily rental rate of luxury cars

Let’s examine the information given to us by the statements:

Statement 1: The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.

This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:

Statement 2: The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday.

This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.

Now, let’s try to tackle both statements together:

The daily rate for luxury cars is \$15 higher than it is for compact cars, and the total rental rates for luxury cars is \$105 higher than it is for compact cars. What constitutes this \$105? It is the higher rental cost of each luxury car (the extra \$15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more  luxury cars were rented, 105 would account for the \$15 higher rent of each luxury car and also for the rent of the extra luxury cars.

Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.

The total extra money collected by renting luxury cars is \$105.

105/15 = 7

Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is \$0 (theoretically), the rent of luxury cars is \$15 and the extra rent charged will be \$105 (7*15 = 105) – this is a valid case.

Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.

Let’s make a slight change to our current numbers to see if they still fit:

Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be \$15*8 = \$120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is \$105 – the \$15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be \$15/9 = \$5/3.

This would mean that the daily rental rate of each luxury car is \$5/3 + \$15 = \$50/3

The total rental cost of luxury cars in this case would be 8 * \$50/3 = \$400/3

The total rental cost of compact cars in this case would be 17 * \$5/3 = \$85/3

The difference between the two total rental costs is \$400/3 – \$85/3 = 315/3 = \$105

Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.

The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:

Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be 10*\$15 = \$150 extra, but it is actually only \$105 extra, a difference of \$45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be \$45/5 = \$9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, \$105. This is a valid case, too.

Hence, there are two strategies we saw in action today:

• Tweak the numbers slightly to see if you will get the same results
• Go for the easy split when choosing numbers to plug in

Getting ready to take the GMAT? We have 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!

# How to Solve “Unsolvable” Equations on the GMAT

The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Statement 2: x^2 – x – 2 < 0

This, we can easily solve:

x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0

We know how to solve this inequality using the method discussed here.

This this will give us -1 < x < 2.

Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.

To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.

Getting ready to take the GMAT? We have 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: A GMAT Quant Question That Troubles Many!

What determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!

In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.

This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.

We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:

† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?

(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089

The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.

The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?

As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”.  Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.

For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:

Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?

Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:

58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.

Along those same lines:

AB = 10A + B
BA = 10B + A

Going back to our original question:

(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)

We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.

We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.

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: Try to Answer This GMAT Challenge Question!

Today, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.

Let’s take a look at the question stem:

 Date of Transaction Type of Transaction June 11 Withdrawal of \$350 June 16 Withdrawal of \$500 June 21 Deposit of x dollars

For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was \$1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was \$1,000, what was the amount of the deposit on June 21?

(A) \$1,000
(B) \$1,150
(C) \$1,200
(D) \$1,450
(E) \$1,600

Think about how you might answer this question:

The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000

Now we have been given the only three transactions that took place:

• A withdrawal of \$350 on June 11 – so on June 11, the account balance goes down to \$650.
• A withdrawal of \$500 on June 16 – so on June 16, the account balance goes down to \$150.
• A deposit of \$x on June 21 – So on June 21, the account balance goes up to 150 + x.

Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) +  150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000

One might then end up doing this calculation to find the value of x:

[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = \$1,450
The answer is D.

But this calculation is rather tedious and time consuming. Can’t we use the deviation method ? After all, we are dealing with large values here! How?

Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:

• The amount from June 11 to June 30 is 350 less.
• The amount from June 16 to June 30 is another 500 less.
• The amount from June 21 to June 30 is x in excess.

Through the deviation method, we can see the shortfall = the excess:

350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)

This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!

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: Evaluating Nasty GMAT Answer Choices

In some Quant questions, we are given big nasty numbers in the answer choices and little else in the question stem. Often in such cases, the starting point is difficult for the test-taker to find, so today, we will discuss how to handle such questions.

The first and only rule with these types of problems is that familiarity helps. Evaluate the answer choices that make sense to you first.

Let’s look at a few questions to understand how to do that:

Which of the following is NOT prime?

(A) 1,556,551
(B) 2,442,113
(C) 3,893,257
(D) 3,999,991
(E) 9,999,991

The first thing that comes to mind when we consider how to find prime numbers should be to “check the number N for divisibility by all prime factors until we get to the √N.” But note that here, we have four numbers that are prime and one number that is not. Also, the numbers are absolutely enormous and, hence, will be very difficult to work with. So, let’s slide down to a number that seems a bit more sane: 3,999,991 (it is very close to a number with lots of 0’s).

3,999,991 = 4,000,000 – 9
= (2000)^2 – 3^2

This is something we recognise! It’s a difference of squares, which can be written as:

= (2000 + 3) * (2000 – 3)
= 2003 * 1997

Hence, we see that 3,999,991 is a product of two factors other than 1 and itself, so it is not a prime number. We have our answer! The answer is D.

Let’s try another problem:

Which of the following is a perfect square?

(A) 649
(B) 961
(C) 1,664
(D) 2,509
(E) 100,000

Here, start by looking at the answer choices. The first one that should stand out is option E, 100,000, since multiples of 10 are always easy to handle. However, we have an odd number of zeroes here, so we know this cannot be a perfect square.

Next, we look at the answer choices that are close to the perfect squares that we intuitively know, such as 30^2 = 900, 40^2 = 1600, 50^2 = 2500. The only possible number whose perfect square could be 961 is 31 – 31^2 will end with a 1 and will be a bit greater than 900 (32^2 will end with a 4, so that cannot be the square root of 961, and the perfect squares of other greater numbers will be much greater than 900).

31^2 = (30 + 1)^2 = 900 + 1 + 2*30*1 = 961

So, we found that 961 is a perfect square and is, hence, the answer!

In case 961 were not a perfect square, we would have tried 1,664 since it is just 64 greater than 1,600. It could be the perfect square of 42, as the perfect square of 42 will end in a 4.

If 1,664 were also not a perfect square (it is not), we would have looked at 2,509. We would have known immediately that 2,509 cannot be a perfect square because it is too close to 2,500. 2,509 ends in a 9, so we may have considered 53 to be its square root, but the difference between consecutive perfect squares increases as we get to greater numbers.

(4^2 is 16 while 5^2 is 25 – the difference between them is 9. The difference between 5^2 and 6^2 will be greater than 9, and so will the difference between the perfect squares of any pair of consecutive integers greater than 6. Hence, the difference between the squares of 50 and 53 certainly cannot be 9.)

Therefore, our answer is B. Let’s try one more question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

(A) 1,296
(B) 1,369
(C) 1,681
(D) 1,764
(E) 2,500

This question is, again, on perfect squares. We can use the same concepts here, too.

30^2 = 900
31^2 = 961 (=(30+1)^2 = 900 + 1 + 2*30)

40^2= 1,600
41^2 = 1,681 (=(40+1)^2 = 1,600 + 1 + 2*40)

50^2 = 2,500
51^2 = 2,601 (=(50+1)^2 = 2,500 + 1 + 2*50)

We know that the difference between consecutive squares increases as we go to greater numbers: going from 30^2 to 31^2 is a difference of 61, while jumping from 40^2 to 41^2 is a difference of 81.

All the answer choices lie in the range from 900 to 2500. In this range, the difference between consecutive squares is between 60 and 100. So, when you add 148 to a perfect square to get another perfect square in this range, we can say that the numbers must be 2 apart, such as 33 and 35 or 42 and 44, etc. Also, the numbers must lie between 30 and 40 because twice 61 is 122 and twice 81 is 162 – 148 lies somewhere in between 122 and 162.

A and B are the only two possible options.

Consider option A – it ends in a 6, so the square root must end in a 6, too. If you add 148, then it will end with a 4 (the perfect square of a number ending in 8 will end in 4). So this answer choice works.

Consider option B – it ends in a 9, so the square root must end in a 3 or a 7. When you add 148, it ends in 7. No perfect square ends in 7, so this option is out. Our answer is, therefore, A.

We hope you see how a close evaluation of the answer choices can help you solve questions of this type. Go get ’em!

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!

# Evaluating “Useful to Evaluate” Critical Reasoning Questions – Part II

Last week we looked at how to handle “useful to evaluate” questions in the Verbal section, and we left you with a tricky “useful to evaluate except” question. Let’s take a look at that problem today. “Except” questions are usually more difficult to deal with since we need to find four “correct” options (which we are not as used to). So, let’s take a look at this question:

Following several years of declining advertising sales, the Greenville Times reorganized its advertising sales force two years ago. Before the reorganization, the sales force was organised geographically, with some sales representatives concentrating on city center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ businesses by having each sales representative deal with only one type of industry or of retailing. After the reorganization, advertising sales increased.

In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be helpful to find out each of the following EXCEPT:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?
(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?
(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

Let’s first break down what the argument says:

• Advertising sales were declining.
• The paper reorganized the advertising sales team two years back.
• Advertising sales increased after reorganisation.

Now, we want to figure out whether the increase actually happened due to the reorganization; in other words, we need to evaluate what else could have caused the increase in sales, if not the reorganization. Say the lead of the sales team changed two years back – it is possible that he is responsible for the increase in revenue. Four of the five answer choices will raise similar questions, while the leftover option (which will be our answer) will not. Let’s take a look at each of the answer choices:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by the advertising sales?

The proportion of advertising sales as a part of the total revenue is immaterial to us – we only need to evaluate why the advertising sales have increased. It is possible that the revenue from other sources has increased much more than the revenue from advertising sales, and hence, advertising sales could be a smaller proportion of the overall revenue now, however this doesn’t matter at all. This option has nothing to do with the increase in advertising sales, and hence, is the correct answer.

Let’s take a look at all the other options too, just to be safe:

(B) Has the circulation of the Greenville Times increased substantially in the last two years?

This answer choice can be evaluated in two ways:

1. Yes, it has increased – If the circulation increased substantially in the last two years, that could have led to the increase in advertising sales.
2. No, it has not increased – If the circulation hasn’t increased substantially, then there must be another reason for the increase in advertising sales. In that case, the reorganization could be the reason.

These two answers affect the argument differently, and hence, this option will be useful in evaluating the argument.

(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?

Again, the answer choice can be evaluated in two ways:

1. Yes, there has – If there has been a substantial turnover in personnel, it is possible that more capable people have been hired, which could have led to higher advertising sales.
2. No, there hasn’t – If there hasn’t been a substantial turnover in personnel, then there would need to be another reason for the increased advertising sales. In that case, the reorganization could be the reason.

The two answers affect the argument differently, so this option will also be useful in evaluating the argument.

(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?

This option can also be evaluated in two ways:

1. Yes, they did find it difficult – Did reorganization make it easier to keep track of relevant developments? If yes, then the reorganization could be responsible for the increase in sales.
2. No, they did not find it difficult – If they did not find it difficult to keep up with relevant developments, then we cannot say whether the reorganization was responsible for the increase in sales or not.

These two responses affect the argument differently. Hence, this option will be useful in evaluating the argument.

(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

Answer choice E can also be evaluated in two ways:

1. Yes, it has – If the economy has been growing rapidly over the past two years, it could be the reason for higher advertising sales. Then we may not be able to attribute the improvement in advertising sales to the reorganization.
2. No it has not – If there has been no such growth in the economy, then reorganization could be the reason for higher advertising sales.

Again, the two responses affect the argument differently, so this option will also be useful in evaluating the argument.

We see that B, C, D and E are all useful in evaluating the argument. Therefore, our answer is A. We hope you will find it easier to handle such questions in the future!

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!

# Evaluating “Useful to Evaluate” Critical Reasoning Questions on the GMAT

In today’s post, we will look at how to answer “useful to evaluate” Critical Reasoning questions in the Verbal section of the GMAT. Arguably, this is one of the toughest question types for test-takers to tackle (perhaps right after boldfaced questions).

To answer this type of question, all you will need to do is follow these six simple steps:

1) Identify the conclusion.
2) Ask yourself the question raised by answer choice A.
3) Answer it with a “yes” and figure out whether it affects the conclusion.
4) Answer it with a “no” and figure out whether it affects the conclusion.
5) Repeat this for all other answer choices.
6) Only one option will affect the conclusion differently in the two cases – that is your answer.

Let’s illustrate this concept with a problem:

In a certain wildlife park, park rangers are able to track the movements of many rhinoceroses because those animals wear radio collars. When, as often happens, a collar slips off, it is put back on. Putting a collar on a rhinoceros involves immobilizing the animal by shooting it with a tranquilizer dart. Female rhinoceroses that have been frequently re-collared have significantly lower fertility rates than uncollared females. Probably, therefore, some substance in the tranquilizer inhibits fertility.

In evaluating the argument, it would be most useful to determine which of the following?

(A) Whether there are more collared female rhinoceroses than uncollared female rhinoceroses in the park.
(B) How the tranquilizer that is used for immobilizing rhinoceroses differs, if at all, from tranquilizers used in working with other large mammals
(C) How often park rangers need to use tranquilizer darts to immobilize rhinoceroses for reasons other than attaching radio collars
(D) Whether male rhinoceroses in the wildlife park lose their collars any more often than the park’s female rhinoceroses do
(E) Whether radio collars are the only practical means that park rangers have for tracking the movements of rhinoceroses in the park

First, we need to break down the argument to find the premises and the conclusion:

• Many rhinoceroses wear radio collars.
• Often, collars slip.
• When a collar slips, the animal is shot with a tranquilizer to re-collar.
• The fertility of frequently re-collared females is less than the fertility of uncollared females.
• Conclusion: Some substance in the tranquilizer inhibits fertility.

Let’s take a look at each answer choice:

(A) Whether there are more collared female rhinoceroses than uncollared female rhinoceroses in the park.

Even if there are more collared female rhinoceroses than uncollared females, this does not affect the argument’s conclusion. This answer choice talks about collared females vs. uncollared females; we are comparing the fertility of re-collared females with that of uncollared females. Anyway, how many of either type there are doesn’t matter. So, whether you answer “yes” or “no” to this question, it is immaterial.

(B) How the tranquilizer that is used for immobilizing rhinoceroses differs, if at all, from tranquilizers used in working with other large mammals.

This option is comparing the tranquilizers used for rhinoceroses with the tranquilizers used for other large mammals. What the conclusion does, however, is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer “very different” or “not different at all” to this question, in the end, it doesn’t matter.

(C) How often park rangers need to use tranquilizer darts to immobilize rhinoceroses for reasons other than attaching radio collars.

This answer choice can be evaluated in two ways:

• Very Often – Tranquilizers are used very often for uncollared females, too. In this case, can we still say that “tranquilizers inhibit fertility”? No! If they did, fertility in uncollared females would have been low, too.
• Rarely – This would strengthen our conclusion. If tranquilizers are not used on uncollared females, it is possible that something in these tranquilizers inhibits fertility.

(D) Whether male rhinoceroses in the wildlife park lose their collars any more often than the park’s female rhinoceroses do.

This answer choice is comparing the frequency of tranquilizers used on male rhinoceroses with the frequency of tranquilizers used on female rhinoceroses. What the conclusion actually does is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer this question with “more frequently” or “not more frequently,” it doesn’t matter.

(E) Whether radio collars are the only practical means that park rangers have for tracking the movements of rhinoceroses in the park.

This option is comparing radio collars with other means of tracking. What the conclusion does is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer this question with “there are other means” or “there are no other means,” again, it does not matter.

Note that only answer choice C affects the conclusion – if you answer the question it raises differently, it affects the conclusion differently. Option C would be good to know to evaluate the conclusion of the argument, therefore, the answer must be C.

Now try this question on your own:

Following several years of declining advertising sales, the Greenville Times reorganized its advertising sales force two years ago. Before the reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ businesses by having each sales representative deal with only one type of industry or of retailing. After the reorganization, advertising sales increased.

In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be helpful to find out each of the following EXCEPT:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?
(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?
(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

We hope you will find this post useful to evaluate the “useful to evaluate” 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: Using the Deviation Method for Weighted Averages

We have discussed how to use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.

The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:

What is the average of 452, 452, 453, 460, 467, 480, 499,  499, 504?

What would you say the average is here? Perhaps, around 470?

Shortfall:
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.

Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.

Excess:
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.

Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.

The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.

So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.

Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)

This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.

We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

(A) y + 3z
(B) (y +z) / 4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z

Grade 1 milk contains 1% fat. Grade 2  milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.

Shortfall = x*(1.5 – 1)
Excess = y*(2 – 1.5) + z*(3 – 1.5)

Since 1.5 is the actual average, the shortfall = the excess.

x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)
x/2 = y/2 + 3z/2
x = y + 3z

And there you have it – the answer is A.

We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your GMAT toolkit.

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 Negate Assumption Answer Choices on the GMAT

Most GMAT test-takers come across the Assumption Negation Technique at some point in their preparation. It is one of the most effective techniques for assumption questions (which are usually fairly difficult) if you learn to apply it successfully.

We already know that many sentences are invalidated by negating the verb of the dominant clause. For example:

There has been a corresponding increase in the number of professional companies devoted to other performing arts.

becomes

There has not been a corresponding increase in the number of professional companies devoted to other performing arts.

Recently, we got a query on how to negate various modifiers such as “most” and “a majority”. So today, we will examine how to negate the most popular modifiers we come across:

• All -> Not all
• Everything -> Not everything
• Always -> Not always
• Some -> None
• Most -> Half or less than half
• Majority -> Half or less than half
• Many -> Not many
• Less than -> Equal to or more than
• Element A -> Not element A
• None ->  Some
• Never ->  Sometimes

Let’s take a look at some examples with these determiners:

1) “All of the 70 professional opera companies are commercially viable options.”
This becomes, “Not all of the 70 professional opera companies are commercially viable options.”

2) “There were fewer than 45 professional opera companies that had been active 30 years ago and that ceased operations during the last 30 years.”
This becomes, “There were 45 or more professional opera companies that had been active 30 years ago and that ceased operations during the last 30 years.”

3) “No one who is feeling isolated can feel happy.”
This becomes, “Some who are feeling isolated can feel happy.”

4) “Anyone who is able to trust other people has a meaningful emotional connection to at least one other human being.”
This becomes, “Not everyone who is able to trust other people has a meaningful emotional connection to at least one other human being.”

5) “The 45 most recently founded opera companies were all established as a result of enthusiasm on the part of a potential audience.”
This becomes, “The 45 most recently founded opera companies were not all established as a result of enthusiasm on the part of a potential audience.”

6) “Many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.”
This becomes, “Not many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.”

7) “The birds of prey capture and kill every single Spotted Mole that comes above ground.”
This becomes, “Not every single Spotted Mole that comes above ground is captured and killed by the birds of prey.”

8) “At least some people who do not feel isolated are happy.”
This becomes, “No people who do not feel isolated are happy.”

9) “Some land-based mammals active in this region, such as fox, will also hunt and eat the Spotted Mole on a regular basis.”
This becomes, “None of the land-based mammals active in this region, such as fox, will also hunt and eat the Spotted Mole on a regular basis.”

10) “No other animal could pose as significant a threat to the above-ground fruits as could the Spotted Mole.”
This becomes, “Some other animals could pose as significant a threat to the above-ground fruits as could the Spotted Mole.”

We hope the next time you come across an assumption question, you will not face any trouble negating the answer choices!

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: Know Your Subtraction for the GMAT!

Your first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:

The last digit of 12^12 + 13^13 – 14^14 × 15^15 =

(A) 0
(B) 1
(C) 5
(D) 8
(E) 9

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

(i) 100-29
100
-29
071

(ii) 29-100
100
-29
071
(But since the sign of 100 is negative, your answer is actually -71.)

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0
–  pq9
ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

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: Linear Relations in GMAT Questions

We have covered the concepts of direct, inverse and variation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.

We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?

(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)
60 = 24m + c ……(II)

(II) – (I)
30 = 18m
m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c
c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20
R = 48

48 (answer choice C) is our answer.

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer.

Getting ready to take the GMAT? We have 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!

# Advanced Number Properties on the GMAT – Part VI

Most people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:

Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

Say N = 3
The numbers are 5, 6, 7 (any three consecutive numbers)
Their sum is 5 + 6 + 7 = 18
Their product is 5*6*7 = 210
Note that both the sum and the product are divisible by 3 (i.e. N).

Say N = 5
The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 2 + 3 + 4 + 5 + 6 = 20
Their product is 2*3*4*5*6 = 720
Again, note that both the sum and the product are divisible by 5 (i.e. N)

Say N = 4
The numbers are 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 3 + 4 + 5 + 6 = 18
Their product is 3*4*5*6 = 360
Now note that the sum is not divisible by 4, but the product is divisible by 4.

If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.
Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

(Multiple of N),
(Multiple of N) +1,
(Multiple of N) + 2,
… ,
(Multiple of N) + (N-2),
(Multiple of N) + (N-1)

In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:

(Multiple of 3) + 2 = 5
(Multiple of 3)       = 6
(Multiple of 3) + 1 = 7

In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:

(Multiple of 4) + 3 = 3
(Multiple of 4)        = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
etc.

What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will always add up in pairs to give the sum of N:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

So when you add up all the integers, you will get a multiple of N.

What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will add up to give integers of N but one will be leftover:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.

For example, let’s reconsider the previous example in which we had four consecutive integers:

(Multiple of 4)      = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
(Multiple of 4) + 3 = 3

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

Which of the following must be true?
1) The sum of N consecutive integers is always divisible by N.
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!

(A) 1, 4, 5
(B) 3, 4, 5
(C) 4 and 5
(D) 1, 2, 3, 4
(E) only 4

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

5) The product of K consecutive integers is divisible by K!

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts I, II, III, IV and V of this series.)

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: Divisibility by Powers of 2

We know the divisibility rules of 2, 4 and 8:

For 2 – If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2.

For 4 – If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4.

For 8 – If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8.

A similar rule applies to all powers of 2:

For 16 – If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16.

For 32 – If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32.

and so on…

Let’s figure out why:

The generic rule can be written like this: A number M is divisible by 2^n if the last n digits of M are divisible by 2^n.

Take, for example, a division by 8 (= 2^3), where M = 65748048 and n = 3.

Our digits of interest are the last three digits, 048.

48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.

A valid question here is, “What about the remaining five digits? Why do we ignore them?”

Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).

Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.

All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.

In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.

In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.

We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:

What is the remainder when 1990990900034 is divided by 32 ?

(A) 16
(B) 8
(C) 4
(D) 2
(E) 0

Breaking down our given number, 1990990900034 = 1990990900000 + 00034.

1990990900000 ends in five 0’s so it is divisible by 32. 34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. 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 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: What is Your Favorite Number?

Fans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this  number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768
(B) 819
(C) 826
(D) 858
(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If  we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

Getting ready to take the GMAT? We have 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: Attacking Gerunds on the GMAT!

A few weeks back, we talked about participles and how they are used on the GMAT. In that post, we had promised to discuss gerunds more in depth at another time. So today, as promised, we’ll be looking at gerunds. Before we do that, however, let’s examine Verbals.

A Verbal is a verb that acts as a different part of speech – not as a verb.

There are three types of verbals:

• Infinitives – these take the form of “to + verb”
• Gerunds – these are the “-ing” form of the verb
• Participles – these can take the “-ing,” “-ed,” “-en” etc. forms

Gerunds end in “-ing” and act as nouns in the sentence. They can act as a subject, direct object, subject complement or object of a preposition. For example:

Running a marathon is very difficult. – Subject
I love swimming. – Direct object
The activity I enjoy the most is swimming. – Subject complement
She thanked me for helping her. – Object of a preposition

You don’t have to identify the part of speech the gerund represents in a sentence; you just need to identify whether a verb’s “-ing” form is being used as a gerund and evaluate whether it is being used correctly.

A sentence could also use a gerund phrase that begins with a gerund, such as, “Swimming in the morning is exhilarating.”

Let’s take a look at a couple of official questions now:

A recent study has found that within the past few years, many doctors had elected early retirement rather than face the threats of lawsuits and the rising costs of malpractice insurance.

(A) had elected early retirement rather than face
(B) had elected early retirement instead of facing
(C) have elected retiring early instead of facing
(D) have elected to retire early rather than facing
(E) have elected to retire early rather than face

Upon reading the original sentence, we see that there is a gerund phrase here – “rising costs of malpractice insurance” – which is parallel to the noun “threat of lawsuits.”

The two are logically parallel too, since there are two aspects that the doctors do not want to face: rising costs and the threat of lawsuits.

Note, however, that they are not logically parallel to “face.” Hence, the use of the form “facing” would not be correct, since it would put “facing” and “rising” in parallel. So answer choices B, C and D are incorrect.

Actually, “retire” and “face” are logically parallel so they should be grammatically parallel, too. Answer choice E has the two in parallel in infinitive form – to retire and (to is implied here) face are in parallel.

Obviously, there are other decision points to take note of here, mainly the question of “had elected” vs. “have elected.” The use of “had elected” will not be correct here, since we are not discussing two actions in the past occurring at different times. Therefore, the correct answer is E.

Take a look at one more:

In virtually all types of tissue in every animal species dioxin induces the production of enzymes that are the organism’s trying to metabolize, or render harmless, the chemical that is irritating it.

(A) trying to metabolize, or render harmless, the chemical that is irritating it
(B) trying that it metabolize, or render harmless, the chemical irritant
(C) attempt to try to metabolize, or render harmless, such a chemical irritant
(D) attempt to try and metabolize, or render harmless, the chemical irritating it
(E) attempt to metabolize, or render harmless, the chemical irritant

Notice the use of the gerund “trying” in answer choice A. “Organism’s” is in possessive form and acts as an adjective for the noun verbal “trying.” Usually, with possessives, a gerund does not work. We need to use a noun only. With this in mind, answer choices A and B will not work.

The other three options replace “trying” with “attempt” and hence correct this error, however options C and D use the redundant “attempt to try.” The use of “attempt” means “try,” so there is no need to use both. Option E corrects this problem, so it is our correct answer.

Unlike participles, which can be a bit confusing, gerunds are relatively easy to understand and use. Feeling more confident about them now?

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!

# How to Go From a 48 to 51 in GMAT Quant – Part VII

Both a test-taker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a test-taker at the 51 level does skillfully:

1. Uses holistic, big-picture methods to solve Quant questions.
2. Handles questions he or she finds difficult in a timely manner.

We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IVV and VI of this series.)

Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level test-taker – in fact, the 51 scorers often have time left over after attempting all these questions.

Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too.

There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions.

Let’s take an official example:

Pam owns an inventory of unopened packages of corn and rice, which she has purchased for \$17 and \$13 per package, respectively. How many packages of corn does she have ?

Statement 1: She has \$282 worth of packages.

Statement 2: She has twice as many packages of corn as of rice.

A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions.

After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force.

The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own \$282 worth of packages.)

So, the test-taker will start working on finding the first solution (using the method discussed in this post). We are told:

Price of a packet of corn = \$17
Price of a packet of rice = \$13

Say Pam has “x” packets of corn and “y” packets of rice.

Statement 1: She has \$282 worth of packages

Using Statement 1, we know that 17x + 13y = 282.

We are looking for the integer values of x and y.

If x = 0, y will be 21.something (not an integer)
If x = 1, y = 20.something
If x = 2, y = 19.something
If x = 3, y = 17.something

This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.

So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient.

Statement 2: She has twice as many packages of corn as of rice.

Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A.

I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.

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 Prepositional Phrases on the GMAT

In previous posts, we have already discussed participles as well as absolute phrases. Today, let’s take a look at another type of modifier – the prepositional phrase.

A prepositional phrase will begin with a preposition and end with a noun, pronoun, gerund, or clause – the “object” of the preposition. The object of the preposition might have one or more modifiers to describe it.

Here are some examples of prepositional phrases (with prepositions underlined):

• along the ten mile highway…
• with a cozy blanket…
• without worrying…
• about what he likes…

A prepositional phrase can function as an adjective or an adverb. As an adjective, it answers the question, “Which one?” while as an adverb it can answer the questions, “How?” “When?” or “Where?”.

For example:

• The book under the table belongs to my mom. Here, the prepositional phrase acts as an adjective and tells us “which one” of the books belongs to my mom.
• We tried the double cheeseburger at the new burger joint. Here, the prepositional phrase acts as an adverb and tells us “where” we tried the cheeseburger.

Like other modifiers, a prepositional modifier should be placed as close as possible to the thing it is modifying.

Let’s take a look at a couple of official GMAT questions to see how understanding prepositional phrases can help us on this exam:

The nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius in two letters to the historian Tacitus.

(A) The nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius in two letters to the historian Tacitus.
(B) To the historian Tacitus, the nephew of Pliny the Elder wrote two letters, being the only eyewitness accounts of the great eruption of Vesuvius.
(C) The only eyewitness account is in two letters by the nephew of Pliny the Elder writing to the historian Tacitus an account of the great eruption of Vesuvius.
(D) Writing the only eyewitness account, Pliny the Elder’s nephew accounted for the great eruption of Vesuvius in two letters to the historian Tacitus.
(E) In two letters to the historian Tacitus, the nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius.

There are multiple prepositional phrases here:

• of the great eruption of Vesuvius (answers “Which eruption?”)
• in two letters (tells us “where” he wrote his account)
• to the historian Tacitus (answers “Which letters?”)

Therefore, the phrase “to the historian Tacitus” should be close to what it is describing, “letters,” which makes answer choices B and C incorrect.

Also, “in two letters to the historian Tacitus” should modify the verb “wrote.” In options A and D, “in two letters to the historian Tacitus” seems to be modifying “eruption,” which is incorrect. (There are other errors in answer choices B, C and D as well, but we will stick to the topic at hand.)

Option E corrects the prepositional phrase errors by putting the modifier close to the verb “wrote,” so therefore, E is our answer.

Let’s try one more:

Defense attorneys have occasionally argued that their clients’ misconduct stemmed from a reaction to something ingested, but in attributing criminal or delinquent behavior to some food allergy, the perpetrators are in effect told that they are not responsible for their actions.

(A) in attributing criminal or delinquent behavior to some food allergy
(B) if criminal or delinquent behavior is attributed to an allergy to some food
(C) in attributing behavior that is criminal or delinquent to an allergy to some food
(D) if some food allergy is attributed as the cause of criminal or delinquent behavior
(E) in attributing a food allergy as the cause of criminal or delinquent behavior

This sentence has two clauses:

Clause 1: Defense attorneys have occasionally argued that their clients’ misconduct stemmed from a reaction to something ingested,

Clause 2: in attributing criminal or delinquent behavior to some food allergy, the perpetrators are in effect told that they are not responsible for their actions.

These two clauses are joined by the conjunction “but,” and the underlined part is a prepositional phrase in the second clause.

Answer choices A, C and E imply that the perpetrators are attributing their own behaviors to food allergies. That is not correct – their defense attorneys are attributing their behavior to food allergies, and hence, all three of these options have modifier errors.

This leaves us with B and D. Answer choice D uses the phrase “attributed as,” which is grammatically incorrect – the correct usage should be “X is attributed to Y,” rather than “X attributed as Y.” Therefore, option B is our answer.

As you can see, the proper placement of prepositional phrases is instrumental in creating a sentence with a clear, logical meaning.  Since that type of clear, logical meaning is a primary emphasis of correct Sentence Correction answers, you should be prepared to look for prepositional phrases (here we go…) *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: Some GMAT Questions Using the “Like” vs. “As” Concept

Today we will look at some official GMAT questions testing the “like” vs. “as” concept we discussed last week.

(Review last week’s post – if you haven’t read it already – before you read this one for greater insight on this concept.)

Take a look at the following GMAT Sentence Correction question:

As with those of humans, the DNA of grape plants contains sites where certain unique sequences of nucleotides are repeated over and over.

(A) As with those of humans, the DNA of grape plants contains sites where
(B) As human DNA, the DNA of grape plants contain sites in which
(C) As it is with human DNA, the DNA of grape plants, containing sites in which
(D) Like human, the DNA of grape plants contain sites where
(E) Like human DNA, the DNA of grape plants contains sites in which

Should we use “as” or “like”? Well, what are we comparing? We’re comparing the DNA of humans to the DNA of grape plants. Answer choice E compares these two properly – “Like human DNA, the DNA of grape plants…” DNA is singular, so it uses the singular verb “contains”.

All other options are incorrect. Answer choice A uses “those of” for DNA, but DNA is singular, so this cannot be right. B uses “as” to compare the two nouns, which is also incorrect. C is a sentence fragment without a main verb. And D compares “human” to “DNA”, which is not the “apples-to-apples” comparison we need to make this sentence correct. Therefore, our answer must be E.

Let’s try another one:

Like Auden, the language of James Merrill is chatty, arch, and conversational — given to complex syntactic flights as well as to prosaic free-verse strolls.

(A) Like Auden, the language of James Merrill
(B) Like Auden, James Merrill’s language
(C) Like Auden’s, James Merrill’s language
(D) As with Auden, James Merrill’s language
(E) As is Auden’s the language of James Merrill

Here, we’re comparing Auden’s language to James Merrill’s language. Answer choice C correctly uses the possessive “Auden’s” to show that language is implied. “Like Auden’s language, James Merrill’s language …” contains both parallel structure and a correct comparison.

Answer choices A, B and D incorrectly compare “Auden” to “language,” rather than “Auden’s language” to “language,” so those options are out. The structure of answer choice E is not parallel – “Auden’s” vs. “the language of James Merrill”. Therefore, the answer must be C.

Let’s try something more difficult:

More than thirty years ago Dr. Barbara McClintock, the Nobel Prize winner, reported that genes can “jump,” as pearls moving mysteriously from one necklace to another.

(A) as pearls moving mysteriously from one necklace to another
(B) like pearls moving mysteriously from one necklace to another
(C) as pearls do that move mysteriously from one necklace to others
(D) like pearls do that move mysteriously from one necklace to others
(E) as do pearls that move mysteriously from one necklace to some other one

This is a tricky question – it’s perfect for us to re-iterate how important it is to focus on the meaning of the given sentence. Do not try to follow grammar rules blindly on the GMAT!

Is the comparison between “genes jumping” and “pearls moving”? Do pearls really move mysteriously from one necklace to another? No! This is a hypothetical situation, so we must use “like” – genes are like pearls. Answer choices B and D are the only ones that use “like,” so we can eliminate our other options. D uses a clause with “like,” which is incorrect. In answer choice B, “moving from …” is a modifier – “moving” doesn’t act as a verb here, so it doesn’t need a clause. Hence, answer choice B is correct.

Here’s another one:

According to a recent poll, owning and living in a freestanding house on its own land is still a goal of a majority of young adults, like that of earlier generations.

(A) like that of earlier generations
(B) as that for earlier generations
(C) just as earlier generations did
(D) as have earlier generations
(E) as it was of earlier generations

Note the parallel structure of the comparison in answer choice E – “Owning … a house… is still a goal of young adults, as it was of earlier generations.” It correctly uses “as” with a clause.

Answer choice A uses “that” but its antecedent is not very clear; there are other nouns between “goal” and “like,” and hence, confusion arises. None of the other answer choices give us a clear, parallel comparison, so our answer is E.

Alright, last one:

In Hungary, as in much of Eastern Europe, an overwhelming proportion of women work, many of which are in middle management and light industry.

(A) as in much of Eastern Europe, an overwhelming proportion of women work, many of which are in
(B) as with much of Eastern Europe, an overwhelming proportion of women works, many in
(C) as in much of Eastern Europe, an overwhelming proportion of women work, many of them in.
(D) like much of Eastern Europe, an overwhelming proportion of women works, and many are.
(E) like much of Eastern Europe, an overwhelming proportion of women work, many are in.

Another tricky question. The comparison here is between “what happens in Hungary” and “what happens in much of Eastern Europe,” not between “Hungary” and “much of Eastern Europe.” A different sentence structure would be required to compare “Hungary” to “much of Eastern Europe” such as “Hungary, like much of Eastern Europe, has an overwhelming …”

With prepositional phrases, as with clauses, “as” is used. So, we have two relevant options – A and C. Answer choice A uses “which” for “women,” and hence, is incorrect. Therefore, our answer is C.

Here are some takeaways to keep in mind:

• You should be comparing “apples” to “apples”.
• Parallel structure is important.
• Use “as” with prepositional phrases.

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 “Like” vs. “As” on the GMAT Verbal Section

If you have seen the Veritas Prep curriculum, then you know we frequently highlight the strategy of “Think like the Testmaker” to answer GMAT questions. Recently, we had a student question the grammatical validity of this construct – this brought the “like” vs. “as” debate to mind, so we decided to tackle it this week.

When should you use “like” and when should you use “as” in a sentence?

Both words can be used in comparisons, however the structure of the sentence will be different in the two cases. This is because traditionally, “like” is a preposition and “as” is a conjunction – a preposition takes the form of an object while a conjunction takes the form of a clause. Therefore:

Using “like,” we compare nouns/pronouns (including gerunds). Usually, a single verb will be used.

Using “as,” we compare actual actions. There will be two verbs used when we compare using “as.”

So, this is how we are going to compare “like” and “as”:

• He runs like a madman. – A single verb, “runs.”
• He runs as a madman does. – Two verbs, “runs” and “does” (which is equivalent to “does run”).

In the same way, both of the following sentences are correct:

• Think like the Testmaker.
• Think as the Testmaker does.

But beware – “as” used with a noun or pronoun alone does not mean that this usage is incorrect. “As” can also be used to show a role or capacity. For example, in the sentence, “She works as a consultant,” the word “as” means that she works in the capacity of a consultant. There is no comparison here, but the sentence is still grammatically correct.

Also, we usually use “like” in the case of hypothetical comparisons. Take, for instance, the sentence, “She screams like a banshee.” Here, it would be odd to say, “She screams as a banshee does,” because we don’t really know how a banshee screams.

Let’s look at a few GMAT Sentence Correction questions now:

Like many self-taught artists, Perle Hessing did not begin to paint until she was well into middle age.

(A) Like
(B) As have
(C) Just as with
(D) Just like
(E) As did

In this sentence, the word “like” is correctly comparing “Perle Hessing” to “many self taught artists.” There is no clause after “like” and we are using a single verb. Hence, the use of “like” is correct and our answer is A.

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

Based on recent box office receipts, the public’s appetite for documentary films, like nonfiction books, seems to be on the rise.

(A) like nonfiction books
(B) as nonfiction books
(C) as its interest in nonfiction books
(D) like their interest in nonfiction books
(E) like its interest in nonfiction books

This sentence also has a comparison, and it is between “appetite” and “interest” and how they are both are on a rise. Answer choice E compares “appetite” to “interest” using “like” as a single verb. None of the answer choices have “as” with a clause so the answer must be E.

These were two simple examples of “like” vs. “as.” Now let’s look at a higher-level GMAT question:

During an ice age, the buildup of ice at the poles and the drop in water levels near the equator speed up the Earth’s rotation, like a spinning figure skater whose speed increases when her arms are drawn in

(A) like a spinning figure skater whose speed increases when her arms are drawn in
(B) like the increased speed of a figure skater when her arms are drawn in
(C) like a figure skater who increases speed while spinning with her arms drawn in
(D) just as a spinning figure skater who increases speed by drawing in her arms
(E) just as a spinning figure skater increases speed by drawing in her arms

There is a comparison here, but between which two things? Answer choice A seems to be comparing “Earth’s rotation” to “spinning figure skater,” but these two things are not comparable. Option E is the correct choice here – it compares “speed up Earth’s rotation” to “skater increases speed.” 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: Are Official Answers Debatable on the GMAT?

Let’s begin with the bottom line: no, they are not. If you are thinking along the lines of, “This official answer cannot be correct! How can the answer be A? It must be C, or C is at least just as valid as A,” then you are wasting your time. The answer given is never debatable. What you should be thinking instead is, “The answer given is A, but  I thought it was C. I must find out where I made a mistake.”

The point is that since you are going to take GMAT, you must learn to think like the GMAT testmakers. The answers they give for these questions are the correct answers, so need to accept that – this way, the next step of figuring out the gap in your understanding will be far easier. Today, let’s take a look at an official question that is often debated:

The average hourly wage of television assemblers in Vernland has long been significantly lower than that in neighboring Borodia. Since Borodia dropped all tariffs on Vernlandian televisions three years ago, the number of televisions sold annually in Borodia has not changed. However, recent statistics show a drop in the number of television assemblers in Borodia. Therefore, updated trade statistics will probably indicate that the number of televisions Borodia imports annually from Vernland has increased.

Which of the following is an assumption on which the argument depends?

(A) The number of television assemblers in Vernland has increased by at least as much as the number of television assemblers in Borodia has decreased.
(B) Televisions assembled in Vernland have features that televisions assembled in Borodia do not have.
(C) The average number of hours it takes a Borodian television assembler to assemble a television has not decreased significantly during the past three years.
(D) The number of televisions assembled annually in Vernland has increased significantly during the past three years.
(E) The difference between the hourly wage of television assemblers in Vernland and the hourly wage of television assemblers in Borodia is likely to decrease in the next few years.

First, let’s look at the premises of the argument:

• The hourly wage of assemblers in Vernland is much lower than that in Borodia.
• 3 years ago, Borodia dropped all tariffs on TVs imported from Vernland.
• The number of TVs sold annually in Borodia is same.
• However, the number of assemblers in Borodia has decreased.

The conclusion is that the trade statistics will probably indicate that the number of televisions Borodia imports annually from Vernland has increased.

This conclusion might look logical, but it is full of assumptions.

Why does this conclusion seem so logical? Wages in Vernland are lower, so it would seem like TVs should be cheaper here. Borodia dropped all tariffs on imported TVs, which means there will be no artificial inflation of Vernland TV prices. Finally, the number of TVs sold in Borodia has not dropped, but number of assemblers in Borodia has dropped, which makes it look like fewer TVs are getting made in Borodia.

An onlooker might conclude that Borodia is importing more TVs from Vernland because they are cheaper, but here are some assumptions that come to mind:

• The cost of a TV in Vernland is lower because assembler’s wage is lower. What if the raw material cost is higher in Vernland? Or other costs are higher? The cost to produce a Vernland TV could actually be higher than the cost to produce a Borodia TV.
• Fewer TVs are getting made in Borodia, but that does not mean that Borodian assemblers have not become more productive. What if fewer assemblers are needed because they can actually complete the assembly process much faster? The number of TVs sold is the same, however, if each assembler is doing more work, fewer assemblers will be needed. In this case, the number of TVs made in Borodia might not have changed even though the number of producers dropped.

Coming to our question now: Which of the following is an assumption on which the argument depends?

We are looking for an assumption, i.e. a NECESSARY premise. We have already identified some assumptions, so let’s see if any of the answer choices gives us one of those:

(A) The number of television assemblers in Vernland has increased by at least as much as the number of television assemblers in Borodia has decreased.

This is the most popular incorrect answer choice. Test takers keep trying to justify why it makes perfect sense, but actually, it is not required for the conclusion to hold true.

The logic of test takers that pick this answer choice is often on the lines of, “If the number of workers from Borodia decreased, in order for Borodia to show an increased number of imports from Vernland, Vernland must have increased their number of workers by at least as much as the number of workers that left Borodia.”

Note that although this may sound logical, it is not necessary to the argument. There are lots of possible situations where this may not be the case:

Perhaps number of TVs being manufactured in Vernland is the same and, hence, the number of assemblers is the same, too. It is possible that out of the fixed number of TVs manufactured, fewer are getting locally bought and more are getting exported to Borodia. So, it is not necessarily true that number of TV assemblers in Vernland has increased.

(B) Televisions assembled in Vernland have features that televisions assembled in Borodia do not have.

This is also not required for the conclusion to hold – the TVs could actually be exactly the same, but the TVs assembled in Vernland could still be cheaper than the TVs assembled in Borodia due to a potentially lower cost of assembly in Vernland.

(C) The average number of hours it takes a Borodian television assembler to assemble a television has not decreased significantly during the past three years.

This is one of the assumptions we discussed above – we are assuming that the reduction in the number of assemblers must not be due to an increase in the productivity of the assemblers because if the assemblers have got more productive, then the number of TVs produced could be the same and, hence, the number of TVs imported would not have increased.

(D) The number of televisions assembled annually in Vernland has increased significantly during the past three years.

This is not required for the conclusion to hold. Perhaps the number of TVs being sold in Vernland has actually reduced while more are getting exported to Borodia, so the overall number of TVs being made is the same.

(E) The difference between the hourly wage of television assemblers in Vernland and the hourly wage of television assemblers in Borodia is likely to decrease in the next few years.

This is also not required for the conclusion to hold. What happens to the hourly wages of assemblers in Vernland and Borodia in the future doesn’t concern this argument – we are only concerned about what has been happening in the last 3 years.

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!

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 Go From a 48 to 51 in GMAT Quant – Part VI

Today’s post the next part in our “How to Go From 48 to 51 in Quant” series. Again, we will learn a technique that can be employed by the test-taker at an advanced stage of preparation – requiring one to understand the situations in which one can use this simplifying technique.

(Before you continue reading, be sure to check out parts I, II, III, IV, and V of this series.)

We all love to use the plug-in method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one that satisfies the equation.

But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options:

If |4x−4|=|2x+30|, which of the following could be a value of x?

(A) –35/3
(B) −21/2
(C) −13/3
(D) 11/5
(E) 47/5

This question is an ideal candidate for the “plug-in” method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than “plug-in”, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in.

Method 1:
|4x – 4| = |2x + 30|

4 * |x – 1| = 2 * |x + 15|

2 * |x – 1| = |x + 15|

This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from -15.

——————(-15) —————————————————(0)——(1)——————

There are two ways to find the value of x:

Case 1: x could be between -15 and 1 such that the distance between them is split in the ratio 2:1.

or

Case 2: x could be to the right of 1 such that the distance between x and -15 is twice the distance between x and 1.

Let’s examine both of these cases in further detail:

Case 1: The distance from -15 to 1 is of 16 units – this can be split into 3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from -15 two times 16/3, i.e. 32/3.

So, x should be at a point 16/3 away from 1 toward the left.

x = 1 – 16/3 = -13/3

This is one of our answer choices and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found -13/3 in the answer options, we would have moved on to Case 2:

Case 2: The distance between -15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and -15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17. If you would not have found -13/3 in the answer choices, then you would have found 17.

Now let’s move on to see how we can make the plug-in method work for us in this case by examining each answer choice we are given:

Method 2:
|4x – 4| = |2x + 30|

2 * |x – 1| = |x + 15|

(A) -35/3

It is difficult to solve for x = -35/3 to see if both sides match. Instead, let’s solve for the closest integer, -12.

2 * |-12 – 1| = |-12 + 15|

On the left-hand side, you will get 26, but on the right-hand side, you will get 3.

These values are far away from each other, so x cannot be -35/3.  As the value of x approaches the point where the equation holds – i.e. where the two sides are equal to each other – the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of -35/3 from -12 will bring the two sides to be equal.

(B) -21/2

For this answer choice, let’s solve for the nearest integer, x = -10.

2 * |-10 – 1| = |-10 + 15|

On the left-hand side, you will get 22; on the right-hand side, you will get 5.

Once again, these values are far away from each other and, hence, x will not be -21/2.

(C) -13/3

For this answer choice, let’s solve for x = -4.

2 * |-4 -1| = |-4 + 15|

On the left-hand side, you will get 10; on the right-hand side, you will get 11.

Here, there is a possibility that x can equal -13/3, as the two sides are so close to one another – plug in the actual value of -13/3 and you will see that the left-hand side of the equation does, in fact, equal the right-hand side. Therefore, C is the correct answer.

Basically, we approximated the answer choices we were given and shortlisted the one that gave us very close values. We checked for that and found that it is the answer.

We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have limited applications – only if you are given the actual values of x, can you use it. Method 1 is far more generic for absolute value questions.

Getting ready to take the GMAT? We have 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: Squares and Square Roots on the GMAT

In 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!

Let’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

Today, 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?

(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

We have discussed standard deviation (SD). 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

Let’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

Today 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.

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.

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

We 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
(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

Critics 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.

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!

Getting ready to take the GMAT? We have 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?

We 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 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

Considering a two dimensional figure, a tangent is a line that touches a curve at a single point.  Here are some examples of tangents:

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!

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

We 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!

It 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 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

We 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

We 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

Consider 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

Today 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).

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 running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and!

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!