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Veritas Prep Blog 03 Sep 2021, 11:07
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FROM Veritas Prep Blog: 3 Formats for GMAT Inequalities Questions You Need to Know

As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

  • Must Be True
  • Could Be True
  • Complete Range
Case 1: Must Be True

If |-x/3 + 1| < 2, which of the following must be true?

(A) x > 0

(B) x < 8

(C) x > -4

(D) 0 < x < 3

(E) None of the above

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2

(1/3) * |x – 3| < 2

|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0

Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8

Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4

Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3

Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Case 2: Could Be True

If −1 < x < 5, which is the following could be true?

(A) 2x > 10

(B) x > 17/3

(C) x^2 > 27

(D) 3x + x^2 < −2

(E) 2x – x^2 < 0

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10

x > 5

No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3

x > 5.67

No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27

x^2 – 27 > 0

x > 3*√(3) or x < -3*√(3)

√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

(Details on how to solve such inequalities are discussed here.)

(D) 3x + x^2 < −2

x^2 + 3x + 2 < 0

(x + 1)(x + 2) < 0

-2 < x < -1

No values of x will lie between -2 and -1, so this also cannot be true.

(E) 2x – x^2 < 0

x * (x – 2) > 0

x > 2 or x < 0

If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.

Case 3: Complete Range

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

(A) 0 < |x| < ½

(B) |x| > ½

(C) -½ < x < 0 or ½ < x

(D) x < -½ or 0 < x < ½

(E) x < -½ or x > 0

We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice.

We are given that x^3 – 4x^5 < 0. This inequality can be solved to:

x^3 ( 1 – 4x^2) < 0

x^3*(2x + 1)*(2x – 1) > 0

x > 1/2 or -1/2 < x < 0

This is our universe of the values of x. It is given that all values of x lie in this range.

Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.

We hope these practice problems will help you become able to distinguish between the three cases now.

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

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

The post 3 Formats for GMAT Inequalities Questions You Need to Know appeared first on Veritas Prep Blog.
This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
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Hi VeritasKarishma,

For the last questions where the entire range is asked for, as per your previous blog I should try bringing the inequality in (x-a) (x-b) <0 form and since we have x^3 (1-4x^2) < 0 here, I didn't do that and got the answer D. I realise my mistake here.
But here's my query:

in the first question we have |-x/3 + 1 | <2,
I got this one right using multiple methods, but if we have |x-a| or |a-x| it doesn't matter here right? We don't need to change |a-x| to |x-a| from by multiplying by -1 and flipping the sign, right?

We don't, right? Because |x-a| = |a-x|.
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Veritas Prep Blog 03 Sep 2021, 23:41
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FROM Veritas Prep Blog: A 700+ GMAT Quant Question on Races

This week we will look at the question on races that we gave you last week.

Question 3: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately?

(A)   8, 10

(B)    4, 5

(C)   5, 9

(D)   6, 9

(E)    7, 10

Solution: Now this question is a little tougher than the previous ones we saw last week.

There are two scenarios given:

1 – A gives B a head start of 200 m and beats him by 30 seconds.

2 – A gives B a head start of 3 mins and is beaten by 1000m.

Let’s study both of them and see what we can derive from them.

Scenario 1: A gives B a start of 200m and beats him by 30 seconds.

As we suggested before, we will start by making a diagram.



A runs from the Start line till the finish line i.e. a total distance of 2000 m.

A gives B a head start of 200 m so B starts, not from the starting point, but from 200 m ahead. A still beats him by 30 sec which means that A completes the race while B takes another 30 sec to complete it. So obviously A is much faster than B.

In this race, A covers 2000m. In the same time, B covers the distance shown by the red line. Since B needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance. The green line distance is given by (1/2)*s where s is the speed of B in meters per minute. The distance B has actually covered in the same time as A is the distance shown by the red line. This distance will be (1/2)*s less than  1800 i.e. it will be [1800 - (1/2)*s].

Scenario 2: A gives B a head start of 3mins and is beaten by 1000m.



A gives B a head start of 3 mins means B starts running first while A sits at the starting point. After 3 mins, B covers the distance shown by the red line which we do not know yet. Now, A starts running too. B beats A by 1000 m which means that B reaches the end point while A is still 1000 m away from the end i.e. at the mid point of the 2000 m track.

In this race, A covers a distance of 1000 m only. In that time, B covers the distance shown by the green line. The distance shown by the red line was covered by B in his first 3 mins i.e. this distance is 3*s. This distance shown by the green line is given by (2000 – 3s).

Now you see that in the first race, A covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, B would also have covered half the previous distance.

Distance covered by B in first race = 2*Distance covered by B in second race

1800 – (1/2)*s = 2*(2000 – 3s) (where s is the speed of B in meters/min)

s = 400 meters/min

Time taken by B to run a 2000 m race = Distance/Speed = 2000/400 = 5 min

Only one option has time taken by B as 5 mins and that must be the answer.

If required, you can easily calculate the time required by A too.

Distance covered by B in scenario 1 = 1800 – (1/2)*s = 1600 m

In the same time, A covers 2000 m which is a ratio of A:B = 5:4. Hence time taken by A:B will be 4:5.

Answer (B)

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

I did the following, wanted to run this by you:

so in the first case, A finished the race but B didn't , time taken by both was same

so for A, 2000 = a *t , where a = speed of A

for B 1800-30b = b*t,
because B was given a headstart of 200 m so B had to cover only 1800, but it didn't even cover that much because A reached the finish line 30 sec earlier and that's when the race ended, and since A beat B by 30 seconds, the distance that B had left up to the finish line was 30 seconds * speed of B = 30b, so net distance covered by B was 1800-30b,

for the second case, let time taken by A be t1, so A ran 1000 m at speed a :

so 1000 = a*t1,

now for B, 2000 = b*(t1+ 180 seconds),
I ended up with a relation between t1 and t and then found B's speed as 20/3 mps and hence time taken by B to completed 2000m = 5min.
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FROM Veritas Prep Blog: A 700+ GMAT Quant Question on Races

This week we will look at the question on races that we gave you last week.

Question 3: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately?

(A)   8, 10

(B)    4, 5

(C)   5, 9

(D)   6, 9

(E)    7, 10

Solution: Now this question is a little tougher than the previous ones we saw last week.

There are two scenarios given:

1 – A gives B a head start of 200 m and beats him by 30 seconds.

2 – A gives B a head start of 3 mins and is beaten by 1000m.

Let’s study both of them and see what we can derive from them.

Scenario 1: A gives B a start of 200m and beats him by 30 seconds.

As we suggested before, we will start by making a diagram.



A runs from the Start line till the finish line i.e. a total distance of 2000 m.

A gives B a head start of 200 m so B starts, not from the starting point, but from 200 m ahead. A still beats him by 30 sec which means that A completes the race while B takes another 30 sec to complete it. So obviously A is much faster than B.

In this race, A covers 2000m. In the same time, B covers the distance shown by the red line. Since B needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance. The green line distance is given by (1/2)*s where s is the speed of B in meters per minute. The distance B has actually covered in the same time as A is the distance shown by the red line. This distance will be (1/2)*s less than  1800 i.e. it will be [1800 - (1/2)*s].

Scenario 2: A gives B a head start of 3mins and is beaten by 1000m.



A gives B a head start of 3 mins means B starts running first while A sits at the starting point. After 3 mins, B covers the distance shown by the red line which we do not know yet. Now, A starts running too. B beats A by 1000 m which means that B reaches the end point while A is still 1000 m away from the end i.e. at the mid point of the 2000 m track.

In this race, A covers a distance of 1000 m only. In that time, B covers the distance shown by the green line. The distance shown by the red line was covered by B in his first 3 mins i.e. this distance is 3*s. This distance shown by the green line is given by (2000 – 3s).

Now you see that in the first race, A covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, B would also have covered half the previous distance.

Distance covered by B in first race = 2*Distance covered by B in second race

1800 – (1/2)*s = 2*(2000 – 3s) (where s is the speed of B in meters/min)

s = 400 meters/min

Time taken by B to run a 2000 m race = Distance/Speed = 2000/400 = 5 min

Only one option has time taken by B as 5 mins and that must be the answer.

If required, you can easily calculate the time required by A too.

Distance covered by B in scenario 1 = 1800 – (1/2)*s = 1600 m

In the same time, A covers 2000 m which is a ratio of A:B = 5:4. Hence time taken by A:B will be 4:5.

Answer (B)

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!
This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB

Hi VeritasKarishma,

I did the following, wanted to run this by you:

so in the first case, A finished the race but B didn't , time taken by both was same

so for A, 2000 = a *t , where a = speed of A

for B 1800-30b = b*t,
because B was given a headstart of 200 m so B had to cover only 1800, but it didn't even cover that much because A reached the finish line 30 sec earlier and that's when the race ended, and since A beat B by 30 seconds, the distance that B had left up to the finish line was 30 seconds * speed of B = 30b, so net distance covered by B was 1800-30b,

for the second case, let time taken by A be t1, so A ran 1000 m at speed a :

so 1000 = a*t1,

now for B, 2000 = b*(t1+ 180 seconds),
I ended up with a relation between t1 and t and then found B's speed as 20/3 mps and hence time taken by B to completed 2000m = 5min.
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Nothing wrong with the logic but think about how much time it took you to solve it using all these equations in so many variables. Mind you, almost all GMAT questions can be solved by taking no variable or one variable. Very rarely do you need to take 2.
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FROM Veritas Prep Blog: 3 Formats for GMAT Inequalities Questions You Need to Know

As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

  • Must Be True
  • Could Be True
  • Complete Range
Case 1: Must Be True

If |-x/3 + 1| < 2, which of the following must be true?

(A) x > 0

(B) x < 8

(C) x > -4

(D) 0 < x < 3

(E) None of the above

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2

(1/3) * |x – 3| < 2

|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0

Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8

Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4

Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3

Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Case 2: Could Be True

If −1 < x < 5, which is the following could be true?

(A) 2x > 10

(B) x > 17/3

(C) x^2 > 27

(D) 3x + x^2 < −2

(E) 2x – x^2 < 0

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10

x > 5

No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3

x > 5.67

No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27

x^2 – 27 > 0

x > 3*√(3) or x < -3*√(3)

√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

(Details on how to solve such inequalities are discussed here.)

(D) 3x + x^2 < −2

x^2 + 3x + 2 < 0

(x + 1)(x + 2) < 0

-2 < x < -1

No values of x will lie between -2 and -1, so this also cannot be true.

(E) 2x – x^2 < 0

x * (x – 2) > 0

x > 2 or x < 0

If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.

Case 3: Complete Range

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

(A) 0 < |x| < ½

(B) |x| > ½

(C) -½ < x < 0 or ½ < x

(D) x < -½ or 0 < x < ½

(E) x < -½ or x > 0

We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice.

We are given that x^3 – 4x^5 < 0. This inequality can be solved to:

x^3 ( 1 – 4x^2) < 0

x^3*(2x + 1)*(2x – 1) > 0

x > 1/2 or -1/2 < x < 0

This is our universe of the values of x. It is given that all values of x lie in this range.

Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.

We hope these practice problems will help you become able to distinguish between the three cases now.

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

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

The post 3 Formats for GMAT Inequalities Questions You Need to Know appeared first on Veritas Prep Blog.
This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB

Hi VeritasKarishma,

For the last questions where the entire range is asked for, as per your previous blog I should try bringing the inequality in (x-a) (x-b) <0 form and since we have x^3 (1-4x^2) < 0 here, I didn't do that and got the answer D. I realise my mistake here.
But here's my query:

in the first question we have |-x/3 + 1 | <2,
I got this one right using multiple methods, but if we have |x-a| or |a-x| it doesn't matter here right? We don't need to change |a-x| to |x-a| from by multiplying by -1 and flipping the sign, right?

We don't, right? Because |x-a| = |a-x|.
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Yes, |x - a| is equal to |a - x| in all cases.
|a - x| = |-(x - a)| = |x - a| because |-5| = |5|

Take some values for a and x to understand why.
Hence, whenever you see |a - x|, feel free to flip it to |x - a| for your own comfort without the need to modify anything else.
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Veritas Prep Blog 09 May 2022, 17:57
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VeritasPrepMarisa, can you please explain the logic behind multiplying 7C5 by 5! from FROM VERITAS PREP BLOG: WHEN PERMUTATIONS & COMBINATIONS AND DATA SUFFICIENCY COME TOGETHER ON THE GMAT! blog?
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VeritasPrepMarisa, can you please explain the logic behind multiplying 7C5 by 5! from FROM VERITAS PREP BLOG: WHEN PERMUTATIONS & COMBINATIONS AND DATA SUFFICIENCY COME TOGETHER ON THE GMAT! blog?
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Please link the post here along with the exact step in which you are facing issues.
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FROM Veritas Prep Blog: Using “Few” vs. “A Few” vs. “Quite a Few” in a GMAT Verbal Question

On quite a few occasions, we at Veritas Prep find ourselves explaining the difference between the terms “few” and “a few” – a subtle, but very important distinction which has, on occasion, completely changed the meaning of a sentence. Hence, we realized that a post on this difference is warranted.

“Few”, when used without a preceding “a”, means “very few” or “none at all”. “Few” is a negative, which puts the quantity of what you are describing near zero.

On the other hand, “a few” is used to indicate “not a large number”. “A few” also indicates a small approximate number, but it is positive nonetheless.

The difference between the two is subtle, yet there are instances where the two can mean completely opposite things. For example, “I have a few friends” is the same as saying “I have some friends”. “I have few friends”, however, implies that I have only very few friends (as opposed to many). It can also imply that I don’t feel very well about it, and I wish I had more friends.

Also, note that there is a very common expression, “quite a few”, which looks like it could mean “rather few” or “very few”, but it does not. It actually means the exact opposite: “a large or significant number” or “many”. So saying, “I have quite a few friends,” is the same as saying “I have quite a lot of friends”.



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

The post Using “Few” vs. “A Few” vs. “Quite a Few” in a GMAT Verbal Question appeared first on Veritas Prep Blog.
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This is a great post! I think we cannot compare all of them in one sentence, but would it mean that, if we were to quantify and compare "few", "a few" and "quite a few" --

(few) < (a few) < (quite a few)
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In pretty much every class I teach, at some point I’ll get the algebra vs. strategy question. Which is better? How do you know? I sympathize with the students’ confusion, as we’ll use the two approaches in different scenarios, but there doesn’t seem to be any magic formula to determine which is preferable. In many instances, both approaches will work fine, and the choice will mostly be a matter of taste and comfort for the test-taker.

In other cases, the question seems to have been specifically designed to thwart an algebraic approach. While there’s no official litmus test, there are some predictable structural clues that will often indicate that algebra is going to be nothing short of hemorrhage-inducing.

Here’s my personal heuristic; if an algebraic scenario involves hideously complex quadratic equations, I avoid the algebra. If, on the other hand, algebra leaves me with one or two linear equations to solve, it will almost certainly be a viable option. You might not recognize which category the question falls under until you’ve done a bit of leg-work. That’s fine. The key is not to get too invested in one approach and to have the patience and flexibility to alter your strategy midstream, if necessary.

Let’s look at some scenarios with unusually complex algebra. Here’s a GMATPrep® question:

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200

B. 19,600

C. 20,000

D. 20,400

E. 20,800

Simple enough. Let’s say the sides of this rectangular park are a and b. We know that the perimeter is 2a + 2b, so 2a + 2b = 560. Let’s simplify that to a + b = 280.

The diagonal of the park will split the rectangle into two right triangles with sides a and b and a hypotenuse of 200. We can use the Pythagorean theorem here to get: a^2 +b^2 = 200^2.

So now I’ve got two equations. All I have to do is solve the first and substitute into the second. If we solve the first for a, we get a = 280- b. Substitute that into the second to get: (280 – b)^2 + b^2 = 200^2. And then... we enter a world of algebraic pain. We’re probably a minute in at this point, and rather than flail away at that awful quadratic for several minutes, it’s better to take a breath, cleanse the mental palate, and try another approach that can get us to an answer in a minute or so.

Anytime we see a right triangle question on the GMAT, it’s worthwhile to consider the possibility that we’re dealing with one of our classic Pythagorean triples. If I see root 2? Probably dealing with a 45:45:90. If we see a root 3? Probably dealing with a 30:60:90. Here, I see that the hypotenuse is a multiple of 5, so let’s test to see if this is, in fact, a 3x:4x:5x triangle. If it is, then a + b should be 280.

Because 200 is the hypotenuse it corresponds to the 5x. 5x = 200 à x = 40. If x = 40, then 3x = 3*40 = 120 and 4x = 4*40 = 160. If the other two sides of the triangle are 120 and 160, they’ll sum to 280, which is consistent with the equation we assembled earlier.

And we’re basically done. If the sides are 120 and 160, we can just multiply to get 120*160 = 19,200. (And note that as soon as we see that ‘2’ is the first non-zero digit, we know what the answer has to be.)

Here’s one more from the Official Guide:

A store currently charges the same price for each towel that it sells. If the current price of each towel were to be increased by $1, 10 fewer of the towels could be bought for $120, excluding sales tax. What is the current price of each towel?

  • $1
  • $2
  • $3
  • $4
  • $12
First the algebraic setup. If we want T towels that we buy for D dollars each, and we’re spending $120, then we’ll have T*D = 120.

If the price were increased by $1, the new price would be D+1, and if we could buy 10 fewer towels, we could then afford T -10 towels, giving us (T-10)(D+1) = 120.

We could solve the first equation to get T = 120/D. Substituting into the second would give us (120/D – 10)(D + 1) = 120. Another painful quadratic. Cue hemorrhage.

So let’s work with the answers instead. Start with D. If the current price were $4, we could buy 30 towels for $120. If the price were increased by $1, the new price would be $5, and we could buy 120/5 = 24 towels. But we want there to be 10 fewer towels, not 6 fewer towels so D is out.

So let’s try B. If the initial price had been $2, we could have bought 60 towels. If the price had been $1 more, the price would have been $3, and we would have been able to buy 40 towels. Again, no good, we want it to be the case that we can buy 10 fewer towels, not 20 fewer towels.

Well, if $4 yields a gap that’s too narrow (difference of 6 towels), and $2 yields a gap that’s too large (difference of 20 towels), the answer will have to fall between them. Without even testing, I know it’s C, $3.

This is all to say that it’s a good idea to go into the test knowing that your first approach won’t always work. Be flexible. Sometimes the algebra will be clean and elegant. Sometimes a strategy is better. If the algebra yields a complex quadratic, there’s an easier way to solve. You just have to stay composed enough to find it.

* GMATPrep® questions courtesy of the Graduate Management Admissions Council.

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By David Goldstein, a Veritas Prep GMAT instructor based in Boston.[/rss2posts
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I understand your overall point in not getting too attached to one method of solving a problem, but both your examples can be solved with fairly straightforward algebra.

For Ex 1:
2a+2b = 560, simplify to a+b=280.
Then square both sides:
(a+b)^2=280^2
a^2 + 2ab + b^2 = 78400
a^2+b^2 = 78400 - 2ab

Take the information we received about the diagonal and apply Pythagorean theorem:
a^2 + b^2 = 200^2

Take first equation and substitute for a^2 + b^2

200^2 = 78400 - 2ab

Since we're looking for the area, we're looking for a*b. So just solve the above for ab:
ab = 19200 (A)


For Ex 2:
If we have T*D = 120 for current pricing for towels and (T-10)(D+1) = 120 if prices were increased by $1,

Take the first equation and solve for T -> T = 120/p. Then take second equation and simplify to T-10 = 120/(p+1). Since we know that the difference between the number of towels we can afford is 10, we can re-write the second equation such that T(current) - T(new) = 10

This gives us 120/p - 12/p+1 = 10. With that, it's a simple matter of plugging in the options. With Option C (3) we get:

(120/3) - 120/(3+1) = 30 which satisfies 40-30 = 10a
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