How to Quickly Interpret Ranges of Variables in GMAT Questions

Quarter Wit, Quarter WisdomSometimes a GMAT Quant question will give us multiple ranges of values that apply to a single variable, and when this happens it can really take us for a ride. Evaluating these ranges to arrive at deductions can extremely confusing, so today we will look at some strategies for how to deal with such problems.

To start off, let’s take a look at an example problem:

If it is true that z < 8 and 2z > -4, which of the following must be true?

(A) -8 < z < 4
(B) z > 2
(C) z > -8
(D) z < 4
(E) None of the above

Given that z < 8 and 2z > -4, we know that z > -2. This means -2 < z < 8. z must lie within that range, hence z can take values such as -1, 0, 5, 7.4, etc.

Now, which of the given answer choices would hold true for ALL such values? Let’s examine each option and see:

(A) -8 < z < 4
We know that z may be more than 4, so this range does not hold true for all possible values of z.

(B) z > 2
We know that z may be less than 2, so this also does not hold true for all possible values of z.

(C) z > -8
No matter what value z will take, it will always be more than -8. This range holds true for all values of z.

(D) z < 4
We know that z may be greater than 4, so this does not hold for all possible values of z.

Our answer is C.

To understand this concept more clearly, let’s use a real life example:

We know that Anna’s weight is more than 120 pounds but less than 130 pounds. Which of the following is definitely true about her weight?

(A) Her weight is 125 pounds.
(B) Her weight is more than 124 pounds.
(C) Her weight is less than 127 pounds.
(D) Her weight is more than 110 pounds.

Can we say that her weight is 125 pounds? No – we just know that it is more than 120 but less than 130. It could be anything in this range, such as 122, 125, 127.5, etc.

Can we say that her weight is more than 124 pounds? This may be true, but it might not be true. Knowing our given range, her weight could very well be 121 pounds, instead.

Can we say her weight is less than 127 pounds? Again, this might not necessarily be true. Her weight could be 128 pounds.

Now, can we say that her weight is more than 110 pounds? Yes – since we know Anna’s weight is between 120 and 130 pounds, it must be more than 110 pounds.

This question uses the same concept as the first question! If you look at that question again, it will hopefully make much more sense. Now try solving this example problem:

If 1/55 < x < 1/22 and 1/33 < x < 1/11, then which of the following could be the value of x?

(i) 1/54
(ii) 1/23
(iii) 1/12

(A) Only (i)
(B) Only (ii)
(C) (i) and (ii)
(D) (ii) and (iii)
(E) (i), (ii) and (iii)

In this problem, we are given two ranges of x. We know that 1/55 < x < 1/22 and 1/33 < x < 1/11, so x is greater than 1/55 AND it is greater than 1/33. Since 1/33 is greater than 1/55 (the smaller the denominator, the larger the number), we just need to know that x will be greater than 1/33.

We are also given that x is less than 1/22 AND it is less than 1/11. Since 1/22 is less than 1/11, we really just need to know that x is less than 1/22.

Hence, the range for x should be 1/33 < x < 1/22. x could take all values that lie within this range, such as 1/32, 1/31, 1/24, 1/23, etc.

Looking at the answer choices, we can see that 1/54 and 1/12 (i and iii) are both out of this range. Therefore, our answer is B.

If we go back to our real life example, this is what the question would look like now:

We know that Anna’s weight is more than 110 pounds but less than 130 pounds. We also know that her weight is more than 115 pounds but less than 140 pounds. Which of the following is definitely true about her weight?

(A) Her weight is 112 pounds.
(B) Her weight is 124 pounds.
(C) Her weight is 135 pounds.

We are given that Anna’s weight is more than 110 pounds and also more than 115 pounds. Since 115 is more than 110, we just need to know that her weight is more than 115 pounds. We are also given that Anna’s weight is less than 130 pounds and also less than 140 pounds. Since 130 is less than 140, we just need to know that her weight is less than 130 pounds.

Now we have the following range: 115 pounds < Anna’s weight < 130 pounds. Only answer choice B lies within this range, so that is our answer.

We hope you see that evaluating ranges of numbers on GMAT questions is not difficult when we consider them in terms of a real life example. The same logic that we use in the simple weight problem is also applicable when algebraic data is given.

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: When Can You Divide by a Variable?

Quarter Wit, Quarter WisdomWe have often come across test takers confused about division by a variable. When is it allowed, when is it not allowed? Why is it allowed in some cases and not in others? What are the constraints we need to look out for?

For example:

Is division by x allowed here: x^2 = 10x?
Is division by x allowed here: y = 4x?
Is division by x allowed here: x^2 < 4x?

Let’s take a detailed look at all these questions today.

The basic guidelines:

  1. Division by 0 is not allowed, hence you cannot divide by a variable until and unless we know that it cannot be 0.
  2. In the case of an inequality, when you divide by a negative number, the sign of the inequality flips. So we cannot divide by a variable until and unless we know that it cannot be 0 AND whether it is positive or negative.

Let’s look at the three questions given above and try to solve them using these guidelines:

Is division by x allowed here: x^2 = 10x?

The first thing to find out here is whether or not x can equal 0.

Case 1: If no other information has been given, then x can be 0 and we cannot divide by it. This is how we proceed in that case:

x^2 – 10x = 0
x(x – 10) = 0
x = 0 or 10

Case 2: If the question stem tells us that x is not 0, then we can divide by x.

x^2/x = 10x/x
x = 10

Obviously, we don’t get the second solution (x = 0) in this case, as we already know that x cannot be 0. Now let’s look at the second problem:

Is division by x allowed here: y = 4x?

Again, this is an equation and we need to know whether or not x can equal 0.

Case 1: If x can be 0, you cannot divide by it. In this case, x = 0 and y = 0 is one of the infinite possible solutions.

Case 2: If the question stem states that x cannot be 0, then we can do the following:

y/x = 4

Now let’s look at the final question:

Is division by x allowed here: x^2 > -4x?

Here, we have an inequality. Before deciding whether we can divide by x or not, we need to know not only whether x can be equal to 0, but also whether x is positive or negative.

Case 1: If we know nothing about the possible values that x can take, then this is how we proceed:

x^2 + 4x > 0
x(x + 4) > 0

Now we can use the method discussed in the first problem to arrive at the range of x.

x > 0 or x < -4

Case 2: If we know that x is positive, then we can proceed like this:

x^2/x > -4x/x
x > -4

Since we are given that x is positive, we know that that x > 0 (looking at the two options above).

Case 3: If we know that x is negative, then this is how we will proceed:

x^2/x < -4x/x (we flip the sign of the inequality because we divide by x, which is negative)
x < -4

The results obtained are logical, right? When x can be anywhere on the number line, we get the range as x > 0 or x < -4.

If x has to be positive, the range is x > 0.
If x has to be negative, the range is x < -4.

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

3 Formats for GMAT Inequalities Questions You Need to Know

Quarter Wit, Quarter WisdomAs 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:

  1. Must Be True
  2. Could Be True
  3. 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
> 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 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 V

Quarter Wit, Quarter WisdomWe 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.
QWQW image 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.

QWQW image 1

 

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 III

Quarter Wit, Quarter WisdomA 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:

QWQW

 

 

 

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!