This week, let’s look at some more properties of exponents and roots. Using a high level data sufficiency question, we will see how a number x is related to ?x and to x^3.

Question: Is x > y?

Statement 1: ?x > y

Statement 2: x^3 > y

It is one of those gorgeous questions that seem very simple at first but surprise you later.

The question asks us whether x is greater than y, but the statements tell us the relation between ?x, x^3 and y. If we know that ?x is greater than y, when can we say that x is certainly greater than y too? If we know that x is greater than ?x, then we can say for sure that x is greater than y too. Is x always greater than ?x? No. Look at the diagram given below.

?x is not defined for negative values of x so let’s ignore the section to the left of 0. When the value of x lies between 0 and 1, x is less than ?x (for example: when ?x = 1/2, x = 1/4). When the value of x is greater than 1, x is greater than ?x (for example: when ?x = 2, x = 4).

Similarly, let’s look at the relation between x^3 and x. If we know that x^3 is greater than y, when can we say that x is certainly greater than y too? If we know that x is greater than x^3, then we can say for sure that x is greater than y too. Is x always greater than x^3? No. Look at the graph below.

When the value of x lies between -1 and 0 or in the region greater than 1, x is less than x^3 (for example: when x = 2, x^3 = 8). When the value of x lies in the region less than -1 or between 0 and 1, x is greater than x^3 (for example: when x = 1/2, x^3 = 1/8)

Let’s look at the statements now

Statement 1: ?x > y

Since ?x is not defined for negative x, we get that x >= 0. As we saw in the first graph above, for some values, x is greater than ?x, for others, x is less than ?x. When x is less than ?x, x may not be greater than y. So this statement alone is not sufficient.

Statement 2: x^3 > y

As we saw in the second graph above, for some values, x is greater than x^3, for others, x is less than x^3. When x is less than x^3, x may not be greater than y. So this statement alone is not sufficient.

Using both the statements together, we know that x >= 0.

When x lies between 0 and 1, we know that x >= x^3. Since statement (2) says that x^3 > y, we can say that x > y.

When x is greater than 1, we know that x > ?x. Since statement (1) says that ?x > y, we can deduce that x > y.

Therefore, for all possible values of x, we can say that x > y. Together the statements are sufficient. Answer (C).

It is important to understand these relations. This concept is very useful, especially for GMAT Algebra!

*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 in Detroit, and regularly participates in content development projects such as this blog!*

## 10 thoughts on “Quarter Wit, Quarter Wisdom: Analyzing a 700+ Quant Question”

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Hi Karishma,

I’m still not really understanding your explanation towards the end:

“Therefore, for all possible values of x, we can say that x > y…”

How can combining the two statements together make all possible values of x > y when it only satisfies one of the statements?

For example, if I pick x = 36. After combining the two statements, I know that this satisfies statement one (since 36 > 36^1/2 > 6), however, when 36 is plugged into statement two, it no longer satisfies x > y (i.e., 36 is NOT > 36^3) …So how can you conclude that both statements combined still provide significant information if it still only satisfies one or the other?

Thank you!

Hey Derek,

I will explain again.

We say that we are now going to combine information from both the statements.

We know from the statements that ?x > y and x^3 > y.

The question is: Is x > y?

Since ?x is real, x must be 0 or a positive number. So basically, x can take any value on the number line to the right of 0. x can be 0, 1/10, 1/2, 1, 26, 10987 etc

Between 0 and 1, say when x = 1/4, x^3 = 1/64. 1/64 is less than 1/4. But this 1/64 is greater than y (because we know that x^3 > y). So obviously x (i.e. 1/4) is greater than y too. This is true for any number between 0 and 1. Also, ?x (1/2 here) is greater than y but this statement doesn’t help us much right now since ?x (i.e. 1/2) is greater than x (1/4). It is good to know that 1/2 > y but it doesn’t help us to prove that 1/4 > y too. Both statements are valid of course but the second one is helpful to deduce that between 0 and 1, x is greater than y.

Similarly, if x is greater than 1, say when x = 4, ?x = 2. From statement 1, we know that ?x > y i.e. 2>y. Then obviously, 4>y too. This is true for all numbers greater than (or equal to) 1. Again, here as well, x^3 i.e. 4^3 = 64 is greater than y but that doesn’t tell us whether 4 is greater than y. Statement 2 is valid as well but not very helpful.

Let’s pick x = 36.

We know from statement 1: ?x > y

i.e. 6 > y

We know that 6 is greater than y. Can we say that 36>y? Sure! If 6 is already greater than y, 36 has to be greater than y too.

We know from statement 2: x^3 > y

36^3 > y

If 36^3 is greater than y, can we say necessarily that 36 is greater than y too? No, we cannot. But it doesn’t matter since the first piece of information already proved that 36 is greater than y.

For x = 36 i.e. a number greater than 1, statement 1 helped us prove that x is greater than y.

For positive numbers less than 1, statement 2 helps us prove that x is greater than y. Hence we need both statements to prove that all values that x can take are definitely greater than y.

Hi Karishma,

Ahhhh…thank you for your further explanation. I had some misconceptions in my head around the whole “x>x^(1/2)” and “x>x^3” as a means of testing the sufficiency of the problems separately (and how the concept is applied once the statements are combined).

While it was, for some reason, very difficult for me to grasp the idea of how the 2 statements combined to work together to solve the problem. I now understand it!

Thank you for further explaining and demystifying this problem. Problems with inequalities tend to trip me up quite a bit (give me an absolute value AND inequality problem and I’m screwed! hehe…)

While I am currently enrolled in aVeritas Live Online class right now, can you offer any advice as to how to help me strengthen the concepts of inequalities and absolute values? Or know of any resources (either Veritas-related or not) that can provide me with some good practice problems?

Thank you! So far my experience with Veritas (both in and outside of the classroom) has been wonderful, thanks for adding to that.

Best,

Derek

Hey Derek,

It’s good to know that you are happy with your Veritas association.

And yes, I know inequalities and mods together can be a nightmare for many.

Here is my suggestion:

Step 1. Your Veritas Algebra book discusses the basics of both these topics. Make sure you are comfortable with the explanations and questions there.

Step 2. I have discussed some trickier concepts in my previous posts:

http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-do-what-dumbledore-did/

http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-the-holistic-approach-to-mods/

http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-the-holistic-approach-to-mods-part-ii/

The first post has extensive application.

Step 3. For more questions, you can check out some GMAT forums e.g. Beat the GMAT, GMAT Club etc. If you come across something that you are not very comfortable with, post it in the Veritas forum (which you can access through your TrueTrack account) and we will get back with an explanation.

Best,

Karishma

Hi , I still have the doubt combining. 1 and 2.

Both statements for some condition or the other fails in the interval 0 to 1

Both statements are true for the interval for 1 and above in the scale.

If we combine x is greater than y for the interval 1 and above.

Am i wrong. Please help.

The important thing to remember when you are combining the statements is that both statements are given to be true. So x can take only those values for which ?x > y and x^3 > y too. Now, between 0 and 1, x is greater than x^3 and you know that x^3 > y so x is greater than y and in the range x>1, x is greater than ?x and you know that ?x > y so again x is greater than y.

Say, you know ?x > y and x^3 > y

x = 1/2

x^3 = 1/8

We know for sure than 1/8 is greater than y so obviously x > y too.

x = 4

?x = 2

We know that 2 > y so obviously x > y too.

In either case, one of the statements will ensure that x > y.

For more, check out the explanation I gave to Derek above.

Hi Karishma,

Could you please let me know where I am making an error –

My explanation –

?x can be defined for positive values of x. Hence x must be positive. Now, 2 cases arises –

A) when y is negative

b) when y is positive.

Taking case 1

When y is negative.

Since positive> negative.

i.e ?x > y

hence, x>y (pos> neg) i.e Answer will be YES.

If second case also provides us YES, then sufficiency will be proved.

Taking Case 2

Both are positive.

?x >y (positive > positive)

Since x> ?x

Hence, x> y

First of all, you need to consider 0 too. ?x is defined for all non negative x.

Secondly, how do you figure that x> ?x for all non negative x. Think x = 0 or x = 1/4. ?x = 0 or 1/2 respectively. In both the cases x is not greater than ?x.

Mind you, in the range 0 to 1, numbers behave differently.

Basically, when looking at number properties, the different ranges to consider are

-infinity to -1,

-1 to 0,

0 to 1

and 1 to infinity.

Hi Karishma,

You explained well with number line and behavior of nos in range ‘-1’-‘0’-‘1’.

Would like to know if you have similar explanations for other similar question types (as no properties change with shift in value from -ve to positive & b/w 0 & 1 ,& -1 & 0) with the number line..in addition to links of mods.

Thanks a lot.

Excellent post.

Kindly evaluate my answer on the example problem.

Question: Is x > y?

Statement 1: ?x > y

Statement 2: x^3 > y

My solution:

1. Statement I

In order to prove x > y, we are required to prove x > (x)^(1/2)

The inequality is true for x > 1 for real positive x.

As such this statement alone is not sufficient as the inequality reverses its sign for x between 0 and 1.

2. Statement 2

In order to prove x > y, we are required to prove x > (x)^(3)

This inequality is true for x < 0 and 0 < x (x)^(3) and x > (x)^(1/2) are satisfied.

Hence answer is (E).

As my answer is different than (C) as given above, I must be missing something.

Kindly explain.

Thanks.