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# Is x > y ? (1) x^(1/2) > y (2) x^3 > y

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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. In the area of inequality, the number of equations means the number of inequalities, since the number of equations means the number of conditions.

In the original question, the number of variables is 2 and the number of equation is 0.

Hence, C is most likely the answer.

For the condition (1), we have two cases.
$$x = 1/4$$ and $$y = 1/3$$ satisfies the condition (1), but $$x > y$$ is not true.
$$x = 4$$ and $$y = 1$$ satisfies the condition (1) and $$x > y$$ is true.
Thus, the condition (1) is not sufficient.

For the condition (2), we also have two cases.
$$x = 2$$ and $$y = 3$$ satisfies the condition (2), but $$x > y$$ is not true.
$$x = 3$$ and $$y = 2$$ satisfies the condition (2) and $$x > y$$ is true.
Thus, the condition (2) is not sufficient, either.

For both the conditions (1) and (2) together, we should consider two cases.

Case 1: $$0 \leq x \leq 1$$.
Since $$x^3 \leq x$$, if $$x^3 > y$$ by the condition (2), $$x \geq x^3 > y$$, that is, $$x >y$$.

Case 2: $$x > 1$$
Since $$x > \sqrt{x}$$, if $$\sqrt{x} > y$$ by the condition (1), $$x > \sqrt{x} >y$$, that is, $$x > y$$.

Hence, both the conditions (1) and (2) together are sufficient, as expected earlier.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

One of those gorgeous questions that seem so simple at first but surprise you later...
Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!)

Statement (1): If I can say that $$x >= \sqrt{x}$$ for all values of x, then I can say that x > y. The green line shows me the region where $$x >= \sqrt{x}$$ but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
I also understand from this statement that x >= 0.

Statement (2): If I can say that $$x >= x^3$$ for all values of x, then I can say that x > y. The green lines show me the region where $$x >= x^3$$ but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
Attachment:
Ques.jpg

Using both together, I know x >= 0.
If 0 <= x <= 1, then we know $$x >= x^3$$. Since statement (2) says that $$x^3 > y$$, I can say that x > y.
If x > 1, then we know $$x > \sqrt{x}$$. Since statement (1) says that $$\sqrt{x} > y$$, I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).

great approach, thanks a lot.
But here I have a doubt (excuse me if silly ) -
When we are combining both graphs - its clear that X>Y when 0>x>1 AND 1>x>+.
But what about when the value of x is between "-1 to 0".....as shown with red color in second graph.

regards,
Tamal
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
tamal99 wrote:
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

One of those gorgeous questions that seem so simple at first but surprise you later...
Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!)

Statement (1): If I can say that $$x >= \sqrt{x}$$ for all values of x, then I can say that x > y. The green line shows me the region where $$x >= \sqrt{x}$$ but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
I also understand from this statement that x >= 0.

Statement (2): If I can say that $$x >= x^3$$ for all values of x, then I can say that x > y. The green lines show me the region where $$x >= x^3$$ but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
Attachment:
Ques.jpg

Using both together, I know x >= 0.
If 0 <= x <= 1, then we know $$x >= x^3$$. Since statement (2) says that $$x^3 > y$$, I can say that x > y.
If x > 1, then we know $$x > \sqrt{x}$$. Since statement (1) says that $$\sqrt{x} > y$$, I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).

great approach, thanks a lot.
But here I have a doubt (excuse me if silly ) -
When we are combining both graphs - its clear that X>Y when 0>x>1 AND 1>x>+.
But what about when the value of x is between "-1 to 0".....as shown with red color in second graph.

regards,
Tamal

Stmnt 1 tells you that x >= 0 (since $$\sqrt{x}$$ is not defined for negative values of x). Hence, when you use both stmnts together, negative values of x are to be ignored.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
shrouded1 wrote:
Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$

(2) $$x^3 > y$$

(1) $$\sqrt{x} > y$$
We can say that x is positive, as we deal with square roots of positive numbers only.
$$x\geq \sqrt{x} > y$$......When $$x\geq 1$$
But when 0<x<1, $$\sqrt{x}>x$$ and we can not say anything about placement of y.
$$\sqrt{x}>x>y$$.......$$\sqrt{\frac{1}{4}}>\frac{1}{4}>0$$
$$\sqrt{x}>y>x$$.......$$\sqrt{\frac{1}{4}}>\frac{1}{3}>\frac{1}{4}$$
Insufficient

(2) $$x^3 > y$$
$$x\geq x^3 > y$$......When $$x< 1$$
But when x>1, $$x^3>x$$ and we can not say anything about placement of y.
$$x^3>x>y$$.......$$2^3>2>0$$
$$x^3>y>x$$.......$$2^3>5>2$$
Insufficient

Combined
x is positive and both $$x^3$$ and $$\sqrt{x}$$ are greater than y.
Say x>1......$$x^3>x>\sqrt{x}>y$$
But if x<1....$$\sqrt{x}>x>x^3>y$$
x=1.....$$\sqrt{x}=x=x^3>y$$
Hence, in all cases, x>y

Sufficient

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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
Bunuel wrote:
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

Is x>y?

(1) $$\sqrt{x}>y$$ --> if $$x=4$$ and $$y=1$$ then the answer will be YES but if $$x=\frac{1}{4}$$ and $$y=\frac{1}{3}$$ then the answer will be NO. Two different answers, hence not sufficient.

Note that from this statement we can derive that $$x\geq{0}$$ because an expression under the square root cannot be negative.

(2) $$x^3>y$$ --> if $$x=4$$ and $$y=1$$ then the answer will be YES but if $$x=2$$ and $$y=3$$ then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) From (1) we have that $$x\geq{0}$$. Now, $$\sqrt{x}$$, $$x$$, $$x^3$$ can be positioned on a number line only in 2 ways:

1. For $$1\leq{x}$$: ------$$\sqrt{x}$$----$$x$$----$$x^3$$, so $$1\leq{\sqrt{x}}\leq{x}\leq{x^3}$$ (the case $$\sqrt{x}=x=x^3$$ is when $$x=1$$). $$y$$ is somewhere in green zone (as $$y<\sqrt{x}$$ and $$y<x^3$$), so if we have this case answer is always YES: $$y<x$$.

2. For $$0\leq{x}<1$$: $$0$$----$$x^3$$----$$x$$----$$\sqrt{x}$$----$$1$$, so $$0\leq{x^3}\leq{x}\leq{\sqrt{x}}$$. $$y$$ is somewhere in green zone (as $$y<\sqrt{x}$$ and $$y<x^3$$), so if we have this case answer is always YES: $$y<x$$.

So in both cases $$y<x$$. Sufficient.

Hope it's clear.

Hi Bunuel can y have negative value

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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
1
Kudos
puneetfitness wrote:
Bunuel wrote:
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$
(2) $$x^3 > y$$

Is x>y?

(1) $$\sqrt{x}>y$$ --> if $$x=4$$ and $$y=1$$ then the answer will be YES but if $$x=\frac{1}{4}$$ and $$y=\frac{1}{3}$$ then the answer will be NO. Two different answers, hence not sufficient.

Note that from this statement we can derive that $$x\geq{0}$$ because an expression under the square root cannot be negative.

(2) $$x^3>y$$ --> if $$x=4$$ and $$y=1$$ then the answer will be YES but if $$x=2$$ and $$y=3$$ then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) From (1) we have that $$x\geq{0}$$. Now, $$\sqrt{x}$$, $$x$$, $$x^3$$ can be positioned on a number line only in 2 ways:

1. For $$1\leq{x}$$: ------$$\sqrt{x}$$----$$x$$----$$x^3$$, so $$1\leq{\sqrt{x}}\leq{x}\leq{x^3}$$ (the case $$\sqrt{x}=x=x^3$$ is when $$x=1$$). $$y$$ is somewhere in green zone (as $$y<\sqrt{x}$$ and $$y<x^3$$), so if we have this case answer is always YES: $$y<x$$.

2. For $$0\leq{x}<1$$: $$0$$----$$x^3$$----$$x$$----$$\sqrt{x}$$----$$1$$, so $$0\leq{x^3}\leq{x}\leq{\sqrt{x}}$$. $$y$$ is somewhere in green zone (as $$y<\sqrt{x}$$ and $$y<x^3$$), so if we have this case answer is always YES: $$y<x$$.

So in both cases $$y<x$$. Sufficient.

Hope it's clear.

Hi Bunuel can y have negative value

Posted from my mobile device

Good question. When we consider the two cases together, y can absolutely be some negative number, say -10. If y is negative and since we concluded that x must be positive, then in this case (y = negative) < (x = positive). So, we'd get the same YES answer to the question: x is greater than y.

Does this make sense?
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Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
shrouded1 wrote:
Is $$x > y$$ ?

(1) $$\sqrt{x} > y$$

(2) $$x^3 > y$$

I thought this was a really tough question !

We can use the concept of a number line to solve this question. Attached video solution to the question -

Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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