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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]

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08 Dec 2014, 07:48
tarikk wrote:
gurpreetsingh wrote:
Statement 1: $$\sqrt{x} > y$$ ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient.

Statement 2: $$x^3 > y$$ take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient.

take statement 1 and 2 together. Now the answer could be either C or E.

Either y is -ve or postive. If y is -ve then x >y always holds true.

If y is +ve then x>y^2 and x^3>y => x^6 > y^2

divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y.

Hence C.

The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains.

Hi gurpreetsingh, could you please elaborate on:
gurpreetsingh wrote:
If y is +ve then x>y^2 and x^3>y => x^6 > y^2
. Thank you!

If y is positive, then we can square x^3 > y to get x^6 > y^2.

Check RAISING INEQUALITIES TO EVEN/ODD POWER in Tips on Inequalities.

Hope it helps.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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06 May 2015, 11:59
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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30 Nov 2015, 04:56
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.

Bunuel : Thanks for the great explanation. I have understood completely except for one small query I have about the second case after combining both statements. You have mentioned that Y is anywhere between $$x^3$$ and 0. But cannot Y be ANYWHERE left of $$x^3$$ (in other words, cannot y be negative) ?
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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30 Nov 2015, 05:01
nishantsharma87 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.

Bunuel : Thanks for the great explanation. I have understood completely except for one small query I have about the second case after combining both statements. You have mentioned that Y is anywhere between $$x^3$$ and 0. But cannot Y be ANYWHERE left of $$x^3$$ (in other words, cannot y be negative) ?

If y is negative, then it would naturally be less than x, which from (1) is positive.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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03 Dec 2015, 05:52
1
2
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.

Is x>y ?

(1) x √ >y
(2) x 3 >y

There are 2 variables (x,y) and 2 equations are given by the conditions, so there is high chance (C) will be the answer.
Looking at the conditions together,
if 0<x<1, sqrt (x)>x>x^3>y answers the question 'yes', and if x>1, x^3>x>sqrt(x)>y also answers the question 'yes'

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink] ### Show Tags 18 Mar 2017, 20:32 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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15 Aug 2018, 07:39
I think I have a quick way to solve this one.

1) Square root (x) > y but Can we compare x and square root (x) ? No because if 0<x<1 then square root (x) is larger than x. And if 1<x then square root (x) is smaller x. NOT SUF
2) x cube > y but can we compare x and x cube ? No because if -1<x<0 then x cube is larger than x and if 0<x<1 then x cube is smaller than x.

Together (1) (2) -> Square root (x) is larger than y, x cube is larger than y -> If we power x, x larger than y, if we root x, x still larger than y. Hence, just x must be larger than y.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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28 Aug 2018, 00:47
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]

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28 Aug 2018, 03:36
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]

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28 Aug 2018, 08:08
shrouded1 wrote:
Is $$x > y$$ ?

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

$$x\,\,\,\mathop > \limits^? \,\,\,y$$

$$\left( 1 \right)\,\,\,\sqrt x > y\,\,\,\,\,\left\{ \begin{gathered} \,Take\,\,\,\left( {x;y} \right) = \left( {1\,;\,0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \,Take\,\,\,\left( {x;y} \right) = \left( {\frac{1}{4}\,;\,\frac{1}{3}} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\ \end{gathered} \right.$$

$$\left( 2 \right)\,\,\,{x^3} > y\,\,\,\,\,\left\{ \begin{gathered} \,(re)\,Take\,\,\,\left( {x;y} \right) = \left( {1;0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \,Take\,\,\,\left( {x;y} \right) = \left( {2\,;\,7} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\ \end{gathered} \right.$$

$$\left( {1 + 2} \right)\,\,\,\,\,$$

$$\left. {x = y\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered} \mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,\sqrt x > x\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < \sqrt x < 1\,\,\,\,\,\,\,\,\left( {\sqrt x \geqslant 1\,\,\, \Rightarrow \,\,\,\sqrt x > x\,\,{\text{false,}}\,\,\,{\text{impossible}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\boxed{0 < x < 1} \hfill \\ \mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,{x^3} > x\,\,\,\,\, \Rightarrow \,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \ne y$$

$$x < y\left. {\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered} \mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,x < y < \sqrt x \,\,\,\,\, \Rightarrow \,\,\,\,\,\sqrt x > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{above}}} \,\,\,\,\,\,\boxed{0 < x < 1}\,\,\,\, \hfill \\ \mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,x < y < {x^3}\,\,\,\, \Rightarrow \,\,\,{x^3} > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{again}}} \,\,\,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,{\text{is}}\,\,{\text{false}}$$

Conclusion: $$\left( {1 + 2} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle$$

The above follows the notations and rationale taught in the GMATH method.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y  [#permalink]

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01 Jul 2019, 06:08
Wow, this is really a good question. You don't need to list the possibilities since it could miss some points.

Just draw the lines of y=x, y=sqrt X and y=x^3.

X>Y means the area is below y=x line.
1): sqrt X > Y: means y under this line. But when x<1, it is possible to have value between y=x and y=sqrt x
2): X^3 is only under y=x when 0<x<1, not sufficient

https://images.app.goo.gl/jqpBNJr4JJRgV2Y47

If combien 1 and 2, it means you have to take the area under x^3 when <1 and sqrt x when >1; in those two areas you have y definitely less than y=x

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

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16 Aug 2019, 04:13
my query is related with the graph ...
as I know when ,
1) 1<x = sqrt x< x< x^2<x^3
2) when 0<x<1= X^3<x^2<x< sqrt x

can anyone please explain me this graph when -1<x<0 and X<-1
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Re: Is x > y? (1) sqrt (x) > y (2) x3 > y  [#permalink]

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01 Oct 2019, 04:21
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Re: Is x > y? (1) sqrt (x) > y (2) x3 > y   [#permalink] 01 Oct 2019, 04:21

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