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If x and y are positive, is x^3 > y?

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If x and y are positive, is x^3 > y? [#permalink] New post 30 Aug 2010, 08:38
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If x and y are positive, is x^3 > y?

(1) \sqrt{x} > y
(2) x > y
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Dec 2012, 02:58, edited 2 times in total.
Edited the question.
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Re: Inequality DS [#permalink] New post 30 Aug 2010, 09:31
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Expert's post
udaymathapati wrote:
If x and y are positive, is x^3 > y?
(1) \sqrt{x} > y
(2) x > y

My approach:
1.\sqrt{x}>y ....>x>y^2....NS
2. x>y ....NS

Combine...y^2>x>y....> x^3>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.


If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) \sqrt{x}>y --> if x=1 and y=\frac{1}{2} then the answer will be YES but if x=\frac{1}{4} and y=\frac{1}{5} then the answer will be NO. Two different answers, hence not sufficient.

(2) x>y --> if x=1 and y=\frac{1}{2} then the answer will be YES but if x=\frac{1}{4} and y=\frac{1}{5} then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

Answer: E.

ALGEBRAIC APPROACH:

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), so if we have this case answer is always YES: y<x^3.

But:

For 0<x<1: 0----x^3----x----\sqrt{x}----1, so 0<x^3<x<\sqrt{x}. y is somewhere in green or red zone (as y<\sqrt{x} and y<x), so if we have this case answer is sometimes YES: y<x^3 (if y is in green zone), and sometimes NO: x^3<y (if y is in red zone). In fact in this case y=x^3 is also possible, for example when x=\frac{1}{2} and y=\frac{1}{8}

Answer: E.

Hope it's clear.
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Re: Inequality DS [#permalink] New post 31 Aug 2010, 03:07
When there are squaring or cubing is involved, always remember -ve values as well as the values between -1 and 1
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Re: Inequality DS [#permalink] New post 18 Mar 2011, 06:11
Bunuel wrote:
udaymathapati wrote:
If x and y are positive, is x^3 > y?
(1) \sqrt{x} > y
(2) x > y

My approach:
1.\sqrt{x}>y ....>x>y^2....NS
2. x>y ....NS

Combine...y^2>x>y....> x^3>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.


If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) \sqrt{x}>y --> if x=1 and y=\frac{1}{2} then the answer will be YES but if x=\frac{1}{4} and y=\frac{1}{5} then the answer will be NO. Two different answers, hence not sufficient.

(2) x>y --> if x=1 and y=\frac{1}{2} then the answer will be YES but if x=\frac{1}{4} and y=\frac{1}{5} then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

Answer: E.

ALGEBRAIC APPROACH:

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), so if we have this case answer is always YES: y<x^3.

But:

For 0<x<1: 0----x^3----x----\sqrt{x}----1, so 0<x^3<x<\sqrt{x}. y is somewhere in green or red zone (as y<\sqrt{x} and y<x), so if we have this case answer is sometimes YES: y<x^3 (if y is in green zone), and sometimes NO: x^3<y (if y is in red zone). In fact in this case y=x^3 is also possible, for example when x=\frac{1}{2} and y=\frac{1}{8}

Answer: E.

Hope it's clear.




+1 for u
Great Post Thanks Bunuel
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Re: Inequality DS [#permalink] New post 18 Mar 2011, 20:25
x = 1/4, y = 1/8

From (1)

1/2 > 1/8, but 1/8 = 1/8

but (9)^3 > 2

and sqrt(9) > 2

In (2) also, the same is true.

Answer - E.
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if x and y are positive, is x³>y? [#permalink] New post 31 Jan 2014, 21:05
Hi guys!

i have a question regarding the following DS problem:

if x and y are positive, is x³>y?

(1) \sqrt{'x'}>y
(2) x>y

ok, statement 1 is insufficient, because x³>y when x>1 (e.g. x=4 and y= 1 --> \sqrt{'4'}=2>1 and 4³>1) but for fractional x-values x³<y (e.g. x=\frac{'1}{100'} and y=\frac{'1}{20'} --> \sqrt{\frac{'1}{100'}}=\frac{'1}{10'}>\frac{'1}{20'} but \frac{'1}{100'}³<\frac{'1}{20'}) -----> insufficient

statement 2: the same logic can be applied here, x³>y when x>1 but can be x³<y when x attains a fractional value.

now, taken (1) and (2) together, my reasoning is the following:

we square the first equation to obtain x²>y and divide this by the second equation to get 1>y. now, since x>y it follows that x>=1>y, from which we can conclude that x³>y. my answer was therefore, C, which however seemed to be wrong. where's the error here in my approach?

i'm thankful for any help! :) cheers
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Re: if x and y are positive, is x³>y? [#permalink] New post 01 Feb 2014, 03:40
Expert's post
damamikus wrote:
Hi guys!

i have a question regarding the following DS problem:

if x and y are positive, is x³>y?

(1) \sqrt{'x'}>y
(2) x>y

ok, statement 1 is insufficient, because x³>y when x>1 (e.g. x=4 and y= 1 --> \sqrt{'4'}=2>1 and 4³>1) but for fractional x-values x³<y (e.g. x=\frac{'1}{100'} and y=\frac{'1}{20'} --> \sqrt{\frac{'1}{100'}}=\frac{'1}{10'}>\frac{'1}{20'} but \frac{'1}{100'}³<\frac{'1}{20'}) -----> insufficient

statement 2: the same logic can be applied here, x³>y when x>1 but can be x³<y when x attains a fractional value.

now, taken (1) and (2) together, my reasoning is the following:

we square the first equation to obtain x²>y and divide this by the second equation to get 1>y. now, since x>y it follows that x>=1>y, from which we can conclude that x³>y. my answer was therefore, C, which however seemed to be wrong. where's the error here in my approach?

i'm thankful for any help! :) cheers


Merging similar topics. Please refer to the solutions above.

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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x and y are positive, is x^3 > y? [#permalink] New post 14 Feb 2014, 12:47
Anyone has an alternate approach for this one? Very tough question indeed

Will provide some nice Kudos +10

Thanks!
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Re: If x and y are positive, is x^3 > y? [#permalink] New post 18 Mar 2014, 05:47
My method was a little different.

Considering case 1: sqrt(x)>y. Squaring both sides, we getx>y^2 . Multiplying both sides by x^2 we get x^3>y*(yx^2).

Thus, if yx^2 >1 our original question is true; if yx^2< 1, our original question is false. Insufficient.

2) x>y. This tells us nothing about the original question, nor the value of yx^2. Insufficient.

Answer: E
Re: If x and y are positive, is x^3 > y?   [#permalink] 18 Mar 2014, 05:47
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