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

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Re: Is x > y ?  [#permalink]

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09 Oct 2012, 06:37
1
Jp27 wrote:
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.

Hi few observations please correct me if Im wrong ->

$$\sqrt{x} > y$$ -> cannot square this but I can always cube both sides

$$y > \sqrt{x}$$ - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ??

$$\sqrt{x} > y$$ -> cannot square this but I can always cube both sides - YES (although it won't help much, the square root will stay)
$$y > \sqrt{x}$$ - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ?? YES, but the right hand side always non-negative (it can be 0)
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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05 Nov 2012, 10:00
Is x > y ?

a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.

b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.

C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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05 Nov 2012, 21:58
Maurice wrote:
Is x > y ?

a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.

b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.

C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E

You might want to re-consider what happens when you take both statements together. Check out the solutions on page 1 or check this link:
http://www.veritasprep.com/blog/2011/08 ... -question/
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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06 Nov 2012, 05:49
Maurice wrote:
Is x > y ?

a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.

b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.

C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E

Answer to the question is C, not E. Check the solutions on page 1 and 2. For example: is-x-y-1-x-1-2-y-2-x-3-y-100636.html#p777179

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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16 Jan 2013, 05:36
It totally helped me to setup a chart to test the values out:

y < x
(a) increasing integer
(b) increasing fraction

y > x
(c) decreasing fraction
(d) decreasing integer

Test this out on statement (1) and statement (2)...
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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22 Jan 2013, 01:35
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

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

I notice usually these types test the fractions... if x and y are swapping signs...

1. $$\sqrt{x} > y$$==>$$x > y^2$$

Let x>y: 2 > 1 (It works!)

Let x<y: 1/3 > (1/4)^2 (It works!)
Let x<y: -1/3 ? 1/16 (It doesn't work)

INSUFFICIENT!

2. $$x^3 > y$$

Let x>y: (2)^3 > 1 (It works!)

Let x<y: 1/27 ? 1/4 (It doesn't work!)
Let x<y: -1/27 > -1/4 (It works)

INSUFFICIENT!

Combine: eq in Statement 1 and 2 works for x>y

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05 Jun 2013, 08:48
Manhnip wrote:
Is x > y?

(1) √ x > y

(2) x^3 > y

Please provide an easy method to solve this problem in less than 2 mins

The number properties change when the numbers move beyond the range of 0 and 1. Draw a number line and try to check the values of $$\sqrt{x}$$ and $$x^3$$.
Both the statements explain only about one range, but as soon as one combines both the statements, the range becomes quite clear and hence C becomes the answer.
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Re: Is x > y ?  [#permalink]

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27 Dec 2013, 09:42
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.

Yeah kinda tough indeed

Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2

Like from both statements to the x^6 > y^2?

Thanks

Cheers!
J

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Re: Is x > y ?  [#permalink]

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01 Jan 2014, 23:24
jlgdr 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.

Yeah kinda tough indeed

Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2

Like from both statements to the x^6 > y^2?

Thanks

Cheers!
J

Kudos rain!

If y is positive and x^3>y, it means x^3 is also positive (since it is greater than y which is positive).
If both sides of an inequality are positive, you can square the inequality.

$$x^3>y$$
$$(x^3)^2>y^2$$
$$x^6 > y^2$$
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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18 Apr 2014, 04:04
shrouded1 wrote:
I thought this was a really tough question !

Is $$x > y$$ ?

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

Statement I is insufficient:

x = 4 and y = 3 then x is greater than y
x = 1/4 and y = 1/3 then x is not greater than y

Statement II is insufficient:
x = 4 and y = 2 then x is greater than y
x = -1/4 and y = -1/2 then x is not greater than y

Combining both we can clearly see that:
x^1/2 > y
x^3 > y

If we plug in positive fractions or positive numbers each time x is greater than y. Hence answer is C.
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Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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07 Jul 2014, 00:10
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.

when if x=1/4 and y=1/3 then \sqrt{X} = 1/2, and 1/2 > 1/3. You have chosen wrong plugin data here.
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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07 Jul 2014, 01:19
honchos 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.

when if x=1/4 and y=1/3 then \sqrt{X} = 1/2, and 1/2 > 1/3. You have chosen wrong plugin data here.

Nope.

If $$x=\frac{1}{4}$$ and $$y=\frac{1}{3}$$, then $$(\sqrt{x}=\frac{1}{2})>(y=\frac{1}{4})$$ (statement is satisfied) but $$(x=\frac{1}{4})<(y=\frac{1}{3})$$ giving a NO answer to the question.
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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, 07:37
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!
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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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Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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26 Nov 2015, 02:47
Quote:
Is $$x > y$$ ?

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

Not sure if this approach is correct. By multiplying 1 and 2, we get: $$x^{1.5} > y^2$$ Now we can substitute the values to check if $$x > y$$
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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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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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11 Dec 2016, 20:38
My approach
Two scenario for x (x can't be negative, see above): |x|>1 or |x|<1. Denote by + if you can tell x will be bigger for sure (- otherwise)
table: IF |X|>1 |X|<1
STATEMENT 1 + - INSUF
STATEMENT 2 - + INSUF
Together, we see that each statement tells us that x will be bigger for both conditions. Thus together sufficient. C
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y  [#permalink]

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18 Mar 2017, 11:35
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.

Here is a little bit typo in red part. it will be better for learner if you make correction.
it is awesome. Bunuel, may we solve this problem by using algebra method?
Thank you...
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.”

Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y &nbs [#permalink] 18 Mar 2017, 11:35

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