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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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Bunuel.... you seem to love number theory! I really am amazed at your patience. Good job. You already have enough Kudos. :P
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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interesting question.......forgot to consider fractions first.......now its clear.....
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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That solution is so simple yet I couldn't figure it out. I knew that each statement alone was insufficient, but I couldn't figure out whether both together were sufficient. I kept trying to plug in numbers and it just never worked. Never did it occur to me to put both scenarios on a number line and just use the statements to prove it. I swear the math on the GMAT really makes you think in different ways and the solutions are so easy we make it harder than it needs to be.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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Again there is a wonderfully simple explanation for this one using graphs.

x>y, sqrt(x)>y, x^3>y all three represents the region below the graph for all three cases.

We need to answer is x>y



(1) sqrt(x)>y
You are below the yellow line does not imply you are below the purple line. Insufficient

(2) x^3>y
Now x need not be just positive, but looking at the graph is enough to conclude this is not sufficient. Being below the blue line does not imply being below the purple line

(1+2) Now x>0 since we are using sqrt(x)
You are below the blue line and the yellow line both
To satisfy both, you must always be below the purple line

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

It's just that the way I was taught algebra, there was a lot of focus on graphs. I find it way more intuitive than algebraic manipulation, more straight forward than plugging values and in almost all cases faster especially for very simple functions like these

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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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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.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
Hi Karishma,

Quote:
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).



from the diagrams you have put up , how can we derive that X>=0 cos from both the diagrams the regions do not over lap!
Please help me with this1
Thanks
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Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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shankar245 wrote:
Hi Karishma,

Quote:
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).



from the diagrams you have put up , how can we derive that X>=0 cos from both the diagrams the regions do not over lap!
Please help me with this1
Thanks


We know that x >= 0 because statement 1 tells us that \(\sqrt{x} > y\).
\(\sqrt{x}\) can be defined only if x is non negative.

Originally posted by KarishmaB on 18 Apr 2012, 03:28.
Last edited by KarishmaB on 08 Oct 2022, 21:33, edited 1 time in total.
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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 ??
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Re: Is x > y ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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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]
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 ? (1) x^(1/2) > y (2) x^3 > y [#permalink]
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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]
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.

Answer: C.

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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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.

Answer: C.

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]
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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