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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 \(y=\frac{1}{8}\)

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 \(y=\frac{1}{8}\)

Re: If x and y are positive, is x^3 > y? [#permalink]
30 May 2014, 06:10

1

This post received KUDOS

udaymathapati wrote:

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

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

My method without number plugging:

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

Square both sides. Since x & y are both positive, we know this doesn't change the sign at all (because \(f(x) = x^2\) is a strictly increasing function for \(x \geq 0\)). Thus \(x > y^2\). We're trying to get this in a form that resembles our original question, so we multiply each side by \(x^2\)(which we know is positive). Thus, \(x^3 > x^2y^2\), or that\(x^3 > y (x^2 y).\) Therefore, our question is TRUE if and only if \(x^2y\geq 1\). Since we don't know anything about the behavior of\(x^2y\), we mark 1) as INSUFFICIENT.

2) \(x>y.\)

Multiplying both sides by \(x^2\), we get that \(x^3 > yx^2\). Thus, if \(x^2 \geq 1,\) we get TRUE, otherwise, we get FALSE. Since we have no information about \(x^2\), we mark this as INSUFFICIENT.

Taking 1) and 2) together, we still have no information about \(x^2\) or \(x^2y\), so 1) & 2) are INSUFFICIENT.

Answer: E.

Last edited by speedilly on 30 May 2014, 14:37, edited 1 time in total.

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.

What if the question says x and y are positive INTEGERS ??

What if the question says x and y are positive INTEGERS ??

If they were both integers, then the answer would be D: either of them are sufficient, since we'd know that\(x^2y \geq 1\) and \(x^2 \geq 1\)(see my solution above). The question doesn't state that, however, so we must consider all cases.

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