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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

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 thatx^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 ofx^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 thatx^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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