If x and y are positive, is x^3 y? (1) x^(1/2) y (2) x y

Question

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

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

Bunuel · Answer

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

amitjash · Answer

When there are squaring or cubing is involved, always remember -ve values as well as the values between -1 and 1

Onell · Answer

+1 for u
Great Post Thanks Bunuel

subhashghosh · Answer

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.

GSBae · Answer

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.

ShajibAhmed · Answer

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

GSBae · Answer

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.

BillyZ · Answer

You will get a better understanding by drawing the number line and plugging in the value.

Basshead · Answer

The key is to remember that if 0 < x < 1, x gets smaller when squared or cubed.

So if we're told \(x > y^2\) (as in statement 1), we can't say with certainty that \(x^3 > y\).

Similarly, if we can't say with certainty that \(x > y^2\), then we can't conclude \(x > y\) either.

bumpbot · Answer

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If x and y are positive, is x^3 > y? (1) x^(1/2) > y (2) x > y

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