If x and y are greater than zero, then what is the value of x^2*y ?

[BTGmoderatorDC]

If x and y are greater than zero, then what is the value of x^2*y ?

(1) y = 1/2 + 1/4 + 1/8 + 1/16
(2) x has exactly two distinct positive factors, one of which is even.


OA C

Source: Princeton Review

[deloitte247]

$$Question:\ What\ is\ the\ value\ of\ x^{2\cdot y}=>x^{2y}$$
$$Statement\ 1:\ y=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$$
$$y=\frac{8+4+2+1}{16}=\frac{15}{16}$$
But value of x is unknown, hence, statement 1 is NOT SUFFICIENT

Statement 2: x has exactly two distinct positive factors one of which is even.
The only number that 2 different factors and one of the factors is an even number = 2.
Therefore, x=2, but the value of y is unknown, hence, statement 2 is NOT SUFFICIENT.

Combining both statement:
y = 15/16 and x=2
$$Therefore,\ x^{2y}=x^{2\cdot\frac{15}{16}}=x^{\frac{15}{8}}$$
From the index rule of fractional exponents
$$a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m}$$
$$2^{\frac{15}{8}}=\sqrt[8]{2^{15}}$$
Both statemement together are SUFFICIENT... Answer = option C