If xy^4/3 =√432, is x + y = 5? (1) y is a positive integer. (2) x is

\(x*y^{\frac{4}{3}} = \sqrt[3]{432}\)
We can solve the above equation in various ways, so let us check the statements

1) y is a positive integer.

\(x*y^{\frac{4}{3}} = \sqrt[3]{432}=432^{\frac{1}{3}}= 432^{\frac{1}{3}}* 1^{\frac{4}{3}}\)
\(x= 432^{\frac{1}{3}} \ \ and \ \ y=1\)....No

\(x*y^{\frac{4}{3}} = \sqrt[3]{432}=432^{\frac{1}{3}}= 27^{\frac{1}{3}}* 16^{\frac{1}{3}}= 3* 2^{\frac{4}{3}}\)
\(x= 3 \ \ and \ \ y=2\)....Yes

Insufficient


2) x is an integer.

\(x*y^{\frac{4}{3}} = \sqrt[3]{432}=432^{\frac{1}{3}}= 27^{\frac{1}{3}}* 16^{\frac{1}{3}}= 3* 2^{\frac{4}{3}}\)
\(x= 3 \ \ and \ \ y=2\)....Yes

\(x*y^{\frac{4}{3}} = \sqrt[3]{432}=432^{\frac{1}{3}}= 27^{\frac{1}{3}}* 16^{\frac{1}{3}}= 3* (-2)^{\frac{4}{3}}\)
\(x= 3 \ \ and \ \ y=-2\)....No

Insufficient



Combined
x=3 and y=2
Sufficient

C

Jitu20