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07 Apr 2021, 03:00
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If \(x*y^{\frac{4}{3}} = \sqrt[3]{432}\), is \(x + y = 5\)?
(1) y is a positive integer.
(2) x is an integer.
(1) y is a positive integer.
(2) x is an integer.
07 Apr 2021, 03:22
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xy^4/3=(2^4*3^3)^1/3=2^4/3*3
Statement 1 y is a positive integer.
comparing term we can say that y =2 in that case x must be 3
x+y=5 sufficient
Statement 2 x is an integer.
Comparing term we can say that x=3 y=2 ; x+y=5 Sufficient
Option D is the right choice
Statement 1 y is a positive integer.
comparing term we can say that y =2 in that case x must be 3
x+y=5 sufficient
Statement 2 x is an integer.
Comparing term we can say that x=3 y=2 ; x+y=5 Sufficient
Option D is the right choice
14 Apr 2021, 06:13
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Bunuel
\(x*y^{\frac{4}{3}} = \sqrt[3]{432}\)
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
