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

Question

If  x*y^{\frac{4}{3}} = \sqrt {432} , is  x + y = 5 ?

(1) y is a positive integer.
(2) x is an integer.

sumitkrocks · Answer

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

chetan2u · Answer

x*y^{\frac{4}{3}} = \sqrt {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 {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 {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 {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 {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

Onalethata · Answer

If x=-3, the sum would not be 5, the question only says x is an integer not that x is a positive integer. Am I missing something?

chetan2u · Answer

x=-3 will not satisfy the expression. If x=-3 and y is positive, then x*y^(4/3) will be negative

HoudaSR · Answer

I could tell that x+y=5 even without looking at the statements, So I assumed that each statement alone is sufficient. See picture below.  Bunuel   chetan2u   TargetTestPrep  can you please tell me where I am going wrong?

Bunuel · Answer

It is important to note that the equation xy^{\frac{4}{3}} = 32^{\frac{4}{3}} does not necessarily mean that x = 3 and y = 2 because we don't know whether x and y are integers. In fact, there are infinitely many solutions to this equation for different values of x and y. For any positive value of x, there exists a corresponding value of y that will satisfy the equation. For example, if x = 1, then y = 2*3^(3/4). If x = 2, then y = 2^(1/4)3^(3/4). If x = 1/2, then y = 26^(3/4). These are just a few examples of the infinitely many solutions to the equation. Similar questions: https://gmatclub.com/forum/what-is-the- ... 68417.html https://gmatclub.com/forum/if-x-2y-3-20 ... 92486.html https://gmatclub.com/forum/if-10-a-3-b- ... 22364.html

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

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