chetan2u wrote:
BrentGMATPrepNow wrote:
If\(j\) ,\(k\) ,\(x\) and\(y\) are each greater than\(1\) , and\( j^{2x} = k^{3y} =j^4k^2\) , then what is the value of\(y\) in terms of\(x\) ?
A)\(\frac{3x-6}{2x}\)
B)\(\frac{2x-6}{3x}\)
C)\(\frac{2x}{3x-6}\)
D)\(\frac{3x}{2x-6}\)
E)\(\frac{2x}{x-3}\)
A)\(\frac{3x-6}{2x}\)
B)\(\frac{2x-6}{3x}\)
C)\(\frac{2x}{3x-6}\)
D)\(\frac{3x}{2x-6}\)
E)\(\frac{2x}{x-3}\)
BrentGMATPrepNow , let me try it with slightly different method.
\( j^{2x} = k^{3y} =j^4k^2\)
\( j^{2x} * k^{3y} =j^4k^2*j^4k^2=j^8*k^4\)
Let us equate powers of j and k.
\(2x=8…..x=4\)
\(3y=4……y=\frac{4}{3}\)
Now, substitute x as 4 and check for option that gives 4/3 as the answer.
A)\(\frac{3x-6}{2x}=\frac{3*4-6}{2*4}=\frac{3}{4}\) …NO
B)[m][fraction]
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