Question: Given: \(\frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27}\). What is the value of \(\frac{x}{y}\)?

1. \(\frac{3}{4}\)
2. \(\frac{6}{5}\)
3. \(\frac{2}{3}\)
4. \(\frac{3}{2}\)
5. \(\frac{5}{6}\)

Answer:

Solution with Explanation:

Approach Solution (1):

If \(\frac{a}{b}=\frac{c}{d}\)then, we have: \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

This is the idea of componendo-dividendo.

To validate: \(\frac{6}{4}=\frac{3}{2} \Rightarrow \frac{6+4}{6-4}=\frac{3+2}{3-2}\)

Apply componendo-dividendo in this equation:

\(\frac{x^3+12x}{6x^2+8}=\frac{y^3+27y}{9y^2+27}\)

\(\frac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8}=\frac{y^3+27y+9y^2+27}{y^3+27y-9y^2-27}\)

\(\frac{x^3+6x^2+12x+8}{x^3+-6x^2+12x-8}=\frac{y^3+27y+9y^2+27}{y^3-9y^2+27y-27}\)

\(\frac{(x+2)^3}{(x-2)^3}=\frac{(y+3)^3}{(y-3)^3}\)

Taking cube root:

\(\Rightarrow \frac{(x+2)}{(x-2)} = \frac{(y+3)}{(y-3)}\)

Apply componendo-dividendo again:

\(\Rightarrow \frac{(x+2+x-2)}{(x+2-x+2)}=\frac{(y+3+y-3)}{(y+3-y+3)}\)

\(\Rightarrow \frac{(2x)}{(4)} = \frac{(2y)}{(6)}\)

\(\Rightarrow \frac{(x)}{(4)}=\frac{(y)}{(6)}\)

\(\Rightarrow \frac{x}{y}=\frac{2}{3}\)

Correct option: C

“Given:\(\frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27}\). What is the value of \(\frac{x}{y}\)?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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