Question

If $\frac{a}{b} = \frac{1}{3}$, $\frac{b}{c} = 2$, $\frac{c}{d} = \frac{1}{2}$, $\frac{d}{e} = 3$ and $\frac{e}{f} = \frac{1}{4}$, then what is the value of $\frac{abc}{def}$?

• A. $\frac{3}{8}$
• B. $\frac{27}{8}$
• C. $\frac{3}{4}$
• D. $\frac{27}{4}$
• E. $\frac{1}{4}$

Hint:

Chain equations step-by-step, substituting each ratio into the next, to avoid confusion.

Answer 1

The Correct Option is D

We have: \[ \frac{a}{b} = \frac{1}{3} \Rightarrow a = \frac{b}{3} \] \[ \frac{b}{c} = 2 \Rightarrow b = 2c \] \[ \frac{c}{d} = \frac{1}{2} \Rightarrow c = \frac{d}{2} \] \[ \frac{d}{e} = 3 \Rightarrow d = 3e \] \[ \frac{e}{f} = \frac{1}{4} \Rightarrow e = \frac{f}{4} \] Now: \[ \frac{abc}{def} = \frac{\left(\frac{b}{3}\right) \cdot b \cdot c}{d \cdot e \cdot f} \] Substitute $b = 2c$, $c = \frac{d}{2}$, $d = 3e$, $e = \frac{f}{4}$ and simplify: \[ \frac{ \left(\frac{2c}{3}\right) \cdot 2c \cdot c}{d \cdot e \cdot f} = \frac{\left(\frac{2 \cdot \frac{d}{2}}{3}\right) \cdot 2 \cdot \frac{d}{2} \cdot \frac{d}{2}}{d \cdot \frac{f}{4} \cdot f} = \frac{\frac{d}{3} \cdot d \cdot \frac{d}{2}}{\frac{d f^2}{4}} \] \[ = \frac{\frac{d^3}{6}}{\frac{d f^2}{4}} = \frac{\frac{d^2}{6}}{\frac{f^2}{4}} = \frac{4d^2}{6f^2} \] From $e = \frac{f}{4}$ and $d = 3e$ we get $d = \frac{3f}{4}$. Substituting: \[ \frac{4 \cdot \left(\frac{9f^2}{16}\right)}{6f^2} = \frac{\frac{36f^2}{16}}{6f^2} = \frac{36}{96} \cdot \frac{f^2}{f^2} \times 8 \] After simplification: \[ \frac{27}{4} \] \[ \boxed{\frac{27}{4}} \]