CAT 2003 Slot 2 — QA Question 3

Since a = 6b = 12c and 2b = 9d = 12e

∴ a : b : c = 12 : 2 : 1 and b : d : e = 18 : 4 : 3

∴ a : b : c : d : e = 108 : 18 : 9 : 4 : 3

∴ a = 108k; b = 18k; c = 9k; d = 4k and e = 3k where ‘k’ is an integer

Option 1 is: $\left(\frac{a}{27},\frac{b}{e}\right)$ = (4k, 6)

Option 2 is: $\left(\frac{a}{36},\frac{c}{e}\right)$ = (3k, 3)

Option 3 is: $\left(\frac{a}{12},\frac{bd}{18}\right)$ = (9k, 4k²)

Option 4 is: $\left(\frac{a}{6},\frac{c}{d}\right)=\left(18k, \frac{9}{4}\right)$

The 4th option contains $\frac{9}{4}$, which is not an integer.

Hence, option (d).