CAT 2002 — QA Question 44

Work done by the large pump = 3 units

∴ Work done by the 3 small pumps = 2 × 3 = 6 units

∴ Work done by all 4 pumps together = 9 units

$\therefore \frac{\mathrm{Rate} \mathrm{of} \mathrm{work} \mathrm{of} \mathrm{the} \mathrm{large} \mathrm{pump}}{\mathrm{Rate} \mathrm{of} \mathrm{work} \mathrm{of} \mathrm{all} 4 \mathrm{pumps}}=\frac{3}{9}=\frac{1}{3}$

Rate of work is inversely proportional to time taken for completing the work.

$\therefore \frac{\mathrm{TIme} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{large} \mathrm{pump}}{\mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{all} 4 \mathrm{pumps}}=\frac{3}{1}$

Hence, option (b).