Skip to main contentSkip to solution

BBA-DBE Bangalore 2025 PYQsQuantitative Ability. Free, no login required.

A and C together can complete a certain work in 30 days. B is 20% more efficient than C. A, B and C together can do the same work in 20 days. C alone completed one fourth of the work and the remaining work was completed by A and B together. In how many days was the entire work completed?

Solution

✅ Correct Option: 2

Let the total work be 6060 units (LCM of 3030 and 2020 for clean calculations).


Find the work rates:

(A+C)(A + C) complete 6060 units in 3030 days, so (A+C)=6030=2(A + C) = \dfrac{60}{30} = 2 units/day

(A+B+C)(A + B + C) complete 6060 units in 2020 days, so (A+B+C)=6020=3(A + B + C) = \dfrac{60}{20} = 3 units/day

Therefore, B=(A+B+C)−(A+C)=3−2=1B = (A + B + C) - (A + C) = 3 - 2 = 1 unit/day


Since BB is 20%20\% more efficient than CC: B=1.2CB = 1.2C

1=1.2C1 = 1.2C

C=11.2=56C = \dfrac{1}{1.2} = \dfrac{5}{6} units/day

A=(A+C)−C=2−56=12−56=76A = (A + C) - C = 2 - \dfrac{5}{6} = \dfrac{12 - 5}{6} = \dfrac{7}{6} units/day


Phase 1: CC completes 14\dfrac{1}{4} of the work

Work done by C=14×60=15C = \dfrac{1}{4} \times 60 = 15 units

Time taken by C=1556=15×65=18C = \dfrac{15}{\dfrac{5}{6}} = 15 \times \dfrac{6}{5} = 18 days


Phase 2: AA and BB complete remaining work

Remaining work =60−15=45= 60 - 15 = 45 units

Combined rate of AA and B=76+1=76+66=136B = \dfrac{7}{6} + 1 = \dfrac{7}{6} + \dfrac{6}{6} = \dfrac{13}{6} units/day

Time taken by AA and B=45136=45×613=27013B = \dfrac{45}{\dfrac{13}{6}} = 45 \times \dfrac{6}{13} = \dfrac{270}{13} days


Total time =18+27013=23413+27013=50413=381013= 18 + \dfrac{270}{13} = \dfrac{234}{13} + \dfrac{270}{13} = \dfrac{504}{13} = 38\dfrac{10}{13} days

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question