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A is 40%40 \% less efficient than BB who can do the same work in 20%20 \% less time than CC. If AA and B together can complete 80%80 \% work in 12 days, then in how many days 60%60 \% work can be completed by B\mathrm{B} and C\mathrm{C} together ?

Solution

Correct Option: 3
  1. Given:

A is 40%40\% less efficient than B: A=0.6BA = 0.6B

B does work in 20%20\% less time than C: B=1.25CB = 1.25C or C=0.8BC = 0.8B

A and B complete 80%80\% work in 1212 days

  1. Step 1: Calculate A and B's combined efficiency

Let total work be 100100 units

Work done by A and B in 12 days = 8080 units

Efficiency of (A+B)=8012=203(A + B) = \frac{80}{12} = \frac{20}{3} units/day

Combined efficiency: A+B=0.6B+B=1.6BA + B = 0.6B + B = 1.6B

Therefore:

1.6B=2031.6B = \frac{20}{3}

B=203×1.6=256B = \frac{20}{3 × 1.6} = \frac{25}{6} units/day

A=0.6B=0.6×256=52A = 0.6B = 0.6 × \frac{25}{6} = \frac{5}{2} units/day

  1. Step 2: Calculate B and C's combined efficiency

B+C=B+0.8B=1.8BB + C = B + 0.8B = 1.8B

B+C=1.8×256=456=152B + C = 1.8 × \frac{25}{6} = \frac{45}{6} = \frac{15}{2} units/day

  1. Step 3: Time for 60%60\% work by B and C

Time = WorkEfficiency=60152=60×215=8\frac{\text{Work}}{\text{Efficiency}} = \frac{60}{\frac{15}{2}} = 60 × \frac{2}{15} = 8 days

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