Manoj alone can complete 35% of a work alone in 7 days. Vipin is 20% less efficient than Manoj. Manoj and Vipin together started the work and left after completing 36% of the work. Then Ram joined and worked with 1/8th of his efficiency which is ‘x’ units/day for 19 days. The remaining work is completed by both Manoj and Vipin in 5 days. Find the time taken by Ram alone to complete 80% of the work with his original efficiency.

I. (2x + 8) days

II. (10x – 2) days

III. (22x – 12) days

Solution

Time taken by Manoj to complete the whole work = 7/0.35 = 20 days Time taken by Vipin to complete the whole work = 20/0.80 = 25 days Let the total work = 100 units (L.C.M of 20 and 25) Efficiency of Manoj = 100/20 = 5 units/day Efficiency of Vipin = 100/25 = 4 units/day Time taken by Manoj and Vipin to complete 36% of the work = 0.36 × 100/(5 + 4) = 4 days Amount of work done by Manoj and Vipin in 5 days = (5 + 4) × 5 = 45 units Remaining work = 100 – (36 + 45) = 19 units Therefore, 19x = 19 Or, x = 1 units/day Original efficiency of Ram = 1 × 8 = 8 units/day Time taken by Ram to complete 80% of work = (0.80 ×100)/8 = 10 days For I: (2x + 8) = 1 × 2 + 8 = 10 days Therefore, I can be the answer. For II: (10x – 2) = 7 × 1 – 2 = 8 days Therefore, II cannot be the answer. For III: (22x – 12) = 22 × 1 – 12 = 10 days Therefore, III can be the answer.