Skip to main contentSkip to solution

Sahil can complete a job in 20 days when working alone. Manoj is twice as fast as Sahil and three times as fast as Ashish in the same job. The collaborative work arrangement involves Sahil and Manoj working on the first day, Sahil and Ashish on the second day, and Manoj and Ashish on the third day. This three-day cycle continues till the work is complete.

What fraction of the total work will then be done by Ashish till the job is completed?

Solution

✅ Correct Option: 1

Take total work =60= 60 units (LCM of 2020, 1010 and 3030).

Sahil does the job in 2020 days, so Sahil does 6020=3\dfrac{60}{20} = 3 units/day.

Manoj is twice as fast as Sahil, so Manoj does 66 units/day.

Manoj is three times as fast as Ashish, so Ashish does 63=2\dfrac{6}{3} = 2 units/day.


Day 1 (Sahil + Manoj) =3+6=9= 3 + 6 = 9 units

Day 2 (Sahil + Ashish) =3+2=5= 3 + 2 = 5 units

Day 3 (Manoj + Ashish) =6+2=8= 6 + 2 = 8 units

One 33-day cycle =9+5+8=22= 9 + 5 + 8 = 22 units, of which Ashish does 2+2=42 + 2 = 4 units.


Two full cycles (66 days) =44= 44 units. Remaining =60−44=16= 60 - 44 = 16 units

Day 7 (Sahil + Manoj) =9= 9 units. Remaining =16−9=7= 16 - 9 = 7 units

Day 8 (Sahil + Ashish) =5= 5 units. Remaining =7−5=2= 7 - 5 = 2 units

Day 9 (Manoj + Ashish) do 88 units in a full day, so 22 units need 28=14\dfrac{2}{8} = \dfrac{1}{4} day.

Ashish's work on day 9 =2×14=12= 2 \times \dfrac{1}{4} = \dfrac{1}{2} unit


Ashish's total work =4×2= 4 \times 2 (two cycles) +2+ 2 (day 8) +12+ \dfrac{1}{2} (day 9)

=10.5= 10.5 units

Fraction =10.560= \dfrac{10.5}{60}

=21120= \dfrac{21}{120}

=740= \dfrac{7}{40}

Therefore, Ashish does 740\dfrac{7}{40} of the total work.

Keyboard Shortcuts

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