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A, B and C can do a piece of work in 20, 30 and 40 days, respectively. They started working together, but from the first day itself, A came to work on alternate days. In how many days would the work be completed?
- a)8
- b)12
- c)15
- d)16
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A, B and C can do a piece of work in 20, 30 and 40 days, respectively...
Given, A, B and C can do a piece of work in 20, 30 and 40 days respectively.
Let the total work be 120 units (LCM of 20, 30 and 40).
A can do 6 units of work in a day (120/20).
B can do 4 units of work in a day (120/30).
C can do 3 units of work in a day (120/40).
From the first day itself, A came to work on alternate days.
So, on day 1, A works for 6 units, on day 2, A doesn't work.
On day 3, A works for 6 units, on day 4, A doesn't work.
This pattern continues.
Let's calculate the work done by A, B and C in the first 12 days.
Day 1: A=6, B=4, C=3 (Total work done = 13)
Day 2: B=4, C=3 (Total work done = 7)
Day 3: A=6, B=4, C=3 (Total work done = 13)
Day 4: B=4, C=3 (Total work done = 7)
Day 5: A=6, B=4, C=3 (Total work done = 13)
Day 6: B=4, C=3 (Total work done = 7)
Day 7: A=6, B=4, C=3 (Total work done = 13)
Day 8: B=4, C=3 (Total work done = 7)
Day 9: A=6, B=4, C=3 (Total work done = 13)
Day 10: B=4, C=3 (Total work done = 7)
Day 11: A=6, B=4, C=3 (Total work done = 13)
Day 12: B=4, C=3 (Total work done = 7)
Total work done in 12 days = 104 units.
Remaining work = 16 units.
Now, A works every day until the work is completed.
A can do 6 units of work in a day.
So, A will complete the remaining 16 units in 16/6 = 2.67 days (approximately 3 days).
Therefore, the total number of days taken to complete the work = 12 + 3 = 15 days.
Hence, the correct answer is option (B).
Let the total work be 120 units (LCM of 20, 30 and 40).
A can do 6 units of work in a day (120/20).
B can do 4 units of work in a day (120/30).
C can do 3 units of work in a day (120/40).
From the first day itself, A came to work on alternate days.
So, on day 1, A works for 6 units, on day 2, A doesn't work.
On day 3, A works for 6 units, on day 4, A doesn't work.
This pattern continues.
Let's calculate the work done by A, B and C in the first 12 days.
Day 1: A=6, B=4, C=3 (Total work done = 13)
Day 2: B=4, C=3 (Total work done = 7)
Day 3: A=6, B=4, C=3 (Total work done = 13)
Day 4: B=4, C=3 (Total work done = 7)
Day 5: A=6, B=4, C=3 (Total work done = 13)
Day 6: B=4, C=3 (Total work done = 7)
Day 7: A=6, B=4, C=3 (Total work done = 13)
Day 8: B=4, C=3 (Total work done = 7)
Day 9: A=6, B=4, C=3 (Total work done = 13)
Day 10: B=4, C=3 (Total work done = 7)
Day 11: A=6, B=4, C=3 (Total work done = 13)
Day 12: B=4, C=3 (Total work done = 7)
Total work done in 12 days = 104 units.
Remaining work = 16 units.
Now, A works every day until the work is completed.
A can do 6 units of work in a day.
So, A will complete the remaining 16 units in 16/6 = 2.67 days (approximately 3 days).
Therefore, the total number of days taken to complete the work = 12 + 3 = 15 days.
Hence, the correct answer is option (B).
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Community Answer
A, B and C can do a piece of work in 20, 30 and 40 days, respectively...
Let the total work = W = 120w
Let the work done by A in 1 day = A
Let the work done by B in 1 day = B
Let the work done by C in 1 day = C
So, 20A = 120w, or A = 6w
30B = 120w or B = 4w
40C = 120w or C = 3w
Let A work for d days.
Then, B and C work for 2d days each.
So, dA + 2d(B + C) = 120w
d(6w) + 2d(7w) = 120w
20dw = 120w
Or, d = 6
i.e. A work for 6 days
that means B and C work for 12 days.
Thus, the work is done in 12 days.
Hence, answer option (2) is correct.
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