A and B can do a piece of work in 18 days, B and C can do it in 24 days, A and C can do it in 36 days. In how many days will A, B and C finish it working separately
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1
16 days
(A + B)’s 1 day’s work = \frac{1}{18}
(B + C)’s 1 day’s work = \frac{1}{24}
(C + A)’s 1 day’s work = \frac{1}{36}
On adding,
2(A + B + C)’s 1 day’s work = \frac{1}{18}+\frac{1}{24}+\frac{1}{36} = \frac{4+3+2}{72} = \frac{1}{8}
∴ (A + B + C)’s 1 day’s work = \frac{1}{16}
∴ A, B and C together will complete the work in 16 days.
Hence option [D] is correct answer.
Answered by
0
16 days
(A + B)’s 1 day’s work = \frac{1}{18}
(B + C)’s 1 day’s work = \frac{1}{24}
(C + A)’s 1 day’s work = \frac{1}{36}
On adding,
2(A + B + C)’s 1 day’s work = \frac{1}{18}+\frac{1}{24}+\frac{1}{36} = \frac{4+3+2}{72} = \frac{1}{8}
∴ (A + B + C)’s 1 day’s work = \frac{1}{16}
∴ A, B and C together will complete the work in 16 days.
Hence option [D] is correct answer.
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