In time and work we will learn to calculate and find the time required to complete a piece of work and also find work done in a given period of time. We know the amount of work done by a person varies directly with the time taken by him to complete the work.
(i) Suppose A can finish a piece of work in 8 days.
Then, work done by A in 1 day = ¹/₈ [by unitary method].
(ii) Suppose that the work done by A in 1 day is ¹/₆
Then, time taken by A to finish the whole work = 6 days.
(i) Suppose if a person A can finish a work in n days.
Then, work done by A in 1 day = 1/nᵗʰ part of the work.
(ii) Suppose that the work done by A in 1 day is \(\frac{1}{n}\)
Then, time taken by A to finish the whole work = n days.
Solution:
Time taken by Aaron to finish the work = 12 days.
Work done by Aaron in 1 day = ¹/₁₂
Time taken by Brandon to finish the work = 15 days.
Work done by Brandon in 1 day = ¹/₁₅
Work done by (Aaron + Brandon) in 1 day = ¹/₁₂ + ¹/₁₅ = ⁹/₆₀ = ³/₂₀
Time taken by (Aaron + Brandon) to finish the work = \(\frac{20}{6}\) days, i.e., 6²/₃ days.
Hence both can finish the work in 6²/₃ days.
Solution:
Time taken by (A + B) to finish the work = 15 days.
Time taken by B alone to finish the work 20 days.
(A + B)’s 1 day’s work = ¹/₁₅
and B’s 1 day’s work = ¹/₂₀
A’s 1 day’s work = {(A + B)’s 1 day’s work} - {B’s 1 day’s work}
= (¹/₁₅ - ¹/₂₀) = (4 - 3)/60 = ¹/₆₀
Therefore, A alone can finish the work in 60 days.
Solution:
Time taken by A to finish the work = 25 days.
A’s 1 day’s work = ¹/₂₅
Time taken by B to finish the work = 20 days.
B’s 1 day’s work = ¹/₂₀
(A + B)’s 1 day’s work = (¹/₂₅ + ¹/₂₀) = ⁹/₁₀₀
(A + B)’s 5 day’s work (5 × ⁹/₁₀₀) = 4̶5̶/1̶0̶0̶ = ⁹/₂₀
Remaining work (1 - ⁹/₂₀) = ¹¹/₂₀
Now, ¹¹/₂₀ work is done by B in 1 day
Therefore, ¹¹/₂₀ work will be done by B in (11/2̶0̶ × 2̶0̶) days = 11 days.
Hence, the remaining work is done by B in 11 days.
Solution:
Time taken by (A + B) to finish the work = 18 days.
(A + B)’s 1 day’s work = ¹/₁₈
Time taken by (B + C) to finish the work = 24 days.
(B + C)’s 1 day’s work = ¹/₂₄
Time taken by (C + A) to finish the work = 36 days.
(C + A)’s 1 day’s work = ¹/₃₆
Therefore, 2(A + B + C)’s 1 day’s work = (¹/₁₈ + ¹/₂₄ + ¹/₃₆) = (4 + 3 + 2)/72 = \(\frac{9}{72}\) = ¹/₈
⇒ (A + B + C)’s 1 day’s work = (¹/₂ × ¹/₈) = ¹/₁₆
Therefore, A, B, C together can finish the work in 16 days.
Solution:
Time taken by (A + B) to finish the work = 12 days.
(A + B)’s 1 day’s work = ¹/₁₂
Time taken by (B +C) to finish the work = 15 days.
(B + C)’s 1 day’s work = ¹/₁₅
Time taken by (C + A) to finish the work = 20 days.
(C + A)’s 1 day’s work = ¹/₂₀
Therefore, 2(A + B + C)’s 1 day’s work = (¹/₁₂ + ¹/₁₅ + ¹/₂₀) = \(\frac{12}{60}\) = ¹/₅
⇒ (A + B + C)’s 1 day’s work = (¹/₂ × ¹/₅) = ¹/₁₀
Therefore, A, B, C together can finish the work in 10 days.
Now, A’s 1 day’s work
= {(A + B + C)’s 1 day’s work} - {(B + C)’s 1 day’s work}
= (¹/₁₀ - ¹/₁₅) = ¹/₃₀
Hence, A alone can finish the work in 30 days.
B’s 1 day’s work
{(A + B + C)’s 1 day’s work} - {(C + A)’s 1 day’s work}
(¹/₁₀ – ¹/₂₀) = ¹/₂₀
Hence, B alone can finish the work in 20 days.
C’s 1 days work
= {(A + B + C)’s 1 day’s work} - {(A + B)’s 1 day’s work}
= (¹/₁₀ – ¹/₁₂) = ¹/₆₀
Hence, C alone can finish the work in 60 days.
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