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. **

● **Time and Work**

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**8th Grade Math Practice** **From Time and Work to HOME PAGE**

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