Question

8 men and 12 boys can finish a piece of work in 10 days, while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.​

Answer 1

Let the time taken by one man alone be 'x' and let the time taken by one boy alone to finish the work be 'y'.

Now,

Work done by one man in one day = 1/x

Work done by one boy in one day = 1/y

Given,

8 men and 12 boys can finish a piece of work in 10 days.

So, work done by 8 men and 12 boys in 1 day = ¹/₁₀

We can express this in the form of an equation -:

$\implies \ 8 \times \dfrac{1}{x} + 12 \times \dfrac{1}{y} = \dfrac{1}{10}$

$\implies \dfrac{8}{x} + \dfrac{12}{y} = \dfrac{1}{10}$             - (i)

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Given,

6 men and 8 boys can finish a piece of work in 14 days.

So, work done by 6 men and 6 boys in 1 day = ¹/₁₄

We can express this in the form of an equation -:

$\implies 6 \times \dfrac{1}{x} + 8 \times \dfrac{1}{y} = \dfrac{1}{14}$

$\implies \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{14}$       - (ii)

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Now let 1/x be 'p' and 1/y be 'q'.

$\implies \dfrac{8}{x} + \dfrac{12}{y} = \dfrac{1}{10}$

$\implies 8p + 12q = \dfrac{1}{10}$              - (iii)

$\implies \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{14}$

$\implies 6p + 8q = \dfrac{1}{14}$            - (iv)

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Now we'll solve the two equations using elimination method.

So, multiply 3 to (iii)

⇒ 3 (8p) + 3 (12q) = 3 (¹/₁₀)

⇒ 24p + 36q = ³/₁₀              - (v)

Now, multiply 4 to (iv)

⇒ 4 (6p) + 4 (8q) = 4 (¹/₁₄)

⇒ 24p + 32q = ²/₇            - (vi)

Now, subtract (v) and (vi)

⇒ 24p + 36q - 24p - 32q = ³/₁₀ - ²/₇

⇒ 36q - 32q = ²¹/₇₀ - ²⁰/₇₀

⇒ 4q = ¹/₇₀

⇒ q = ¹/₂₈₀

Substitute 'q' in (i)

⇒ 8p + 12 × ¹/₂₈₀ = ¹/₁₀

⇒ 8p + ³/₇₀ = ¹/₁₀

⇒ 8p = ¹/₁₀ - ³/₇₀

⇒ 8p = ⁴/₇₀

⇒ p = ⁴/₇₀ × ¹/₈

⇒ p = ¹/₁₄₀

∴ p = 1/x = ¹/₁₄₀

∴ x = 140

∴ q = 1/y = ¹/₂₈₀

∴ y = 280

∴ Time taken by one man alone to finish the work is 140 days

∴ Time taken by one boy alone to finish the work is 280 days.

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