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Question
2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.
Solution
Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man’s one day’s work = `1/x`
And, one boy’s one day’s work = `1/y`
2 men and 5 boys can finish the work in 4 days.
∴ (2 men’s one day’s work) + (5 boys’ one day’s work) =` 1/4`
`⇒ 2/x + 5/y = 1/4`
⇒ 2u + 5v = `1/4` …….(i) Here, `1/x= u and 1/y`= v
Again, 3 men and 6 boys can finish the work in 3days.
∴ (3 men’s one day’s work) + (6 boys’ one day’s work) = `1/3`
`⇒ 3/x + 6/y = 1/3`
⇒ 3u + 6v = `1/3` …….(ii) Here, `1/x = u and 1/y = v`
On multiplying (iii) from (iv), we get:
`3u = (5/3− 6/4) = 2/12 = 1/6`
`⇒ u = 1/(6 × 3) = 1/18 ⇒ 1/x = 1/18 ⇒ x = 18`
On substituting u = `1/1`8 in (i), we get:
`2 × 1/18 + 5v = 1/4 ⇒ 5v = (1/4− 1/9) = 5/36`
`⇒ v = (5/36 × 1/5) = 1/36 ⇒ 1/y = 1/36 ⇒ y = 36`
Hence, one man alone can finish the work is 18days and one boy alone can finish the work in 36 days.
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