Question

4 men and 6 boys can finish a piece work in 5 days while 3 men and 4 boys can finish it in 7 days. find the time taken by 1 man alone or that by 1 boy alone.

Answer 1

Number of days taken by 1 man alone and 1 boy alone are 35 days and 70 days respectively.

Explanation:

Let, the work completed by 1 man in 1 day = x and the work completed by 1 boy in 1 day = y.

So, 4 men and 6 boy's 1 day work is
`(4)/(x)` and
`(6)/(y)`

And, 3 men and 4 boy's 1 day work is
`(3)/(x)` and
`(4)/(y)`

Since, 4 men and 6 boys complete the work in 5 days and 3 men and 4 boys complete the work in 7 days.

So, we have,

`(4)/(x)+(6)/(y)=(1)/(5)`

`(3)/(x)+(4)/(y)=(1)/(7)`

Let,
`u=(1)/(x)` and
`v=(1)/(y)`

So, the equations are,

`4u+6v=(1)/(5)` ................(1)

`3u+4v=(1)/(7)` ................(2)

Multiply (1) by 3 and (2) by 4 and subtract the equations, we get,

`2v=(1)/(35)` i.e.
`v=(1)/(70)` i.e.
`(1)/(y)=(1)/(70)` i.e. y = 70.

Substitute value of 'v' in (1) gives,

`4u+(6)/(70)=(1)/(5)` i.e.
`4u=(-6)/(70)+(1)/(5)` i.e.
`4u=(8)/(70)` i.e.
`u=(8)/(70* 4)` i.e.
`u=(1)/(35)` i.e.
`(1)/(x)=(1)/(35)` i.e. x = 35

So, the number of days taken by 1 men alone and 1 boy alone are 35 days and 70 days respectively.