Question

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

Answer 1

Let 1 man alone can finish the work in X days and 1 boy alone can finish it in Y .

Then,

1 man's 1 days work = 1/X And,

1 boy's 1 day's work = 1/Y

(4 men's 1 day's work )+(6 boys 1 day's work )= 1/5

=> 4/X + 6/Y = 1/5

=> 4U + 6V = 1/5 [ Where 1/X = U and 1/Y =V]

=> 4U + 6V = 1/5 --------(1)

Again,

( 3 men's 1 day's work ) + ( 4 boys 1 day's work ) = 1/7

=> 3/X + 4/Y = 1/7

=> 3U + 4V = 1/7 -------(2)

On multiplying (1) by 3 and ,(2) by 4 we get,

12U + 18V = 3/5 --------(3)

And,

12U + 16 V = 4/7 --------(4)

Subtracting (3) and (4) we get,

2V = ( 3/5 - 4/7)

2V = 1/35

V = 1/35 ×2

V = 1/70

1/Y = V

1/Y = 1/70

Y = 70 days

Putting V = 1/70 in equation (1) we get,

4U + 6V = 1/5

4U = ( 1/5 - 6V )

4U = ( 1/5 - 6/70 )

4U = ( 14 - 6 /70)

4U = ( 8/70)

U = 8/70 × 1/4

U = 1/35

1/X = U

1/X = 1/35

X = 35 days

Therefore,

One man alone can finish the work in 70 days and One boy alone can finish the work in 35 days.