Math, asked by Shredhky, 1 month ago

2 men and 7 boys can do a piecee of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it alone?​

Answers

Answered by Anonymous
2

HII!!

Time taken by one man = x days , and that by 1 boy = y days

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

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

work done by 2 men in one day = 2/x

work done by 7 boys in one day = 7/y

so ,

2/x + 7/y = 1/4 -----> [ 1]

================================

Similarly ,

work done by 4 men in one day = 4/x

work done by 4 boys in one day = 4/y

4/x + 4/y = 1/3 ----> [2]

Multiplying equation 1 by , 2 ,

2(2/x + 7/y = 1/4 )

= 4/x + 14/y = 1/2 ---> [3]

solving equations 2 and 3

4/x + 4/y = 1/3

- 4/x + 14/y = 1/2

==============

-10/y = 2-3/6

-10/y = -1/6

-1/6y = -10

y = -10 × 6/-1

= 60

2/x + 7/y = 1/4

2/x + 7/60 = 1/4

7/60 - 1/4 = -2/x

-8/60 = -2/x

-120 = -8x

x = -120/-8

x = 15

=================================

Time taken by 1 man to do the whole job = 15 days

time taken by 1 boy to do the whole job = 60 days

time taken by one boy and one man to do one day's work = 1/60 + 1/15

= 1/60 + 4/60

= 5/60 = 1/12

∴ Together they can do it in 12 days !

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