Question

2 men and 7 boys can do a piece of work in 4 days.It is done by 4 men and 4 boys in 3 days.Find time taken by 1 men and 1 boy to complete the work seperately?

Answer 1

here is the answer for your question

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]

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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

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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

Thank you

Answer 2

let suppose
one men can finish work in x days
one boy can finish work in y days
2x+7y=4. 1
4x+4y=3 2
multiply eq 1 by 2
eq 1 becomes
4x+14y=8
4x+4y=3
10y=5
y=1/2
put value of y in eq 2
4x+2=3
4x=1
x=1/4