Question

33. 2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men
and 6 boys. How long would it take one man and one boy to do it? (15,60)​

Answer 1

Answer:

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 !

Answer 2

Answer:

≈ 14days

Step-by-step explanation:

Man can do work in m days,

• in one day a man can do = 1/m of work

Boy can do work in b days,

• in one day a boy can do= 1/b of work

We have:

• 2/m+7/b=1/4 => mb= 4(2b+7m) => mb=8b+28m
• 4/m+6/b=1/3 => mb= 3(4b+6m) => mb= 12b+18m

Comparing above 2 equations:

• 8b+28m=12b+18m => 10m=4b => m=2/5b

Replacing m in first equation to find b:

• 2/m+7/b=1/4
• 2/(2/5 b)+7/b=1/4
• 5/b+7/b=1/4
• 12/b=1/4
• b=48     boy needs 48 days to complete work
• m=2/5 b= 96/5  man needs 19 1/5 days to complete work

In one day a boy and a man can complete:

• 1/b+1/m= 1/48+5/96= 7/96

They need to compete work:

• 96/7= 13 5/7  ≈ 14days