Question

If 6 men and 8 boys do a piece of work in 10 days while 26 men and 48boys can do the same work in 2 days then what is the time taken by 15 men and 20 days to complete the same type of work

Answer 1

bonjour dear!

Units of work done is measured in terms of man-days, boy-days etc. you will have to find our the number of man-days and boy-days required to do the job and equate it.

10(6M+ 8B) = 2 (26M + 48B) since the same work is done they will be equal

60M + 80B = 52M + 96B

8M = 16B or M = 2B substitute 2B = M and you will get the total man-days required to do the job and it is

10( 6M + 4M) = 100 man-days

Similarly substitute 20B as 10M to find out the answer

x(15M+10M) = 100M

solving this x = 4 so 15 men and 20 boys will have to work for 4 days to complete the job

hope this helps

keep smiling

:)

Answer 2

Answer:

let men be x and boys be y

so, 6 x+ 8 y= 1/10 ..........i

26x+48 y = 1/2..............ii

both are one day work

then by solving both equations

We get x=1/100 and y= 2/100

hence,

15 men+20 boys = 15/100+20/200= 1/4

so,answer is 4 days