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
Answers
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:
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