question is based on time and work relation.

Question

2 men and 7 boys can do a piece of work in14 days, while 3 men and 8 boys can do it in 11 days. Then find in how many days would 8 men and 6 boys do the work three time as big as the first?

Answer 1

Let x = work per day by each man, i.e. how many pieces of work each man does per day.

Let y = work per day by each boy

 

Then 14*2*x + 14*7*y = 1

and 11*3*x + 11*8*y = 1

 

This is a system of two linear equations with two unknowns, hence 0 solutions, exactly 1 solution, or infinitely-many solutions.

 

28x + 98y = 1

33x + 88y = 1

 

-5x + 10y = 0 after subtracting the second equation from the first

x = 2y             solving for x in terms of y

28*2*y + 98*y = 1

154y = 1

y = 1/154

x = 1/77

That is the solution to the system of linear equations.

 

Finally let d = number of days it takes 8 men and 6 boys to do a job three times as big. So

 

d * 8 * x + d * 6 * y = 3

d * 8/77 + d * 6/154 = 3

d *( 16 + 6 )/154 = 3

22d = 462

d = 21. It would take 21 days.

 

The trick is choosing good variables at the beginning.