If x paints a house in 2.5 days and x+y can paint the same house in 1.5 days, how long would it take y to paint the same house alone?

Answer 1

In 2.5 dys, x can paint a house.

So x paints at the rate of:

1 / 2.5 house per dys

Together, they can paint at the rate of:

1 / x + 1 / y

1 / 2.5 + 1 / y house per dys or:

1 / 2.5 + 1 / y = 1 / 1.5 dys

1 / y = 1 / 1.5 - 1 / 2.5

1 / 1.5 = (1 / 1 ) / ( 3 / 2 ) = 2 / 3

1 / 2.5 = ( 1 / 1 ) / ( 5 / 2 ) = 2 / 5

1 / y = 1 / 1.5 - 1 / 2.5 =

2 / 3 - 2 / 5 =

10 / 15 - 6 / 15 =

4 / 15

y paints at the rate of 4 / 15 house per dys

y paint the same house alone for 15 / 4 = 3.75 dys

Answer 2

let the time taken by y be t days
x's rate = 1/2.5 = 2/5
y's rate = 1/t
combined rate is 2/5 + 1/t = (2t+5)/(5t)

so combined time = 1/[(2t+5)/)5t)] = 5t/(2t+5)
but 5t/(2t+5) = 3/2
10t = 6t+15
4t = 15
t = 15/4 or 3.75 days