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?
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
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
To solve this problem, let's break it down step by step.
Let's assume that painting the entire house is considered as one complete job.
Given that x can paint the house in 2.5 days, we can say that x's work rate is 1/2.5 or 2/5 of the whole house per day.
Now, let's consider the combined work rate of x and y working together. We are given that x+y can paint the house in 1.5 days. Since they are working together, their combined work rate is 1/1.5 or 2/3 of the whole house per day.
To find y's work rate, we need to subtract x's work rate from the combined work rate of x and y. So, the work rate of y alone is (2/3 - 2/5) or 4/15 of the whole house per day.
Now, we can calculate how long it would take y to paint the house alone by taking the reciprocal of y's work rate. So, y would take 15/4 or 3.75 days to paint the same house alone.
Therefore, it would take y approximately 3.75 days to paint the same house alone.