Question

two men painted a room together in 3 hours. How long will it take each one working alone if the faster one can do the job in 2 1/2 hours less than the other one?

Answer 1

thanks in advance for those who will answer :))

Answer 2

The way in general to approach these problems is to think in terms of rates of doing something.

Man 1 can paint r1 of a room in one hour; Man 2 paints at the rate r2. This means that Man 1 by himself can paint a room in 1/r1 hours and Man 2 in 1/r2 hours.

So together they paint a room at a rate of r1 + r2. This means that the time required to paint one room is 1/(r1 + r2)

From the problem we know that it takes 3 hours for them to paint a room together. hence

1/(r1 + r2) = 3.

We also know that

1/r1 - 1/r2 = 2.5 [we have assumed Man 1 paints slower than Man 2]

So now you have two equations in terms of r1 and r2 and it's a question of algebra to get them back in linear form so you can solve them.

Can you take it from here?

Answer 3

let A and B arae 2 men

let A can finish in x hours

so B can finish in x-5/2 hours

in 1 hour x can do 1/x part of work

in 1 hour B can do 1/(x+2.5) part of work

togather in one hour they can do

1/x + 1/(x+2.5) part of work

they finished togather in 3 hours

so 3[1/x + 1/(x+2.5)] = 1

(3/x)+(3/(x+2.5))=1

3(x+2.5)+3(x)=(x)(x+2.5)

3x+7.5+3x=x^2+2.5x

x^2-3.5x-7.5=0

2x^2-7x-15=0

(2x+3)(x-5)=0

x=-3/2 and x=5

time cannot be negative

so x=5

x+2.5=7.5

Answer 4

hope it was helpful!