Question

P does a piece of work alone in 3 days.
P and Q together do the work in 2 days.
How
many days will Q take to do the
work alone?
​

Answer 1

Answer:

Q, alone will do it in 18.39 days

Explanation:

Let P+Q, together, take x days to do the work.

P alone will do the same work in (x+12) days while Q does it in (x+3) days.

P does [1/(x+12)]th of the work in 1 day.

Q does [1/(x+3)]th of the work in 1 day.

So P and Q together do [1/(x+12)]+[1/(x+3)] or [x+3+x+12]/[(x+12)(x+3)] =(2x+15)/[(x+12)(x+3)]th part of the work in 1 day.

So P and Q will take [(x+12)(x+3)]/(2x+15) days which is the same as x.

Or, [(x+12)(x+3)]/(2x+15)=x, or

x^2+15x+36 = 2x^2+30, or

x^2–15x-6=0

x = [15+(225+24)^0.5]/2

= [15+15.78]/2

= 15.39

So P does the work, alone, in 27.39 days while Q, alone will do it in 18.39 days

Answer 2

Answer:

6 days

Explanation:

Let the work done by Q in one day be x

Time taken by P to complete work= 3 days

Work done by P in one day= 1/3

Time taken by P and Q together to complete work= 2 days

Work done by P and Q together in one day= 1/2

1/2 = 1/3 + x

x= 1/2 - 1/3

x= 3/6 - 2/6

x= 1/6

Time taken by Q to complete the work alone= 6 days (reciprocal)

Thank you, hope it helps :)