Question

A and B together can complete a work in 40/3 days. B can complete the same work working alone in 6 more days than that taken by A. find the number of days taken by B to complete the work alone?

Answer 1

Let no. of days A required be x then no. of days B will required will be x+6
So,
work done by A in a day=1/x
and B=1/(x+6)
so, work done by A and B in a day=3/40
=>1/x + 1/(x+6) = 3/40
=>(x+6+x)/x(x+6) = 3/40
=>(2x+6)/(x^2+6x) = 3/40
=>40(2x+6)=3(x^2+6x)
=>80x+240=3x^2+18x
=>3x^2+18x-80x-240=0
=>3x^2-62x-240=0
=>3x^2-72x+10x-240=0
=>3x(x-24)+10(x-24)=0
=>(x-24)(3x+10)

So, x =24 as days cant be negative..
so, no of days taken by B to complete the work alone=30 days..