Question

a and b can do a piece of a work in 12 days,b and c in 15 days, b and a in 20 days . how much time will alone take it to finish the work?

Answer 1

time taken by a + b to finish the work=12 days
Work done by a+b in 1 day=1/12
Time taken by b+c to to finish the work =15 days
Work done by b+c in 1 day=1/15
Time taken by c+a to finish the work =20 days
Work done by c+a in 1 day=1/20
Time taken by (a+b) (b+c) (c+a) =1/12+1/15+1/20=(5/60+4/60+3/60)=1/5
Time taken by2(a+b+c) in 1 day=1/5
Time taken by (a+b+c) in 1 day=1/5*2=1/10
Time taken by (a+b+c) =1/1/10=10 days

Answer 2

Given:

A and B can do a piece of work in = 12 days

Work done by A and B in 1 day = 1/12

B and C can do a piece of work in = 15 days

Work done by B and C in 1 day = 1/15

A and C can do a piece of work in = 20 days

Work done by A and C in 1 day = 1/20

Solution:

On adding A, B and C we get,

2(A+B+C)’s one day work

= 1/12 + 1/15 + 1/20

= (5+4+3)/60 (by taking LCM for 12, 15 and 20 which is 60)

= 12/60

= 1/5

A+B+C one day work = 1/(5×2) = 1/10

Knowing that:

Work A can do in 1 day

= (A+B+C)’s 1 day work – (B+C)’s 1 day work

= 1/10 – 1/15

= (3-2)/30 (by taking LCM for 10 and 15 which is 30)

= 1/30

Answer:

∴ A alone can finish the work in = 1/(1/30) = 30day.