Question

A and B can do a piece of work in 12 days B and C in 15 days and C and A in 20 days How much time will A alone take to finish the job? ​

Answer 1

Answer:

A needs  30 days to finish the work.

Step-by-step explanation:

Given A and B do piece of work in 12 days and B and C in 15 days and C and A in 20 days

LCM of 12,15,20 = 60

One day work of A and B = $\frac{60}{12}$= 5 units

One day work of B and C = $\frac{60}{15}$ = 4 units

One day work of C and A = $\frac{60}{20}$= 3 units

Now if we add all three we get:-

2(A+B+C)=12 units

So one day work of A, B and C put together is $\frac{12}{2}$ = 6 units

To find one day work of A:-

One day work of A, B and C together - one day work of B and C

6–4=2 units

If the total work is of 60 units and the one day work of A is 2 unit, then alone A needs $\frac{60}{2}$ = 30 days to finish the work.

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) = 30days.