Question

A and B can do a piece of work in 30 days. While B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work ?​

Answer 1

Answer:

$\huge\underline\bold {Answer:}$

(A + B)’s one days’ work = 1/30

(B + C)’s one days’ work = 1/24 .....(1)

(C + A)’s one days’ work = 1/20

Therefore (A + B + C)’s one days’ work

$\frac{1}{2} ( \frac{1}{30} + \frac{1}{24} + \frac{1}{20} ) \\ = \frac{1}{2} \times ( \frac{20 + 25 + 30}{600} ) \\ = \frac{75}{1200} \\ = \frac{1}{16} \: \: \: \: \: \: \: ...(2)$

(A + B + C)’s 10 days’ work

= 10/6 = 5/8

From (1) and (2), A's one days’ work

= 1/16 – 24

= 1/48

Therefore remaining 3/8 of the work is done by A alone in 3/8 × 48

= 18 days.

Answer 2

Answer:

$\boxed{Answer}$

A & B do the work in 30 days

So in 1 day A & B do 1/30 of the work

Similarly in 1 day B & C do 1/24 of the work

And in 1 day C & A do 1/20 of the work

So 2 ( A + B + C) = 1/30+1/24 +1/20 = (4 +5 +6) /120 =15/120 =1/8

So A , B & C together take 16 days to complete the work

In 1 day they do 1/16

In 10 days they do 10/16 = 5/8 of the work

Work remaining = 3/8

Amount of work A does in 1 day = 1/16 - 1/24 = 1/48

So A alone takes 48 days to do the complete work

So 3/8 will take A 3/8 x 48 = 18 more days