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

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

(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

Given :

• 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

To find :

• How many days more will A take to finish the work ?

Solution :

(A + B)'s one day work = 1/30

(B + C) 's one day work = 1/24 ........(1)

(C + A) 's one day work = 1/20

∴ (A + B + C)'s one day work

=> ½(1/38 + 1/24 + 1/20)

=> ½ x (38 + 24 + 20/600)

=> 75/1200

=> 1/16 ...........(2)

(A + B + C)'s ten day work

=> 10/16 = 5/8

from (1) and (2) A's one days work

=> 1/16 - 1/24

=> 1/48

∴ remaining ⅜ of the work done is by A alone in

=> 3/8 x 48

=> 18 Days