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:

1. A + B= 1/30 of the job or A=1/30-B

• A + C= 1/20 of the job or A=1/20-C
• C + B= 1/24 of the job or C=1/24-B
• 1/30-B =1/20-1/24 + B
• B = (1/30 - 1/20 + 1/24 ) / 2
• So B = 1/ 80
• Then we find C = 1/24–1/80 = (20 - 6) /480 = 14/480
• C = 7/240
• then we find A=1/30-B = 1/30 -1/ 80
• A = 5 /240
• Now with their rendiment in ten days they do together:
• = 10 × ( 5/240 + 1/80 + 7/ 240)
• = 15/24
• What is left to do is by A only is:
• 1- 15/24 = 3/8 of the job
• Since A does 5/ 240 per day we calculate:
• X = (3/8) / (5/240)
• X= 18

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