Question

1. A can do a piece of work in 30 days. B and C together can
do the same work m 20 days. A and B worked together and
completed 2/3 of the work in 12 days. How many days will be
required by C to complete the remaining part of the work?​

Answer 1

Answer:

18 days

Step-by-step explanation:

Let the number of days taken by A, B & C to complete the work individually be a, b & c respectively.

Given, A and B can do a piece of work in 30 days

=> 1/a + 1/b = 1/30

Given, B and C can do the same work 24 days

=> 1/b + 1/c = 1/24

Given, A and C can do the same work in 20 days

=> 1/c + 1/a = 1/20

Adding all the three equations,

=> 2(1/a +1/b + 1/c) = 1/30 + 1/24 + 1/20 = (20+25+30)/600 = 75/600 = 1/8

=> 1/a + 1/b + 1/c = 1/16

=> When A, B & C work together, the amount of work done in one day is 1/16

(1/a + 1/b + 1/c) - (1/b + 1/c) = 1/16 - 1/24 = (3–2)/48 = 1/48

=> 1/a = 1/48

=> The work done by A individually in one day = 1/48

Given that A, B & C work together for 10 days,

=> Work done by them together for 10 days = 10 * (1/16) = 10/16 = 5/8

=> The remaining work to be done by A alone = 1–5/8 = 3/8

The number of days taken by A alone to complete the remaining work = (3/8)/(1/48) = 18 days

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Answer 2

Answer:

36

Step-by-step explanation:

A can do a piece of work in 30 days.

A 1day work = 1/30

(B+C) can do the same work m 20 days.

(B+C) 1day work= 1/20

(A+B+C) 1day work = 1/30+1/20

(A+B+C) 1day work = 2/60+3/60

(A+B+C) 1day work = 5/60

(A+B+C) 1day work = 1/12

(A+B) can completed 2/3 of the work in 12 days.

(A+B) can complete 1work in 12×3/2

18days

(A+B) 1day work= 1/18

C 1day work = 1day work of(A+B+C)-(A+B)

= 1/12-1/18

= 3/36 - 2/36

= 1/36

C can complete whole work in 36 days.

........666.......