Question

A can do a piece of particular work in 25 days and B can finish it in 20 days. They work together for 10 days and then A leaves. In how many days will B finish the remaining work?​

Answer 1

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☆ 11 Days ☆.

METHOD 1

A works only for 5 days completing 5/25 = 1/5 part of work.

Remaining 4/5 work is done by B in 4/5 x 20 = 16 days.

=> B works for 11 days after A leaves.

METHOD 2

Let B take m days after A leaves.

Since A & B work together for 5 days, we have :

5/25 + (5 + m)/20 = 1

Taking LCM as 100,

20+ 5(5 + m) = 100

=> 20 + 25 + 5m = 100

5m = 100 - 45 = 55

m = 55/5= 11 days.

Answer 2

Answer:

2 days

Step-by-step explanation:

One day work of A = 1/25

One day work of B = 1/20

One day work of A and B together = 1/25 + 1/20 = 9/100

10 days work of A and B together = 9×10/100 = 9/10

Remaining work = 1 - 10 day work of A and B together = 1 - 9/10 = 1/10

NOW, Time taken by B to do the remaining work = remaining work divided by one day work of B

Time taken = 1/10/1/20 = 20/10 = 2 days

hope this helps

thank you ☺️