Math, asked by prince65469, 2 months ago

A can do 7/10 of a piece of work in 14 days and with the help of B he is able to finish the same work in 10 days. What time will B alone take to finish the work ?​

Answers

Answered by Anonymous
1

Answer:

B will require 10 days to complete the whole work alone.

A does 7/10th of the work in 14 days.

So he will require 20 days to complete the whole work.

After 14 days 3/10th of the work is remaining.

Of this when they work together A does 1/20x2 or 1/10 th of the work and the balance 2/10 of the work is done by B.

To do 1/5th of the work B requires 2 days.

To do the whole work he will require 10 days.

Answered by Itzxblackloverxx
0

Answer:

A completes 7/10 of a work in 14 days, means A's daily performance is...

7/10 divided by 14 = 1/20 part of the work or A can complete the work in 20 days.

After 14 days the unfinished work left is...

1 - 7/10 = 3/10 part and A and B jointly completed this part in 2 days.

A had contributed 1/20 x 2 = 2/20 or 1/10 part and B's contribution is... The rest i.e..

3/10 - 1/10 =2/10 in 2 days or B's daily contribution is 1/10 part of the work. Therefore B Alone could have completed the work in...

10 days. Answer.

Note : For A the reamaining 3/10 part of the work would have taken 3/10 by 1/20 = 6 days more. Whereas by the help of B this part was completed in 2 days and rate of performance of B is obviously DOUBLE of A. Hence B could do the work within HALF of Timeframe for A, i.e 20/2=10 answer.

Step-by-step explanation:

hope it's helpful

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