Math, asked by ankanikar, 6 months ago

A and B together can do a piece of work in 25 days. If
B works alone for the last 10 days, it is completed in
30 days. In
how
many days A alone can do it?

Answers

Answered by satvikum2006
4

Answer:

Step-by-step explanation:

A and B can together complete a work in 25 days.

In 1 day A and B does: 1/25 work.

Its said, the work is completed in 30 days, if B works alone for last 10 days.

Means, A and B together worked for 20 days. Then A alone worked to complete the work in 30 days.

A and B in 20 days do: 30* 1/25 = 1.5 work

Work remaining = 1 - 1.5 = -0.5 work

-0.5 work done by A in 10 days

-0.5== 10 days

1== X

X = 60 days.

A can complete the work alone in 60 days.

In 1 day A does: 1/60 work.

Let in 1 day B does 1/b work

1/25 = 1/60 + 1/b

1/b = 1/25 - 1/60 = 0.00066666666

b = 90 days.

Hence B can complete the work alone in 90 days.

Hope it helps.

SHORT CUT

A+B=== 25 days

A completed the work alone in last 10 days and the work was competed in 40 days.

WORK REMAINING:

A+B                         A

6 days                     10 days

L.C.M(6.10) = 30 = Remaining work

A+B effic = 30/6 =5

A's effic = 30/10 = 3

Hence, B's effic = 5-3 = 2 [efficiencies are constant hence doesn't matter whether its determined from total work or remaining work)

TOTAL WORK = TOTAL EFFICIENCY * DAYS

TOTAL WORK = 5*36 = 180

B's time(days) = Work/ffic = 180/2 = 90 days.

Answered by MightyGuy1409
0

Answer:

See, The answer should be 12 days

As If there r 2 persons A..B so if both can do a piece of work in 25 days. One can do 12 (25/2)

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