Math, asked by gamer12, 6 months ago

A and B can do a piece of work in 12 days, B and C in 15 days, C and A in 20 days. How much time will A alone take to finish the work

Answers

Answered by OKP12
2

Answer:

30 days.

Step-by-step explanation:

2(A+B+C) 1 DAY WORK=1 UPON 12 + 1 UPON 15 +1 UPON 20.

2(A+B+C)1 DAY WORK=12 UPON 60=1 UPON 5.

(A+B+C)1 DAY WORK=1 UPON 5INTO2=1 UPON 10.

A'S 1 DAY WORK=(A+B+C)1 DAY WORK-(B+C)1 DAY WORK.

A'S 1 DAY WORK=1 UPON 10-1 UPON 15.

A'S 1 DAY WORK=1 UPON 30.

NO. OF DAYS=1 UPON 1 DAY WORK.

NO. OF DAYS=1/1 UPON 30.

NO. OF DAYS=30 DAYS.

ANS=A ALONE WILL COMPLETE THE WORK IN 30 DAYS.

THANK YOU.

Answered by TheUntrustworthy
0

Given:

A and B can do a piece of work in = 12 days

Work done by A and B in 1 day = 1/12

B and C can do a piece of work in = 15 days

Work done by B and C in 1 day = 1/15

A and C can do a piece of work in = 20 days

Work done by A and C in 1 day = 1/20

Solution:

On adding A, B and C we get,

2(A+B+C)’s one day work

= 1/12 + 1/15 + 1/20

= (5+4+3)/60 (by taking LCM for 12, 15 and 20 which is 60)

= 12/60

= 1/5

A+B+C one day work = 1/(5×2) = 1/10

Knowing that:

Work A can do in 1 day

= (A+B+C)’s 1 day work – (B+C)’s 1 day work

= 1/10 – 1/15

= (3-2)/30 (by taking LCM for 10 and 15 which is 30)

= 1/30

Answer:

∴ A alone can finish the work in = 1/(1/30) = 30days.

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