Math, asked by doreamon5962, 1 year ago

A can complete a work in 10 days b in 15 days and c in 12 days. All of them begin the work together butahad to leave the work after two days of the start and c 3 days before the completation of the work. How long did the work last?/

Answers

Answered by Anonymous
11

Answer:

let the total work done by them in the end of 15 days be 1 work

Work done by A in 1 day=1/10

Work done by Bin 1 day=1/15

Work done by C in 1 day=1/12

a,b and c's  work together for 2 days

total work done by them in one day is=(1/10+1/12+1/15)=1/4

work done by them in two days=2*1/4=1/2

work done by b in the last three days(15-12)=1/15*3=1/5

[NOTE : 2days work done TOGETHER +3 days of  B's WORK =5days]

[NOTE: NOW WE HAVE TO FIND OUT NUMBER OF DAYS IN WHICH REMAINING WORK IS COMPLETED  BY C and B]

remaining work = (number of days required) * (WORK IS COMPLETED

BY C and B)

3/10 = x* (1/12 +1/15)

x= 2

TOTAL DAYS = (2days work done by them together) + (3 days of b's work) + (2 days B+c work)

thus 2+2+3=7.

Total work was done in 7 days.

hope it helps

please mark as brainliest

Step-by-step explanation:

Answered by amreshsahu6
0

Step-by-step explanation:

A-10d

B-15d

c-12d

LCM of ABC working time is 60

it means ,

A works - 6d

B works - 4d

C works - 5d

now ABC works together for 1 day= 6+4+5=15 units

so that ,

for 2 days = 15×2=30

now ramain work is 60-30=30 units

if C will work last 3 day with B = 5×3 = 15

now remain work will be = 30+15= 45 units

now B+C works= 4+5=9 unit/day

45/9= 5 days

so total days for total work is 2+5 =7 days..

this is the sort trik ..

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