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
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
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Step-by-step explanation:
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 ..