A and B can do a piece of work in 12 days, which B and C can do in 16 days. After A has been working for 5 days and B for 7 days, C finishes it in 13 days. In how many days C alone can finish the work?
Answers
Answer:
Lcm of (12,16) = 48 units <--total work
So A+B = 48/12 = 4 units per day
B+C = 48/16 = 3 units per day
Now,
5A+7B+13C = 48 units
5 (A+B)+2 (B+C)+11C = 48
5 (4)+2 (3)+11C = 48
20+6+11C = 48
11C = 22
C = 2 units
So C does 2 units of work daily
So total time required to complete work alone is 48/2 = 24 days
Step-by-step explanation:
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Step-by-step explanation:
A+B=12days.
(A=5days,B=days)
(A + B) one day work = 100/12 = 8.33% per day
B+C=16days
(B + C) one day work = 100/16 = 6.25% per day.
(A + B)'s 5 days work + (B + C)'s 2 days work + C's 11 days work = 1
therefore.5 days work of B has been included with A and 2 das work of C has been included with B
=(8.33 * 5 )+ (2 *6.25) + C's 11 days work = 100%
=C's 13 days work = 100 - 41.66 - 12.5 = 45.84%
=one day work of C = 45.84/11 = 4.16%
=C can done work = 100/4.16 = 24 days
therefore, C finish it's work in 24 days