A can do a piece of work in 30 days. He worked for 6 days and B finished the remaining work in 48 days. B alone can do the whole work in how many days? Options: 1) 60
2) 55
3) 75
4) 70
Answers
Explanation:
Total time required by A to complete the work= 30 days
He worked for 6 days
Remaining work
24 days of work of A is completed by A and B both in 9 days
A×24 =(A+B)×9
A/9 = (A+B)/24
That means B/15 ( the values in denominators are their per day speeds with which they work)
So to total work = A×6 +(A+B)×9
=9×6 + 24*9
= 270 units
Time taken by B= 270/15 = 18 days
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Answer:
A- 30 days for full work; 1 day: 1/30 work.
6 days: 6/30 work done; 6/30 is 1/5.
Remaining work:4/5
Together done in 9 days.
Together in one day: (4/5)÷9=4/45=8/90
Done by A in one day: 1/30=3/90
Done by B in one day=(8/90)-(3/90)=(8–3)/90=5/90=1/18
B does 1/18 work per day: so 18 days required to complete work alone.
Ans: 18 days.
Alternate method: 6, 9 and 30 are numbers used in problem description. So, let us take common multiple; 90 is one such number.
We consider full work as 90 units.
A does 90 units in 30 days.; so 3 units per day.
In 6 days A does6×3=18 units.
Remaining work =72 units done by A and B together in 9 days.; 8 units per day.
Out of 8 units per day together, A does 3 units.
So, B does 5 units per day.
At 5 units per day, B needs 18 days to complete 90 units of work (full work).
Ans: 18 days.