A and B together can do a piece
of work in 19 days. A is thrice as
efficient as B. If A and B together
works for 7 days, then in how many
days the remaining work will be
completed by B alone?
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Answer:
A finishes 10 days before B, so time taken for A, TaTa , is time taken by B, TbTb less 10 days :
Ta=Tb−10Ta=Tb−10
A is 3 times as efficient as B, and so time taken for A, TaTa is 1313 of TbTb :
3Ta=Tb3Ta=Tb
Ta=3Ta−10Ta=3Ta−10
2Ta=102Ta=10
∴Ta=5∴Ta=5
∴Tb=3T∴Tb=3T a=15a=15
A does 1515 or 315315 of the job each day.
B does 115115 of the job each day.
A & B do 315+115=415315+115=415 each day.
3×415=12153×415=1215
4×415=16154×415=1615
So they’ll finish on Day 4.
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