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- 16. A and B can do a piece of work in 12 days and 24days. Both starts the work together for some time, but B leaves the job 6 days before the work is completed. Find the time in which work is finished.
A.10 days
B.12 days
C.14 days
D.5 days
Answer & Explanation
Answer: Option B 
A & B completes the work in 12*24/12+24 = 12*24/36 = 8 days
2 days work = 2+1/8 = ¼ work
Remaining work = 1-1/4 = ¾
A completes ¾ work in ¾*16 = 12 days
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- 17. A can do a piece of work in 15 days. A does the work for 5 days only and leaves the job. B does the remaining work in 6 days.In how many days B alone can do the work?
A.5days
B.7 days
C.12 days
D.9 days
Answer & Explanation
Answer: Option D
A’s 5 day work = 5*1/15=1/3
Remaining work = 1-1/3=2/3
B completes 2/3 work in 6 days
B alone can do in x days
2/3*x=6
X=9 days
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- 18. 36 workers can reap a field in 6 days. If the work is to be completed in 4 days, the extra workers required are
A.10
B.25
C.18
D.15
Answer & Explanation
Answer: Option C
M1D1 = M2D2
36(6)=X(4)
=54
Extra required workers are 54-36=18
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- 19. A does work in 20 days and B does work in 30 days. In how many days they together will do the same work?
A.6
B.12
C.16
D.20
Answer & Explanation
Answer: Option B
A’s one day work is 1/20
B’s one day work is 1/30
A and B work together 1/20+1/30=5/60=12 days
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- 20. A tyre has two punctures. The first puncture alone would have made the tyre flat in 10 min and second alone would have done it in 20min. If air leaks out at a constant rate how long does it take both the punctures together to make it flat?
A.6 22/3
B.6 2/3
C.6 1/3
D.6 4/3
Answer & Explanation
Answer: Option B
1st puncture:1/10
2nd puncture:1/20
both punctures make the tyre flat in: 1/10 + 1/20 =3/20
Time = 20/3 =6 2/3 min
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