Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Time&Work)

Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Time&Work):

Dear Readers, Important Practice Aptitude Questions for IBPS Exams 2017 was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this material.

1.A can do a piece of work in 60 days while B can do it in 75 days. They begin together but 20 days before the completion of the work, A leaves off. How many days B finishes the remaining work?

1).A’s one day work = 1/60
B’s one day work = 1/75
(A+B)’s one day work = (1/60+1/75) = 3/100
Both A and B complete the work in 20 days = 20 × (3/100) = 3/5
∴ remaining (1-3/5 = 2/5) 2/5th work is done by B = 2/5 × 75 = 30 days.
Answer: D

2.A and B can do a job in 15 days, B and C can do a job in 20 days and C and A can do it in 12 days. How long would C take to do that job?

2).(A+B)’s one day work = 1/15
(B+C)’s one day work = 1/20
(C+A)’s one day work = 1/12
The work done by (A+B+C) is= 2(A+B+C) = 1/15+1/20+1/12 = 1/5
∴ (A+B+C)’s one day work = ½ × 1/5 = 1/10
Hence, C can do a job in one day = 1/10 – 1/15 = 5/150 = 1/30
∴ C can finish the job in 30 days.
Answer: B

3.A contractor undertook to finish a certain work in 144 days and employed 120 men. After 64 days, he found that he had already done 2/3 of the work. How many men can be discharged now, so that the work may finish in time?

3).Work = Men × Day
W1/W2 = (M1 × D1) / (M2 × D2)
[(2/3) W] / [(1/3) W] = (120×64) / (M2 × 80)
2×M2 × 80 = 120× 64
M2 = 48
Hence, 120 – 48 = 72 men can be discharged.
Answer: C

4.A, B and C can do a piece of work in 36, 48 and 72 days respectively. They started the work but A left 9 day before the completion of the work while B left 12 days before the completion of the work. How many days C worked?

4).Let C can complete the work in X days
(X-9)/36 + (X-12)/48 + (X/72) = 1 work
(12(X-9) + 9(X-12) + 6X)/432 = 1
12X -108 +9X -108 +6X = 432
∴ X= 648/27 = 24
Hence, C worked for 24 days.
Answer: B

5.A ,B and C can do a piece of work in 40 days, 30 days and 60 days respectively. A works continuously, on every third day A is assisted by B and C. In how many days work is completed?

5). Day1 Day2 Day3 … A A A+B+C The above cycle is repeated for the next days.
The work done by A, B and C in one day is 1/30, 1/40 and 1/60
Hence, 3 (3A+B+C) = 3 [1/40] + [1/30] + [1/60]
= 3(3) +4 +2 /120 = 15/120 = 1/8
∴ (3A+B+C) = 1/8×3 = 1/24
Hence, the work can be completed in 24 days.
Answer: D

6.A work is started by 15 men. After 4 days, 3 more men accompanied them to finish the work in next 10 days. How many men should have started the work to finish it in 8 days?

6). Work = Men × Days
(15×4) + (15+3) ×10 = M × 8
60 + 180 = M × 8
∴ M = 240/8 = 30
Hence, 30 men are required to finish the work in 8 days.
Answer: C

7.A can finish a work in 8 days. After working for 3 days, B is joined and the remaining work got finished in next one day. If B works alone, how many days will be required to finish the work?

7).Total work = A × 8 = A ×3 + (A+B) ×1
8A = 3A + A + B
∴ B = 4A
Number of days to finish the work by B = ¼ (No. of days of A)
= ¼ × 8 = 2 days.
Answer: A

8.Two taps A and B fill an overhead tank in 10 hrs and 15 hrs respectively. Both the taps are opened for 4 hrs and then B is turned off. How much time will A take to fill the remaining tank?

8).Tank filled in 1 hr by tap A = 1/10
Tank filled in 1 hr by tap B = 1/15
Both taps filled in 4 hrs = 4(1/10 + 1/15)
= 4 (1/6) = 2/3
∴ remaining 1-2/3 = 1/3 of the tank is filled by tap A = 1/3 ×10
= 3/10 = 3 1/3 hrs.
Answer: B

9.A boiler can be filled by two taps in 20 min and 30 min respectively and can be emptied by a third tap in 48 min. If they are all turned on at once, when will the boiler be half full?

9).A boiler is filled by 2 taps= 1/20 + 1/30 = 5/60 = 1/12
∴ half full the boiler = ½ B = 1/12 – 1/48
= 3/ 48 = 1/16
B = 2/16 =1/8
Hence, 8 mins required to half fill the boiler.
Answer: D

10.If 8 women or 12 children can do a piece of work in 16 days, what is the number of days required to complete the work by 20 women and 6 children?

10).8 women = 12 children
1 W = 12/8 C = 3/2 C
∴ 20 W = 20 × 3/2 C = 30 C
(12 C) ×16 = (20 W + 6 C) × D
(12 C) ×16 = (30 C + 6 C) × D
D = 12 C × 16 / 36 C
= 16 /3 = 5 1/3 days.
Answer: C