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Quantitative Aptitude Set -278

Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.

1) A, B and C can do a piece of work in 14, 28 and 32 days respectively. They all begin to working together. A work continuously till it is finished, B leaves the

work 3 days before its completion and C leaves the work 2 days before its completion. In what time is the work finished?

A.7(20/31) days

B.8(14/31) days

C.9(14/31) days

D.7(20/31) days

E.none of these

2) A and B together can complete the job in 16 days, B and C together can complete the job in 20 days. First A did the job for 6 days, B did the job for 9 days and C completed the remaining job in 22 days. Find the number of days in which C alone can complete the job?

A.40 days

B.48 days

C.30 days

D.36 days

E.None of these

3) A and B together can complete 33(1/3) % of the work in 10 days, C and D together can complete 85% of the work in 17 days and A, B and C together can complete half of the work in 7.5 days. A and D together can complete 66(2/3) % of the work in 16 days. In how many days B alone complete 40% of the work?

A.32 days

B.36 days

C.45 days

D.48 days

E.None of these

4) A can complete the 75% of the work in 9 days and A and B together can complete 60% of the work in 4.8 days. If A got Rs. 5000 for doing the whole work and then what is the difference between the amount received by A and B?

A.Rs.2000

B.Rs.2500

C.Rs.3000

D.Rs.1500

E.None of these

5) A alone complete four – ninth of the work in 20 days, B alone complete half of the work in 15 days and C alone complete four – fifth of the work in 36 days. A and B started the work and after 5 days C joined with them, in how many days the work will be completed?

A.13(2/7) days

B.14(2/7) days

C.15(2/7) days

D.16(2/7) days

E.None of these

Answers :

1) Answer: B

Let the work finished in x days

A’s one day work = 1/14

B’s one day work = 1/28

C’s one day work = 1/32

x/14 + (x-3)/28 + (x-2)/32 = 1

=> (16x + 8x – 24 + 7x – 14)/224 = 1

=> 31x = 224 + 38

=> x = 262/31

=> x = 8 (14/31) days

2) Answer: A

LCM of 16 and 20 = 80 units

(A + B)’s one day work = 5 units

(B + C)’s one day work = 4 units

A’s 6 days work + B’s 9 days work + C’s 22 days work = 80

(A + B)’s 6 days work + (B + C)’s 3 days work + C’s 19 days work = 80

(5*6) + (4*3) + C’s 19 days work = 80

C’s 19 days work = 80 – 30 – 12 = 38

C’s one day work = 38/19 = 2 units

C alone can complete the work in, (80/2) = 40 days

3) Answer: D

A + B = 300/100 * 10 = 30 days

C + D = 100/85 * 17 = 20 days

A + B + C = 2/1 * 7.5 = 15 days

C = 1/15 – 1/30 = 1/30

D = 1/20 – 1/30 = 1/60

A + D = 300/200 * 16 = 24 days

A = 1/24 – 1/60 = (5 – 2)/120 = 1/40

B = 1/30 – 1/40 = 1/120

B complete 40% of the work = 40/100 * 120 = 48 days

4) Answer: B

A = 4/3 * 9 = 12 days

A + B = 100/60 * 4.8 = 8 days

B = 1/8 – 1/12 = 3 – 1/24 = 1/24

Ratio of the amount received by A and B = 1/12:1/24 = 2:1

Required difference = 1/2 * 5000 = 2500

5) Answer: B

A = 9/4 * 20 = 45

B = 2/1 * 15 = 30

C = 5/4 * 36 = 45

(x + 5/45) + (x + 5/30) + (x/45) = 1

2x + 10 + 3x + 15 + 2x = 90

7x = 65

x = 65/7 = 9(2/7)

Total time = 5 + 9(2/7) = 14(2/7) days

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