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Quantitative Aptitude DI

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

Study the following graph carefully and answer the given questions.

The table shows the number of days taken by different persons to complete the whole work alone.

1) A and C started to working together, after 3 days the person who has equal efficiency of A and C working together was joins with them. And 4 days after, the person who has 100% less efficiency than the person E completed the remaining work. In how many days required to complete the whole work? 

a) 7

b) 6

c) 9

d) 12

e) 10

2) C and F both started the work together but F left after some days. Find the number of days required to complete the whole work?

Statement I: The number of days C and F worked to complete the whole work is in the ratio of 9: 5

Statement II: If F gets Rs. 1800 from the total wage of Rs. 4500.

a) Only I

b) Only II

c) Both I and II

d) Neither I nor II

e) Either I or II

3) D and C can complete the work individually working 10 hours a day. Work is to be done in two shifts. Morning shift lasts for 6 hours and evening shift last for 4 hours. On the first day D works in the morning shift while C works in the evening shift. Next day D works in the evening shift while C works in the morning shift and so on. It means they work alternatively with respect to their shifts. Thus they work on this pattern till the work is completed. On which day the work got completed?

a) 15th

b) 13th

c) 12th

d) 14th

e) None of these

4) E and B started working together for 5 days, after which B was replaced by P. If the work was finished in next 2 days, then the number of days in which P alone could do the work?

a) 20 days

b) 24 days

c) 32 days

d) 28 days

e) None of these

5) Quantity I: A, B and E started working together but A works with 100% more efficiency. A left 5 days before the work was completed and B left 2 days after A had left. Find the number of days taken to complete the whole work?

Quantity II: C, D and E started working together but C works with 25% more efficiency. D and E left the work after 3 days and 5 days of the work started. Find the number of days required to complete the whole work

a) Quantity I > Quantity II

b) Quantity I < Quantity II

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II (or) Relationship cannot be determined

Answers :

1) Answer: C

(A+C)’s one day work = 1/20 + 1/30

= 5/60 = 1/12

The person E has equal efficiency of A and C working together

(A+C)’s 3 days work = 3*(1/12)

= ¼

Remaining = 1 – ¼ = ¾

A, C and E’s 4 days work = 4*(1/20 + 1/30 + 1/12)

= 4 (10/60)

= 2/3

Remaining work = ¾ – 2/3

= (9-8)/12

= 1/12

The person D has 100% less efficiency than by E,

D completes the remaining work,

Number of days taken by D to compete remaining 1/12 work,

= 1/12 * 24 = 2 days

Total number of days to complete the whole work = 3+4+2 = 9 days

2) Answer: E

C’s per day work = 1/30

F’s per day work = 1/25

LCM of 30 and 25 = 150

Total work = 150 units

C’s per day work = 150/30 = 5 units

F’s per day work = 150/25 = 6 units

From statement I,

Number of days C and F worked is 9x and 5x respectively

According to the question,

9x*5 + 5x * 6 = 150

45x + 30x = 150

75x = 150

= > x = 150/75 = 2

Required number of days

= (9*2 – 5*2) + 5*2

= 18 – 10 + 10 = 18 days

Hence, statement I alone is sufficient to answer the given question.

From statement II,

Wage per unit = 4500/150 = Rs.30

Number of units completed by F = 1800/30 = 60 units

Number of days required by F to complete 60 units = 60/6 = 10 days

Number of units completed by C = 150 – 60 = 90 units

Number of days required by C to complete 90 units = 90/5 = 18 days

Hence, statement II alone is sufficient to answer the given question.

3) Answer: D

LCM of 24 and 30 = 120

Total work = 120 units

D’s per day work = 120/24 = 5 units

C’s per day work = 120/30 = 4 units

Day – 1 = 5 + 4 = 9 units

Day – 2 = 4 + 5 =9 units

1 cycle (2 days) = 18 units

6th cycle (12 days) = 18 * 6 = 108 units completed

Remaining = 120 – 108 = 12 units

9 units completed in 13th day

Remaining units completed in 14th day        

4) Answer: B

E and B worked together

1/12 + 1/15 = (12 + 15)/(12*15) = 3/20

E and B’s 5 days work = (3/20)*5 = (3/4)

Remaining work 1/4 done by E and P

E and P finished it in 4 days

(1/4)*(E + P)’s whole work = 2

(E + P)’s whole work = 8

P’s one day work = (1/8) – (1/12) = 1/24

P alone can complete the work in 24 days

5) Answer: A

Quantity I:

LCM of 20, 15 and 12 = 120

Total work = 120 units

A’s per day work = 120/20 = 6 units

A’s 100% more efficiency = 6 * 200/100 = 12 units

B’s per day work = 120/15 = 8 units

E’s per day work = 120/12 = 10 units

A’s 5 days work = 12 * 5 = 60 units

E’s (5 – 2 = 3) days work = 3 * 10 = 30 units

Required number of units to complete

= 120 + 60 + 30 = 210 units completed by A, B and E together

Required number of days = 210/(12+8+10)

= 210/30 = 7 days

Quantity II: C, D and E started working together but C works with 25% more efficiency. D and E left the work after 3 days and 5 days of the work started. Find the number of days required to complete the whole work

LCM of 30, 24 and 12 = 120

Total work = 120 units

C’s per day work = 120/30 = 4 units

C’s 25% more efficiency = 4 * 125/100 = 5 units

D’s per day work = 120/24 = 5 units

E’s per day work = 120/12 = 10 units

D’s 3 days work = 5*3 = 15 units

E’s 5 days work = 10*5 = 50 units

Number of units completed by all the three persons together = 120 – 50 – 15

= 55 units

Required number of days = 55/(5+5+10)

= 55/20 = 2 (15/20)

= 2 (3/4) days

Hence, Quantity I > Quantity II