# IBPS RRB Clerk 2019 Mains: Quant Test Day 13

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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5): There are 3 number series you have to find the value of a, b, c and then establish relation among a, b, c.**A) 12, 16, 7, (a), -2, 34

B) 4, 4, 8, (b), 96, 480

C) 2, 7, (c) 77, 238, 723Correct**Answer: 5) a <b=c**

**Explanation:**

A)

12 + 2^{2}= 16

16-3^{2}= 7

7+4^{2}=**23 (a)**

23-5^{2}= -2

-2+6^{2}= 34

B)

4*1 = 4

4*2 = 8

8*3 =**24 (b)**

24*4=96

96*5 = 480

C)

2*3+1 = 7

7*3+3 =**24 (c)**

24*3+5=77

77*3+7 = 238

238*3+9 = 723Incorrect**Answer: 5) a <b=c**

**Explanation:**

A)

12 + 2^{2}= 16

16-3^{2}= 7

7+4^{2}=**23 (a)**

23-5^{2}= -2

-2+6^{2}= 34

B)

4*1 = 4

4*2 = 8

8*3 =**24 (b)**

24*4=96

96*5 = 480

C)

2*3+1 = 7

7*3+3 =**24 (c)**

24*3+5=77

77*3+7 = 238

238*3+9 = 723 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5): There are 3 number series you have to find the value of a, b, c and then establish relation among a, b, c.**A) 512, 512, 256, (a), 8, 0.5

B) 1, 8, 9, (b), 25, 216

C) 6, 7, 15, 42, (c), 231Correct**Answer:****2) a =b****<c**

**Explanation:**

A)

512/1 = 512

512/2 = 256

256/4 =**64 (a)**

64/8 = 8

8/16 = 0.5

B)

1^{2}=1

2^{3}=8

3^{2}=9

4^{3}=**64 (b)**

5^{2}=25

6^{3}=216

C)

6+1^{3}=7

7+2^{3}=15

15+3^{3}=42

42+4^{3}=**106(c)**

106+5^{3}=231Incorrect**Answer:****2) a =b****<c**

**Explanation:**

A)

512/1 = 512

512/2 = 256

256/4 =**64 (a)**

64/8 = 8

8/16 = 0.5

B)

1^{2}=1

2^{3}=8

3^{2}=9

4^{3}=**64 (b)**

5^{2}=25

6^{3}=216

C)

6+1^{3}=7

7+2^{3}=15

15+3^{3}=42

42+4^{3}=**106(c)**

106+5^{3}=231 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5): There are 3 number series you have to find the value of a, b, c and then establish relation among a, b, c.**A) 5, 12, 24, 36, 52, (a)

B) 1, 6, 30, (b), 360, 720,

C) 105, 110, 100, 115, 95, (c)Correct**Answer: 4) a <b=c**

**Explanation:**

A)

2+3 = 5

5+7 =12

11+13 = 24

17+19 = 36

23+29=52

31+37 =**68 (a)**

B)

1*6=6

6*5=30

30*4=**120 (b)**

120*3=360

360*2=720

C)

105+5 = 110

110-10 = 100

100+15 = 115

115-20 = 95

95+25 =**120 (c)**Incorrect**Answer: 4) a <b=c**

**Explanation:**

A)

2+3 = 5

5+7 =12

11+13 = 24

17+19 = 36

23+29=52

31+37 =**68 (a)**

B)

1*6=6

6*5=30

30*4=**120 (b)**

120*3=360

360*2=720

C)

105+5 = 110

110-10 = 100

100+15 = 115

115-20 = 95

95+25 =**120 (c)** - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative AptitudeA) 48, 24, 36, (a), 315, 1417.5

B) 24, 35, 48, 65, 84, (b)

C) 101, 103, (c), 109, 113, 127Correct**Answer: 4) a <b=c**

**Explanation:**

A)

48*0.5 = 24

24*1.5 = 36

36*2.5 =**90 (a)**

90*3.5 = 315

315*4.5 = 1417.5

B)

24+11 = 35

35+13 = 48

48+17=65

65+19 = 84

84+23 =**107 (b)**

C) Prime Numbers: 101, 103,**107(c)**, 109, 113, 127Incorrect**Answer: 4) a <b=c**

**Explanation:**

A)

48*0.5 = 24

24*1.5 = 36

36*2.5 =**90 (a)**

90*3.5 = 315

315*4.5 = 1417.5

B)

24+11 = 35

35+13 = 48

48+17=65

65+19 = 84

84+23 =**107 (b)**

C) Prime Numbers: 101, 103,**107(c)**, 109, 113, 127 - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative AptitudeA) 3, 14, (a), 385, 1536, 4605

B) 1, 10, 0.5, (b), 0.375, 18.75

C) 5, 3, 2, 1.5, (c), 1.125Correct**Answer: 3) a >b>c**

**Explanation:**

A)

3*7-7 = 14

14*6-6 =**78 (a)**

78*5-5 = 385

385*4-4 = 1536

1536*3-3 = 4605

B) 1*10 = 10

10÷20 = 0.5

0.5*30 =**15 (b)**

15÷40 = 0.375

0.375*50 = 18.75

C)

5*0.5 + 0.5 = 3

3*0.5+0.5 = 2

2*0.5+0.5 = 1.5

1.5*0.5+0.5 =**1.25 (c)**

1.25*0.5+0.5 = 1.125Incorrect**Answer: 3) a >b>c**

**Explanation:**

A)

3*7-7 = 14

14*6-6 =**78 (a)**

78*5-5 = 385

385*4-4 = 1536

1536*3-3 = 4605

B) 1*10 = 10

10÷20 = 0.5

0.5*30 =**15 (b)**

15÷40 = 0.375

0.375*50 = 18.75

C)

5*0.5 + 0.5 = 3

3*0.5+0.5 = 2

2*0.5+0.5 = 1.5

1.5*0.5+0.5 =**1.25 (c)**

1.25*0.5+0.5 = 1.125 - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (6-10): Data below represent the number of people required for the various project to be completed in various days:**

**If 3 people were working for Project A, how many days did it take to complete the work?**Correct**Answer: 2) 4 days**

Explanation:

Project A:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

(X+1)*X = (X-1)*(X+3)

X2+X = X2+3X-X-3

X = 2X-3

X = 3

Let the work = (3+1)*3 = 12 units

Workers*Days = 12

3*Days = 12

Days = 4**Common Explanation:**

**Project A:**

Workers*Days (Situation 1) = Workers*Days (Situation 2)

(X+1)*X = (X-1)*(X+3)

X^{2}+X = X^{2}+3X-X-3

X = 2X-3

X = 3

**Project B:**

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

**Project C:**

Workers*Days (Situation 1) = Workers*Days (Situation 2)

M^{3}*27 = 8*N^{3}

3M = 2NIncorrect**Answer: 2) 4 days**

Explanation:

Project A:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

(X+1)*X = (X-1)*(X+3)

X2+X = X2+3X-X-3

X = 2X-3

X = 3

Let the work = (3+1)*3 = 12 units

Workers*Days = 12

3*Days = 12

Days = 4**Common Explanation:**

**Project A:**

Workers*Days (Situation 1) = Workers*Days (Situation 2)

(X+1)*X = (X-1)*(X+3)

X^{2}+X = X^{2}+3X-X-3

X = 2X-3

X = 3

**Project B:**

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

**Project C:**

Workers*Days (Situation 1) = Workers*Days (Situation 2)

M^{3}*27 = 8*N^{3}

3M = 2N - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (6-10): Data below represent the number of people required for the various project to be completed in various days:**

**Find the number of people required to finish the project B in 8Y days.**Correct**Answer: 4) 5**

Explanation:

Project B:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

8Y = 8*2 = 16

Let the work = (Y+3)*16 = (2+3)*16 = 80 units

People*Days = 80

People*8Y = 80

People*16 = 80

People = 5Incorrect**Answer: 4) 5**

Explanation:

Project B:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

8Y = 8*2 = 16

Let the work = (Y+3)*16 = (2+3)*16 = 80 units

People*Days = 80

People*8Y = 80

People*16 = 80

People = 5 - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (6-10): Data below represent the number of people required for the various project to be completed in various days:**

**If M=4 then find the number of days required to complete the project C in situation 2.**Correct**Answer: 3) 216**

**Explanation:**

Project C:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

M^{3}*27 = 8*N^{3}

3M = 2N

3*4 = 2N

N = 6

N^{3}= 6^{3}= 216Incorrect**Answer: 3) 216**

**Explanation:**

Project C:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

M^{3}*27 = 8*N^{3}

3M = 2N

3*4 = 2N

N = 6

N^{3}= 6^{3}= 216 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude

**2*N*X is what percent of 3*M*Y?**Correct**Answer: 5) 150%**

Explanation:

Project A:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

(X+1)*X = (X-1)*(X+3)

X2+X = X2+3X-X-3

X = 2X-3

X = 3

Project B:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

Project C:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

M3*27 = 8*N3

3M = 2N

2*N*X = (2N)*3 = 3M*3 = 9M

3*M*Y = 3*M*2 = 6M

Required percentage = (9M/6M)*100 = (3/2)*100 = 150%Incorrect**Answer: 5) 150%**

Explanation:

Project A:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

(X+1)*X = (X-1)*(X+3)

X2+X = X2+3X-X-3

X = 2X-3

X = 3

Project B:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

Project C:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

M3*27 = 8*N3

3M = 2N

2*N*X = (2N)*3 = 3M*3 = 9M

3*M*Y = 3*M*2 = 6M

Required percentage = (9M/6M)*100 = (3/2)*100 = 150% - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude

**4 workers started working for project B and after 8 days they were replaced by 10 children. The efficiency of a child is 40% of the efficiency of a worker. Find the total time taken to finish the project B.**Correct**Answer: c) 20 days**

Explanation:

Project B:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

Let the work = (Y+3)*16 = (2+3)*16 = 80 units

8 workers completed the work in [(3Y+4) = (3*2+4)] = 10 days

Work completed by 8 workers in 1 day = 80/10 = 8 units

Work done by 4 workers in 1 day will be 4 units

Work done by 1 worker in 1 day = 4/4 = 1 unit

Work done by 4 workers in 8 days will be 4*8 = 32 units

Work left = 80-32 = 48units

Work done by 1 child in one day = 0.4*1 = 0.4 unit (efficiency of a child is 40% of the efficiency of worker)

Work done by 10 children in one day will be 0.4*10 = 4 units

Remaining 48 units will be completed by 10 children in 48/4 = 12 days

Total time taken = 8 +12 = 20 daysIncorrect**Answer: c) 20 days**

Explanation:

Project B:

Workers*Days (Situation 1) = Workers*Days (Situation 2)

8*(3Y+4) = (Y+3)*16

24Y+32 = 16Y+48

8Y = 16

Y = 2

Let the work = (Y+3)*16 = (2+3)*16 = 80 units

8 workers completed the work in [(3Y+4) = (3*2+4)] = 10 days

Work completed by 8 workers in 1 day = 80/10 = 8 units

Work done by 4 workers in 1 day will be 4 units

Work done by 1 worker in 1 day = 4/4 = 1 unit

Work done by 4 workers in 8 days will be 4*8 = 32 units

Work left = 80-32 = 48units

Work done by 1 child in one day = 0.4*1 = 0.4 unit (efficiency of a child is 40% of the efficiency of worker)

Work done by 10 children in one day will be 0.4*10 = 4 units

Remaining 48 units will be completed by 10 children in 48/4 = 12 days

Total time taken = 8 +12 = 20 days

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