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

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Table below shows the number of people required for the various projects to be completed in various days.**

**The amount of hard work that goes while doing project B is double to that of project A. What is the value of X?**Correct**Answer- 2) 40**

Explanation-

According to question, we have,

X × 55 = (X + 10) × 88/2

55X = 44X + 440

11X = 440

X = 40Incorrect**Answer- 2) 40**

Explanation-

According to question, we have,

X × 55 = (X + 10) × 88/2

55X = 44X + 440

11X = 440

X = 40 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Table below shows the number of people required for the various projects to be completed in various days.**

**Team P = (1.5Y + 15) workers and it can complete project C in 21 days. In how many days team Q will complete the project C if team Q consists of (Y + 10) workers?**Correct**Answer- 5) 31.5 days**

Explanation-

We have,

Y(Y – 3) = (1.5Y + 15) × 21

Y2 – 3Y = 31.5Y + 315

Y2 – 34.5Y – 315 = 0

By solving, we have,

Y = 42

Let the number of days in which team Q will complete the work are x. Then,

(42 + 10) × x = 42 × 39

x = 42 × 39/52 = 31.5Incorrect**Answer- 5) 31.5 days**

Explanation-

We have,

Y(Y – 3) = (1.5Y + 15) × 21

Y2 – 3Y = 31.5Y + 315

Y2 – 34.5Y – 315 = 0

By solving, we have,

Y = 42

Let the number of days in which team Q will complete the work are x. Then,

(42 + 10) × x = 42 × 39

x = 42 × 39/52 = 31.5 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Table below shows the number of people required for the various projects to be completed in various days.**

**By using the information given in Q2, calculate the number of days in which 37 workers will complete the project D?**Correct**Answer- 3) 47**

Explanation-

Let the required number of days are x

Then,

x × 37 = (42 + 5)(42 – 5)

37x = 47 × 37

x = 47Incorrect**Answer- 3) 47**

Explanation-

Let the required number of days are x

Then,

x × 37 = (42 + 5)(42 – 5)

37x = 47 × 37

x = 47 - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude

**What is the value of M, if 108 workers can do a project which is three times as big as project F in 392 days?**Correct**Answer- 4) 14**

Explanation-

We have,

M2 × 72 = 108 × 392/3

M2 = 196, M = 14Incorrect**Answer- 4) 14**

Explanation-

We have,

M2 × 72 = 108 × 392/3

M2 = 196, M = 14 - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude

**What is the value of Z if Z4 no of workers can do project E in 10 days?**Correct**Answer- 1) 2**

**Explanation-**

We have,

Z^{2}× 40 = Z^{4}× 10

Z^{2}= 4, Z = 2Incorrect**Answer- 1) 2**

**Explanation-**

We have,

Z^{2}× 40 = Z^{4}× 10

Z^{2}= 4, Z = 2 - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). There are five number series. You have to find the value of a, b and c and then establish relation among a, b and c.**(i). 723, 727, 736, 752, (a)

(ii). 23, 24, 50, 153, (b), 3085

(iii). 110, 165, 330, (c), 2475, 8662.5Correct**Answer- 3) b < a < c**

**Explanation-**(i). Difference is 4, 9, 16 and 25.

So, a = 777(ii). Sequence is,

23 × 1 + 1 = 24

24 × 2 + 2 = 50

50 × 3 + 3 = 153

153 × 4 + 4 = 616

So, b = 616(iii). Sequence is,

110 × 1.5 = 165

165 × 2 = 330

330 × 2.5 = 825

So, c = 825

Therefore, b < a < c.Incorrect**Answer- 3) b < a < c**

**Explanation-**(i). Difference is 4, 9, 16 and 25.

So, a = 777(ii). Sequence is,

23 × 1 + 1 = 24

24 × 2 + 2 = 50

50 × 3 + 3 = 153

153 × 4 + 4 = 616

So, b = 616(iii). Sequence is,

110 × 1.5 = 165

165 × 2 = 330

330 × 2.5 = 825

So, c = 825

Therefore, b < a < c. - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). There are five number series. You have to find the value of a, b and c and then establish relation among a, b and c.**(i). 56, (a), 66, 77, 90

(ii). 42, (b), 60, 78, 102

(iii). 14336, 896, 112, (c), 14Correct**Answer- 1) a > b > c**

**Explanation-**(i). Sequence is,

56 + 3 = 59

59 + 7 = 66

66 + 11 = 77

77 + 13 = 80

80 + 17 = 97

Therefore, a = 59(ii). Sequence is,

42 + 6 = 48

48 + 12 = 60

60 + 18 = 78

78 + 24 = 102

Therefore, b = 48(iii). Sequence is,

14336/16 = 896

896/8 = 112

112/4 = 28

Therefore, c = 28

Therefore, a > b > cIncorrect**Answer- 1) a > b > c**

**Explanation-**(i). Sequence is,

56 + 3 = 59

59 + 7 = 66

66 + 11 = 77

77 + 13 = 80

80 + 17 = 97

Therefore, a = 59(ii). Sequence is,

42 + 6 = 48

48 + 12 = 60

60 + 18 = 78

78 + 24 = 102

Therefore, b = 48(iii). Sequence is,

14336/16 = 896

896/8 = 112

112/4 = 28

Therefore, c = 28

Therefore, a > b > c - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). There are five number series. You have to find the value of a, b and c and then establish relation among a, b and c.**(i). 151, 154, (a), 190, 271, 514

(ii). 50, 60, 78, 109.2, (b), 262.08

(iii). 105, 112, 126, (c), 175Correct**Answer- 2) a < b > c**

**Explanation-**(i)Sequence is

151 + 3 = 154

154 + 9 = 163

163 + 27 = 190

190 + 81 = 271

271 + 243 = 514

Therefore, a = 163(ii) Sequence is,

50 × 1.2 = 60

60 × 1.3 = 78

78 × 1.4 = 109.2

109.2 × 1.5 = 163.8

Therefore, b = 163.8(iii) Sequence is,

105 + 7 = 112

112 + 14 = 126

126 + 21 = 147

147 + 35 = 175

Therefore, c = 147

Therefore, a < b > cIncorrect**Answer- 2) a < b > c**

**Explanation-**(i)Sequence is

151 + 3 = 154

154 + 9 = 163

163 + 27 = 190

190 + 81 = 271

271 + 243 = 514

Therefore, a = 163(ii) Sequence is,

50 × 1.2 = 60

60 × 1.3 = 78

78 × 1.4 = 109.2

109.2 × 1.5 = 163.8

Therefore, b = 163.8(iii) Sequence is,

105 + 7 = 112

112 + 14 = 126

126 + 21 = 147

147 + 35 = 175

Therefore, c = 147

Therefore, a < b > c - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude(i) 725, 750, (a), 900, 1100

(ii) 50000, 10000, 2000, (b), 80

(iii) 200, 300, (c), 2625, 11812.5Correct**Answer- 2) a > c > b**

**Explanation-**(i)Sequence is:

725 + 25 = 750

750 + 50 = 800

800 + 100 = 900

900 + 200 = 1100

Therefore, a = 800(ii) Sequence is:

50000/5 = 10000

10000/5 = 2000

2000/5 = 400

400/5 = 80

Therefore, b = 400(iii) Sequence is:

200 × 1.5 = 300

300 × 2.5 = 750

750 × 3.5 = 2625

2625 × 4.5 = 11812.5

Therefore, c = 750

Therefore, a > c > bIncorrect**Answer- 2) a > c > b**

**Explanation-**(i)Sequence is:

725 + 25 = 750

750 + 50 = 800

800 + 100 = 900

900 + 200 = 1100

Therefore, a = 800(ii) Sequence is:

50000/5 = 10000

10000/5 = 2000

2000/5 = 400

400/5 = 80

Therefore, b = 400(iii) Sequence is:

200 × 1.5 = 300

300 × 2.5 = 750

750 × 3.5 = 2625

2625 × 4.5 = 11812.5

Therefore, c = 750

Therefore, a > c > b - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude(i) 5, 15, 90, (a), 9720

(ii) 26, 41.5, 72.5, (b), 537.5

(iii) 9, 45, (c), 3375, 42187.5Correct**Answer- 5) a > c > b**

**Explanation-**(i)Sequence is,

5 × 3 = 15

15 × 6 = 90

90 × 9 = 810

810 × 12 = 9720

Therefore, a = 810(ii) Sequence is,

26 + 15.5 = 41.5

41.5 + 31 (15.5 × 2) = 72.5

72.5 + 93 (31 × 3) = 165.5

165.5 + 372 (93 × 4) = 537.5

Therefore, b = 165.5(iii) Sequence is,

9 × 5 = 45

45 × 7.5 = 337.5

337.5 × 10 = 3375

3375 × 12.5 = 42187.5

Therefore, c = 337.5

Therefore, a > c > bIncorrect**Answer- 5) a > c > b**

**Explanation-**(i)Sequence is,

5 × 3 = 15

15 × 6 = 90

90 × 9 = 810

810 × 12 = 9720

Therefore, a = 810(ii) Sequence is,

26 + 15.5 = 41.5

41.5 + 31 (15.5 × 2) = 72.5

72.5 + 93 (31 × 3) = 165.5

165.5 + 372 (93 × 4) = 537.5

Therefore, b = 165.5(iii) Sequence is,

9 × 5 = 45

45 × 7.5 = 337.5

337.5 × 10 = 3375

3375 × 12.5 = 42187.5

Therefore, c = 337.5

Therefore, a > c > b

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