SBI Clerk and SBI PO Prelims exam are going to be held in the upcoming months. We have already provided you with the PDFs of many topics of Quantitative Aptitude like Simplification/Approximation, Number Series,datainterpretation.important Arithmetic questions to prepare for SBI Clerk and PO Pre exam. Practicing these questions will help you to know about the level of the questions. To increase your speed and accuracy, enhance your calculations.

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__START TEST__1.In each of these questions, one term in the given number series is wrong. Find out the wrong term.

5, 15, 81, 479, 3360, 26897

A. 479

B. 26879

C. 3360

D. 81

E. None of these

2.23, 46, 76, 110, 152, 192

A. 76

B. 152

C. 46

D. 110

E. None of these

3.What value should come at the place of question mark in the following number series?

10, 15, 30, 65, 155 ?

A. 411.5

B. 415.5

C. 417.5

D. 419.5

E. None of these

4.215, 335, 485, 665, 875, ?

A. 1125

B. 1115

C. 1005

D. 995

E. None of these

5.44, 72, 104, 140, 180, ?

A. 210

B. 205

C. 315

D. 224

E. None of these

6.In each of the following questions, two equations are given. You have to solve these equations to find the relation between x and y.

8x2 – 15x + 7 = 0

2y2 – 15y + 28 = 0

A.if x < y

B.if x > y

C.if x ≤ y

D.if x ≥ y

E.if relationship between x and y cannot be determined

7.

2x2 – 7x + 6 = 0

y2 + 13y + 42 = 0

A. if x < y

B. if x > y

C. if x ≤ y

D. if x ≥ y

E. if relationship between x and y cannot be determined

8.

2x2– 9x + 9 = 0

8y2+ 34y + 21 = 0

A. if x < y

B. if x > y

C. if x ≤ y

D. if x ≥ y

E. if relationship between x and y cannot be determined

9.

2x2 – 13x + 15 = 0

3y2 + 28y + 65 = 0

A. if x < y

B. if x > y

C. if x ≤ y

D. if x ≥ y

E. if relationship between x and y cannot be determined

10.

x2 = 2401

y3 = 117649

A. if x < y

B. if x > y

C. if x ≤ y

D. if x ≥ y

E. if relationship between x and y cannot be determined

11.Study the following information carefully and answer the questions given below:

There are total 1600 students in a class. Each likes one among the four breakfasts viz. Idli, Dosa, Upma and Puri Bhaji. Respective ratio of boys and girls among them is 5:3. 30% of the boys like Dosa. 25% of the girls like Puri Bhaji. 20% of the total number of students likes Idli. Respective ratio of number of boys who like Idli and number of boys who like Dosa is 2:5. Respective ratio of number of girls who like Puri Bhaji and number of girls who like Upma is 5:4. Number of boys who like Upma is 25% more than the number of girls who like Upma

Number of boys who like Upma is what percent of the number of girls who like Idli?

A. 55%

B. 75%

C. 50%

D. 60%

E. None of these

12.Find the difference between total number of students who like Dosa and total number of students who like Puri Bhaji.

A. 80

B. 150

C. 120

D. 100

E. None of these

13.Find the respective ratio of number of girls who like Puri Bhaji and total number of students who like Upma.

A. 3:5

B. 4:9

C. 5:9

D. 2:5

E. None of these

14.Find the average of the number of boys who like Puri Bhaji and number of girls who like Upma.

A. 235

B. 250

C. 240

D. 275

E. None of these

A. Rs.60000

B. Rs.52000

C. Rs.64000

D. Rs.48000

E. None of these

*SOLUTIONS:*1. Answer: C

5 x 4 – 5 = 15

15 x 5 + 6 = 81

81 x 6 – 7 = 479

479 x 7 – 8 = 3361 (not 3360)

3361 x 8 + 9 = 26897

2.Answer: D

4 x 6 – 1 = 23

6 x 8 – 2 = 46

8 x 10 – 4 = 76

10 x 12 – 8 = 112 (Not 110)

12 x 14 – 16 = 152

14 x 16 – 32 = 192

3.Answer: C

10 x 0.5 + 10 = 15

15 x 1 + 15 = 30

30 x 1.5 + 20 = 65

65 x 2 + 25 = 155

155 x 2.5 + 30 = 417.5

4.Answer: B

63 – 13 = 215

73 – 23 = 335

83 – 33 = 485

93 – 43 = 665

103 – 53 = 875

113 – 63 = 1115

5.Answer: D

4 x 11 = 44

6 x 12 = 72

8 x 13 = 104

10 x 14 = 140

12 x 15 = 180

14 x 16 = 224

6.Answer: A

I.8x2 – 15x + 7 = 0

=> 8x2 – 8x – 7x + 7 = 0

=> 8x(x – 1) – 7(x – 1) = 0

=> (8x – 7)(x – 1) = 0

=> x = 7/8, 1

II.2y2 – 15y + 28 = 0

=> 2y2 – 8y – 7y + 28 = 0

=> 2y(y – 4) – 7(y – 4) = 0

=> (2y – 7)(y – 4) = 0

=> y = 7/2, 4

Hence, x < y

7.Answer: B

I.2x2 – 7x + 6 = 0

=> 2x2 – 4x – 3x + 6 = 0

=> 2x(x – 2) – 3(x – 2) = 0

=> (2x – 3)(x – 2) = 0

=> x = 3/2, 2

II.y2 + 13y + 42 = 0

=> y2 + 6y + 7y + 42 = 0

=> y(y + 6) + 7(y + 6) = 0

=> (y + 7)(y + 6) = 0

=> y = -7, -6

Hence, x > y

8.Answer: B

I.2x2– 9x + 9 = 0

=> 2x2 – 6x – 3x + 9 = 0

=> 2x(x – 3) – 3(x – 3) = 0

=> (2x – 3)(x – 3) = 0

=> x = 3/2, 3

II.8y2+ 34y + 21 = 0

=> 8y2 + 28y + 6y + 21 = 0

=> 4y(2y + 7) + 3(2y + 7) = 0

=> (4y + 3)(2y + 7) = 0

=> y = -3/4, -7/2

Hence, x > y

9.Answer: B

I.2x2 – 13x + 15 = 0

=> 2x2 – 10x – 3x + 15 = 0

=> 2x(x – 5) – 3(x – 5) = 0

=> (2x – 3)(x – 5) = 0

=> x = 3/2, 5

II.3y2 + 28y + 65 = 0

=> 3y2 + 15y + 13y + 65 = 0

=> 3y(y + 5) + 13(y + 5) = 0

=> (3y + 13)(y + 5) = 0

=> y = -13/3, -5

Hence, x > y

10.Answer: C

I.x2 = 2401

=> x = ± √2401

=> x = ± 49

II.y3 = 117649

=> y = 3√117649

=> y = 49

Hence, x ≤ y

11.

Answer: B

Required percentage = 150/200 x 100 = 75%

12.Answer: B

Required difference = 580 – 430 = 150

13.Answer: C

Required ratio = 150 : 270 = 5:9

14.Answer: D

Required average = (430 + 120)/2

= 550/2

= 275

15.Answer: B

Number of products sold by company P in 2018 = 42/100 x 250000 = 105000

Number of products sold by company P in 2017 = 105000 x 80/100 = 84000

Number of products sold by company Q in 2018 = 58/100 x 320000 = 185600

Required ratio = 84000 :185600 = 105 : 232