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SBI PO Prelims Quantitative Aptitude Question Paper 2021

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  • Last Updated : 25 Mar, 2022
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Directions (01-05): In the given number series find out the answer in place of the question mark ‘?’.

1. Question

118    119    111    ?    74    199

a) 112

b) 92

c) 138

d) 128

e) 101

Answer: C

Explanation:

118 + 13 = 119

119 – 23  = 111

111 + 33 = 138

138 – 43  = 74

74 + 53   = 199

Thus, the required number is = 138.

 

2. Question

74   128   102   156   ?    212   270

a) 118

b) 130

c) 142

d) 158

e) 172

Answer: D

Explanation:

74 + 28  = 102

128 + 28 = 156

102 + 56 = 158

156 + 56 = 212

158 + 112 = 270

Thus, the required number is = 156.

 

3. Question

4    5    14    51    220    ?

a) 440

b) 880

c) 1025

d) 1125

e) 1250

Answer: D

Explanation:

4 × 1 + 1^2  = 5

5 × 2 + 2^ 2 = 14

14 × 3 + 3^2 =51

51 × 4 + 4^2 = 220

220 × 5 + 5^2 =1125

Thus the required number is = 1125.

 

4. Question

7    28    63    126    215    ?

a) 299

b) 344

c) 312

d) 327

e) 409

Answer: B

Explanation:

23 – 1 = 7

33 + 1 = 28

43 – 1 = 63

53 + 1 =126

63 – 1 = 215

73 + 1 = 344

Thus, the required number is = 344.

 

5. Question

25    24    27    25    29    ?     31    27

a) 25

b) 26

c) 27

d) 28

e) 30

Answer : B

Explanation:-

25 + 2 = 27

24 + 1 = 25

27 + 2 = 29

25 + 1 = 26

29 + 2 = 31

Thus, the required number is = 26.

 

Directions: (06 -10) Study the following information carefully and answer the questions given beside.

Indian footballer Sunil Chetri scored goals against different countries in three different years. 

 (NOTE: Total goals scored in a year= Bangladesh + England + Denmark )

 2010:

The total goals scored in 2010 were 1200. The goals scored against England were 1/3rd of the goals against Denmark in 2011. The average number of goals scored against Bangladesh and England was 300. 

 2011:

The total number of goals scored against Bangladesh and Denmark was 1200. The ratio of the total goals scored against Denmark in 2010 to that of the total goals scored against Denmark in 2011 is 4:3. The total goals scored against England in 2011 were equal to the total goals scored against England in 2012. 

 2012: The sum of the total goals scored against Bangladesh and England is equal to the total goals scored against Denmark. The total goals scored in 2012 were 1400. The total runs scored against Bangladesh were twice of the goals scored against England in 2010.

6. Question

What were the total goals scored in 2011?

a) 1000

b) 1200

c) 1400

d) 1600

e) 1500

Answer: D

Explanation:

2010: 

The total goals scored in 2010 were 1200. 

The average number of goals scored against Bangladesh and England was 300. So, the total runs scored against Bangladesh and England was 600. 

The total goals scored against others = 1200 – 600 = 600 

2011: 

The ratio of the total goals scored against Denmark in 2010 to that of the total goals scored against Denmark in 2011 is 4 : 3.  

So, the total goals against Denmark in 2011

= (600 × 3)/4

= 450

The total goals scored against Bangladesh and Others = 1200 

The total goals scored against Bangladesh = 1200 – 450 = 750 

2012: 

The total goals scored in 2012 were 1400. 

The sum of the total goals scored against Bangladesh and England is equal to the total goals scored against Denmark. It means the total goals scored against Denmark is half i.e 700 goals and the sum of the total goals scored against Bangladesh and England was 700. 

Years Bangladesh England Denmark

2010

 

 

600

2011

750

 

450

2012

 

 

700

2010:

The goals scored against England were 1/3rd of the runs against Denmark in 2012.

So the total goals scored against England were  =450 × 1/3= 150

The total goals scored against Bangladesh = 600 – 150 = 450

2012:

The total goals scored against Bangladesh were twice of the goals scored against England in 2010.

The total goals scored against Bangladesh = 150 × 2= 300

The total goals scored against England = 700 – 300=400

2011:

The total goals scored against England in 2011 were equal to the total goals scored against England in 2012.

The total goals scored against England = 400

Year Bangladesh England Denmark

2010

(1200) 

450

150

600

2011

(1600) 

750

400

450

2012

(1400) 

300

400

700

(Note: The above explanation for question no 06 – 10)

The total goals scored in 2011 were 1600.

Here, the correct option is D.

7. Question

What is the ratio of the total goals scored against Bangladesh in 2010 to that of the total goals scored against England in 2012?

a) 9:7

b) 7:9

c) 9:8

d) 11:7

e) 11:13

Answer: C

Explanation:

The total goals scored against Bangladesh in 2010 = 450 

 The total goals scored against England in 2012 = 400 

 So, required ratio = 450 : 400 = 9 : 8 

Hence, option C is correct.

 

8. Question.

What is the sum of the goals scored against England in all three years?

a) 800

b) 850

c) 950

d) 1050

e) 1100

Answer: C

Explanation:

The total goals scored against England in all three years

= (150 + 400 + 400 )

= 950

Hence, option C is correct.

 

9. Question.

The total goals scored against Bangladesh in 2011 is what percentage of the total goals scored against Bangladesh in 2012?

a) 100%

b) 125%

c) 150%

d) 200%

e) 250%

Answer: E

Explanation:

The total goals scored against Bangladesh in 2011 = 750

The total goals scored against Bangladesh in 2012 = 300

So, reqd. % = 750/300 × 100 % = 250%

Hence, option E is correct.

 

10. Question

What is the difference between the total goals scored against Denmark in 2010 to the total goals scored against Denmark in 2011?

a) 300

b) 250

c) 200

d) 150

e) 100

Answer: D

Explanation:

The total goals scored against Denmark in 2010 = 600 

 The total goals scored against Denmark in 2011 = 450 

 So, required difference = (600 – 450) = 150

Hence, option D is correct.

 

Direction (11-15): Study the bar chart carefully and answer the questions given beside.

A number of workers from Bihar, Odisha, and Jharkhand are working in a production factory in Mumbai city. Given bar graph shows the percentage of employees from Bihar, Odisha, and Jharkhand in Mumbai city over the different months. Study the bar graph and answer the following questions:

 

 

11. Question

If the ratio of the total workers who are doing their jobs in the month January and April is 3:5 and the total workers from Bihar in April are 4125, then find the total workers from Jharkhand in the month of January in Mumbai city.

a) 1575

b) 1690

c) 2250

d) 1920

e) 1780

Answer: A

Explanation:

Total workers doing their jobs in Mumbai in the month of April is

= (4125 × 100 )/27.5 = 15000

Total workers doing their jobs in Mumbai in the month of January

= (15000 × 3)/5 = 9000

Total workers from Jharkhand in the month of January in Mumbai

= 17.5% of 9000

= 1575

Thus, the total number of workers from Jharkhand in the month of January is = 1575.

 

12. Question

If total workers from Bihar in Mumbai in the month of February are equal to the total workers from Odisha in Mumbai in the month of May, then find the ratio of workers from Jharkhand in Mumbai in the respective months.

a) 2 : 5

b) 2 : 3

c) 1 : 5

d) 3 : 5

e) 5 : 7

Answer: C

Explanation:

Let workers on February and May are ‘x’ and ‘y’ respectively.

 According to the question,

 45% of x = 30% of y

 x : y  = 2 : 3

Required ratio-

15% of x : 50% of y = 15x : 50y

= 1 : 5

 

13. Question

If total workers in March are three times the total workers from Jharkhand in April and also the total workers from Bihar in the month of March is 5670, then find that the total workers from Bihar in the month of March are how much percent more than the total workers from Jharkhand in the month April.

a) 5%

b) 6%

c) 4%

d) 3%

e) 7%

Answer: A

Explanation:

Total workers on March= (5670 × 100 )/35 = 16200

Total workers from Jharkhand on April = 16200/3 = 5400

Required per cent = (5670 – 5400 )/5400 × 100 = 5%

Thus, workers from Bihar in the month of March are 5 % more than the total workers from Jharkhand in April.

14. Question

If the total number of workers from Odisha in the month of January is 2250 and the ratio of workers from Odisha in the months January and February is 45:88, then find the total number of workers in Mumbai city in the months January and February together.

a) 21000

b) 22000

c) 23000

d) 25000

e) 20000

Answer: B

Explanation:

Total workers in the month January =( 2250 × 100) / 25= 9000

Total workers from Odisha  is in the month February =( 2250 × 88 ) / 45= 4400

Total workers in the month February =( 4400 × 100 )/40 = 11000

Total workers in the Mumbai city in the month January and February together =( 9000 + 11000 )= 20000

Thus, the total workers in Mumbai city in the month of January and February together = 20000.

15. Question

If the ratio of total workers in all the given months from January to May is 2 : 3 : 3 : 4 : 5, then find the ratio of total workers from Bihar in Mumbai over all the given months from January to May.

a) 23 : 27 : 21 : 22 : 20

b) 22 : 29 : 23 : 20 : 21

c) 13 : 21 : 23 : 31 : 11

d) 24 : 27 : 21 : 29 : 25

e) None of these

Answer: A

Explanation:

Let the total number of workers in the given months be 2x, 3x, 3x, 4x and 5x respectively.

 Required ratio:

 = 57.5% of 2x : 45% of 3x : 35% of 3x : 27.5% of 4x : 20% of 5x

 = 57.5 × 2 : 45 × 3 : 35 × 3 : 27.5 × 4 : 20 × 5

 = 115 : 135 : 105 : 110 : 100

 = 23 : 27 : 21 : 22 : 20

Thus the ratio over all the given months from January to May is

= 23 : 27 : 21 : 22 : 30

 

16. Question

What approximation value will come in place of ‘ x ‘ in the given question ( find approximate value)

143.99 – (80 × 19.99 ) /39.99  + x = 69.99

a) 27

b) 34

c) 49

d) 21

e) 56

Answer: B

Explanation:

143.99 – ( 80 × 19.99 ) /39.99 + ?  = 69.99

144 – ( 80 × 20 ) /40 + x = 70

144 – 40 -70 = x

x = 34

Thus, the value of x is = 34.

 

17. Question

16 cm and 20 cm are the sides of a triangle. The angle included between the two sides is 30°. Then find the area of the triangle.

a) 80

b) 70

c) 90

d) 100

e) 150

Answer: A

Explanation:

Let  a = 16 cm

       b= 20 cm

Area = 1/2 × ab Sinθ

        =1/2 × 16 ×20 × Sin30°

        =1/2 ×16 × 20 ×1/2

        =80 cm2

Thus, the area of the triangle is 80 cm2.

 

18. Question

In a classroom, a madam multiplied a number by 4/5 instead of 5/4. What is the error percentage in the calculation?

a) 64 %

b) 25 %

c) 36 %

d) 48 %

e) None of these

Answer: C

Explanation:

4/5  ,  5/4

LCM = 20

4/5 × 20 = 16

5/4 × 20 = 25

Required percentage =( 25 – 16  ) / 25 × 100 %

                                    = 36%

Thus, the error is = 36 %

 

19. Question

A boy did a piece of work in 3 days. That piece of work was done by a woman in 4 days. If the boy and woman worked together, they got total wages of Rs. 3500. How much did the woman get?

a) 1500

b) 2000

c) 1000

d) 1200

e) None of these

Answer: A

Explanation:

 

Boy

Woman

Time

3

4

Efficiency

4

3

(Time and efficiency are inversely proportional)

Woman get = 3500 ×  3/7

                    = 1500

Thus, the woman gets the wages of Rs 1500.

 

20. Question

Pritam covers a certain distance between his uncle’s house and the shopping mall by walking. With an average speed of 15 km/hr, he is late by 10 mins. If he increases his speed to 20 km/hr, he reaches the gym 5 min earlier. What is the distance between his uncle’s house and the shopping mall?

a) 10 km

b) 12 km

c) 15 km

d) 18 km

e) 30 km

Answer: C

Explanation:

Let the distance = x

Difference between time = (10 + 5 ) min = 15/60 hrs

According to question,

x / 15 – x /20 = 1/4

4x – 3x = 15

x = 15

Thus, the distance is = 15 km

 

Directions (21-25): In the following question contains two equations as I and II. You have to solve both the equations and determine the relationship between them and give answers as,

a)  x > y

b)  x ≥ y

c)  x < y

d)  x ≤ y

e)  x = y or the relation cannot establish.

21. Question

I. 7x^2 + 10x + 3 = 0

II. 2y^2 – 3y – 44 = 0

Answer: E

Explanation:  

7x2 + 10x + 3 = 0

7x2 + 7x + 3x + 3 = 0

(x + 1) (7x + 3) = 0

x =  -3/7 , -1/3

 

2y2 -3y – 44 = 0

2y2 +8y – 11y – 44 = 0

(y + 4) (2y – 11)  = 0

y = 11/2 , – 4

Thus, the relationship can not be established.

 

22. Question

I. x2 – 11x + 28 = 0

II. y2 – 45y + 324 = 0

Answer :- C

Explanation :-

x2 – 11x + 28 = 0

x2 – 4x – 7x + 28 = 0

(x – 7) (x – 4) = 0

x = 4 , 7

 

y2 – 45y + 324 = 0

y2 – 36y – 9y + 324 = 0

(y – 9) (y – 36) = 0

y = 9 , 36

x < y

Thus, the correct option is C.

 

23.Question

I. x2 +14x + 60 =15

II. y2 = y + 30  

Answer: D

Explanation:

x2 + 14x +60 = 15

x2 + 14x + 45 = 0

x2 + 5x + 9x + 45 = 0

(x + 5) (x + 9) = 0

x = – 5 , – 9

 

y2 = y + 30

y2 – y – 30 = 0

y2 – 6y + 5y – 30 = 0

(y – 6) (y + 5) = 0

y = 6 , -5

x ≤ y

Thus, the correct option is D.

 

24.Question

I.  21x2 + 2x – 3 = 0

II. 12y2 – y – 6 = 0

Answer: E

Explanation:

21x2 + 2x – 3 = 0

21x2 – 7x + 9x – 3 = 0

(7x + 3) (3x – 1) = 0

x = 1/3 , – 3/7

 

12y2 – y – 6 = 0

12y2 – 9y + 8y – 6 = 0

(3y + 2) (4y – 3) = 0

y = 3/4 , – 2/3

Thus, the relationship can not be established

So the correct option is E.

 

25. Question

I. 2x2 – 288 = 0

II. y2 + 24x + 144 = 0

Answer: B

Explanation:

2x2 – 288 = 0

x2 – 144 = 0

(x + 12) (x – 12) = 0

x = 12 , – 12

 

y2 + 24x + 144 = 0

(y + 12)2 = 0

y + 12 = 0

y = -12

x ≥ y

Thus, the correct option is B

 

Directions (26-30):

Read the following information carefully and answer the questions. The given information shows the total number of AC and Cooler sold by four different seller P, Q, R, S.

AC: The ratio of the number of AC sold by sellers P and Q is 1 : 2 respectively. The difference between the number of AC sold by P and S is 90. The number of AC sold by R is 20 % less than that of shop S. Number of AC sold by R is 60 % of the number of AC sold by Q.

Cooler: Average number of Cooler sold by all sellers is 175 and the number of cooler sold by sellers P is two times the number of coolers sold by R. Ratio of the number of coolers sold by P and S is 5 : 4 respectively. The number of coolers sold by Q is 80 more than that of S. 

26. Question

Seller P’s selling of cooler is what percentage of total selling of cooler by four seller together?

a) 200/7 %

b) 100/3 %

c) 300/7 %

d) 400/17 %

e) None of these

Answer: B

Explanation:

For AC,

P : Q = 1 : 2

P – S = 90  

R : S = 80 : 100 = 4 : 5   (× 3)                      = 12 : 15

R : Q = 60 : 100 = 3 : 5 (×2) =6 : 10   (× 2)  = 12 : 20

P : Q = 1 : 2  (×5) =5 : 10   (×2)                    = 10 : 20    

P : Q : R : S = 10 : 20 : 12 : 15    

P – S = (15 – 10 ) = 5

5 unit = 90

1 unit = 90 ÷ 5 = 18

P = 18 × 10 = 180

Q = 18 × 20 = 360

R = 18 × 12 = 216

S = 18 × 15 = 270

For Cooler,

P + Q + R + S =175 × 4 = 700

P = 2R

P : R = 2 : 1  (×5) = 10 : 5

P : S = 5 : 4   (×2) =10 : 8

P : R : S = 10x : 5x : 8x

Q = (S + 80)

   = (8x + 80)

P + Q + R + S = 10x + 8x + 80 + 5x + 8x

                       = 31x + 80

31x + 80 = 700

31x = 620

x = 20

:. P   = 10 × 20      = 200

:. Q   = 8 ×20 + 80 = 240

:. R   = 5 × 20         = 100

:. S    = 8 ×20         = 160

( Above explanation is for question 26 to 31 )

Required % = 200 / 700 × 100 %

                    = 200/7 %

:. The required percentage = 200/7 %

27. Question:

Find out the ratio of total selling of cooler by four sellers to the selling of AC by seller P, Q and S together.

a) 40: 41

b) 51 : 61

c) 70 : 81

d) 1 : 3

e) 3 : 5

Answer : C

Explanation :-

(P + Q +S) ‘s selling of AC = (180 + 360 + 270 )

                                             = 810

Required Ratio = 700 : 810

                          = 70 : 81

:. The ratio between selling of Cooler by all and selling of AC by P, Q and S = 70 : 81

28. Question

Find out the ratio between the (P + Q)’s selling of AC and ( R+Q) ‘s selling of cooler.

a) 13 : 17

b) 19 :23

c) 33 :23

d) 27 : 22

e) None of these

Answer 😀

Explanation :

AC : Cooler =( 180 + 360 ) : ( 100 + 240)

                    = 540 : 440

                     = 27 : 22

:. The ratio between (P + Q) ‘s selling of AC and (R + Q) ‘s selling of Cooler is = 27 : 22

29. Question

Find the difference between the selling of AC and Cooler by seller (R + S) together.

a) 226

b) 200

c) 250

d) 175

e) 225

Answer :A

Explanation :-

AC – Cooler = (216 + 270) – ( 100 + 160)

                    = 486 – 260

                    = 226

:. The difference between the selling of AC and Cooler by seller ( R + S) = 226

30. Question

Total AC and Cooler selling by Q is what percentage of total selling of Cooler by the four sellers.

a) 300/7 %

b) 500/3

c) 600/7

d) 800/3

e) None of these

Answer :C

Explanation :

Required % = (240 + 360 ) /700 × 100%

                    = 600/7 %

:. Thus the correct answer is = 600/7 % ( option C)

31. Question

Which seller’s selling of AC and Cooler together is maximum?

a) P

b) Q

c) R

d) S

Answer :B

Explanation :

P   = 180 + 200      = 380

Q   = 360 + 240      = 600

R   = 216 + 100      = 316

S   = 270 + 160      = 430

:. The maximum selling by the seller = Q

32. Question

Sakshi and Simran started a business investing amounts of 18500 and 22500 respectively. If Simran’s share in the profit earned by them is 2700, then what is the total profit earned by them together?

a) 4000

b) 4530

c) 4780

d) 4920

e) 5200

Answer : D

Explanation :-

Investment ratio of  Sakshi : Simran = 18500 : 22500

                                                               = 37 : 45

According to question,

45x = 2700      ( investment ratio = profit ratio)

x = 60

:. Profit of Sakshi = 37 × 60

                              = 2220

:. Total profit = (2220 + 2700 )

                      = 4920  

:.The total profit earned by together = 4920

33. Question

Mother’s age 9 years ago and daughter’s age 6 years hence is equal. The average age of mother, daughter, and another boy is 27 years. The Boy is 9 years younger than his mother. Find the age of Daughter 10 years hence.

a) 30 years

b) 25 years

c) 28 years

d) 32 years

e) 35 years

Answer :  A

Explanation :

According to question,

Mother – Daughter = 15

Mother + Daughter + Boy = 3 × 27 = 81 years

Let, Daughter ‘s age = x years

Mother’s age = x + 15 years

Boy’s age =( x+15) – 9 years = (x + 6) years

x + (x + 15) + (x + 6) = 81

3x + 21 = 81

3x = 60

x = 20

:. Daughter’s age 10 years hence = (20 + 10 ) years

                                                        = 30 years

34.Question

Find the value of,

6000 – 999(1/7) – 999(2/7) – 999(3/7) – 999(4/7) – 999(5/7) – 999(6/7) .

a) 5999

b) 5997

c) 9

d) 3

e) none of these

Answer :- C

Explanation :-

6000 – 999(1/7) – 999(2/7) – 999(3/7) – 999(4/7) – 999(5/7) – 999(6/7)  

= 6000 – {(999 + 999 +999 + 999 +999 + 999) +(1/7 + 2/7 + 3/7 + 4/7 + 5/7 + 6/7) }

= 6000 – (6 × 999) + 21/7

= 6000 – 6 ( 1000 – 1)  + 3

= 6000 – 6000 + 6 + 3

= 9

35.Question

In the given fractions, find out the sum of the largest and the smallest fractions.

1/2 , 5/7 , 3/4 , 4/9 , 6/11

Answer :-

Explanation :-

1/2 = 0.50

5/7 = 0.71

3/4 = 0.75

4/9 = 0.44

6/11 = 0.54

Largest fraction  = 3/4

Smallest fraction = 1/2

Sum = 3/4 + 1/2  

        = 5/4

        = 1.25

:.The sum of the largest and smallest fractions is = 1.25


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