SBI PO Prelims Quantitative Aptitude Question Paper 2021
Directions (0105): 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 + 1^{3} = 119
119 – 2^{3} = 111
111 + 3^{3} = 138
138 – 4^{3} = 74
74 + 5^{3} = 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:
2^{3} – 1 = 7
3^{3} + 1 = 28
4^{3} – 1 = 63
5^{3} + 1 =126
6^{3} – 1 = 215
7^{3} + 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 (1115): 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 cm^{2}
Thus, the area of the triangle is 80 cm^{2}.
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 (2125): 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:
7x^{2} + 10x + 3 = 0
7x^{2} + 7x + 3x + 3 = 0
(x + 1) (7x + 3) = 0
x = 3/7 , 1/3
2y^{2} 3y – 44 = 0
2y^{2} +8y – 11y – 44 = 0
(y + 4) (2y – 11) = 0
y = 11/2 , – 4
Thus, the relationship can not be established.
22. Question
I. x^{2} – 11x + 28 = 0
II. y^{2} – 45y + 324 = 0
Answer : C
Explanation :
x^{2} – 11x + 28 = 0
x^{2} – 4x – 7x + 28 = 0
(x – 7) (x – 4) = 0
x = 4 , 7
y^{2} – 45y + 324 = 0
y^{2} – 36y – 9y + 324 = 0
(y – 9) (y – 36) = 0
y = 9 , 36
x < y
Thus, the correct option is C.
23.Question
I. x^{2} +14x + 60 =15
II. y^{2} = y + 30
Answer: D
Explanation:
x^{2} + 14x +60 = 15
x^{2} + 14x + 45 = 0
x^{2} + 5x + 9x + 45 = 0
(x + 5) (x + 9) = 0
x = – 5 , – 9
y^{2} = y + 30
y^{2} – y – 30 = 0
y^{2} – 6y + 5y – 30 = 0
(y – 6) (y + 5) = 0
y = 6 , 5
x ≤ y
Thus, the correct option is D.
24.Question
I. 21x^{2} + 2x – 3 = 0
II. 12y^{2} – y – 6 = 0
Answer: E
Explanation:
21x^{2} + 2x – 3 = 0
21x^{2} – 7x + 9x – 3 = 0
(7x + 3) (3x – 1) = 0
x = 1/3 , – 3/7
12y^{2} – y – 6 = 0
12y^{2} – 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. 2x^{2} – 288 = 0
II. y^{2} + 24x + 144 = 0
Answer: B
Explanation:
2x^{2} – 288 = 0
x^{2} – 144 = 0
(x + 12) (x – 12) = 0
x = 12 , – 12
y^{2} + 24x + 144 = 0
(y + 12)^{2} = 0
y + 12 = 0
y = 12
x ≥ y
Thus, the correct option is B
Directions (2630):
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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