SBI Clerk Quantitative Aptitude Questions 2024

The SBI Clerk exam is a key entry-level position in the banking and finance sector, representing an important starting point in a career with the State Bank of India. One of the most important topics in this exam is Quantitative Aptitude, which is a significant part of the paper. Out of a total of 100 questions, 35 are devoted to Quantitative Aptitude. In this article, we will share a collection of 2024 SBI Clerk Quantitative Aptitude Questions 2024 to help you prepare.

Here, we present the main topics of Quantitative Aptitude, which is an important section of the SBI Clerk 2024 exam. Understanding these topics not only enhances problem-solving skills but also increases accuracy and confidence, making it an important part of your preparation journey.

• Simplification
• Approximation
• Missing Series
• Quadratic Equation
• Data Interpretation (Bar, Line, Pie, Tabular)
• Data Sufficiency
• Wrong Series
• Time & Work, Pipes & Cisterns
• Problems on Ages
• Average, Ratio, Percentage, Profit & Loss
• Simple Interest & Compound Interest
• Speed, Distance & Time
• Permutation & Combination
• Boat & Stream
• Mensuration
• Volumes
• Probability
• Partnership
• Mixture & Allegation

SBI Clerk Quantitative Aptitude Questions 2024

Directions (1 – 10): What should come in place of question mark (?) in the following questions?

Q1. (7294 – 3241 + 716) – (3267 + 2425 – 961) = ?
(a) 35
(b) 48
(c) 43
(d) 38
(e) 47

Q2. 1217 + 841 – 724 + 819 = ? + 1843
(a) 210
(b) 310
(c) 360
(d) 270
(e) 410

Q3. 0.06 × 0.84 = ? × 1.2 × 0.015
(a) 8.2
(b) 6.4
(c) 2.6
(d) 3.8
(e) 2.8

Q4. 8.41 + 6.25 + 0.79 = ? – 0.55
(a) 17
(b) 14.9
(c) 13.9
(d) 16
(e) 14.7

Q5. 616 + 472 – 811 + 317 = ? + 576
(a) 28
(b) 16
(c) 24
(d) 18
(e) 14

Q6. Average of present age of A and C is 29 years and C is 8 years younger than B. Ratio of present age of A to present age of B is 6:5. Find present age of B?
(a) 32 years
(b) 36 years
(c) 28 years
(d) 30 years
(e) 16 years

Q7. A boat can travel with the speed of 17 kmph in upstream. If the speed of river is 3 kmph, then find the speed of boat in downstream in the same river.
(a) 23 kmph
(b) 20 kmph
(c) 25 kmph
(d) 19 kmph
(e) 21 kmph

Q8. If a shopkeeper marks an article 20% above cost price and sold it at 5% discount and get Rs. 28 profit. Find difference between C.P. and market price of article?
(a) Rs. 50
(b) Rs. 45
(c) Rs. 32

(d) Rs. 40
(e) Rs. 44

Q9. The ratio of speed of two trains which are running in the same direction is 4 : 5. The train having higher speed crosses the second train in 30 seconds and a pole in 4 seconds respectively. Find the ratio of their lengths.
(a) 1 : 2
(b) 2 : 3
(c) 3 : 5
(d) 4 : 5
(e) 6 : 7

Q10. A train crosses a man, who is running in the same direction of train at the speed of 2m/sec. in 10 seconds. The same train crosses a tunnel in 54 seconds. If speed of train is 72 km/h then what is the length of tunnel?
(a) 850 m
(b) 800 m
(c) 900 m
(d) 750 m
(e) 650 m

Q11. 4, 7, 13, 24, 42, 69, ?
(a) 91
(b) 107
(c) 112
(d) 98
(e) 102
Q12. 7, 13, 19, 29, 37, 43, ?
(a) 61
(b) 57
(c) 53
(d) 51
(e) 55
Q13. ?, 5, 11, 23, 43, 73, 115
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Q14. 32, ?, 24, 60, 210, 945, 5197.5
(a) 12
(b) 8
(c) 15
(d) 18
(e) 16
Q15. 5, 3, 3, 5, 15, 69, ?
(a) 319
(b) 287
(c) 361
(d) 407
(e) 443

Solutions

S1. Ans.(d)

Sol.
4769–4731=?
? = 38

S2. Ans.(b)
Sol. ? = 1217 + 841 + 819 – 724 – 1843
= 2877 – 2567
? = 310

S3. Ans.(e)
Sol.
? = 2.8

S4. Ans.(d)
Sol.
?= 15.45 + 0.55
? = 16

S5. Ans.(d)
Sol.
? = 1405 – 811 – 576
? = 18

S6. Ans.(d)
Sol.
Total age of A and C together = 58 years
Let present age of A and B be 6x and 5x years respectively
Therefore, present age of C = 5x-8
ATQ,
5x-8 + 6x = 58
x= 6
present age of B = 5 ×6 = 30 years

S7. Ans.(a)
Sol.
Speed of boat in upstream = 17 kmph
Speed of river water = 3 kmph
So speed of boat in still water = 17 + 3 = 20 kmph
So speed of boat in downstream = 20 + 3 = 23 kmph

S8. Ans(d)
Sol. let C.P. of article = Rs.100x
Then, M.R.P. of article = x
S.P. of article =
ATQ

x = 2
So, required difference = Rs.40

S9. Ans.(a)
Sol.
Let length of slower train = ℓ 1
Length of faster train = ℓ 2
4x = speed of slower train
5x = speed of faster train
…(i)

And, ℓ 2 = 5x × 4
= 20 x
= 10x

S10. Ans.(c)
Sol.
Speed of man = 2 m/sec
Speed of train = = 20 m/sec
∴ Length of train = (20 – 2) × 10 = 180 m
∴ Length of tunnel = 54 × 20 – 180 = 900 m

S11. Ans. (b)
Sol.
The pattern of the series is –
4+(1²+2)=7
7+(2²+2)=13
13+(3²+2)=24
24+(4²+2)=42
42+(5²+2)=69
69+(6²+2)=107

S12. Ans. (c)
Sol.
The series is set of alternate prime numbers.
7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53

S13. Ans. (c)
Sol.
The pattern of the series is –
3+(1×2)=5
5+(2×3)=11
11+(3×4)=23
23+(4×5)=43
43+(5×6)=73
73+(6×7)=115

S14. Ans. (e)
Sol.
The pattern of the series is –
32×0.5=16
16×1.5=24
24×2.5=60
60×3.5=210
210×4.5=945
945×5.5=5197.5

S15. Ans. (d)
Sol.
The pattern of the series is –
5×1-2=3
3×2-3=3
3×3-4=5
5×4-5=15
15×5-6=69
69×6-7=407

S16. Ans.(d)
Sol.
I. 2x² – 17x+ 36 = 0
2x² – 8x – 9x + 36 = 0
2x (x – 4) – 9 (x – 4) = 0
(2x – 9) (x- 4) = 0
x=9/2, 4
II. 2y² – 19y + 45 = 0
2y² – 10y – 9y + 45 = 0
2y (y- 5) – 9 (y- 5) = 0
(2y- 9) (y- 5) = 0
y=9/2,5
∴ y ≥ x

S17. Ans.(e)
Sol.
I. x² – 25x + 154 = 0
x² – 14x – 11x + 154 = 0
x (x – 14) – 11 (x- 14) = 0
(x – 11) (x- 14) = 0
x = 11, 14
II. y² – 28y + 195 = 0
y² – 13y – 15y + 195 = 0
y (y- 13) – 15 (y -13) =0
(y- 13) (y – 15) = 0
y = 13, 15
∴ no relation

S18. Ans.(a)
Sol.
I. 10/x-24/x² =1
Multiplying by x² on both side
10x – 24 = x²
x² – 10x + 24 = 0
x² – 6x -4x+ 24 = 0
x(x – 6) – 4 (x- 6) = 0
(x – 4) (x- 6) = 0
x= 4, 6
II. 5/y-6/y² =1
Multiplying by y² on both side
5y – 6 = y²
y² – 5y + 6 = 0
y² – 3y – 2y + 6 = 0
y (y- 3) – 2 (y- 3) = 0
(y – 2) (y- 3) = 0
y = 2, 3
∴ x > y

S19. Ans.(d)
Sol.
I. 3x² – 10x – 8 = 0
3x² – 12x + 2x – 8 = 0
3x (x – 4) + 2 (x- 4) = 0
(3x+ 2) (x- 4) = 0
x= -2/3,4
II. 2y²-23y+60=0
2y² – 8y- 15y + 60 = 0
2y (y- 4) -15(y-4) = 0
(y- 4) (2y- 15) = 0
y=4,15/2
∴ y ≥ x

S20. Ans.(a)
Sol.
I. 12x – 16y +16 = 0
3x – 4y + 4 = 0 …(i)
II. 17y- 13x = 12 …(ii)
By multiplying equation (i) by 13 & equation (ii) by 3
39x – 52y = -52
-39x + 51y = 36
y = 16 & x = 20
∴ x > y

S21. Ans.(a)
Sol.
Female Senior citizen of age group (50 – 60) yrs
=3/8×2400=900
Female senior citizen of age group (61 – 70) yrs
=1/4×3200
= 800
Required percentage =(900-800)/800×100
= 12.5% more

S22. Ans.(c)
Sol.
Required average =1/2×(3/4×3200+2/3×6000)
=1/2×6400
= 3200

S23. Ans.(a)
Sol.
Remaining senior citizens
=(100-100/3)% of 6000
= 4000
∴ Required percentage =4000/(3200+4800)×100
=1/2×100
= 50%

S24. Ans.(c)
Sol.
Non–pensioner males =(100-20)% of 7/12 of 4800
=80/100×7/12×4800
= 2240

S25. Ans.(d)
Sol.
Required difference =(5/8×2400+3/4×3200)-(3/8×2400+1/4×3200)
= 1500 + 2400 – 900 – 800
= 2200

S26. Ans.(b)
Sol.
Total rooms booked in Oberai on Tuesday and Thursday = 280 + 520 = 800
Total rooms books in Grand on Monday and Thursday = 280 + 720 = 1000
Required percentage = (1000–800)/1000×100 = 20%

S27. Ans.(d)
Sol.
Total rooms booked in Oberai, Lodhi and Taj on Monday
= 360 + 260 + 640 = 1260
Total rooms booked in Taj, Grand and Eros on Thursday
= 375 + 720 + 275 = 1370
Required difference = 1370 – 1260 = 110

S28. Ans.(a)
Sol.
Total room booked in Eros on Wednesday and Thursday = 265 + 275 = 540
Total rooms booked in Lodhi on Thursday and Friday = 215 + 305 = 520
Required ratio = 540 : 520
= 27 : 26

S29. Ans.(c)
Sol.
Average of room booked in Eros on Monday, Wednesday and Friday
= (155+265+315)/3
= 245
Average of room booked in ‘Grand’ on Monday & Friday
= (280+220)/2
= 250
Required sum = 245 + 250 = 495

S30. Ans.(e)
Sol.
Required percent = (480 –360)/360×100 = 33 1/3%

Solution
S31. Ans.(d)
Sol.
15×25/100×140=? ×15
525/15=?
35=?

S32. Ans.(e)
Sol.
1570+2190-4/9 of 198=?/4
3760-88=?/4
14688=?

S33. Ans.(d)
Sol.
80/100×560+361- ∛4913
448+361-17=?
792=?

S34. Ans.(d)
Sol.
44/4×9/3+3= √(?)
√(?) =11×3+3
36 = √(?)
? =1296

S35. Ans.(b)
Sol.
√(40%×375+75%×60) = ?
? = √(150+45)
? = √195
? = 14

S36. Ans(d)
Sol. 56/100×550+252= ?2 – √256
308 + 252 + 16 = ?2
576 = ?2
? = 24

S37. Ans(a)
Sol. (112/100×700)/(?×7) + 343 = 175 ×2
784/(?×7) = 350 – 343
784/(?×7) = 7
? = 784/49
? = 16

S38. Ans(e)
Sol. 4 + 15/100 ×1280= ?2
4 + 192 = ?2
? = 14

S39. Ans(d)
Sol. (240+25/100×720+280)/70= √(3&?)
(240+180+280)/70 =√(3&?)
700/70=√(3&?)
10 = √(3&?)
? = 1000

S40. Ans(d)
Sol. 25 ÷ 22 × 27 = 2?
2(5-2) ×27 = 2
210 = 2?
? = 10

S41. Ans (b)
Sol. (A+B+C)=(2×(29000+37000+34000))/2
A+B+C=100000
So, require income =100000-(2×34000)
=100000-68000=Rs 32000

S42. Ans.(d)
Sol.
Total age of A and C together = 58 years
Let present age of A and B be 6x and 5x years respectively
Therefore, present age of C = 5x-8
ATQ,
5x-8 + 6x = 58
x= 6
present age of B = 5 ×6 = 30 years

S43. Ans. (c)
Sol. Number of boys in the class = 72×5/9
= 40
Number of girls in the class = 72×4/9
= 32
Required average = ((60×72)-(40×80))/32
= (4320-3200)/32
= 35 kg

S44. Ans (c)
Sol. Total score of Rohit sharma in 11 matches =11×85=935
Total score in first four matches =4×72=288
Total score in next six matches =6×90=540
So, required runs =935-288-540=107

S45. Ans (d)
Sol. let present age of Deepak, Sanjay and Harish be x, y and z years respectively.
ATQ,
⟹ (x-4)/(y-4)=3/4
4x-3y=4 …….(i)
⟹ x+y+z=26×3=78 …..(ii)
⟹ z=y-11 …..(iii)
From (i), (ii) and (iii)
x=25,y=32,z=21
So, present age of Sanjay =y=32 years

S46. Ans. (a)
Sol.
Total number of food packets not received by villages B & D together = 125+ 210 = 335

S47. Ans. (e)
Sol.
Required percentage = (1500-250)/1500×100=83 1/3%

S48. Ans. (e)
Sol.
Required percentage = 25/5×100=500%

S49. Ans. (d)
Sol.
Required average = (60+125+25)/3=70

S50. Ans. (b)
Sol.
Required ratio = (60+125+25) : (200+50) = 210:250=21:25