SBI PO Quantitative Questions with Solutions & Explanation

The SBI PO Quantitative Aptitude section in the 2026 Prelims exam tests a candidate's speed, accuracy, and problem-solving ability across 35 questions to be solved in 20 minutes, making it one of the most challenging sections of the paper.

Most Important SBI PO Quantitative Questions

Practising the top 30 SBI PO Quantitative Aptitude questions is essential for every serious aspirant, as these questions cover the most frequently tested topics including Number Series, Simplification and Approximation, Data Interpretation (Bar Graph, Line Graph, Tabular, and Pie Chart), Quadratic Equations, Arithmetic Word Problems (covering Percentage, Profit and Loss, Simple and Compound Interest, Time and Work, Speed Distance and Time, and Ratio and Proportion), and Missing Number Series. 

Q1. Sushma invested some amount in two schemes A and B in the ratio of 5:7 respectively for two years. Scheme A offers 20% p.a. compound interest and Scheme B offers 15% p.a. simple interest. Difference between the interests earned from both the schemes is Rs.1520. How much amount invested in scheme A?

(A). Rs. 72200
(B). Rs. 74200
(C). None of these
(D). Rs. 75500
(E). Rs. 76000

Answer: E

Solution: Given: Scheme A offers 20% p.a. compound interest and Scheme B offers 15% p.a. simple interest.
​Difference between the interests earned from both the schemes is Rs.1520. 

Q2.  A vessel P contains mixture of alcohol and water in the ratio of 4: 7 and vessel Q contains mixture of alcohol and water in the ratio of 4: 5. Both vessels mixed in big vessel X. if total quantity of liquid in vessel X is 236 lit and quantity of water in vessel P and Q was same, then find the quantity of alcohol in vessel P.

(A). 20 lit 
(B). 56 lit 
(C). 36 lit 
(D). 40 lit 
(E). 70 lit 

Answer: D

Solution: Given: P contains mixture of alcohol and water in the ratio of 4: 7
Q contains mixture of alcohol and water in the ratio of 4: 5.

Q3. What will come in the place of question (?) mark in following number series.
+  = 51.98% of 400.01

(A). 4
(B). 12
(C). 8
(D). 32
(E). 16

Answer: E

Solution: 

624/? + 169 = 52/100 X 100
624/? = 208 - 169
? = 624/39
?= 16

Q4. Ravi can do three fourth of a work in 27/2 hours while Hira can do two third of the same work in 8 hours. If both started working together then in how much time the work will be completed?

(A). 8h
(B). 7.2h
(C). 8.4 h
(D). 9 h
(E). 9.2 h

Answer: b

Solution:

Information: Time taken by Ravi to complete three fourth of a work = 27/2 hours
Time taken by Hira to complete two third of the work = 8 hours

Formula Used: Total work = efficiency x time

Explanation: 3/4 th work can be done by Ravi in =27/2 hrs
∴ whole work completed by Ravi =4/3×27/2
= 18 h
And,
Whole work completed by Hira in = 3/2×8=12h
∴ Required time =(18×12)/(18+12)= 7.2 h

Q5. The lengths of train A and train B are in the ratio 5 : 4 respectively. Find the length of train A (in meter).
Statement (I): Train A and Train B are running in opposite directions, and the speed of Train A is 108 km/h. They cross each other in 20 seconds.
Statement (II): Train B is running at a speed of 72 km/h and crosses a 240 m long platform in 24 seconds.

(A). I alone
(B). II alone
(C). Both I & II together
(D). Either I or II
(E). Neither I nor II

Answer: b

Solution:
​​
Let: Length of Train A = 5x meters
Length of Train B = 4x meters
From I: Train A and Train B are running in opposite directions. Speed of Train A = 108 km/h = 108 × 5 / 18 = 30 m/s
They cross each other in 20 seconds.
When two trains move in opposite directions,
Relative speed = Sum of their speeds
Let speed of train B be ‘b’ m/sec
ATQ, 9x = 20(30 + b)
We have two unknowns - x and b - and only one equation. So, Statement (I) alone is
not sufficient.
Statement (II): Train B is running at 72 km/h = 20 m/s And it crosses a 240 m platform in 24 seconds .
When crossing a platform: Distance = Length of train + Length of platform = 20
4x + 240 = 480
4x = 240
x = 60
So, length of Train A = 5x = 5 × 60 = 300 meter
So, Statement (II) aloneis sufficient.

Q6. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m³. The emptying capacity of the pump is 10 m³ per minute more than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump (in m³ per minute)?

(A). 54
(B). 60
(C). 50
(D). 45
(E). 65

Answer: c

Solution:

Given: Capacity of the tank = 2400 m³
Formula Used: Work = efficiency x time 
Explanation: Let the filling capacity of the pump be x m³ per minute
Then, emptying capacity of the pump = (x + 10) m³ per minute
So,
=> x² + 10x – 3000 = 0
=> (x – 50) (x + 60) = 0
=> x = 50 m³ per minute

Q7. What will come in the place of question (?) mark in following number series.
? ×(20.01% of 580.01 + √(196.01 )) = + 17.01 ×3.03

(A). 6
(B). 4
(C). 2
(D). 8
(E). 12

Answer: a

Solution:

? X (20/100 X 580 + 14) = 729 + 51
? X 130 = 780
? = 6

Q8. Ritika marked up a jacket 20% above its cost price. Find the cost price (in Rs) of the jacket.
Statement (I): Ritika gave a discount of Rs 40 on the marked price and earned 10% profit.
Statement (II): If Ritika had given two successive discounts of 15% and 10%, then she would have incurred a loss of Rs 32.8 ​​

(A). I alone
(B). II alone
(C). Both I & II together
(D). Either I or II
(E). Neither I nor II

Answer: d

Solution:

Let the cost price be Rs 100x
Marked price = 100x × 120/100​​=120x Rs
From I: Selling price = 120x - 40
ATQ, 120x - 40 = 100x× 110/100
120x - 40 = 110x
120x - 110x = 40
10x = 40
x = 4
The cost price = 100x = Rs 400
From I: Selling price = 120x x 85% x 90%​​=91.8x Rs
ATQ, 100x - 91.8x = 32.8
8.2x = 32.8
x = 4
The cost price = 100x = Rs 400
So, Either I or II

Read the information and answer the following question.

The table shows the total number of females and ratio of male to females in five different companies.

Q9. In company F, total number of females are 20% more than females in A and males are 20% less than that of females. Find the difference between total employees in F and E.

(A). 56
(B). 48
(C). 29
(D). 40
(E). 57

Answer: A

Solution:

In A, Males: females = 7:5
5x = 25
5 = x
7x =35 = males
Total = 12x = 60
Similarly,
 

Females in F = 120% of 25 = 30
Males in F = 80% of 30 = 24
Required answer = 110 – (30 +24) = 56

Q10. Find the average number of males in D and B.

(A). 39
(B). 38
(C). 35
(D). 40
(E). 37

Answer: A

Solution:

In A,
Males: females = 7:5
5x = 25
5 = x
7x =35 = males
Total = 12x = 60
Similarly,

Required answer = (48+30)/2=39

Q11. The ratio of promoted to non-promoted from total employees in D is 2:3. Total females who got promoted in D is 25. Find the females who got promoted in D is what percentage of non-promoted females in D.

(A). 11.11
(B). 22.22
(C). 33.33
(D). 44.44
(E). 55.55

Answer: E

Solution:

In A,
Males: females = 7:5
5x = 25
5 = x
7x =35 = males
Total = 12x = 60
Similarly,

Let the promoted and non-promoted employees be 2x and 3x respectively.
5x = 100
20 = x
Promoted employees = 40
non- promoted employees = 60
promoted females = 25
promoted males =40 – 25 = 15
non – promoted females = 70 – 25 = 45
non – promoted males = 60- 45 = 15
required answer 25/45×100=55.55%

Q12. Find the ratio of total employees in A to males in B & D together.

(A). 5:6
(B). 4:9
(C). 10:13
(D). 14:13
(E). 5:7

Answer: C

Solution:

In A,
Males: females = 7:5
5x = 25
5 = x
7x =35 = males
Total = 12x = 60
Similarly,

Required answer = 60:48+30 = 60:78 = 20:26 = 10:13

Q13. Find the difference between the total number of males and females in all the companies.

(A). 102
(B). 108
(C). 109
(D). 106
(E). 100

Answer: D

Solution:

In A,
Males: females = 7:5
5x = 25
5 = x
7x =35 = males
Total = 12x = 60
Similarly,

Required answer = 312 – 206 = 106

Q14. ​The ratio of intern to permanent employees in C is 1:1 and out of that 25 are female intern. Find the permanent male employees in C.​

(A). 1
(B). 2
(C). 3
(D). 4
(E). 5

Answer: E

Solution:

In A,
Males: females = 7:5
5x = 25
5 = x
7x =35 = males
Total = 12x = 60
Similarly,

Intern : permanent = 1:1 = 1x and 1x
2x = 136
68 = x
Total Intern = 68
Females intern = 25
Males intern = 68 – 25 = 43
Permanent male employees = 48 – 43 = 5

Q15. If the average of 8 numbers is 39.5 and when two new numbers added, then the average of the numbers decreased by 2.5. If the ratio of the two new numbers is 2: 1, then find the value of greater number in the two new numbers?

(A). 28
(B). 18
(C). 27
(D). 36
(E). None of these

Answer: d

Q16. A sum of Rs. x was invested at 10% simple interest for 3 years. If the same sum was invested at 4% more for same period, then it would have fetched Rs. 120 more. Find the value of 5x. (in Rs.)

(A). 5000
(B). 4800
(C). 3600
(D). 5500
(E). 4000

Answer: a

Solution:

Information Given:
Principal = Rs. x
Rate 1 = 10%, Time = 3 years

Rate 2 = 14% (4% more), Time = 3 years
Difference in interest = Rs. 120

Formula Used: Simple Interest (SI) = (P × R × T) / 100
Difference = SI at 14% - SI at 10%

Explanation:
Difference = x × 3 × (14 - 10)/100 = 120
3x × 4 /100 = 120
12x/100 = 120
x = 120 × 100/12 = 1000
5x = 5 × 1000 = 5000

Q17. What will come in the place of question (?) mark in following number series.
1248.01 +  ​ = 96.01 % of 1525.01

(A). 6
(B). 5
(C). 4
(D). 10
(E). 12

Answer: a

Read the information carefully and answer the following questions. The pie chart shows the percentage of total students (male and females) in four colleges and another pie chart shows the number of females in these colleges.

Q18. Find the average number of males in B, C & D.

(A). 240
(B). 120
(C). 360
(D). 100
(E). 150

Answer: A

Solution:

Total females in B = 800 – (130+220+150)= 300=X

Required average = (200+30+490)/3=240

Q19. Find the ratio of males in college C and D together to females in A and C together.

(A). 24:25
(B). 52:35
(C). 35:36
(D). 9:10
(E). 15:14

Answer: B

Solution:

Total females in B = 800 – (130+220+150)= 300=X

Required ratio = 30+490: 130+220 = 520:350 = 52:35

Q20. Find the total females in C and B together is what percentage of total students in D.

(A). 84.25%
(B). 82%
(C). 81.25%
(D). 100%
(E). 81.2%

Answer: C

Solution:

Total females in B = 800 – (130+220+150)= 300=X

Required answer = (300+220)/640×100=81.25%

Q21. In college E, total number of students is 3X, out of that 35% are females. Find the males in E is what percentage more/less than total students in B?

(A). 15%
(B). 12%
(C). 16%
(D). 10%
(E). 17%

Answer: E

Solution:

Total females in B = 800 – (130+220+150)= 300=X

Total students in E = 3X = 3 ×300=900
Females = 35% of 900 = 315
Males = 900 – 315 = 585
Required answer (585-500)/500×100=17%

Q22. Find the difference between total students in A & B together and twice the males in B.

(A). 740
(B). 720
(C). 760
(D). 710
(E). 750

Answer: D

Solution:

Total females in B = 800 – (130+220+150)= 300=X

Required difference = (610+500)- 2×200= 710

Read the following information carefully and answer the questions given below.

A, B, and C started a business. The investment of A and B is in the ratio of 11:9, respectively. The investment of C is 33.33% more than that of B. The time periods of A, B, and C are 8 months, 6 months, and 12 months, respectively.

Q23. If the initial investment of A is Rs 1650, then find the initial investment of C (in Rs).

(A). 1500
(B). 1350
(C). 1800
(D). 1650
(E). 2100

Answer: C

Solution:

Let the investments of A and B be R 11x and Rs 9x respectively
Investment of C =4/3​​×9x=12x Rs              (33.33% =1/3​​)
The profit-sharing ratio of A, B and C = 11x×8 :9x×6 :12x×12
= 44 : 27 : 72
The initial investment of A = Rs 1650
The initial investment of C = 1650 × 12/11​​=1800 Rs

Q24. At the end of the year the total profit of Rs 4290, then find the profit share of B (in Rs)

(A). 860
(B). 950
(C). 810
(D). 740
(E). 1020

Answer: C

Solution:

Let the investment of A and B be R 11x and Rs 9x respectively
Investment of C =4/3​​×9x=12x Rs              (33.33% =1/3​​)
The profit-sharing ratio of A, B and C = 11x×8 :9x×6 :12x×12
= 44 : 27 : 72
The profit share of B = 4290 ×27/ (44+27+72)
= 4290 × 27/143
Rs 810

Q25. If the profit share of C is Rs 1080, then find the profit share of A (in Rs).

(A). 710
(B). 840
(C). 920
(D). 600
(E). 660

Answer: E

Solution:

Let the investment of A and B be R 11x and Rs 9x respectively
Investment of C =4/3​​×9x=12x Rs              (33.33% =1/3​​)
The profit-sharing ratio of A, B and C = 11x×8 :9x×6 :12x×12
= 44 : 27 : 72
The profit share of A = 1080 × ​= Rs 660

Q26. The length of a rectangle is 4 cm more than the breadth and area of the rectangle is 96 cm². If side of a square is 6 cm more than length of the rectangle, then find the area (in cm²) of the square.

(A). 256
(B). 324
(C). 225
(D). 361
(E). 144

Answer: B

Q27. P, Q and R started a business with certain amount. P invested 40% of the amount for 8 months. Q invested 20% of total amount for 4 months and remaining amount invested by R for 12 months. At the end of a year, P received Rs 5760 as profit share. Find the total profit of the business (in Rs,)?

(A). 15960
(B). 14480
(C). 15840
(D). 13840
(E). 15560

Answer: c

Solution:

Information Given:
Ratio of investment of P, Q and R = 40% : 20% : 100% - (40%+20%) = 2 : 1 : 2
Time for P invested = 8 months
Time for Q invested = 4 months
Time for R invested = 12 months
At end of 12 months, A received profit = 5760 Rs

Formula Used:
Profit sharing ratio = Investment × time

Explanation:
Let total investment = 5x
Investment of P = 5x × 2/5=2x Rs
Investment of Q = 5x × 1/5=x Rs
Investment of R = 5x × 2/5=2x Rs
Profit sharing ratio of P, Q and R = 2x×8 : x ×4 : 2x ×12 = 16x : 4x : 24x
= 4 : 1 : 6
Required profit = 5760/4×11 = 15840 Rs

Q28. Instruction:
A spends 20% of his monthly salary on house rent and 25% of the remaining monthly salary on travelling. He spends his remaining monthly salary on food and children's education in the ratio of 3:5 respectively. If the difference between the amount spent on children's education and house rent is Rs.700, then find the monthly salary of A (in Rs).

(A). 2500
(B). 4000
(C). 4500
(D). 5000
(E). 2000

Answer: b

Solution:

Given:
​Difference between the amount spent on children's education and house rent is
Rs.700

Formula used:
Concept of Successive percentage

Explanation:
Let monthly salary of A be Rs.100x
Expenditure on house rent =Rs. 20x
Expenditure on travelling = (100x-20x)×25/100=Rs.20x
Expenditure on food = (100x-20x-20x)×3/8=Rs.22.5x
Expenditure on children's education = (100x-20x-20x)×5/8=Rs.37.5x
ATQ,
37.5x-20x=700
x=40
Monthly salary of A = 40×100=Rs.4000

Q29. A boat covers D km downstream in 2T hours and it takes T hours when the boat covers (D – 400) km in still water. If the speed of boat in still water is six time the speed of current, then find the value of D.

(A). 750
(B). 710
(C). 740
(D). 730
(E). 700

Answer: E

Solution:

Let the speed of current be x km/hr.
The speed of boat in = 6x km/hr.

Q30. Two years ago Raju’s age was 75% of his sister, Rita’s age at that time. After two years, Rita’s age will be 33 1/3% of her father’s age. Average age of Rita’s father and mother is 31 yrs. If Rita’s mother’s age is 28 yrs then what is the present age of Raju?

(A). 10 yrs
(B). 6 yrs
(C). 8 yrs
(D). 12 yrs
(E). 14 yrs

Answer: c

Solution:

Information:

The age of Raju two years ago = 75% of Rita’s age at that time. After two years, Rita’s age will be 33 1/3% of her father’s age. Average age of Rita’s father and mother = 31 yrs Rita’s mother’s age =28 yrs

Explanation:

Rita’s father’s age = 31 × 2 – 28
= 34 yrs
Rita’s age after two yr =100/300×(36)
= 12 yr
∴ Rita’s present age = 10 yr
∴ Raju’s present age =(10-2)×75/100+2
= 8 yr