SBI Clerk 2026 Quant Memory Based Paper: Prelims Most Asked Questions & Answers

The Quantitative Aptitude section of the SBI Clerk examination tests speed, conceptual clarity, and calculation efficiency. Excelling in this paper requires a strong command over arithmetic fundamentals such as percentages, ratios, and averages, alongside quick data interpretation skills. Understanding memory-based question patterns helps candidates identify high-weightage topics, avoid common traps, and build efficient calculation strategies for maximum accuracy.

1. Compound Interest: Incremental Interest Concept

Problem Statement

A principal sum of ₹7,000 is invested for 2 years at an interest rate of 20% per annum compounded annually. Find the difference between the compound interest earned in the second year and the interest earned in the first year.

Also Check -  SBI Clerk Prelims Exam Analysis 2026, 27th September Shift 1.

Mathematical Derivation & Solution

• Principal (P): Rs. 7,000

• Rate (R): 20% per annum

1. First-Year Interest (CI1):
   CI1 = 20% of 7000 = Rs. 1,400

2. Second-Year Interest (CI2):
   Compound interest applies interest on interest.
   CI2 = CI1 + (20% of CI1)
   CI2 = 1400 + (0.20 * 1400) = 1400 + 280 = Rs. 1,680

3. Difference (CI2 - CI1):
   Difference = 1680 - 1400 = Rs. 280

The difference between the second-year interest and the first-year interest reflects only the interest earned on the previous period (Memory Tip: The difference between the compound interest of the second year and the first year is simply the interest calculated on the first year's interest: Difference = R%  CI1*).

2. Profit And Loss: Base Value Equivalence

Problem Statement

The cost price of an article is given as CP = 2800 + X. A shopkeeper sells the article at a 15% profit. If the selling price is increased by ₹700, the profit percentage becomes 35%. Find the value of X.

Mathematical Derivation & Solution

• Initial Profit = 15% of CP

• New Profit = 35% of CP

• Change in Profit Percentage = 35% - 15% = 20%

Since profit percentage is always calculated with the Cost Price (CP) as the base:

• 20% of CP = Rs. 700

• 10% of CP = Rs. 350

• 100% of CP = Rs. 3,500

Given CP = 2800 + X:

3500 = 2800 + X

X = 700

Also Check - SBI Clerk Prelims Exam Analysis 2026, 26 September: Shift 3 Review

3. Mixtures And Proportions: Fractional Removal Rule

Problem Statement

A vessel contains 75 liters of a milk and water mixture. The quantity of milk is 45 liters more than that of water. If 15 liters of the mixture is removed, determine the remaining quantity of milk.

Mathematical Derivation & Solution

1. Initial Distribution:

• Total = 75 L

• Difference = 45 L

• Remaining quantity to distribute equally = 75 - 45 = 30 L

• Initial Water = 30 / 2 = 15 L

• Initial Milk = 15 + 45 = 60 L

2. Removal Principle:

• Fraction of mixture removed = 15 / 75 = 1/5

• When a fraction of a homogeneous mixture is withdrawn, each individual component decreases by the exact same fraction (1/5).

3. Final Milk Quantity:

• Milk removed = (1/5) * 60 = 12 L

• Remaining Milk = 60 - 12 = 48 L

4. Speed, Time, And Distance: Ratio Method

Problem Statement

Car A covers a distance of 126 km in 2 hours. The ratio of the speed of Car A to Car B is 7 : 6. Find the time taken by Car B to cover a distance of 135 km.

Mathematical Derivation & Solution

1. Speed of Car A (SA):
   SA = Distance / Time = 126 / 2 = 63 km/h

2. Speed of Car B (SB):
   SA / SB = 7 / 6
   63 / SB = 7 / 6
   SB = (63 * 6) / 7 = 54 km/h

3. Time Required for Car B (TB):
   TB = 135 / 54 = 2.5 hours

5. Mensuration: 2D Geometric Equivalence

Problem Statement

The length of a rectangle is 18 units and its area is 180 sq. units. The perimeter of the rectangle equals the perimeter of a square. Find the area of the square.

Mathematical Derivation & Solution

1. Dimensions of the Rectangle:

• Length (l) = 18

• Area (A) = l * b = 180

• Breadth (b) = 180 / 18 = 10

2. Perimeter Equivalence:

• Perimeter of Rectangle = 2 * (l + b) = 2 * (18 + 10) = 56

• Perimeter of Square = 4 * side = 56

• Side of Square (a) = 14

3. Area of the Square:
   Area = a^2 = 14^2 = 196 sq. units

6. Successive Discounts And Equivalent Markup

Problem Statement

A shopkeeper offers two successive discounts of 20% and 25% on the Marked Price (MP) of a laptop, yet achieves a profit of 20%. If the Selling Price (SP) is ₹36,000, find the total discount amount.

Mathematical Derivation & Solution

1. Unit Base Method (MP = 100 units):

• After 1st discount (20%): Value = 80 units

• After 2nd discount (25% of 80 = 20 units): SP = 80 - 20 = 60 units

• Total Discount = MP - SP = 100 - 60 = 40 units

2. Value Scaling:

• 60 units = Rs. 36,000

• 1 unit = Rs. 600

• Total Discount Amount = 40 * 600 = Rs. 24,000

7. Averages: Deviation And Exit Analysis

Problem Statement

The average height of 15 people is 180 cm. Three individuals (A, B, C) leave the group, causing the average height of the remaining people to decrease by 5 cm. If two of the departing individuals have heights of 170 cm and 195 cm, find the height of the third individual.

Mathematical Derivation & Solution

1. Baseline Departure Sum:
   Baseline Sum = 3 * 180 = 540 cm

2. Deviation Contribution:
   The departure causes the remaining 12 individuals to lose 5 cm each. This deficit was carried away by the departing group:
   Extra Sum Carried Away = 12 * 5 = 60 cm
   Total Height of (A + B + C) = 540 + 60 = 600 cm

3. Third Individual's Height (h3):
   170 + 195 + h3 = 600
   365 + h3 = 600
   h3 = 235 cm

8. Partnership: Variable Capital And Duration

Problem Statement

A and B start a business with equal initial investment X. After 5 months, A withdraws ₹1,000, while B maintains the initial investment for the entire year (12 months). At the end of the year, the total profit is ₹11,300, and B's profit share is ₹6,000. Find the initial investment X.

Mathematical Derivation & Solution

1. Profit Allocation Ratios:

• Total Profit = Rs. 11,300

• Profit of B = Rs. 6,000

• Profit of A = 11,300 - 6,000 = Rs. 5,300

• Ratio (Profit A / Profit B) = 5,300 / 6,000 = 53 / 60

2. Investment-Time Equivalent:

• Equivalent Investment of B = X * 12

• Equivalent Investment of A = (X * 5) + ((X - 1000) * 7) = 12X - 7000

3. Equating Ratios:
   (12X - 7000) / (12X) = 53 / 60
   (12X - 7000) / X = 53 / 5
   5 * (12X - 7000) = 53X
   60X - 35000 = 53X
   7X = 35000
   X = Rs. 5,000

9. Data Interpretation: Tabular Form (Subject Scores)

Data Overview

Also Check -  SBI Clerk Prelims Exam Analysis 2026, 26 September, Shift 2 Review

Key Worked Problems

• Average Total Marks: 696 / 5 = 136.8

• Difference (Total Marks of A and D): 140 - 134 = 6

• Ratio of Total Marks (Student B to Student E): 144 / 138 = 24 : 23

• Average Economics Marks (Students A, B, E): (64 + 86 + 74) / 3 = 224 / 3 = 74.67

• Geography Score Extension:

• Geography marks of A = 16 + (64 + 70) = 150

• Ratio of Geography marks (A : B) = 5 : 4 -> Geography of B = (150 / 5) * 4 = 120

• Difference for Student B between Total and Geography = 144 - 120 = 24

10. Data Interpretation: Line Graph (Workshop Attendance)

Extracted Values

• P: 48

• Q: 64

• R: 24

• S: 40

Key Worked Problems

• Percentage Comparison (Q vs R):
  Percentage More = [(64 - 24) / 24] * 100% = (40 / 24) * 100% = 166.66%

• Gender Segregation:

• Females in P = 25% of 48 = 12 -> Males in P = 48 - 12 = 36

• Females in S = 20% of 40 = 8 -> Males in S = 40 - 8 = 32

• Total Males (P + S) = 36 + 32 = 68

• Group Partitioning:

• Total in (P + S) = 48 + 40 = 88

• Group 1 equals R = 24

• Group 2 = 88 - 24 = 64

• Aggregate Percentage:
  [(P + Q) / (R + S)] * 100% = (112 / 64) * 100% = 175%

11. Time And Work: Efficiency Conversion

Problem Statement

4 men and 3 women can complete a work piece in 12 days. The efficiency of a man is twice that of a woman (1 Man = 2 Women). Determine the time required for 2 women to complete the identical task.

Mathematical Derivation & Solution

1. Workforce Conversion:
   4 Men = 4 * 2 = 8 Women
   Total Initial Workforce = 8 Women + 3 Women = 11 Women

2. Total Work:
   Total Work = 11 Women * 12 Days = 132 Woman-Days

3. Time for 2 Women:
   Time = 132 / 2 = 66 days

12. Boats And Streams: Speed And Ratio Dynamics

Problem Statement

The ratio of time taken by a boat to cover a fixed distance downstream vs. upstream is 3 : 5. The speed of the stream is 4 km/h. Find the time taken to travel 160 km downstream.

Mathematical Derivation & Solution

1. Speed and Time Inversion:
   Ratio of Time (Downstream : Upstream) = 3 : 5
   Ratio of Speed (Downstream : Upstream) = 5 : 3

2. Speed Components:

• Downstream Speed = 5k, Upstream Speed = 3k

• Speed of Stream = (5k - 3k) / 2 = 1k

• Given Stream Speed = 4 km/h -> k = 4

• Downstream Speed = 5 * 4 = 20 km/h

3. Time Calculation:
   Time Downstream = 160 / 20 = 8 hours

Also Check - SBI Clerk Prelims Exam Analysis 2026, 26 September: Shift 2