SBI PO 2026 Quantitative Aptitude: Full Paper Analysis and Tricks

SBI PO 2026 Quantitative Aptitude section in the SBI PO 2026 Prelims is designed to test a candidate’s numerical ability, problem-solving skills, and speed in calculations. 

With 30 questions to solve in just 20 minutes, this section challenges both accuracy and time management. It covers a mix of topics including Speed Maths, Data Interpretation, Caslet DI, and Arithmetic Word Problems, each requiring a strategic approach. 

Success in this section depends not only on mastering formulas and concepts but also on knowing which questions to attempt first, how to skip time-consuming problems, and how to apply shortcuts and approximation techniques. A strong foundation in basic mathematics combined with daily practice can help aspirants maximize their score in this high-pressure, fast-paced exam.

SBI PO 2026 Quantitative Aptitude 

SBI PO Prelims Quantitative Aptitude section tests problem-solving and numerical ability. This analysis provides insights into the exam pattern, question distribution, and effective strategies for the 2026 cycle. Understanding how to manage time, prioritize questions, and practice consistently are vital for success.

SBI PO Prelims Quantitative Aptitude Exam Pattern

The Quantitative Aptitude section comprises 30 questions to be completed in 20 minutes, totaling 30 marks. The paper typically includes:

Important Note: Do NOT try to skip any arithmetic chapters. While attempting every question isn't necessary, a complete syllabus understanding is vital for maximizing scores.

Exam Attempt Strategy and The Art of Skipping

The 20-minute time limit makes solving all 30 questions nearly impossible. A realistic and safe target is around 20+ questions.

The most important skill for banking exams is knowing which questions to skip. Many students fail not due to lack of study, but because they get stuck on difficult or time-consuming problems.

Core Principles:

1. Never Get Attached to a Question: Even if a question is from a strong topic, if it appears too calculative or complex, skip it immediately.

2. Avoid Time Sinks: The exam includes 2-3 intentionally difficult or lengthy arithmetic questions. Identify and bypass these to save time for easier, scoreable questions.

3. Heed the Instructions: Examiners advise: "Do not waste too much time on a single question." Taking this seriously is key to success.

Topic-Wise Strategy and Daily Practice

• Speed Maths (Approximation, Number Series, Quadratic Equations): This is a highly scoring section. Daily practice allows solving these 5 questions in under 2 minutes.

• Data Interpretation (DI): One of two DI sets is generally doable. Success requires strong language comprehension and proficiency in Percentage, Ratio, and Average.

• Caslet DI: Can be easy with good language command and practice, but otherwise time-consuming.

• The Power of Daily Practice: Consistent practice guarantees mastery. Solving just 5 questions daily for a challenging topic (like Number Series) ensures solving at least 4 out of 5 questions in the exam. This applies to Approximation, Quadratic Equations, and DI.

Prerequisites for Speed: To achieve required speed, especially in Speed Maths, memorize:

• Squares up to 50

• Cubes up to 20

Worked Examples: Approximation

Question 1

Problem: x² + 479.89 - 80.01% of 899.91 = 8.02% of 2500.12

Solution:

Approximate values: x² + 479 - 80% of 900 = 8% of 2500.

x² + 479 - 720 = 200

x² - 241 = 200

x² = 441

x = 21

Question 2

Problem: 35.89 / 8.98 * 35.12 + 244.14 - 28.01% of 75.14 * 14.98 = x

Solution:

Approximate values: (36 / 9) * 35 + 244 - 28% of 75 * 15 = x.

(Memory Tip: Calculate a% of b as b% of a for easier computation.)

28% of 75 is 75% of 28 = (3/4) * 28 = 21.

4 * 35 + 244 - 21 * 15 = x

140 + 244 - 315 = x

384 - 315 = x

x = 69

Question 3

Problem: 25.01% of 399.98 + 300.01% of 69.91 = x - 90.11

Solution:

Approximate values: 25% of 400 + 300% of 70 = x - 90.

100 + 210 = x - 90

310 = x - 90

x = 400

Question 4

Problem: 32.01 * 11.01 + √63.98 + 8.99² = x

Solution:

Approximate values: 32 * 11 + √64 + 9² = x.

352 + 8 + 81 = x

360 + 81 = x

x = 441

Worked Example: Data Interpretation Set 1 (Pie Chart)

Problem Statement: Pie chart shows percentage distribution of male employees in five companies (P, Q, R, S, T). Total males (P+Q+R+S+T) is 100%. In each company, the average number of male and female employees is 50. Ratio of male to female employees in company P is 1:4.

Step-by-Step Solution:

1. Solve for 'a': Sum of all male percentages (20% + (a+2)% + (3a+1)% + (2a-1)% + (4a-2)%) must be 100%.
   10a + 20 = 100 => 10a = 80 => a = 8.

2. Calculate Male Percentage for Each Company:
   P: 8+2 = 10%, Q: 3(8)+1 = 25%, R: 2(8)-1 = 15%, S: 4(8)-2 = 30%, T: 20%.

3. Deduce Total Employees: "Average male and female is 50" means (Male + Female)/2 = 50, so Total Employees = 100 per company.

4. Determine Total Male Count: For Company P, Male:Female is 1:4. With total 100 employees, Males = 20, Females = 80. Since P has 10% of total males, and 10% corresponds to 20 employees, then 1% of total males = 2 employees.

5. Construct Final Data Table: Use 1% = 2 males to find male counts, then calculate females (100 - males).

Worked Example: Data Interpretation Set 2 (Table DI)

Problem Statement: Table provides data for three types of ships (Space, Water, Air) in five countries. Given: Total Ships, Ratio of (Space : Water) ships, and Average of (Water + Air) ships.

Framing the Data Table:

1. Sum (Water + Air): Average * 2.

2. Space ships: Total Ships - Sum (Water + Air).

3. Water ships: Use Space : Water ratio with calculated Space ships.

4. Air ships: Sum (Water + Air) - Water ships.

Example Calculation for Country A:

1. Sum (Water + Air): 55 * 2 = 110.

2. Space ships: 200 (Total) - 110 = 90.

3. Water ships: Ratio Space : Water = 9 : 7. Since 9 units = 90, 1 unit = 10. Water ships = 7 * 10 = 70.

4. Air ships: 110 - 70 = 40.

Final Data Table (Fully Solved):

Worked Example: Arithmetic (Word Problems)

Question 1: Rectangle and Square

Problem: Length of a rectangle is 10 cm more than its breadth. Area is 375 cm². If breadth is increased by x%, it becomes a square. Find x.

Solution:

Let breadth b. Length L = b + 10. Area b(b + 10) = 375.

Solving b² + 10b - 375 = 0 gives b = 15 (breadth cannot be negative).

Length L = 15 + 10 = 25 cm.

For a square, new breadth must be 25 cm. Increase needed = 25 - 15 = 10 cm.

Percentage increase x = (10 / 15) * 100 = (2/3) * 100 = **66.67%**.

Question 2: Pipes and Cisterns

Problem: Pipe A fills a tank in 10 minutes. Pipe B empties it in 6 minutes. Tank is 2/5th full. If both pipes open, in what time will the tank be emptied?

Solution:

LCM of 10 and 6 = 30 units (tank capacity).

Efficiency A = +30/10 = +3 units/min.

Efficiency B = -30/6 = -5 units/min.

Initial water = (2/5) * 30 = 12 units.

Combined efficiency = 3 - 5 = -2 units/min (tank empties).

Time to empty = 12 units / 2 units/min = **6 minutes**.

Question 3: Population Growth

Problem: City population after 3 years will be 26,620. Growth rate is 10% per year. Find present population.

Solution:

Let present population be P. Growth factor (1 + 10/100) = 11/10.

P * (11/10)³ = 26,620

P * (1331/1000) = 26,620

P = 26,620 * (1000/1331)

P = 20 * 1000 = **20,000**.

Question 4: Ratios and Proportions

Problem: Students in classes A, B, C are in ratio 8:9:11. Total students 140. When a + 5 students are added to each class, ratio becomes 11:12:14. Find 'a'.

Solution:

Total ratio parts = 8+9+11 = 28. 28 units = 140 students. 1 unit = 5 students.

Initial students: A = 8*5 = 40, B = 9*5 = 45, C = 11*5 = 55.

After adding a+5:

(40 + a + 5) / (45 + a + 5) = 11 / 12

(45 + a) / (50 + a) = 11 / 12

12(45 + a) = 11(50 + a)

540 + 12a = 550 + 11a

a = 10.

Question 5: Simple vs. Compound Interest

Problem: Equal sum invested in scheme P (SI 20% p.a. for 2 years) and scheme Q (CI 10% p.a. for 2 years). Total interest is ₹3050. Find amount invested in each scheme.

Solution:

Let principal be 100%.

Scheme P (SI): 20% * 2 = 40% interest.

Scheme Q (CI): Effective rate 10 + 10 + (10*10)/100 = 21% interest.

Total interest = 40% + 21% = 61% of principal.

61% = ₹3050

1% = ₹3050 / 61 = ₹50

Principal (100%) = 50 * 100 = **₹5,000**.

Question 6: Age-related Problems

Problem: Ratio of present ages of Amit and Akhlak is 5:7. Difference between Akhlak's present age and Amit's age after 6 years is 2. Find sum of their present ages.

Solution:

Amit's present age = 5a. Akhlak's present age = 7a.

Akhlak's present age - Amit's age after 6 years = 2.

7a - (5a + 6) = 2

2a - 6 = 2

2a = 8

a = 4.

Sum of present ages = 5a + 7a = 12a = 12 * 4 = **48 years**. 

Step-by-Step Data Framing:

1. Balls A and C: Balls by A is 33.33% (1/3) more than C. So, A:C balls ratio is 4:3.

2. Runs Condition: Average runs of (A and B) = average runs of (C and D). This means Runs(A) + Runs(B) = Runs(C) + Runs(D).

3. Player B's Data: Faced 26 balls. Scored 2 runs/ball => 26 * 2 = 52 runs.

4. Player A's Data: Played 40 balls. From A:C ratio (4:3), if 4 units = 40, then 1 unit = 10. So, C played 3 * 10 = 30 balls.

5. Player D's Balls: Played 25% of total balls (A+B+C+D). Thus, A+B+C played 100% - 25% = 75% of total balls.
   Balls (A+B+C) = 40 + 26 + 30 = 96.
   If 75% = 96, then 25% = 96 / 3 = 32. So, D played 32 balls.

6. Player D's Runs: No runs on first 2 balls. 1 run on each subsequent ball.
   Scoring balls = 32 - 2 = 30. Runs by D = 30 * 1 = 30.

7. Runs for A and C: Total runs (A+C) = 78. Let Runs(A) = r. Runs(C) = 78 - r.
   Using the condition Runs(A) + Runs(B) = Runs(C) + Runs(D):
   r + 52 = (78 - r) + 30
   r + 52 = 108 - r
   2r = 56 => r = 28.
   So, Runs(A) = 28. Runs(C) = 78 - 28 = 50.

Final Framed Data Table: