IBPS PO Previous Year Question Paper (Prelims & Mains) – Quant Analysis with Solutions

If you are preparing for IBPS PO, the IBPS PO previous year question paper is the single most useful practice material you can get. It shows you the real difficulty level, the real pattern and the topics that keep coming back year after year. In this post, we cover how to use previous year papers properly, the Prelims and Mains pattern, the topic-wise weightage of Quant, and 10 solved Quant questions in the IBPS PO style with step-by-step solutions.

IBPS PO Previous Year Question Paper: Overview

IBPS PO (Probationary Officer) is conducted by the Institute of Banking Personnel Selection for recruitment in participating public sector banks. The selection has three stages: Prelims, Mains and Interview.

Previous year papers are available for both Prelims and Mains of earlier years. Some years have multiple shifts, so you get several papers from the same year. The best way to use them is to treat each paper as a full mock test, not as a question bank to skim.

Important: IBPS does not officially release question papers shift by shift right after the exam. The papers you see online right after an exam are usually memory-based papers, which are unofficial and are prepared from questions recalled by candidates. A few questions may be recalled slightly wrong, so use them for pattern and practice, not as an exact record.

Why You Should Solve IBPS PO Previous Year Papers

1. You see the real level. Mock tests from different providers vary a lot in difficulty. Previous year papers show you what IBPS actually asks.
2. You spot repeating topics. Simplification, Data Interpretation, Quadratic Equations and Arithmetic show up in some form almost every year.
3. You learn time management. Prelims gives you only 60 minutes for 100 questions, so deciding which questions to skip is a skill you build only with timed practice.
4. You find your weak areas. After a few papers, you will see which topics eat your time and which cost you marks.
5. You reduce exam fear. When the paper in the exam looks familiar, your mind stays calmer.

IBPS PO Prelims Exam Pattern

Section	Questions	Time
English Language	30	20 minutes
Quantitative Aptitude	35	20 minutes
Reasoning Ability	35	20 minutes
Total	100	60 minutes

Each section has its own time limit. There is negative marking of one fourth of the marks for every wrong answer. Prelims is a qualifying round, so your marks here do not count in the final merit.

IBPS PO Mains Exam Pattern

Section	Questions	Time
Reasoning and Computer Aptitude	45	60 minutes
English Language	35	40 minutes
Data Analysis and Interpretation	35	45 minutes
General, Economy and Banking Awareness	40	35 minutes
Objective Total	155	180 minutes
English Letter and Essay (descriptive)	2	30 minutes

Mains marks count in the final merit along with the interview. Since you are targeting Mains level, the Data Analysis and Interpretation section deserves extra practice, because the calculations are heavier than in Prelims.

Please always cross-check the latest pattern with the official IBPS notification, because the pattern and number of questions can change.

IBPS PO Previous Year Question Paper PDF: Year-wise Links

Year	Prelims	Mains
2025	Oliveboard	Oliveboard
2024	Adda247	Adda247

IBPS PO 2026 Memory-Based Papers

The IBPS PO Prelims 2026 was held on 22 and 23 August 2026, and memory-based papers for the different shifts are available on various preparation websites. Remember, these are unofficial. Use them to understand the latest trend, and also keep solving older papers, because the repeating concepts are easier to see across many years.

Quant Topic-wise Weightage in IBPS PO Papers

The exact split changes from shift to shift, but the Quant section usually follows this pattern:

Topic	What to expect
Simplification / Approximation	5 to 10 questions in Prelims, quick scoring
Quadratic Equations	5 questions, relation-based (x > y, x ≤ y and so on)
Data Interpretation	5 to 10 questions, tables, bar graphs, line graphs, caselets
Number Series	5 questions, missing number or wrong number
Arithmetic (Percentage, Profit and Loss, Ratio, Time and Work, TSD, Mixture, SI and CI, Average)	8 to 12 questions, mixed level
Permutation, Combination and Probability	1 to 3 questions, usually medium level

If you are short on time, finish Simplification, Quadratic Equations and Number Series first, because these are the fastest marks. Then attempt DI and Arithmetic.

10 Solved Quant Questions (IBPS PO Style)

These questions are written in the style of IBPS PO papers. Try each one yourself for a few minutes before reading the solution.

Question 1 (Simplification)

18.5% of 640 + 27.5% of 480 − 12.5% of 320 = ?

Solution:

• 18.5% of 640 = 0.185 × 640 = 118.4
• 27.5% of 480 = 0.275 × 480 = 132
• 12.5% of 320 = 0.125 × 320 = 40
• Total = 118.4 + 132 − 40 = 210.4

Question 2 (Quadratic Equations)

I. x² − 11x + 30 = 0 II. y² − 13y + 42 = 0

Find the relation between x and y.

Solution:

• From I: x² − 11x + 30 = (x − 5)(x − 6) = 0, so x = 5 or 6
• From II: y² − 13y + 42 = (y − 6)(y − 7) = 0, so y = 6 or 7
• Compare every pair: 5 < 6, 5 < 7, 6 = 6, 6 < 7
• In no case is x greater than y, and x equals y in one case
• So the answer is x ≤ y

Question 3 (Percentage)

The price of sugar rises by 25%. By what percent must a family reduce its consumption so that its expenditure stays the same?

Solution:

• Let the old price be 100 and old consumption be 1 unit. Old expenditure = 100
• New price = 125
• New consumption = 100 ÷ 125 = 0.8 units
• Reduction = (1 − 0.8) × 100 = 20%
• Shortcut: reduction = 25 ÷ (100 + 25) × 100 = 20%

Question 4 (Profit and Loss)

A shopkeeper marks an article 40% above its cost price. He gives two successive discounts of 10% and 5%. Find his profit percent.

Solution:

• Let the cost price be 100. Marked price = 140
• After the first discount of 10%: 140 × 0.9 = 126
• After the second discount of 5%: 126 × 0.95 = 119.7
• Profit = 119.7 − 100 = 19.7
• Profit percent = 19.7%

Question 5 (Ratio and Proportion)

The ratio of A to B is 3 : 5. If 12 is added to each, the ratio becomes 5 : 7. Find the value of A.

Solution:

• Let A = 3x and B = 5x
• (3x + 12) ÷ (5x + 12) = 5 ÷ 7
• Cross multiply: 7(3x + 12) = 5(5x + 12)
• 21x + 84 = 25x + 60
• 4x = 24, so x = 6
• A = 3 × 6 = 18 (and B = 30)
• Check: (18 + 12) : (30 + 12) = 30 : 42 = 5 : 7 ✓

Question 6 (Time, Speed and Distance: Boats and Streams)

The speed of a boat in still water is 15 km/h and the speed of the stream is 5 km/h. The boat goes to a place and comes back in a total of 6 hours. Find the distance of the place.

Solution:

• Downstream speed = 15 + 5 = 20 km/h
• Upstream speed = 15 − 5 = 10 km/h
• Let the distance be d km
• d ÷ 20 + d ÷ 10 = 6
• (d + 2d) ÷ 20 = 6
• 3d = 120, so d = 40 km

Question 7 (Time and Work)

A can complete a work in 12 days and B can complete it in 18 days. They work together for 4 days, then A leaves. In how many days in total will the work be finished?

Solution:

• Take total work = 36 units (LCM of 12 and 18)
• A’s one day work = 3 units, B’s one day work = 2 units
• Together in 4 days = 4 × 5 = 20 units
• Remaining work = 36 − 20 = 16 units
• B alone finishes 16 units in 16 ÷ 2 = 8 days
• Total time = 4 + 8 = 12 days

Question 8 (Mixture and Alligation)

A vessel contains 60 litres of a milk and water mixture in the ratio 2 : 1. How much water must be added so that the ratio of milk to water becomes 1 : 2?

Solution:

• Milk = 60 × 2/3 = 40 litres, water = 20 litres
• Milk does not change, so for ratio 1 : 2, water must be 2 × 40 = 80 litres
• Water to be added = 80 − 20 = 60 litres

Question 9 (Simple and Compound Interest)

The difference between compound interest and simple interest on a certain sum for 2 years at 10% per annum is ₹150. Find the sum.

Solution:

• For 2 years, CI − SI = P × (r ÷ 100)²
• 150 = P × (10 ÷ 100)² = P × 0.01
• P = 150 ÷ 0.01 = ₹15,000

Question 10 (Probability)

A bag contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are of the same colour.

Solution:

• Total ways to pick 2 balls from 10 = 10C2 = 45
• Both red = 4C2 = 6 ways
• Both blue = 6C2 = 15 ways
• Favourable ways = 6 + 15 = 21
• Probability = 21 ÷ 45 = 7/15

Topics Wise Questions

Topic 1: Simplification / Approximation

Q1. 24.98% of 1199.8 + 15.02² − √2024.9 = ?

Solution:

• 24.98% of 1199.8 is about 25% of 1200 = 300
• 15.02² is about 15² = 225
• √2024.9 is about √2025 = 45
• 300 + 225 − 45 = 480

Answer: 480

Q2. (7.98³ − 5.01³) ÷ (7.98 − 5.01) + 11.03% of 1799.7 = ?

Solution:

• Round: 7.98 to 8 and 5.01 to 5
• (8³ − 5³) ÷ (8 − 5) = (512 − 125) ÷ 3 = 387 ÷ 3 = 129
• 11.03% of 1799.7 is about 11% of 1800 = 198
• 129 + 198 = 327

Answer: 327

Q3. [(3/4 of 2/3 of 1296) ÷ 9] × (5/6) + 15² − 13² = ?

Solution:

• 2/3 of 1296 = 864
• 3/4 of 864 = 648
• 648 ÷ 9 = 72
• 72 × 5/6 = 60
• 15² − 13² = 225 − 169 = 56
• 60 + 56 = 116

Answer: 116

Q4. 1599.8 ÷ 15.98 × 3.99 + 19.97% of 2504.6 − 7.99² = ?

Solution:

• 1599.8 ÷ 15.98 is about 1600 ÷ 16 = 100
• 100 × 3.99 is about 400
• 19.97% of 2504.6 is about 20% of 2500 = 500
• 7.99² is about 64
• 400 + 500 − 64 = 836

Answer: 836

Q5. (0.75 × 3.2 + 1.25 × 4.8) ÷ 0.6 − √1.44 × 5 = ?

Solution:

• 0.75 × 3.2 = 2.4
• 1.25 × 4.8 = 6.0
• Sum = 8.4, and 8.4 ÷ 0.6 = 14
• √1.44 = 1.2, and 1.2 × 5 = 6
• 14 − 6 = 8

Answer: 8

Topic 2: Quadratic Equations

In each question, solve both equations and find the relation between x and y.

Q1. I. 2x² − 13x + 21 = 0 II. 3y² − 17y + 20 = 0

Solution:

• I: 2x² − 6x − 7x + 21 = 2x(x − 3) − 7(x − 3) = (2x − 7)(x − 3) = 0, so x = 3.5 or 3
• II: 3y² − 12y − 5y + 20 = 3y(y − 4) − 5(y − 4) = (3y − 5)(y − 4) = 0, so y = 4 or 5/3 (1.67)
• x = 3.5 is less than 4 but more than 1.67, so no single relation holds

Answer: No relation

Q2. I. x² + 5x − 66 = 0 II. y² − 20y + 96 = 0

Solution:

• I: (x + 11)(x − 6) = 0, so x = −11 or 6
• II: (y − 8)(y − 12) = 0, so y = 8 or 12
• Largest x is 6, smallest y is 8, so every x is less than every y

Answer: x < y

Q3. I. 6x² + 7x − 20 = 0 II. 4y² − 4y − 15 = 0

Solution:

• I: 6x² + 15x − 8x − 20 = 3x(2x + 5) − 4(2x + 5) = (3x − 4)(2x + 5) = 0, so x = 4/3 or −5/2
• II: 4y² − 10y + 6y − 15 = 2y(2y − 5) + 3(2y − 5) = (2y + 3)(2y − 5) = 0, so y = −3/2 or 5/2
• x = 4/3 is more than y = −1.5 but less than y = 2.5, so no fixed relation

Answer: No relation

Q4. I. x² − 17x + 72 = 0 II. 4y² − 60y + 224 = 0

Solution:

• I: (x − 8)(x − 9) = 0, so x = 8 or 9
• II: divide by 4: y² − 15y + 56 = (y − 7)(y − 8) = 0, so y = 7 or 8
• Compare all pairs: 8 ≥ 7, 8 = 8, 9 > 7, 9 > 8, so x is never less than y

Answer: x ≥ y

Q5. I. 5x² − 18x + 16 = 0 II. 2y² − 5y + 3 = 0

Solution:

• I: discriminant = 324 − 320 = 4, so x = (18 ± 2) ÷ 10 = 2 or 1.6
• II: (2y − 3)(y − 1) = 0, so y = 1.5 or 1
• Smallest x is 1.6, largest y is 1.5, so every x is greater than every y

Answer: x > y

Topic 3: Data Interpretation

Q1. The table shows the sales (in lakh rupees) of four shops.

Shop	2023	2024
P	480	540
Q	600	570
R	720	810
S	360	468

By what percent (to 2 decimal places) did the total sales of all four shops increase from 2023 to 2024?

Solution:

• Total 2023 = 480 + 600 + 720 + 360 = 2160
• Total 2024 = 540 + 570 + 810 + 468 = 2388
• Increase = 2388 − 2160 = 228
• Percent increase = 228 ÷ 2160 × 100 = 10.56%

Answer: 10.56%

Q2. The table shows the number of students and the percentage of boys in five classes.

Class	Total students	Boys (%)
A	60	40%
B	80	62.5%
C	50	56%
D	90	60%
E	70	50%

18 boys of class D are shifted to class E, and 9 girls of class E are shifted to class D. What is the new ratio of boys to girls in class E?

Solution:

• Class E originally: boys = 35, girls = 35
• Class E after the shift: boys = 35 + 18 = 53, girls = 35 − 9 = 26
• Ratio = 53 : 26

Answer: 53 : 26

Q3. A school has 1200 students. 35% of them study in the primary section and the rest in the secondary section. In the primary section the ratio of boys to girls is 3 : 2, and in the secondary section it is 5 : 7. What is the difference between the total number of girls and the total number of boys in the school?

Solution:

• Primary = 35% of 1200 = 420, secondary = 780
• Primary: boys = 420 × 3/5 = 252, girls = 168
• Secondary: boys = 780 × 5/12 = 325, girls = 780 × 7/12 = 455
• Total boys = 252 + 325 = 577, total girls = 168 + 455 = 623
• Difference = 623 − 577 = 46

Answer: 46

Q4. The table shows the income and expenditure (in lakh rupees) of two companies.

Year	Income A	Expense A	Income B	Expense B
2020	450	360	520	400
2021	540	450	650	500
2022	600	480	780	600

In which year was the combined profit of A and B the highest, and what was the combined profit percent in that year (to the nearest whole number)?

Solution:

• Profit 2020: A = 90, B = 120, combined = 210
• Profit 2021: A = 90, B = 150, combined = 240
• Profit 2022: A = 120, B = 180, combined = 300, which is the highest
• 2022 combined income = 600 + 780 = 1380, combined expense = 480 + 600 = 1080
• Profit percent = 300 ÷ 1080 × 100 = 27.78%, which is about 28%

Answer: 2022, about 28%

Q5. A person’s monthly salary is ₹80,000. His expenditure is: Rent 25%, Food 20%, Education 15%, Transport 10%, Miscellaneous 12%, and the rest is saved. Next year his salary rises by 15%. Rent rises by 10%, food by 20%, education by 25%, miscellaneous by 5%, and transport stays the same. Find his new savings (in ₹) and the new savings as a percent of the new salary.

Solution:

• Old amounts: rent 20,000; food 16,000; education 12,000; transport 8,000; miscellaneous 9,600
• New salary = 80,000 × 1.15 = 92,000
• New rent = 22,000; food = 19,200; education = 15,000; transport = 8,000; miscellaneous = 9,600 × 1.05 = 10,080
• Total expenditure = 22,000 + 19,200 + 15,000 + 8,000 + 10,080 = 74,280
• New savings = 92,000 − 74,280 = 17,720
• Savings percent = 17,720 ÷ 92,000 × 100 = 19.26% (it was 18% before)

Answer: ₹17,720 and about 19.26%

Topic 4: Number Series

Q1. 6, 13, 28, 59, 122, ?

Solution:

• 6 × 2 + 1 = 13
• 13 × 2 + 2 = 28
• 28 × 2 + 3 = 59
• 59 × 2 + 4 = 122
• 122 × 2 + 5 = 249

Answer: 249

Q2. 3, 4, 10, 33, 136, ?

Solution:

• 3 × 1 + 1 = 4
• 4 × 2 + 2 = 10
• 10 × 3 + 3 = 33
• 33 × 4 + 4 = 136
• 136 × 5 + 5 = 685

Answer: 685

Q3. Find the wrong number: 5, 6, 14, 45, 186, 925, 5556

Solution:

• Pattern: ×1 + 1, ×2 + 2, ×3 + 3, ×4 + 4, ×5 + 5, ×6 + 6
• 5 × 1 + 1 = 6, 6 × 2 + 2 = 14, 14 × 3 + 3 = 45
• 45 × 4 + 4 = 184, not 186
• 184 × 5 + 5 = 925 and 925 × 6 + 6 = 5556 both fit with 184

Answer: 186 (it should be 184)

Q4. 2, 3, 7, 25, 121, ?

Solution:

• Add 1 to a factorial each time: 1! + 1 = 2, 2! + 1 = 3, 3! + 1 = 7, 4! + 1 = 25, 5! + 1 = 121
• Next term = 6! + 1 = 720 + 1 = 721

Answer: 721

Q5. 11, 12, 20, 47, 111, ?

Solution:

• Differences are cubes: 12 − 11 = 1³, 20 − 12 = 2³, 47 − 20 = 3³, 111 − 47 = 4³
• Next difference = 5³ = 125
• 111 + 125 = 236

Answer: 236

Topic 5: Time and Work

Q1. A and B together can do a work in 12 days, B and C in 15 days, and C and A in 20 days. In how many days can A alone do the work?

Solution:

• Add the three rates: 2(A + B + C) = 1/12 + 1/15 + 1/20 = (5 + 4 + 3)/60 = 1/5
• A + B + C = 1/10 per day
• A = (A + B + C) − (B + C) = 1/10 − 1/15 = 1/30
• A alone takes 30 days

Answer: 30 days

Q2. Pipes A and B can fill a tank in 12 hours and 15 hours, and pipe C can empty it in 20 hours. They are opened one at a time for one hour each, in the order A, B, C, A, B, C and so on. In how much time will the empty tank be completely filled?

Solution:

• Take the capacity = 60 units (LCM). A fills 5 units/h, B fills 4 units/h, C empties 3 units/h
• Net in one 3-hour cycle = 5 + 4 − 3 = 6 units
• After 9 cycles (27 hours) = 54 units
• Hour 28 (pipe A): 54 + 5 = 59 units
• Hour 29 (pipe B): only 1 unit is needed at 4 units/h, so it takes 1/4 hour
• Total time = 28 + 0.25 = 28.25 hours

Answer: 28 hours 15 minutes

Q3. A is 50% more efficient than B. Working together they finish a work in 12 days. They start together, but B leaves after 5 days and A completes the remaining work alone. In how many days in total is the work finished?

Solution:

• Efficiency A : B = 3 : 2. Total work = (3 + 2) × 12 = 60 units
• Work done in 5 days together = 5 × 5 = 25 units
• Remaining = 35 units, A does 3 units/day, so 35 ÷ 3 = 11 2/3 days
• Total = 5 + 11 2/3 = 16 2/3 days

Answer: 16 2/3 days (16 days 16 hours)

Q4. A, B and C can do a work in 10, 15 and 30 days respectively. They take up the work for ₹9,000. A and C both leave 3 days before the work is completed and B works till the end. Find the total time taken and B’s share of the money.

Solution:

• Let the total time be T days. A and C work (T − 3) days, B works T days
• (T − 3)/10 + T/15 + (T − 3)/30 = 1
• Multiply by 30: 3(T − 3) + 2T + (T − 3) = 30, so 6T − 12 = 30, which gives T = 7
• Work shares (in 30ths): A = 4 days × 3 = 12, B = 7 × 2 = 14, C = 4 × 1 = 4
• Ratio = 12 : 14 : 4 = 6 : 7 : 2, so one part = 9000 ÷ 15 = 600
• B’s share = 7 × 600 = 4,200

Answer: 7 days and ₹4,200

Q5. 6 men and 8 women can complete a work in 10 days. 26 men and 48 women can complete the same work in 2 days. How many days will 15 men and 20 women take?

Solution:

• Let one man’s and one woman’s daily work be m and w
• 6m + 8w = 1/10, multiply by 5: 30m + 40w = 1/2
• 26m + 48w = 1/2
• Equate: 30m + 40w = 26m + 48w, so 4m = 8w and m = 2w
• 6(2w) + 8w = 20w = 1/10, so w = 1/200 and m = 1/100
• 15m + 20w = 15/100 + 20/200 = 0.25 per day, so 4 days

Answer: 4 days

Topic 6: Mixture and Alligation

Q1. A vessel has 80 litres of pure milk. 16 litres are taken out and replaced with water. This is done two more times (three times in all). Find the ratio of milk to water at the end.

Solution:

• Milk left = 80 × (1 − 16/80)³ = 80 × (4/5)³ = 80 × 64/125 = 40.96 litres
• Water = 80 − 40.96 = 39.04 litres
• Ratio = 40.96 : 39.04 = 4096 : 3904 = 64 : 61

Answer: 64 : 61

Q2. Vessel P has milk and water in the ratio 5 : 3 and vessel Q has them in the ratio 7 : 1. In what ratio should the mixtures be taken so that the final mixture has milk and water in the ratio 2 : 1?

Solution:

• Milk fraction in P = 5/8 = 15/24, in Q = 7/8 = 21/24, wanted = 2/3 = 16/24
• P : Q = (21 − 16) : (16 − 15) = 5 : 1
• Check: 5 parts of P and 1 part of Q give milk = 5 × 15/24 + 21/24 = 96/24 = 4 in 6 parts, which is 2/3 ✓

Answer: 5 : 1

Q3. Tea costing ₹180 per kg is mixed with tea costing ₹240 per kg. The mixture is sold at ₹264 per kg and gives a profit of 20%. In what ratio were the two teas mixed?

Solution:

• Cost price of the mixture = 264 ÷ 1.2 = ₹220 per kg
• Cheaper : costlier = (240 − 220) : (220 − 180) = 20 : 40 = 1 : 2
• Check: 1 kg at 180 + 2 kg at 240 = 660 for 3 kg = 220 per kg ✓

Answer: 1 : 2

Q4. A container has 60 litres of a milk and water mixture with 70% milk. A certain quantity is removed and replaced with water. The same quantity is again removed and replaced with water. The milk is now 25.2% of the mixture. How many litres were removed each time?

Solution:

• Let x litres be removed each time. Milk fraction = 70% × (1 − x/60)²
• 0.7 × (1 − x/60)² = 0.252, so (1 − x/60)² = 0.36
• 1 − x/60 = 0.6, so x = 24
• Check: milk left = 60 × 0.252 = 15.12 litres

Answer: 24 litres

Q5. A vessel has 40 litres of wine and water in the ratio 3 : 2. How many litres of the mixture must be removed and replaced with pure wine so that the ratio of wine to water becomes 9 : 1?

Solution:

• Initially wine = 24 L, water = 16 L
• Final water = 40 × 1/10 = 4 L, and water is lost only by removal
• Each litre of mixture removed takes out 0.4 L of water, so 16 − 0.4x = 4 and x = 30
• Check: removing 30 L takes 18 L wine and 12 L water, leaving 6 L wine and 4 L water. Adding 30 L wine gives 36 : 4 = 9 : 1 ✓

Answer: 30 litres

Topic 7: Simple and Compound Interest

Q1. A sum of money becomes ₹8,400 in 3 years and ₹9,000 in 5 years at simple interest. Find the sum and the rate of interest.

Solution:

• Interest for 2 years = 9000 − 8400 = 600, so yearly interest = 300
• Interest for 3 years = 900, so principal = 8400 − 900 = 7,500
• Rate = 300 ÷ 7500 × 100 = 4%

Answer: ₹7,500 at 4% per annum

Q2. The difference between compound interest and simple interest on a sum for 3 years at 10% per annum is ₹3,100. Find the sum.

Solution:

• For 3 years, CI − SI = P × (r/100)² × (3 + r/100)
• = P × 0.01 × 3.1 = 0.031P
• 0.031P = 3100, so P = 1,00,000
• Check: CI = 33,100 and SI = 30,000, difference = 3,100 ✓

Answer: ₹1,00,000

Q3. ₹40,000 is lent in two parts, one at 6% and the other at 10% simple interest per annum. After 2 years the total interest is ₹6,880. How much was lent at 10%?

Solution:

• Let x be lent at 6%, so (40,000 − x) is lent at 10%
• Interest for 2 years = x × 0.12 + (40,000 − x) × 0.20 = 8,000 − 0.08x
• 8,000 − 0.08x = 6,880, so x = 14,000
• Amount at 10% = 40,000 − 14,000 = 26,000
• Check: 14,000 × 0.12 = 1,680 and 26,000 × 0.20 = 5,200, total = 6,880 ✓

Answer: ₹26,000

Q4. Find the compound interest on ₹20,000 for 1½ years at 10% per annum, compounded half-yearly.

Solution:

• Rate per half-year = 5%, number of half-years = 3
• Amount = 20,000 × (1.05)³ = 20,000 × 1.157625 = 23,152.50
• CI = 23,152.50 − 20,000 = 3,152.50

Answer: ₹3,152.50

Q5. A sum amounts to ₹7,986 in 3 years and ₹8,784.60 in 4 years at compound interest (compounded annually). Find the rate and the sum.

Solution:

• The 4th year interest = 8,784.60 − 7,986 = 798.60, and this is interest on 7,986
• Rate = 798.60 ÷ 7986 × 100 = 10%
• Sum = 7,986 ÷ (1.1)³ = 7,986 ÷ 1.331 = 6,000

Answer: 10% and ₹6,000

Topic 8: Permutation, Combination and Probability

Q1. In how many ways can the letters of the word BANKING be arranged so that the two vowels are not together?

Solution:

• 7 letters with N repeated twice: total arrangements = 7! ÷ 2! = 2,520
• Vowels A and I together: treat them as one unit, so 6 units with N repeated twice = 6! ÷ 2! = 360, and the vowels can swap in 2! ways, giving 720
• Not together = 2,520 − 720 = 1,800

Answer: 1,800

Q2. A committee of 5 is to be formed from 6 men and 5 women. In how many ways can it be formed with at least 2 women and at least 1 man?

Solution:

• 2 women + 3 men: 5C2 × 6C3 = 10 × 20 = 200
• 3 women + 2 men: 5C3 × 6C2 = 10 × 15 = 150
• 4 women + 1 man: 5C4 × 6C1 = 5 × 6 = 30
• 5 women + 0 man is not allowed
• Total = 200 + 150 + 30 = 380

Answer: 380

Q3. Two dice are thrown together. Find the probability that the sum of the numbers is a multiple of 4 or the product of the numbers is 12.

Solution:

• Sum is 4: (1,3), (2,2), (3,1) = 3 ways. Sum is 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 ways. Sum is 12: (6,6) = 1 way. Total = 9
• Product is 12: (2,6), (3,4), (4,3), (6,2) = 4 ways
• Both conditions: (2,6) and (6,2) = 2 ways
• Favourable = 9 + 4 − 2 = 11, so probability = 11/36

Answer: 11/36

Q4. Two cards are drawn at random from a pack of 52 cards. Find the probability that both are face cards of different suits.

Solution:

• There are 12 face cards (3 in each suit)
• Total ways = 52C2 = 1,326
• Both face cards = 12C2 = 66. Both from the same suit = 4 × 3C2 = 12
• Different suits = 66 − 12 = 54
• Probability = 54/1326 = 9/221

Answer: 9/221

Q5. Bag A has 3 red and 5 blue balls. Bag B has 4 red and 4 blue balls. A bag is chosen at random and two balls are drawn without replacement. Find the probability that both balls are red. Also find the probability that it was bag A, given that both balls are red.

Solution:

• Bag A: 3C2 ÷ 8C2 = 3/28. Bag B: 4C2 ÷ 8C2 = 6/28
• P(both red) = 1/2 × 3/28 + 1/2 × 6/28 = 9/56
• P(bag A given both red) = (1/2 × 3/28) ÷ (9/56) = (3/56) ÷ (9/56) = 1/3

Answer: 9/56 and 1/3

How to Use IBPS PO Previous Year Papers: A 3-Step Strategy

Step 1: Solve without a timer (first 2 papers). Understand the pattern, the level and the type of questions. Note down every concept you did not know.

Step 2: Solve with a strict timer (next papers). Sit at a desk, set the exact section time (20 minutes per section for Prelims) and attempt it like the real exam. No phone, no breaks.

Step 3: Analyse every paper properly. After each paper, write down three things: questions you got wrong because of a concept gap, questions you got wrong because of a silly mistake, and questions that took more than 90 seconds. Revise those topics before the next paper. A paper you do not analyse is a paper half wasted.

Repeat the important papers after a few weeks. As a good rule, you should be able to sense the pattern without looking at the solutions.

Common Mistakes to Avoid

• Solving only the latest year and ignoring older papers
• Checking the answer key too early instead of trying the question again
• Skipping the analysis and moving straight to the next paper
• Attempting every question in Prelims instead of choosing the easy ones first
• Treating memory-based questions as official without cross-checking

FAQs

Q1. Where can I get the IBPS PO previous year question paper PDF? You can get Prelims and Mains papers from the official IBPS website and from trusted preparation platforms. Always check that the paper is complete before using it.

Q2. Is the IBPS PO 2026 memory-based paper official? No. It is an unofficial paper prepared from questions recalled by candidates after the exam.

Q3. How many previous year papers should I solve? Try to solve at least the last 5 years of Prelims and Mains papers, and repeat the important ones.

Q4. Is IBPS PO Prelims marks counted in the final merit? No. Prelims is a qualifying stage. Final merit is based on Mains and the interview.

Q5. Which Quant topics should I prepare first? Start with Simplification, Quadratic Equations, Number Series and Data Interpretation, then move to Arithmetic topics like Percentage, Ratio, Profit and Loss, and Time and Work.

Final Words

Previous year papers are not just practice, they are a map of what the exam wants from you. Solve them with a timer, analyse every mistake, and come back to the same papers after a few weeks. For more topic-wise practice, check our other series on BrainQuro: Percentage, Ratio and Proportion, and Time Speed and Distance.

Comment Challenge

Pipe A can fill a tank in 20 minutes, pipe B can fill it in 30 minutes, and pipe C can empty the full tank in 15 minutes. A and B are opened together. After 10 minutes, pipe C is also opened. In how many minutes in total will the tank be completely filled?

Write your answer in the comments.