IPMAT Arithmetic Questions with Detailed Solutions

Overview: Crack the pattern of IPMAT arithmetic questions, and the rest of the paper won't feel half as tricky.

The IPM Arithmetic Questions section forms the foundation of the Quantitative Aptitude paper in the Integrated Program in Management Aptitude Test (IPMAT), conducted by IIM Indore and IIM Rohtak.

Arithmetic is one of the most important and scoring areas of the exam, as it tests a student's conceptual understanding and problem-solving ability.

This article explains all major topics, types of IPMAT Arithmetic Questions, essential formulas, shortcuts, and preparation strategies to help students achieve high accuracy and speed in the exam.

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IPMAT Arithmetic Questions & Answers with Detailed Solutions

Q1. A product's price is increased by 20% and then later reduced by 25%. What is the net percentage change in price?

Solution: Take initial price = 100 (convenient).

After +20%: 100×1.20=120.100 \times 1.20 = 120.100×1.20=120.

After −25%: 120×0.75=90.120 \times 0.75 = 90.120×0.75=90.

Net change = 90−100=−10.90 - 100 = -10.90−100=−10. → net decrease 10%.

Shortcut: Multiply factors (1.20 \times 0.75 = 0.90) → net −10%.

Answer: 10% decrease.

Q2. Two items are sold for ₹1,000 each. On one item, the seller makes a 20% profit and on the other a 10% loss. Overall, did the seller gain or lose - and by what % (w.r.t total cost)?

Solution:  For item1: SP = 1000, gain 20% → CP1 = (1000/1.20 = 833.\overline{3}.)

For item2: SP = 1000, loss 10% → CP2 = (1000/0.90 = 1111.\overline{1}.)

Total CP = (833.\overline{3} + 1111.\overline{1} = 1944.\overline{4} = 17500/9.)

Total SP = (2000.) Net = (2000 - 1944.\overline{4} = 55.\overline{5} = 500/9.)

Percentage w.r.t total CP = (\dfrac{500/9}{17500/9}\times100 = \dfrac{500}{17500}\times100 = \dfrac{1}{35}\times100 \approx 2.857%.)

Answer: Overall profit ≈ 2.857% (≈ 2 6/7 %).

Q3. For 2 years at 12% p.a., the difference between compound interest (annual compounding) and simple interest on a sum is ₹252. Find the principal.

Solution: SI for 2 years = (P \times 12% \times 2 = 0.24P.)

Amount with CI after 2 yrs = (P(1.12^2) = P(1.2544).)

CI = (P(1.2544 - 1) = 0.2544P.)

Difference = CI − SI = (0.2544P - 0.24P = 0.0144P.)

Given (0.0144P = 252) → (P = 252 / 0.0144 = 252 \times (10000/144) )

But compute directly: (0.0144 = 144/10000) so (P = 252 \times 10000/144 = 252 \times 69.444\ldots) do exact: divide 252/144 = 1.75. So (P = 1.75 \times 10000 = 17500.)

Answer: ₹17,500.

Check: SI = (0.24\times17500 = 4200.) CI = (0.2544\times17500 = 4452.) Difference = 252 ✓

Q4. A and B have incomes in ratio 7:5. Their expenditures are in ratio 4:3. If A saves ₹7,000 and B saves ₹3,000, find their incomes.

Solution:  Let incomes: (A=7x,\ B=5x.)

Expenditures: (A_{exp}=4y,\ B_{exp}=3y.)

Savings: (7x - 4y = 7000) … (1) and (5x - 3y = 3000) … (2)

Multiply (2) by 4 and (1) by 3 to eliminate y: (2)*4: (20x - 12y = 12000.) (1)*3: (21x - 12y = 21000.)

Subtract: ((21x - 12y) - (20x - 12y) = 21000 - 12000 \Rightarrow x = 9000.)

So incomes: (A = 7x = 63000,\ B = 5x = 45000.)

Answer: A = ₹63,000; B = ₹45,000.

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Q5. A can complete a job in 10 days, B in 15 days and C in 20 days. They all start working but A leaves after 3 days and B leaves after 6 days (counting from start). How many more days will C take to finish the remaining work?

Solution: Rates: A = 1/10, B = 1/15, C = 1/20 (work/day).

Work done in first 3 days by all three: (3\times(1/10 + 1/15 + 1/20).)

Compute LCM 60: rates sum = ((6+4+3)/60 = 13/60.)

So first 3 days work = (3\times13/60 = 39/60 = 13/20.)

After 3 days, A leaves; B and C work together for next 3 days (from day 3 to day 6): combined rate = (1/15 + 1/20 = (4+3)/60 = 7/60.)

Work in next 3 days = (3\times7/60 = 21/60 = 7/20.) Total done by day 6 = (13/20 + 7/20 = 20/20 = 1.)

So the whole work is finished exactly at 6 days - C takes 0 more days.

Answer: Work completes exactly at 6 days (C requires 0 extra days).

Tip: Always sum incremental contributions; sometimes tasks finish exactly.

Q6. A vessel contains 100 L of milk. 20 L is removed and replaced by water. The operation is repeated two more times (total 3 such replacements). Find the final quantity of milk remaining.

Solution:  Fraction of milk remaining after one replacement = (\dfrac{100 - 20}{100} = 0.8.)

After 3 replacements: milk fraction = (0.8^3 = 0.512.) So milk left = (100 \times 0.512 = 51.2) L.

Answer: 51.2 L of milk.

Shortcut: Use ((1 - \frac{r}{V})^n) where r replaced each time.

Q7. A cyclist rides 30 km at 15 km/h, then 40 km at 20 km/h, and finally 50 km at 25 km/h. Find the average speed for the whole journey.

Solution:  Total distance = (30 + 40 + 50 = 120) km.

Times: t1 = (30/15 = 2) h; t2 = (40/20 = 2) h; t3 = (50/25 = 2) h.

Total time = 6 h. Average speed = total distance / total time = (120/6 = 20) km/h.

Answer: 20 km/h.

Observation: Here, times happen to be equal; average = arithmetic mean of speeds when times equal.

Q8. The average of 8 numbers is 42. If two numbers, 36 and 54, are removed, the average of the remaining numbers becomes 43. Find the sum of the remaining numbers and the number of remaining numbers.

Solution:  Sum of 8 numbers =8×42=336.= 8 \times 42 = 336.=8×42=336.

Sum of remaining 6 numbers =6×43=258.= 6 \times 43 = 258.=6×43=258.

Sum of two removed numbers =336−258=78.= 336 - 258 = 78.=336−258=78.

One removed number is 36, so the other =78−36=42.= 78 - 36 = 42.=78−36=42.

Answer: The other removed number is 42. (Removed numbers: 36 and 42.)

Note: IPMAT questions sometimes intentionally give values that force you to check consistency - watch reading.

Q9. A shopkeeper mixes two varieties of tea: one costing ₹120/kg and another ₹200/kg. He mixes them to sell at ₹170/kg, making a 15% profit on cost. What are the proportions of the two teas in the mixture?

Solution:  Let the mixture cost per kg = C.

He sells at SP = 170, profit 15% → cost = (170 / 1.15 = 148.\overline{26}) (exact: (170 \times 100/115 = 17000/115 = 3400/23 ≈148. 260869...)).

Keep fraction exact: cost = (3400/23). Let x fraction of ₹120 tea and (1−x) fraction of ₹200 tea → cost per kg = (120x + 200(1-x) = 200 − 80x.) Set = (3400/23.) Solve: (200 - 80x = 3400/23.)

Multiply 23: (4600 - 1840x = 3400.) So (4600 - 3400 = 1840x → 1200 = 1840x → x = 1200/1840 = 120/184 = 30/46 = 15/23.) So ratio of 120-cost tea : 200-cost tea = (15/23 : 1 - 15/23 = 8/23) → ratio = (15 : 8.)

Answer: 15 : 8 (₹120 tea : ₹200 tea).

Shortcut: Find mixture cost from desired SP and profit, then alligation.

Q10. Rahul invests ₹40,000 at 6% p.a. compounded annually. What is the amount after 3 years? Also compute the compound interest earned.

Solution: Amount = (40000 \times (1.06)^3.) Compute: (1.06^2 = 1.1236.)

Multiply by 1.06 → (1.191016.)

So amount = (40000 \times 1.191016 = 47640.64.) CI = (47640.64 - 40000 = 7640.64.)

Answer: Amount = ₹47,640.64; Compound interest ≈ ₹7,640.64.

Tip: For currency answers, round to paise if needed.

Q11. A and B start a business. A invests ₹90,000 and leaves after 6 months. B invests ₹60,000 and continues for 12 months. C joins after 3 months with ₹30,000 and stays till end (12 months total). Find profit share of A : B : C.

Solution: Compute capital×time (months): A: (90000 \times 6 = 540000.) B: (60000 \times 12 = 720000.) C: (30000 \times 9) (since C joins after 3 months and stays for the remaining 9 months) -

But the statement said "stays till end (12 months total)" is ambiguous; clarifying: if business duration = 12 months and C joined after 3 months, C invested for 9 months → (30000 \times 9 = 270000.)

So ratio = (540000 : 720000 : 270000 = 54 : 72 : 27) divide by 9 → (6 : 8 : 3.)

Answer: 6:8:3.

Q12. Two trains 200 km apart start moving towards each other at the same time. One at 50 km/h, the other at 70 km/h. A bird starts from the first train and flies to the second at 100 km/h, instantly turns back and continues until the trains meet. What is the total distance flown by the bird?

Solution: Time till trains meet = distance / relative speed = (200 / (50 + 70) = 200 / 120 = 5/3) hours = 1 hour 40 minutes.

Bird flies at 100 km/h for that entire time, so distance = (100 \times 5/3 = 500/3 = 166.\overline{6}) km.

Answer: (166.\overline{6}) km (i.e., 500/3 km).

Shortcut: Bird's total distance = bird speed × meeting time.

Q13. A 3-digit number is such that the difference between the number and the number formed by reversing its digits is 495. Find the number(s).

Solution:  Let number = 100a+10b+c100a + 10b + c100a+10b+c (with aaa nonzero).

Reversed = 100c+10b+a.100c + 10b + a.100c+10b+a.

Difference =99(a−c)=495.= 99(a - c) = 495.=99(a−c)=495.

So a−c=49599=5.a - c = \dfrac{495}{99} = 5.a−c=99495​=5.

Therefore hundreds digit aaa is 5 greater than units digit ccc.

Possible aaa values are 5,6,7,8,95,6,7,8,95,6,7,8,9 with corresponding c=0,1,2,3,4.c = 0,1,2,3,4.c=0,1,2,3,4. Middle digit bbb can be any digit 000-999.

Count of solutions = 5 choices for aaa × 10 choices for bbb = 50 numbers. Example numbers: 500, 510, …, 590; 611, 621, …, 691; etc.

Answer: All 3-digit numbers where hundreds digit − units digit = 5; there are 50 such numbers.

Q14. The average of 10 consecutive integers is 102. What is the sum of the first three integers?

Solution:  For an odd number of consecutive integers, the average equals the middle term. For 9 numbers the middle is the 5th term, so the 5th term = 102.

Therefore the first term =102−4=98.= 102 - 4 = 98.=102−4=98. (Because the sequence is 98,99,100,101,102,…98,99,100,101,102,\dots98,99,100,101,102,….)

Sum of first three =98+99+100=297.= 98 + 99 + 100 = 297.=98+99+100=297.

Answer: 297.

Q15. A worker is paid ₹200 per day for first 10 days, ₹240 per day for next 10 days, and then 10% more than previous rate for each of next 10 days respectively (i.e., day 21 wage = 1.1×240, day 22 wage = 1.1×(1.1×240), etc.). Find his total earning over 30 days (round to two decimals).

Solution: Earnings for days 1-10: 10×200=2000.10 \times 200 = 2000.10×200=2000.

Earnings for days 11-20: 10×240=2400.10 \times 240 = 2400.10×240=2400.

For days 21-30 (10 days) wages form a geometric progression with first term a=240×1.1=264a = 240 \times 1.1 = 264a=240×1.1=264 and common ratio r=1.1.r = 1.1.r=1.1.

Sum of last 10 days =ar10−1r−1=264×1.110−10.1.= a \dfrac{r^{10} - 1}{r - 1} = 264 \times \dfrac{1.1^{10} - 1}{0.1}.=ar−1r10−1​=264×0.11.110−1​.

Compute 1.110≈2.5937424601.1.1^{10} \approx 2.5937424601.1.110≈2.5937424601. Then 1.110−1≈1.5937424601.1.1^{10} - 1 \approx 1.5937424601.1.110−1≈1.5937424601.

Divide by 0.10.10.1 gives 15.937424601.15.937424601.15.937424601. Multiply by 264: 264×15.937424601≈4207.48.264 \times 15.937424601 \approx 4207.48.264×15.937424601≈4207.48.

Total earning =2000+2400+4207.48=8607.48.= 2000 + 2400 + 4207.48 = 8607.48.=2000+2400+4207.48=8607.48.

Answer: ₹8,607.48 (approximately).

Shortcut: For geometric pay, use GP sum formula and known powers of 1.1 (1.1^10 ≈ 2.5937424601).

IPMAT Arithmetic Previous Year Questions PDF

Before jumping into new problems, it's crucial to see how the real exam frames its arithmetic questions.

That's why we've compiled IPMAT arithmetic previous year questions.

IIM Indore IPMAT Previous Year Arithmetic Question Papers PDF

Download IPMAT 2024 Question Paper

Download IPMAT 2023 Question Paper

Download IPMAT 2022 Question Paper

Download IPMAT 2021 Question Paper

IIM Rohtak IPMAT Previous Year Arithmetic Question Papers PDF

Download IPMAT Rohtak Question Paper 2020

Download IPMAT Rohtak Question Paper 2019

Important Topics Covered in IPMAT Arithmetic Questions

The following table lists the most frequently tested areas in IPMAT number theory arithmetic questions along with the level of difficulty and concepts involved:

Topic	Key Concepts	Difficulty Level
Percentages	Successive changes, discounts, percentage increase/decrease	Easy
Profit, Loss and Discount	Cost price, selling price, successive discounts	Easy to Moderate
Simple and Compound Interest	Rate, time, principal, and CI-SI difference	Moderate
Ratio and Proportion	Direct and inverse variation, mixing quantities	Easy
Averages	Weighted averages, incremental method	Easy
Mixtures and Alligation	Concentration and composition adjustment	Moderate
Time, Speed and Distance	Relative speed, trains, circular track problems	Moderate
Time and Work / Pipes and Cisterns	Efficiency, work-time relation, combined work	Moderate
Boats and Streams	Upstream and downstream calculations	Moderate
Partnerships	Profit sharing based on time and capital	Easy

These are the foundation topics for IPMAT arithmetic questions, and mastering them is essential before moving on to advanced areas such as algebraic applications or data interpretation.

Read: How to Crack the Arithmetic Section in IPMAT Exam 2026?

Concept Integration in IPM Arithmetic Questions

In recent years, IPMAT arithmetic questions have shown a clear trend of combining two or more concepts in a single problem. For example:

1. Time, Speed and Distance + Ratio: Comparing speeds when one participant gets a head start.
2. Simple and Compound Interest: Comparing the total amounts under both types of interest.
3. Mixture and Ratio: Finding the amount of water to be added to reach a desired concentration.

Such integrated IPMAT arithmetic questions require conceptual clarity and the ability to connect multiple ideas logically.

Tricks and Shortcuts for Solving IPMAT Arithmetic Questions

To perform well in arithmetic IPMAT questions, students must develop both accuracy and speed. The following are some essential formulas and techniques:

• Break the question into stories. Picture people moving, money growing, or tanks filling up. It makes numbers come alive.
• Use ratios instead of fractions whenever you can - it's faster and easier to simplify.
• Practice mental math. You don't need to be a genius - just train your brain to see 20% as one-fifth instantly.
• Set mini timers. Challenge yourself: "Let's solve this in 50 seconds." You'll learn to think faster under pressure.

And remember - even if a question looks messy, there's always a simple path hidden inside. Arithmetic is about finding that clean line through the chaos.

How to Prepare for IPMAT Arithmetic Section 2026

A systematic approach is crucial for mastering IPMAT arithmetic questions.

1. Build Conceptual Clarity: Revise the basic formulas from NCERT Class 9 and 10. Understanding the logic behind every formula is more important than memorizing it.
2. Practice Chapter-wise: Start with one topic at a time. Solve 30-40 questions per chapter to strengthen fundamentals.
3. Create a Formula Booklet: Maintain a small notebook summarizing key formulas, ratios, and shortcuts from IPMAT arithmetic questions.
4. Take Topic-wise Tests: After revising each topic, attempt a short quiz or sectional test to identify weak areas.
5. Analyze Mistakes: After every IPMAT mock test, review each incorrect IPMAT arithmetic question carefully to understand whether the error was due to concept confusion or calculation oversight.
6. Manage Time Effectively: In IPMAT, arithmetic problems should ideally be solved within one minute to leave sufficient time for higher-level quantitative and logical reasoning questions.

Weightage of IPMAT Arithmetic Questions

The following table shows the approximate distribution of arithmetic topics based on recent papers:

Topic	IPMAT Rohtak	IPMAT Indore
Percentages and Ratio	3-4 Questions	2-3 Questions
Time, Speed and Distance	2-3 Questions	2 Questions
Time and Work	1-2 Questions	1 Question
Profit and Loss	2 Questions	1-2 Questions
Simple and Compound Interest	2 Questions	1-2 Questions
Averages and Mixtures	2 Questions	1 Question
Miscellaneous	1-2 Questions	1-2 Questions

This weightage confirms that a strong understanding of arithmetic questions significantly boosts overall performance in the IPMAT quantitative aptitude section.

IPMAT arithmetic questions form the backbone of the quantitative aptitude section and are essential for a strong overall score.

By focusing on conceptual clarity, consistent practice, and time management, students can achieve high accuracy and efficiency in this section.

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Key Takeaways

• Arithmetic questions form the backbone of the IPMAT quantitative aptitude section - they're your best chance to score consistently.
• The questions test clarity and logic, not just memorized formulas.
• Solving arithmetic questions from IPMAT previous years is the smartest way to understand patterns and difficulty levels.
• Focus on accuracy first, then build speed with IPMAT sample papers daily.
• Regularly review your IPMAT arithmetic questions and answers to strengthen weak areas.
• Treat arithmetic like a puzzle - once you learn how each piece fits (percentages, averages, ratios), you'll solve faster and more confidently.