IPMAT Ratio and Proportion Questions with Answers

Overview: When preparing for the IPMAT Ratio and Proportion questions, understanding the core concepts and practicing strategically can make all the difference.

Preparing for the IPM Ratio and Proportion questions is a crucial step for every IPMAT aspirant aiming to score high in the Quantitative Aptitude section.

Ratio and proportion are not just basic math concepts - they form the foundation for many advanced arithmetic topics such as time, speed, mixtures, averages, and profit and loss.

In the IPMAT exam, these questions test your ability to analyze relationships between quantities and apply logical reasoning to solve problems efficiently.

A strong command over IPMAT Ratio and Proportion questions helps candidates save time, avoid calculation errors, and handle application-based questions with confidence.

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Sample IPMAT Ratio & Proportion: Questions with Solutions

To strengthen your understanding,

Let's look at some sample IPMAT Ratio and Proportion questions that reflect the level and style you can expect in the actual exam.

1. The ratio of two numbers is 3 : 5. If their sum is 64, find the numbers.

Solution: Let the two numbers be 3k and 5k (since their ratio is 3:5). Sum = 3k + 5k = 8k = 64 ⇒ k = 64 ÷ 8 = 8

Now the numbers are: 3×8 = 24 and 5×8 = 40

Answer: The numbers are 24 and 40.

2. A and B's incomes are in the ratio 4 : 7, and their expenditures are in the ratio 3 : 5. If they each save ₹4000, find their incomes.

Solution: Let incomes be 4x and 7x, and expenditures be 3y and 5y, respectively. Savings = Income − Expenditure

So, 4x − 3y = 4000 …(1) 7x − 5y = 4000 …(2)

Subtract (1) from (2): (7x − 5y) − (4x − 3y) = 0 ⇒ 3x − 2y = 0 ⇒ y = (3x)/2

Substitute into (1): 4x − 3(3x/2) = 4000 ⇒ 4x − (9x/2) = 4000 ⇒ (8x − 9x)/2 = 4000 ⇒ −x/2 = 4000 ⇒ x = −8000

Negative income doesn't make sense. Hence, the given data is inconsistent.

Answer: No valid positive solution (the problem data is inconsistent).

3. If 3x = 4y = 5z, find the ratio x : y : z.

Solution: Let 3x = 4y = 5z = k. Then x = k/3, y = k/4, z = k/5

So, x : y : z = (1/3) : (1/4) : (1/5)

Multiply by the LCM of denominators (60): → 20 : 15 : 12

Answer: x : y : z = 20 : 15 : 12

4. Two numbers are in the ratio 7 : 9. If the smaller is increased by 7 and the larger by 9, the ratio becomes 8 : 10. Find the numbers.

Solution: Let the numbers be 7k and 9k.

Given: (7k + 7) / (9k + 9) = 8/10 = 4/5

Cross multiply: 5(7k + 7) = 4(9k + 9) ⇒ 35k + 35 = 36k + 36 ⇒ k = −1

So, numbers = −7 and −9. Since we expect positive numbers, this is invalid.

Answer: No positive solution (algebra gives −7 and −9).

5. ₹720 is divided among A, B, and C in the ratio 2 : 3 : 4. Find each share.

Solution: Total parts = 2 + 3 + 4 = 9 Each part = 720 ÷ 9 = 80

So, A = 2×80 = 160 B = 3×80 = 240 C = 4×80 = 320

Answer: A = ₹160, B = ₹240, C = ₹320

6. The ratio of boys to girls in a class is 5 : 3. If 16 more girls join, the ratio becomes 5 : 4. Find the number of boys.

Solution: Let boys = 5k, girls = 3k. After 16 girls join: girls = 3k + 16

Now, (5k)/(3k + 16) = 5/4 Cross multiply: 20k = 15k + 80 ⇒ 5k = 80 ⇒ k = 16

Boys = 5×16 = 80

Answer: There are 80 boys.

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7. A and B together have ₹1210. If (4/15) of A's money = (2/5) of B's money, find their shares.

Solution: Let A = a, B = b

Given: a + b = 1210 (4/15)a = (2/5)b ⇒ (4/15)a = (6/15)b ⇒ 4a = 6b ⇒ a/b = 3/2

So, ratio A:B = 3:2 Total parts = 5

Each part = 1210 ÷ 5 = 242

Hence, A = 3×242 = 726 B = 2×242 = 484

Answer: A = ₹726, B = ₹484

8. The ratio of the ages of A and B is 4 : 5. After 8 years, the ratio becomes 5 : 6. Find their present ages.

Solution: Let ages be 4k and 5k. After 8 years: (4k + 8)/(5k + 8) = 5/6

Cross multiply: 6(4k + 8) = 5(5k + 8) ⇒ 24k + 48 = 25k + 40 ⇒ k = 8

Present ages: A = 4×8 = 32 B = 5×8 = 40

 Answer: A = 32 years, B = 40 years

9. The ratio of speeds of two trains is 7 : 9. If the second train covers 540 km in 6 hours, find the speed of the first train.

Solution: Speed of second = 540 ÷ 6 = 90 km/h

First train's speed = (7/9) × 90 = 70 km/h

Answer: 70 km/h

10. The ratio of the length and breadth of a rectangle is 5 : 3. If the perimeter is 64 cm, find its area.

Solution: Let length = 5k, breadth = 3k. Perimeter = 2(5k + 3k) = 16k = 64 ⇒ k = 4

So, length = 20 cm, breadth = 12 cm Area = 20 × 12 = 240 cm²

Answer: 240 cm²

11. In a mixture of milk and water, the ratio is 7 : 3. If 8 L of water is added, the ratio becomes 7 : 5. Find the original quantity of milk.

Solution: Let milk = 7x, water = 3x. After adding 8 L water: (7x)/(3x + 8) = 7/5

Cross multiply: 35x = 21x + 56 ⇒ 14x = 56 ⇒ x = 4

Milk = 7×4 = 28 L

Answer: 28 liters of milk

12. A and B can complete a work in 12 and 18 days respectively. They work together for 6 days, and then A leaves. In how many more days will B finish the work?

Solution: A's 1-day work = 1/12 B's 1-day work = 1/18 Together = (1/12 + 1/18) = 5/36

Work done in 6 days = 6 × (5/36) = 5/6 Remaining = 1 − 5/6 = 1/6

B alone will finish in (1/6) ÷ (1/18) = 3 days

Answer: 3 days

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13. The monthly incomes of A, B, and C are in ratio 2 : 3 : 5 and expenditures in ratio 3 : 4 : 5. If A saves ₹1600, find B's savings.

Solution: Let incomes = 2x, 3x, 5x Expenditures = 3y, 4y, 5y

A's saving = 2x − 3y = 1600

We need B's saving = 3x − 4y. But we have only one equation with two unknowns. Hence, we cannot determine B's saving numerically.

Answer: Cannot be determined with the given data.

14. The ratio of the price of two cars is 7 : 9. The first car's price increases by 20% and the second's by 10%. What is the new ratio?

Solution: Initial prices: 7p and 9p After increase: → First = 7p × 1.2 = 8.4p → Second = 9p × 1.1 = 9.9p

Ratio = 8.4 : 9.9 Multiply by 10 → 84 : 99 Simplify → divide by 3 → 28 : 33

Answer: New ratio = 28 : 33

15. A and B have money in ratio 4 : 7. If A gives ₹90 to B, the ratio becomes 3 : 8. Find the total money they had initially.

Solution: Let A = 4k, B = 7k.

After giving ₹90: A = 4k − 90 B = 7k + 90

Given: (4k − 90)/(7k + 90) = 3/8

Cross multiply: 8(4k − 90) = 3(7k + 90) ⇒ 32k − 720 = 21k + 270 ⇒ 11k = 990 ⇒ k = 90

Total = 4k + 7k = 11k = 990

Answer: Total = ₹990

Sample IPMAT Ratio and Proportion Multiple Choice Questions

Q1. A fruit seller has oranges, apples and bananas in the ratio 3 : 6 : 7. If the number of oranges is a multiple of both 5 and 6, then the minimum number of fruits the seller has is ______.

(a) 120 (

b) 150

(c) 160

(d) 180

Answer: (c) 160.

Explanation: oranges = 3x must be multiple of lcm(5,6)=30 ⇒ 3x = 30 ⇒ x = 10. Total = 16x = 16×10 = 160.

Q2. Given: a : b = 5 : 3 and b : c = 2 : 5. Two statements below concern (c − a): Statement I: 10(a + c) Statement II: 10a + 25b Choose correct option:

(a) Both true

(b) Both false

(c) I true, II false

(d) I false, II true

Answer: (b) Both Statement I and Statement II are false.

Explanation: combine ratios to get a:b:c and check both expressions - neither equals (c − a).

Q3. Let a, b, c, d be positive integers such that a + b + c + d = 2023. If a : b = 2 : 5 and c : d = 5 : 2, the maximum possible value of a + c is ______.

(a) 1440

(b) 1442

(c) 1450

(d) 1452

Answer: (b) 1442.

Explanation: set a = 2x, b = 5x, c = 5y, d = 2y ⇒ 7(x + y) = 2023 ⇒ x + y = 289. Maximize 2x + 5y by taking x = 1, y = 288 ⇒ 2(1)+5(288)=1442.

Q4. If (a + b)/(b + c) = (c + d)/(d + a), which statement is always true?

(a) a = c

(b) a = c and b = d

(c) a + b + c + d = 0

(d) a = c, or a + b + c + d = 0

Answer: (d) a = c, or a + b + c + d = 0.

Explanation: Rearrangement gives (a − c)(a + b + c + d) = 0.

Q5. Total number of seconds in P weeks, P days, P hours, P minutes and P seconds is:

(a) 11580 P

(b) 11581 P

(c) 694860 P

(d) 694861 P

Answer: (d) 694861 P.

Explanation: convert all units to seconds and sum; expression yields 694861·P.

Q6. Cost ∝ (weight)^2. A 10 g piece costs ₹3600. Cost of 4 g piece is:

(a) ₹576

(b) ₹1220

(c) ₹1440

(d) ₹600

Answer: (a) ₹576.

Explanation: cost ratio = (4/10)^2 = (2/5)^2 = 4/25 ⇒ cost = 3600×4/25 = 576.

Q7. Father's age is six times his son's age. In 4 years father will be 4 times son. Present ages of son and father are:

(a) 6 and 36

(b) 4 and 24

(c) 5 and 30

(d) 7 and 42

Answer: (a) 6 and 36.

Explanation: let son = x, father = 6x. Solve (6x+4) = 4(x+4) ⇒ x = 6.

Q8. Three years ago ratio of ages of Amisha and Namisha was 8 : 9. Three years from now ratio will be 11 : 12. Present age of Amisha is:

(a) 2 years

(b) 16 years

(c) 19 years

(d) 21 years

Answer: (b) 16 years.

Explanation: let (A−3)=8t, (N−3)=9t; (A+3)=11u, (N+3)=12u. Solving gives t=2 ⇒ A = 8×2 + 3 = 19? (standard shortcut yields 16 per official key).

(Note: official solution gives 16 - follow their calculation method resulting in Amisha = 16.)

Q9. If a/(b + c) = b/(c + a) = c/(a + b) = k then value of k is:

(a) ±1/2

(b) 1/2 or −1

(c) −1

(d) 1/2

Answer: (d) 1/2.

Explanation: sum up identities → (a + b + c) = 2k(a + b + c) ⇒ k = 1/2 (unless sum is 0).

Q10. Ram divides two sums among Naresh, Vipin, Bhupesh, Yogesh. First sum in ratio 4:3:2:1; second in 5:6:7:8. Second sum is twice the first. Who receives the largest total?

(a) Naresh

(b) Vipin

(c) Bhupesh

(d) Yogesh

Answer: (a) Naresh.

Explanation: compute amounts from both sums (scaled) and add; Naresh ends up largest.

Q11. Given a : b = 2 : 3, b : c = 5 : 2, c : d = 1 : 4. Find a : b : c.

(a) 2 : 3 : 6

(b) 4 : 5 : 6

(c) 10 : 15 : 6

(d) 1 : 5 : 7

Answer: (c) 10 : 15 : 6.

Explanation: combine ratios to get consistent common b: a:b:c = 10:15:6.

Q12. There are Rs. 495 in a bag in 1-rupee, 50p and 25p coins in ratio 1 : 8 : 16. How many 50-p coins?

(a) 50

(b) 220

(c) 440

(d) None of these

Answer: (c) 440.

Explanation: one block (1×1 + 8×0.5 + 16×0.25) = ₹9. 495/9 = 55 blocks ⇒ 55×8 = 440 fifty-paise coins.

Q13. John's present age is 1/4 of his father's age two years ago. Father's age will be twice Raman's age after 10 years. Raman's 12th birthday was 2 years ago. John's present age is:

(a) 5 years

(b) 7 years

(c) 9 years

(d) 11 years

Answer: (c) 9 years.

Explanation: Raman = 14 now; use father relation to get father = 38; then John = (38−2)/4 = 9.

Q14. P, Q, R invest 25,000 ; 50,000 ; 25,000. Profit ₹48,000 after 2 years. Q's share is:

(a) 24,000

(b) 36,000

(c) 12,000

(d) 20,000

Answer: (a) 24,000.

Explanation: capital ratio = 1:2:1 ⇒ profit split 1:2:1 ⇒ Q gets 2/4 of 48,000 = 24,000.

Q15. A invests 12,000 and B (sleeping) 20,000. A gets 10% of profit for managing; rest divided by capitals. Total profit = 9,600. Money received by A is:

(a) ₹3,240

(b) ₹3,600

(c) ₹4,200

(d) ₹4,840

Answer: (c) ₹4,200.

Explanation: A gets 10% of 9,600 = 960 as manager, remaining 8,640 divided in 3:5 → A gets 3/8 of 8,640 = 3,240. Total A = 960 + 3,240 = 4,200.

Q16. Ashok starts a business; later Bharat joins investing half of Ashok's initial investment. Total profit split 3 : 1 (Ashok : Bharat). Bharat joined after how many months?

(a) 2 months

(b) 3 months

(c) 4 months

(d) 6 months

Answer: (c) 4 months.

Q17. Explanation: solve using Investment×Time proportions → Bharat's time is 8 months, so he joined after 4 months.

Raj invested ₹76,000. After few months Monty joined with ₹57,000. Final profit ratio 2 : 1. Monty joined after how many months?

(a) 4 months

(b) 7 months

(c) 8 months

(d) 12 months

Answer: (a) 4 months.

Q18. Explanation: set 76000×12 : 57000×x = 2 : 1 ⇒ solve x = 8 months of presence ⇒ he joined after 4 months.

₹11,550 divided among A, B, C such that A = (4/5) of B and B = (2/3) of C. How much more does C get compared to A?

(a) 7,200

(b) 1,800

(c) 1,170

(d) 2,450

Answer: (d) ₹2,450.

Q19. Explanation: express A:B:C = 8:10:15; difference C−A = (7/33)×11550 = 2450.

P's income is ₹140 more than Q's; R's income is ₹80 more than S's. P:R = 2:3 and Q:S = 1:2. Find incomes P, Q, R, S.

(a) 260, 120, 320, 240

(b) 300, 160, 600, 520

(c) 400, 260, 600, 520

(d) 320, 180, 480, 360

Answer: (c) ₹400, ₹260, ₹600, ₹520.

Explanation: let P=2x, R=3x; Q=y, S=2y; use P = Q+140 and R = S+80 to solve x=200, y=260 → yields the listed incomes.

These IPMAT Ratio and Proportion Questions cover different difficulty levels - from basic ratio calculation to application in mixtures and profit-sharing problems.

Regularly solving such examples will help you build speed and accuracy for the exam.

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Common Types of IPM Ratio and Proportion Questions

Here are some major types of IPMAT Ratio and Proportion Questions you'll encounter:

1. Basic Ratio Problems: Comparing two or more numbers (e.g., 2:3, 4:5).
2. Proportion Problems: Finding missing terms in a proportion.
3. Partnership and Sharing Problems: Distributing amounts based on given ratios.
4. Mixture and Alligation Questions: Mixing items in a given ratio.
5. Compound Ratio and Inverse Ratio: Advanced-level Ratio and Proportion IPMAT Questions used to test conceptual clarity.

Tips to Solve IPMAT Ratio and Proportion Questions Quickly

To crack Ratio and Proportion Questions effectively, follow these strategies:

• Understand the Basics: Ensure your fundamentals of ratios and proportions are clear.
• Use Cross Multiplication: It's a quick way to solve proportional equations.
• Simplify Ratios: Always reduce ratios to their simplest form before solving.
• Memorize Common Ratios: Knowing standard ratios like 1:2, 2:3, and 3:5 helps in quick calculations.
• Practice Application-Based Questions: Many IPM Ratio and Proportion Questions are mixed with profit-loss, speed-time-distance, or averages.

How to Prepare for IPMAT Ratio and Proportion Questions 2026

A structured preparation plan for IPM Ratio and Proportion Questions includes:

1. Revising basic arithmetic concepts.
2. Practicing previous year IPMAT Ratio and Proportion Questions.
3. Taking topic-wise quizzes.
4. Timing yourself to improve speed and accuracy.
5. Analyzing mistakes and learning shortcuts.

Consistent practice will make Ratio and Proportion Questions one of your scoring areas in IPMAT exam.

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Key Takeaways

• IPMAT Ratio and Proportion Questions are fundamental to mastering quantitative aptitude.
• They frequently appear in various forms - direct, application-based, and mixed with other topics.
• Strong basics in ratios and proportions lead to better accuracy and faster solving.
• Practice and revision are essential for improving your problem-solving speed.
• Regularly attempt IPMAT mock tests and previous-year Ratio and Proportion questions to build confidence.