What Are IPMAT Linear Equations Questions?

Overview: Linear equations are one of the most predictable and high-utility areas in the IPMAT Quantitative Ability section. IPMAT linear equations questions test how comfortably a candidate can interpret relationships, form equations, and solve them logically within a limited time.

Linear equations form a key part of the Quantitative Ability section in IPMAT.

These questions test how well a student can apply basic algebra to real situations, often appearing in topics like ratios, ages, or time and work.

IPMAT linear equations questions are usually simple in concept but require clarity and speed in execution.

Since they appear almost every year and are quick to solve, they're considered one of the most reliable scoring areas in the exam.

This article highlights the pattern and approach for solving these questions, along with sample problems and solutions to strengthen your preparation.

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Importance of IPMAT Linear Equations Questions

Why should you focus on Linear Equations during your IPMAT preparation?

1. They are conceptually simple yet frequently asked.
2. They appear as independent algebra problems and within word-based arithmetic questions.
3. Solving IPMAT Linear Equations Questions helps you quickly translate English statements into mathematical form.
4. Mastery here boosts performance in related topics like Ratio-Proportion, Ages, and Mixtures.

Expected Weightage: On average, 2-4 questions in every IPMAT Quant section are based directly or indirectly on Linear Equations.

Sample IPMAT Linear Equations Questions And Answer

The following are carefully curated IPMAT linear equations questions created to match the difficulty and pattern of IPMAT Indore and IPMAT Rohtak exams.

Each set includes a mix of direct, word-based, and application-based linear problems.

Multiple Choice Questions (MCQs)

Q1. If 5x−7=185x - 7 = 185x−7=18, the value of xxx is:

A) 4

B) 5

C) 6

D) 7

Answer: B) 5

Explanation: 5x=25⇒x=55x = 25 \ ⇒x = 55x=25⇒x=5

Q2. If 2x+3=5x−92x + 3 = 5x - 92x+3=5x−9, then x=?x = ?x=?

A) 2

B) 3

C) 4

D) 5

Answer: C) 4

Explanation: 3+9=3x⇒x=43 + 9 = 3x \⇒x = 43+9=3x⇒x=4

Q3. The sum of two numbers is 35 and their difference is 7. The greater number is:

A) 14

B) 21

C) 22

D) 24

Answer: B) 21

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Explanation: x+y=35,x−y=7⇒2x=42⇒x=21x + y = 35, x - y = 7 \⇒2x = 42 \⇒x = 21x+y=35,x−y=7⇒2x=42⇒x=21

Q4. If 3x+2y=123x + 2y = 123x+2y=12 and 4x−y=54x - y = 54x−y=5, what is the value of xxx?

A) 1

B) 2

C) 3

D) 4

Answer: B) 2

Explanation: Substituting y=4x−5y = 4x - 5y=4x−5 gives 3x+8x−10=12⇒x=23x + 8x - 10 = 12 \⇒x = 23x+8x−10=12⇒x=2

Q5. If x3+x4=14\frac{x}{3} + \frac{x}{4} = 143x​+4x​=14, find xxx:

A) 12

B) 18

C) 24

D) 28

Answer: C) 24

Explanation: LCM = 12 → 7x/12=14⇒x=247x/12 = 14 \⇒x = 247x/12=14⇒x=24

Q6. A's present age is twice B's. Ten years ago, A was four times B's age. Find B's present age.

A) 10

B) 12

C) 15

D) 20

Answer: C) 15

Explanation: 2x−10=4(x−10)⇒x=152x - 10 = 4(x - 10) \⇒x = 152x−10=4(x−10)⇒x=15

Q7. Two numbers are in the ratio 4:5, and their sum is 72. Find the numbers.

A) 24, 48

B) 32, 40

C) 36, 45

D) 28, 35

Answer: C) 36, 45 Explanation: 4k+5k=72⇒k=8⇒32,404k + 5k = 72 \⇒k = 8 \⇒32, 404k+5k=72⇒k=8⇒32,40

Q8. If 7x−2(3x−4)=167x - 2(3x - 4) = 167x−2(3x−4)=16, find xxx:

A) 6

B) 7

C) 8

D) 9

Answer: C) 8

Explanation: 7x−6x+8=16⇒x=87x - 6x + 8 = 16 \⇒x = 87x−6x+8=16⇒x=8

Q9. The perimeter of a rectangle is 40 cm, and its length is 4 cm more than its breadth. The breadth is:

A) 7 cm

B) 8 cm

C) 9 cm

D) 10 cm

Answer: B) 8 cm

Explanation: 2(l+b)=40⇒l+b=20⇒b=8,l=122(l + b) = 40 \⇒l + b = 20 \⇒b = 8, l = 122(l+b)=40⇒l+b=20⇒b=8,l=12

Q10. A and B together have ₹1500. A has twice as much as B. What is A's share?

A) ₹400

B) ₹600

C) ₹900

D) ₹1000

Answer: C) ₹1000

Explanation: A=2B⇒2B+B=1500⇒B=500,A=1000A = 2B \⇒2B + B = 1500 \⇒B = 500, A = 1000A=2B⇒2B+B=1500⇒B=500,A=1000

Short Answer Type Questions (SA)

Q1. Solve for xxx: 4x−3=134x - 3 = 134x−3=13

Solution: 4x=16⇒x=44x = 16 \⇒x = 44x=16⇒x=4

Answer: x=4x = 4x=4

Q2. If 2x+y=102x + y = 102x+y=10 and x−y=2x - y = 2x−y=2, find yyy.

Solution:

From the second equation: x=y+2x = y + 2x=y+2

Substitute in the first: 2(y+2)+y=10⇒3y+4=10⇒y=22(y + 2) + y = 10 \⇒3y + 4 = 10 \⇒y = 22(y+2)+y=10⇒3y+4=10⇒y=2

Answer: y=2y = 2y=2

Q3. The cost of 5 pens and 3 pencils is ₹45. The cost of 3 pens and 2 pencils is ₹29. Find the cost of each pen.

Solution:

Let the cost of a pen = ppp, and pencil = ccc

Equations: 5p+3c=455p + 3c = 455p+3c=45 …(1) 3p+2c=293p + 2c = 293p+2c=29 …(2)

Multiply (2) by 3 and (1) by 2 for elimination: 15p+9c=13515p + 9c = 13515p+9c=135 6p+4c=586p + 4c = 586p+4c=58

Subtract → 9p+5c=779p + 5c = 779p+5c=77

Now solve directly → p=7.5p = 7.5p=7.5

Answer: ₹7.5 per pen

Q4. If 7(x+2)=3(2x+5)7(x + 2) = 3(2x + 5)7(x+2)=3(2x+5), find xxx.

Solution:

Expand both sides: 7x+14=6x+157x + 14 = 6x + 157x+14=6x+15 ⇒7x−6x=15−14⇒x=1\⇒7x - 6x = 15 - 14 \⇒x = 1⇒7x−6x=15−14⇒x=1

Answer: x=1x = 1x=1

Q5. If the sum of two consecutive odd numbers is 56, find the numbers.

Solution:

Let the numbers be xxx and x+2x + 2x+2. Then x+x+2=56⇒2x=54⇒x=27x + x + 2 = 56 \⇒2x = 54 \⇒x = 27x+x+2=56⇒2x=54⇒x=27.

Hence, the numbers are 27 and 29.

Answer: 27 and 29

Q6. A train travels 300 km at a uniform speed. If the speed were 10 km/h more, it would take 1 hour less. Find the original speed.

Solution:

Let the speed = xxx km/h.

Time = 300/x300/x300/x.

New speed = x+10x + 10x+10, new time = 300/(x+10)300/(x + 10)300/(x+10).

Given: 300x−300x+10=1\frac{300}{x} - \frac{300}{x + 10} = 1x300​−x+10300​=1

Simplify: 300[(x+10−x)/(x(x+10))]=1⇒3000=x(x+10)300[(x + 10 - x)/(x(x + 10))] = 1 \⇒3000 = x(x + 10)300[(x+10−x)/(x(x+10))]=1⇒3000=x(x+10) ⇒x2+10x−3000=0\⇒x^2 + 10x - 3000 = 0⇒x2+10x−3000=0 ⇒(x+60)(x−50)=0⇒x=50\⇒(x + 60)(x - 50) = 0 \⇒x = 50⇒(x+60)(x−50)=0⇒x=50

Answer: 50 km/h

Q7. A and B can complete a piece of work together in 8 days. A person alone can do it in 12 days. In how many days can B alone finish the work?

Solution:

Work done per day: A = 1/121/121/12, A + B = 1/81/81/8. So, 1/B=1/8−1/12=(3−2)/24=1/241/B = 1/8 - 1/12 = (3 - 2)/24 = 1/241/B=1/8−1/12=(3−2)/24=1/24.

Answer: B alone = 24 days

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Q8. If 3(x+4)=5(x−2)3(x + 4) = 5(x - 2)3(x+4)=5(x−2), find xxx.

Solution: 3x+12=5x−103x + 12 = 5x - 103x+12=5x−10 ⇒22=2x⇒x=11\⇒22 = 2x \⇒x = 11⇒22=2x⇒x=11

Answer: x=11x = 11x=11

Q9. The sum of three consecutive even numbers is 78. Find the numbers.

Solution: Let the numbers be xxx, x+2x + 2x+2, and x+4x + 4x+4.

Then x+x+2+x+4=78⇒3x+6=78⇒x=24x + x + 2 + x + 4 = 78 \⇒3x + 6 = 78 \⇒x = 24x+x+2+x+4=78⇒3x+6=78⇒x=24.

Hence, the numbers are 24, 26, and 28.

Answer: 24, 26, 28

Q10. If x+1x=5x + \frac{1}{x} = 5x+x1​=5, find x2+1x2x^2 + \frac{1}{x^2}x2+x21​.

Solution: Square both sides: (x+1x)2=52⇒x2+1x2+2=25(x + \frac{1}{x})^2 = 5^2 \⇒x^2 + \frac{1}{x^2} + 2 = 25(x+x1​)2=52⇒x2+x21​+2=25. ⇒x2+1x2=23\⇒x^2 + \frac{1}{x^2} = 23⇒x2+x21​=23

Answer: 23

Types of IPMAT Linear Equations Questions

Let's understand the significant patterns of IPMAT Linear Equations Questions that appear in the exam:

Type	Description	Example
Type 1 - Direct Linear Equations	Simple equations in one variable.	5x−3=225x - 3 = 225x−3=22
Type 2 - Simultaneous Equations	Two equations with two variables to solve simultaneously.	3x+2y=12,4x−y=53x + 2y = 12, 4x - y = 53x+2y=12,4x−y=5
Type 3 - Word Problems	Translate real-life situations into equations.	"A's age is twice B's; ten years ago, A was four times B."
Type 4 - Ratio or Fractional Forms	Linear relations are expressed using ratios or fractions.	x3+x4=14\frac{x}{3} + \frac{x}{4} = 143x​+4x​=14
Type 5 - Application-Based Equations	Linear relations in work, time, or mixture problems.	"A and B together can finish work in 10 days…"

Each type tests conceptual clarity and equation-forming ability - both crucial for scoring well in IPMAT Linear Equations Questions.

Trends of IPMAT Linear Equations Questions (2021-2025)

Here's how Linear Equations have appeared across recent IPMAT exams:

Year	Exam	No. of Questions	Difficulty	Common Pattern
2025	IPMAT Indore &Rohtak	2-3	Moderate	Direct, word-based, and application-based
2024	IPMAT Indore	2-3	Easy	Word-based + Ratio form
2023	IPMAT Rohtak	3	Easy-Moderate	Simultaneous equations
2022	JIPMAT	2	Easy	Direct + Fractional
2021	IPMAT Indore	2	Moderate	Age and Mixture Problems

Analysis: Across all papers, IPMAT Linear Equations Questions remain among the most predictable and high-accuracy areas for candidates.

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Key Takeaways

• Consistent Presence: Linear Equations appear every year in IPMAT, usually contributing 2-4 questions.
• Conceptual Clarity Pays: Questions are rooted in school-level algebra but demand quick logical translation.
• Question Variety: Includes single-variable, two-variable, and applied word-based equations.
• Quick to Solve: Most can be completed in under a minute with proper practice.
• Reliable Scoring Area: Accuracy is high, and mistakes are minimal when concepts are strong.
• Preparation Focus: Strengthen basics, practice word problems daily, and revise standard equation forms.