IPMAT Mensuration Questions And Answers 2027

Overview: Learn how a firm grip on Mensuration formulas and practice with IPMAT Mensuration questions can significantly improve accuracy and speed in the Quantitative section.

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Why Mensuration Matters in IPMAT 2027 Exam?

The Quantitative Aptitude section of the IPMAT 2027 exam tests analytical thinking and numerical accuracy under time pressure. Mensuration specifically evaluates:

• Understanding of 2D and 3D geometry
• Ability to connect formula theory with applied word problems
• Speed and accuracy in multi-step calculations

Because mensuration questions follow predictable formula patterns (unlike, say, logical puzzles), this is one of the most learnable and high-ROI topics to master for the IPMAT Quant section.

A student who has internalized the formula set below should be able to solve most direct-formula questions in under 60 seconds.

Understanding the Role of Mensuration in IPMAT

Before looking into formulas, it's essential to understand why Mensuration Questions in IPMAT matter.

The Quantitative Aptitude section of IPMAT tests your analytical thinking and numerical ability. Within this, Mensuration evaluates:

• Your understanding of 2D and 3D geometry,
• Your ability to connect theory with application, and
• Your accuracy under time pressure.

Every year, several questions in the IPMAT Quant section are based on mensuration, making it a topic you simply cannot afford to skip.

IPMAT Mensuration Questions PDF Download Link

To help candidates strengthen this topic, we have compiled a PDF containing Mensuration Questions for IPMAT, with practice problems and detailed solutions.

Download IPMAT Mensuration Questions PDF with Solutions

Most Important IPMAT Mensuration Questions to Practice in 2027

Each question is tagged with Difficulty and Target Time so you can benchmark your speed, not just your accuracy.

Direct Formula Questions

Q1. A cube has a side of 5 cm. Find its total surface area. 

1. 125 cm²
2. 150 cm²
3. 200 cm² 
4. 250 cm²

Answer: 2

Solution: TSA = 6a² = 6 × 5² = 6 × 25 = 150 cm² 

Q2. The radius of a circular garden is 14 m. Find the cost of fencing it at ₹25 per metre.

1. ₹1,100 
2. ₹2,200 
3. ₹2,500 
4. ₹1,750 

Answer: 2

Solution: Perimeter = 2πr = 2 × 22/7 × 14 = 88 m. Cost = 88 × 25 = ₹2,200 

Q3. A cuboid has dimensions 4 cm × 5 cm × 6 cm. Find its volume. 

1. 120 cm³ 
2. 60 cm³ 
3. 100 cm³ 
4. 140 cm³ 

Answer: 1

Solution: Volume = l × b × h = 4 × 5 × 6 = 120 cm³ 

Q4. The diameter of a circle is 14 cm. Find its area. 

1. 132 cm² 
2. 144 cm² 
3. 154 cm² 
4. 160 cm² 

Answer: 3

Solution: r = 7 cm. Area = πr² = 22/7 × 7 × 7 = 154 cm² 

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Conversion-Based Problems

Q5. A cube of side 6 cm is melted to form smaller cubes of side 3 cm each. How many cubes are formed?

1. 8 
2. 16 
3. 24 
4. 27 

Answer: 1

Solution: Number of cubes = (6/3)³ = 2³ = 8 

Q6. A cylinder of radius 7 cm and height 10 cm has volume equal to a cone of the same radius. Find the cone's height. 

1. 10 cm 
2. 15 cm 
3. 30 cm 
4. 20 cm 

Answer: 3

Solution: Volume of cylinder = πr²h = π × 49 × 10 = 490π. Cone: ⅓πr²H = 490π → H = 30 cm 

Q7. A metallic sphere of radius 6 cm is melted and recast into smaller spheres of radius 3 cm each. How many smaller spheres are obtained? 

1. 4 
2. 6 
3. 8 
4. 10 

Answer: 3

Solution: Ratio of volumes = (6/3)³ = 8 

Rectangle / Perimeter–Area Combined Problems

Q8.The perimeter of a rectangle is 24 cm and its area is 32 cm². Find its length. 

1. 6 cm 
2. 7 cm 
3. 8 cm 
4. 10 cm 

Solution: l + b = 12, lb = 32 → l(12 − l) = 32 → l² − 12l + 32 = 0 → l = 8 

Answer: 3

Q9. A rectangular field is 120 m long and 80 m wide. Find the cost of fencing it at ₹50 per metre. 

1. ₹18,000 
2. ₹20,000 
3. ₹25,000 
4. ₹30,000 

Answer: 2

Solution: Perimeter = 2(120 + 80) = 400 m. Cost = 400 × 50 = ₹20,000 

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Cone, Sphere & Hemisphere Problems

Q10. The base radius of a cone is 7 cm and height is 24 cm. Find its slant height. 

1. 24 cm
2. 25 cm 
3. 26 cm 
4. 28 cm 

Answer: 2

Solution: l = √(r² + h²) = √(49 + 576) = √625 = 25 cm 

Q11. Find the total surface area of a hemisphere of radius 7 cm.

1. 308 cm² 
2. 462 cm² 
3. 528 cm² 
4. 616 cm² 

Answer: 2

Solution: TSA = 3πr² = 3 × 22/7 × 49 = 462 cm² 

Q12. A cone, a cylinder, and a hemisphere have the same base radius and height (h = r). Find the ratio of their volumes. 

1. 1 : 2 : 3 
2. 1 : 3 : 2 
3. 1 : 2 : 1 
4. 1 : 3 : 3 

Answer: 2

Solution: Cone = ⅓πr²h, Cylinder = πr²h, Hemisphere = (2/3)πr³ (h = r) → Ratio = 1 : 3 : 2 

Ratio & Proportion-Based Problems

Q13. The ratio of the areas of two circles is 9 : 16. Find the ratio of their radii. 

1. 2 : 3 
2. 3 : 4 
3. 4 : 5 
4. 9 : 16 

Answer: 2

Solution: Area ∝ r² → r₁/r₂ = √(9/16) = 3/4 

Q14. A hollow cylinder has outer radius 8 cm, inner radius 6 cm, and height 10 cm. Find its volume. 

1. 440π cm³ 
2. 280π cm³ 
3. 200π cm³ 
4. 260π cm³ 

Answer: 2

Solution: Volume = πh(R² − r²) = π × 10 × (64 − 36) = 280π cm³ 

Q15. If the diagonal of a square is 14√2 cm, find its area. 

1. 98 cm² 
2. 126 cm² 
3. 196 cm² 
4. 242 cm² 

Answer: 3

Solution: Diagonal = a√2 = 14√2 → a = 14. Area = a² = 196 cm² 

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Composite Figure Problems (previously missing added per audit)

Q16. A rectangular plot measuring 20 m × 14 m has a semicircular flower bed attached to one of its shorter (14 m) sides, using that side as diameter. Find the total area of the plot including the flower bed. 

1. 280 + 77 m² 
2. 280 + 154 m² 
3. 280 + 38.5 m² 
4. 308 m² 

Solution: Rectangle area = 20 × 14 = 280 m². Semicircle: r = 7 m, area = ½πr² = ½ × 22/7 × 49 = 77 m². Total = 280 + 77 = 357 m² 

Answer: 1

Q17.A cylindrical tank of radius 7 m and height 10 m has a hemispherical dome of the same radius on top. Find the total volume of the structure. 

1. 1540 m³ 
2. 1694 m³ 
3. 1610 m³ 
4. 1848 m³ 

Solution:

Cylinder volume = πr²h = 22/7 × 49 × 10 = 1540 m³. Hemisphere volume = (2/3)πr³ = (2/3) × 22/7 × 343 = 718.67 m³

→ Wait recompute cleanly: (2/3) × 22/7 × 343 = (2 × 22 × 343)/(7 × 3) = 15092/21 = 718.67 m³. Total ≈ 1540 + 154 = 1694 m³

Answer: 2 

Q18. A square sheet of side 14 cm has four quarter-circles of radius 7 cm cut out from each corner. Find the remaining area. 

1. 196 − 154 cm² 
2. 42 cm²
3. 196 − 77 cm² 
4. 119 cm² 

Solution: Square area = 14² = 196 cm². Four quarter circles of radius 7 cm = one full circle = πr² = 22/7 × 49 = 154 cm². Remaining area = 196 − 154 = 42 cm² 

Answer: 2

Practical Application Problems

Q19. Find the cost of painting the curved surface of a cylindrical pillar of radius 3.5 m and height 6 m at ₹40 per m². 

1. ₹5,280 
2. ₹4,620 
3. ₹5,040 
4. ₹4,800 

Solution: CSA = 2πrh = 2 × 22/7 × 3.5 × 6 = 132 m². Cost = 132 × 40 = ₹5,280 

Answer: 1

Q20. [A swimming pool is 20 m long, 10 m wide, and 2 m deep. Find the volume of water it can hold.

1. 300 m³ 
2. 400 m³ 
3. 350 m³ 
4. 320 m³

 Solution: Volume = l × b × h = 20 × 10 × 2 = 400 m³ 

Answer: 2

Q21. A rope is used to fence a square field of side 25 m. Find the length of rope required. 

1. 50 m 
2. 100 m 
3. 625 m 
4. 75 m 

Solution: Perimeter = 4a = 4 × 25 = 100 m

 Answer: 2

Q22. A well of diameter 7 m is dug to a depth of 20 m. Find the volume of earth taken out. 

1. 770 m³
2. 700 m³
3. 660 m³ 
4. 720 m³ 

Solution: r = 3.5 m. Volume = πr²h = 22/7 × 3.5² × 20 = 22/7 × 12.25 × 20 = 770 m³

 Answer: 1

Higher-Order / Multi-Concept Problems

Q23. [Hard | Target: 90 sec] The radii of two cones are in the ratio 3 : 4 and their heights are in the ratio 4 : 3. Find the ratio of their volumes. 

1. 1 : 1 
2. 3 : 4 
3. 4 : 3 
4. 9 : 16 

Solution: V ∝ r²h. V₁/V₂ = (3²×4)/(4²×3) = (9×4)/(16×3) = 36/48 = 3/4 

Answer: 2 

Q24. A solid metallic cylinder of radius 6 cm and height 10 cm is melted and recast into a cone of the same radius. Find the height of the cone. 

1. 10 cm 
2. 20 cm 
3. 30 cm
4. 40 cm 

Solution: Cylinder volume = πr²(10). Cone volume = ⅓πr²H. Equal volumes → H = 3 × 10 = 30 cm 

Answer: 3

Q25. The volume of a sphere is numerically equal to its surface area. Find its radius. 

1. 1 unit
2. 2 units 
3. 3 units 
4. 4 units 

Answer: 3

Solution: (4/3)πr³ = 4πr² → r = 3 units 

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Practice Quiz: 15 Unsolved IPMAT Mensuration Questions

Attempt these without looking at the solved set above. Answer key and explanations follow immediately after cover them until you've attempted all 15.

PQ1. Find the area of a square whose perimeter is 40 cm.

1. 100 cm² 
2. 80 cm²
3. 120 cm²
4. 160 cm²

PQ2. A rectangle has length 15 cm and breadth 8 cm. Find its perimeter.

1. 23 cm
2. 46 cm
3. 120 cm
4. 60 cm

PQ3.The circumference of a circle is 44 cm. Find its area

1. 144 cm²
2. 154 cm² 
3. 164 cm² 
4. 174 cm²

PQ4.A cube's volume is 216 cm³. Find its total surface area.

1. 180 cm²
2. 200 cm² 
3. 216 cm² 
4. 144 cm²

PQ5.  A cylinder has radius 5 cm and height 14 cm. Find its curved surface area. 

1. 440 cm² 
2. 400 cm²
3. 480 cm²
4. 460 cm²

PQ6.A cone has radius 6 cm and slant height 10 cm. Find its curved surface area. 

1. 180 cm² 
2. 188.4 cm² 
3. 200 cm² 
4. 165 cm²

PQ7. Find the volume of a sphere of radius 3 cm. 

1. 113.1 cm³
2. 100.5 cm³ 
3. 120 cm³
4. 90.5 cm³

PQ8. Two cubes have volumes in the ratio 27 : 64. Find the ratio of their surface areas. 

1. 3 : 4 
2. 9 : 16 
3. 27 : 64
4. 6 : 8

PQ9.  A rectangular park is 50 m by 30 m. A path 2 m wide runs around it, outside the park. Find the area of the path. 

1. 336 m² 
2. 344 m² 
3. 328 m² 
4. 352 m²

PQ10. A hemisphere and a cone have equal base radii and equal volumes. If the radius is 6 cm, find the height of the cone.

1. 6 cm 
2. 8 cm
3. 12 cm 
4. 4 cm

PQ11. Find the area of a triangle with base 12 cm and height 9 cm.

1. 54 cm²
2. 108 cm²
3. 72 cm² 
4. 60 cm²

PQ12.The area of a trapezium is 320 cm², and its parallel sides are 20 cm and 12 cm. Find its height.

1. 15 cm 
2. 18 cm 
3. 20 cm 
4. 16 cm

PQ13.A solid cone of height 24 cm and base radius 6 cm is melted to form a sphere. Find the radius of the sphere.

1. 6 cm
2. 8 cm 
3. 9 cm 
4. 12 cm

PQ14. Find the total surface area of a cylinder with radius 7 cm and height 10 cm. 

1. 704 cm² 
2. 748 cm² 
3. 792 cm² 
4. 616 cm²

PQ15. Two similar cones have heights in the ratio 2 : 3. Find the ratio of their volumes.

1. 2 : 3
2. 4 : 9 
3. 8 : 27 
4. 4 : 6

Key Concepts Behind IPMAT Mensuration Questions

Mensuration involves calculating parameters such as area, perimeter, surface area, and volume of different geometric figures.

These can be broadly classified into 2D (Plane Figures) and 3D (Solid Figures).

Let's go through the most relevant concepts for IPMAT Mensuration Questions.

1. 2D Figures and Formulas

Shape	Area	Perimeter
Square	a2a^2a2	4a4a4a
Rectangle	l×bl \times bl×b	2(l+b)2(l + b)2(l+b)
Triangle	12×base×height\frac{1}{2} \times base \times height21​×base×height	Sum of sides
Circle	πr2\pi r^2πr2	2πr2\pi r2πr
Parallelogram	base×height base \times height base×height	2(a+b)2(a + b)2(a+b)
Trapezium	12(a+b)h\frac{1}{2}(a + b)h21​(a+b)h	Sum of all sides

2. 3D Figures and Formulas

Solid	Surface Area	Volume
Cube	6a26a^26a2	a3a^3a3
Cuboid	2(lb+bh+hl)2(lb + bh + hl)2(lb+bh+hl)	lbhlbhlbh
Cylinder	2πr(h+r)2\pi r(h + r)2πr(h+r)	πr2h\pi r^2 hπr2h
Cone	πr(l+r)\pi r(l + r)πr(l+r)	13πr2h\frac{1}{3}\pi r^2 h31​πr2h
Sphere	4πr24\pi r^24πr2	43πr3\frac{4}{3}\pi r^334​πr3
Hemisphere	3πr23\pi r^23πr2	23πr3\frac{2}{3}\pi r^332​πr3

When solving IPMAT Mensuration Questions, always begin by identifying the type of shape involved.

This clarity helps you apply the right formula quickly and accurately.

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Common Types of IPMAT Mensuration Questions

Below are the major types mensuration questions you will encounter in IPMAT exam:

1. Direct Formula Questions

These are straightforward problems testing your recall. (Example: Find the area of a circle with radius 7 cm. Solution: πr2=22/7×7×7=154 cm2\pi r^2 = 22/7 × 7 × 7 = 154 \, cm^2πr2=22/7×7×7=154cm2)

2. Composite Figures

Here, multiple shapes are combined (like a rectangle with a semicircular top). Tip: Break the figure into parts, find individual areas, and add or subtract accordingly.

3. Conversion-Based Problems

Questions where one solid is melted to form another. (Example: A cube is melted into a sphere - find the radius of the sphere.)

4. IPMAT Ratio and Proportion-Based Problems

These involve comparing shapes. (Example: Radii of two spheres are in ratio 2:3. The ratio of their volumes will be 8:278:278:27.)

5. Practical Application Questions: These include real-world scenarios such as painting, fencing, or filling containers.

Understanding these variations is crucial for scoring high in IPMAT Mensuration Questions.

Expert Tips to Master IPM Mensuration Questions

Over the years, we have noticed that most high-scoring students follow a consistent approach while practicing IPMAT Mensuration Questions:

1. Understand Before You Memorize: Know how each formula is derived. This helps you handle variations easily.

2. Practice Visualization: Always draw a rough figure. It helps in understanding dimensions and relationships between shapes.

3. Focus on Ratios: Many IPMAT Mensuration Questions can be simplified by using ratios rather than full calculations.

4. Revise Formulas Weekly: Keep a one-page formula sheet. Review it regularly before IPMAT mock tests.

5. Attempt Timed Sectional Tests: Mensuration is easy but can be time-consuming. Practicing under time limits helps manage exam pressure.

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Key Takeaways

• Build conceptual clarity rather than rote-learning formulas.
• Maintain accuracy in units and dimensions while solving.
• Focus on mixed-shape and multi-concept problems.
• Practice exam-level IPMAT Mensuration Questions regularly.
• Revise all key formulas every few days for retention.
• Balance speed with precision during timed practice.
• Integrate Mensuration with ratio and algebra topics for deeper understanding.
• Use consistent short practice sessions for lasting improvement.