IPMAT Mensuration Questions And Answers 2026

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.

Mensuration is one of the most structured and reliable scoring areas in the Quantitative Aptitude section of the IPMAT exam.

Every year, students face questions that test their understanding of two-dimensional and three-dimensional figures, such as circles, triangles, cubes, cylinders, and cones.

A solid grasp of formulas, combined with logical application, can help candidates solve these problems quickly and accurately.

Practicing a wide range of IPMAT Mensuration Questions helps aspirants to identify patterns, improve calculation speed, and develop confidence before the exam.

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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.

Sample IPMAT Mensuration Questions with Solutions

Let us review a few representative problems to understand how questions appear in the IPMAT exam.

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

A. 125 cm²

B. 150 cm²

C. 200 cm²

D. 250 cm²

Solution: Surface area of cube = 6a2=6×52=6×25=150 cm26a^2 = 6 × 5^2 = 6 × 25 = 150 \, cm^26a2=6×52=6×25=150cm2.

Correct Answer: B (150 cm²)

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

A. ₹1,100

B. ₹2,200

C. ₹2,500

D. ₹1,750

Solution: Perimeter = 2πr=2×22/7×14=88 m2\pi r = 2 × 22/7 × 14 = 88 \, m2πr=2×22/7×14=88m.

Cost = 88×25=₹2,20088 × 25 = ₹2,20088×25=₹2,200.

Correct Answer: B (₹2,200)

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

A. 120 cm³

B. 60 cm³

C. 100 cm³

D. 140 cm³

Solution: Volume = l×b×h=4×5×6=120 cm3l × b × h = 4 × 5 × 6 = 120 \, cm^3l×b×h=4×5×6=120cm3.

Correct Answer: A (120 cm³)

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

A. 132 cm²

B. 144 cm²

C. 154 cm²

D. 160 cm²

Solution: Radius r=7 cmr = 7 \, cmr=7cm. Area = πr2=22/7×7×7=154 cm2\pi r^2 = 22/7 × 7 × 7 = 154 \, cm^2πr2=22/7×7×7=154cm2.

Correct Answer: C (154 cm²)

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

A. 8

B. 16

C. 24

D. 27

Solution: Number of cubes = (6/3)3=23=8(6/3)^3 = 2^3 = 8(6/3)3=23=8.

Correct Answer: A (8 cubes)

Q6. A cylinder of radius 7 cm and height 10 cm has its volume equal to that of a cone. Find the height of the cone if its radius is also 7 cm.

A. 10 cm

B. 15 cm

C. 30 cm

D. 20 cm

Solution: Volume of cylinder = πr2h=π×72×10=490π\pi r^2 h = \pi × 7^2 × 10 = 490\piπr2h=π×72×10=490π.

Volume of cone = 13πr2H\frac{1}{3}\pi r^2 H31​πr2H.

Equating: 13π×72×H=490π⇒H=30 cm\frac{1}{3}\pi × 7^2 × H = 490\pi \Rightarrow H = 30 \, cm31​π×72×H=490π⇒H=30cm.

Correct Answer: C (30 cm)

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Q7. The perimeter of a rectangle is 24 cm and its area is 32 cm². Find its length.

A. 6 cm

B. 7 cm

C. 8 cm

D. 10 cm

Solution: 2(l+b)=24⇒l+b=122(l + b) = 24 \Rightarrow l + b = 122(l+b)=24⇒l+b=12. lb=32lb = 32lb=32. l(12−l)=32⇒l2−12l+32=0⇒l=8l(12 - l) = 32 \Rightarrow l^2 - 12l + 32 = 0 \Rightarrow l = 8l(12−l)=32⇒l2−12l+32=0⇒l=8.

Correct Answer: C (8 cm)

Q8. The base radius of a cone is 7 cm and height is 24 cm. Find its slant height.

A. 24 cm

B. 25 cm

C. 26 cm

D. 28 cm

Solution: Slant height l=r2+h2=72+242=625=25 cml = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{625} = 25 \, cml=r2+h2​=72+242​=625​=25cm.

Correct Answer: B (25 cm)

Q9. Find the total surface area of a hemisphere of radius 7 cm.

A. 308 cm²

B. 462 cm²

C. 528 cm²

D. 616 cm²

Solution: Total surface area = 3πr2=3×22/7×7×7=462 cm23\pi r^2 = 3 × 22/7 × 7 × 7 = 462 \, cm^23πr2=3×22/7×7×7=462cm2.

Correct Answer: B (462 cm²)

Q10. A rectangular field is 120 m long and 80 m wide. Find the cost of fencing the field at ₹50 per meter.

A. ₹18,000

B. ₹20,000

C. ₹25,000

D. ₹30,000

Solution: Perimeter = 2(l+b)=2(120+80)=400 m2(l + b) = 2(120 + 80) = 400 \, m2(l+b)=2(120+80)=400m. Cost = 400×50=₹20,000400 × 50 = ₹20,000400×50=₹20,000.

Correct Answer: B (₹20,000)

Q11. 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?

A. 4

B. 6

C. 8

D. 10

Solution: 4/3π×634/3π×33=(6/3)3=8\frac{4/3\pi × 6^3}{4/3\pi × 3^3} = (6/3)^3 = 84/3π×334/3π×63​=(6/3)3=8.

Correct Answer: C (8 spheres)

Q12. If the diagonal of a square is 14√2 cm, find its area.

A. 98 cm²

B. 126 cm²

C. 196 cm²

D. 242 cm²

Solution: Diagonal =a2=142⇒a=14= a\sqrt{2} = 14\sqrt{2} \Rightarrow a = 14=a2​=142​⇒a=14. Area = a2=196 cm2a^2 = 196 \, cm^2a2=196cm2.

Correct Answer: C (196 cm²)

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

A. 2 : 3

B. 3 : 4

C. 4 : 5

D. 9 : 16

Solution: Area ∝ r2r^2r2. So, (r1/r2)2=9/16⇒r1/r2=3/4(r_1 / r_2)^2 = 9/16 \Rightarrow r_1 / r_2 = 3/4(r1​/r2​)2=9/16⇒r1​/r2​=3/4.

Correct Answer: B (3 : 4)

Q14. A hollow cylinder has an outer radius of 8 cm and inner radius of 6 cm. Its height is 10 cm. Find its volume.

A. 440π cm³

B. 280π cm³

C. 200π cm³

D. 260π cm³

Solution: Volume = πh(R2−r2)=π×10(82−62)=10π(64−36)=280π cm3\pi h (R^2 - r^2) = \pi × 10 (8^2 - 6^2) = 10\pi (64 - 36) = 280\pi \, cm^3πh(R2−r2)=π×10(82−62)=10π(64−36)=280πcm3.

Correct Answer: B (280π cm³)

Q15. A cone, a cylinder, and a hemisphere have the same base radius and height. Find the ratio of their volumes.

A. 1 : 2 : 3

B. 1 : 3 : 2

C. 1 : 2 : 1

D. 1 : 3 : 3

Solution: Volume of cone = 13πr2h\frac{1}{3}\pi r^2 h31​πr2h Volume of cylinder = πr2h\pi r^2 hπr2h Volume of hemisphere = 23πr3\frac{2}{3}\pi r^332​πr3 (for same h = r) Ratio = 1:3:21 : 3 : 21:3:2.

Correct Answer: B (1 : 3 : 2)

These examples reflect the diversity of Mensuration Questions in IPMAT and show how conceptual clarity directly translates into speed and accuracy.

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.

This resource includes:

• 50+ carefully curated Mensuration Questions
• Key formulas and derivations
• Level-wise practice: Basic, Moderate, and Advanced
• Complete solutions for self-assessment

You can download download previous year IPMAT questions PDF

Year	Download Link
IPMAT 2022	Download Now
IPMAT 2023	Download Now
IPMAT 2024	Download Now
IPMAT 2025	Download Now

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×heightbase \times heightbase×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.

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.