Mensuration, 3D and Optimization Previous Year Questions (PYQs) for CAT: 1+ Solved Questions with Step-by-Step Solutions

Solve 1+ Mensuration, 3D and Optimization previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Mensuration, 3D and Optimization

CAT QA Geometry

Mensuration, 3D and Optimization

A short chapter where formulas become fast decision tools.

chapter PYQs

①

📦 t1 — Three-Dimensional Mensuration

You master cuboids, spheres, surface area, volume, space diagonal, and 3D enclosure logic.

1 own-course PYQ Selected topic Formula + algebra

②

📐 t2 — Area Optimization in Plane Figures

You learn how geometric restrictions force a maximum area.

1 own-course PYQ

③

💰 t3 — Perimeter and Cost Optimization

You convert side lengths and area constraints into minimum-cost decisions.

1 own-course PYQ

End goal: identify whether the question is asking for a 3D measure, a maximum area, or a minimum cost, then apply the shortest formula path.

Topic Hero: Three-Dimensional Mensuration

Geometry → Mensuration → t1

Three-Dimensional Mensuration

One-line hook: in 3D, the hidden diagonal often controls the whole solid.

What you'll learn here

Cuboid surface area, volume, and total edge length.
Sphere radius, surface area, and volume basics.
Space diagonal of a rectangular box.
Why an inscribed box has diagonal equal to sphere diameter.
How CAT combines formulas using algebraic identities.

Mensuration, 3D and Optimization: Solved Questions with Step-by-Step Explanations (1 Problems)

Question 1 · Quantitative Ability MCQ

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

A.
120000
B.
90000
C.
100000
D.
160000

Correct Answer:

Step-by-Step Solution

This is a cost-optimization question with unequal fencing rates. The trigger is that one side costs a different price per foot than the other three — this means we must minimize actual rupee cost, not just minimize perimeter.

Step 1: Set up variables. Let the side that has the expensive fencing on one edge have length $x$ , and the perpendicular side have length $y$ . The rectangle has two sides of length $x$ (one costing ₹200/ft, the opposite one costing ₹100/ft) and two sides of length $y$ (both costing ₹100/ft, since only "one side" is singled out for the higher rate).

Step 2: Write the total cost expression:

$Cost = 200 x + 100 x + 100 y + 100 y = 300 x + 200 y$

Step 3: Use the fixed area condition. Since area is 60000 sq ft:

$x y = 60000 \Rightarrow y = \frac{60000}{x}$

Step 4: Substitute into the cost expression to get cost purely in terms of $x$ :

$Cost (x) = 300 x + \frac{200 \times 60000}{x} = 300 x + \frac{12000000}{x}$

Step 5: Minimize this using AM-GM (since both terms are positive for $x > 0$ ):

$300 x + \frac{12000000}{x} \geq 2 300 x \times \frac{12000000}{x} = 2 3, 600, 000, 000$

$= 2 \times 60000 = 120000$

Step 6: The minimum is achieved when the two terms are equal:

$300 x = \frac{12000000}{x} \Rightarrow x^{2} = 40000 \Rightarrow x = 200$

Then $y = 60000/200 = 300$ , and cost $= 300 (200) + 200 (300) = 60000 + 60000 = 120000$ , confirming the AM-GM result.

Answer: the lowest possible cost is ₹120000, option A.

Common trap: students often assume the cheapest rectangle is always a square (which minimizes plain perimeter), but here the unequal costs mean the optimal shape is NOT a square — you must minimize the actual cost expression using AM-GM, not just the perimeter.

A question from this chapter

Question 1

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

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