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CSAT

CSAT · 20 questions

Geometry & mensuration

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Geometry & mensuration questions per year: 2016: 6, 2017: 1, 2018: 3, 2019: 0, 2020: 3, 2021: 2, 2022: 2, 2023: 2, 2024: 1, 2025: 0, 2026: 0 Asked in 8 of 11 years · most in 2016 (6)

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” See the full syllabus →

A piece of tin is in the form of a rectangle having length 12 cm and width 8 cm. This is used to construct a closed cube. The side of the cube is:

Answer & explanation

Answer: (c) 4 cm

The whole sheet of 12 × 8 = 96 cm² becomes the surface of the cube. A closed cube has six square faces, so 6s² = 96, s² = 16 and each side is 4 cm.

  1. Area of the sheet = 12 × 8 = 96 cm².
  2. Closed cube: 6s² = 96, so s² = 16 and s = 4 cm.
  3. Check: 6 × 4 × 4 = 96.

Remember · Sheet-to-solid problems: the sheet's area equals the surface area of the solid made from it.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

A person climbs a hill in a straight path from point ‘O’ on the ground in the direction of north-east and reaches a point ‘A’ after travelling a distance of 5 km. Then, from the point ‘A’ he moves to point ‘B’ in the direction of north-west. Let the distance AB be 12 km. Now, how far is the person away from the starting point ‘O’?

Answer & explanation

Answer: (b) 13 km

North-east and north-west are at right angles, so the path turns through 90° at A. OB is the hypotenuse of a right triangle with arms 5 km and 12 km: √(5² + 12²) = 13 km.

  1. North-east and north-west are perpendicular, so angle OAB = 90°.
  2. OA = 5 km and AB = 12 km are the arms of the right angle.
  3. OB = √(5² + 12²) = √169 = 13 km.

Remember · Directions 90° apart (NE and NW, N and E) form a right angle — use Pythagoras; 5-12-13 is a triple worth knowing.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

An agricultural field is in the form of a rectangle having length X₁ meters and breadth X₂ meters (X₁ and X₂ are variable). If X₁ + X₂ = 40 meters, then the area of the agricultural field will not exceed which one of the following values?

Answer & explanation

Answer: (a) 400 sq m

When the length and breadth add up to 40 m, the area is greatest when the rectangle is a square of side 20 m. So the area can never exceed 20 × 20 = 400 sq m.

  1. For a fixed sum X₁ + X₂ = 40, the product X₁ × X₂ is largest when X₁ = X₂ = 20.
  2. Maximum area = 20 × 20 = 400 sq m.
  3. Check: 19 × 21 = 399 and 10 × 30 = 300, both below 400.

Remember · For a fixed sum, a product is largest when the parts are equal.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

AB is a vertical trunk of a huge tree with A being the point where the base of the trunk touches the ground. Due to a cyclone, the trunk has been broken at C which is at a height of 12 meters, broken part is partially attached to the vertical portion of the trunk at C. If the end of the broken part B touches the ground at D which is at a distance of 5 meters from A, then the original height of the trunk is:

Answer & explanation

Answer: (b) 25 m

The standing 12 m of trunk and the 5 m along the ground form a right triangle whose hypotenuse is the broken part: √(12² + 5²) = 13 m. The original height was 12 + 13 = 25 m.

  1. AC = 12 m (standing), AD = 5 m (on the ground), angle CAD = 90°.
  2. Broken part CD = √(12² + 5²) = √169 = 13 m.
  3. Original height = AC + CB = 12 + 13 = 25 m.

Remember · Broken-tree problems: original height = standing part + hypotenuse from the break to where the top touches the ground.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

A cube has all its faces painted with different colours. It is cut into smaller cubes of equal sizes such that the side of the small cube is one-fourth the big cube. The number of small cubes with only one of the sides painted is:

Answer & explanation

Answer: (b) 24

Each edge of the big cube is cut into 4, so every face is a 4 × 4 grid of small cubes. Only the inner 2 × 2 on each face has exactly one painted side: 6 × 4 = 24.

  1. The big cube becomes 4 × 4 × 4 = 64 small cubes.
  2. On each face, the cubes painted on that face alone form the inner (4 − 2) × (4 − 2) = 4.
  3. Six faces: 6 × 4 = 24.

Remember · Cube cut into n per edge: one face painted = 6(n − 2)², two faces = 12(n − 2), three faces = 8.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

A cylindrical overhead tank of radius 2 m and height 7 m is to be filled from an underground tank of size 5.5 m × 4 m × 6 m. How much portion of the underground tank is still filled with water after filling the overhead tank completely?

Answer & explanation

Answer: (a) 1/3

The cylinder holds 22/7 × 2² × 7 = 88 m³, and the full underground tank holds 5.5 × 4 × 6 = 132 m³. After the cylinder is filled, 44 m³ remain — one-third of the underground tank.

  1. Overhead tank: πr²h = 22/7 × 2 × 2 × 7 = 88 m³.
  2. Underground tank (taken as full at the start): 5.5 × 4 × 6 = 132 m³.
  3. Left: 132 − 88 = 44 m³, and 44/132 = 1/3.

Remember · When 7 appears in a radius or height, take π = 22/7; the numbers are usually set to cancel neatly.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·