Geometry & mensuration

Answer: (d) 1, 2 and 3

All three are standard metric equivalences: a cubic metre holds 1000 litres, a metric ton (tonne) is 1000 kg, and a hectare is a 100 m × 100 m square, i.e. 10000 m².

1. 1 m³ = 100 cm × 100 cm × 100 cm = 10,00,000 cm³, and 1 litre = 1000 cm³, so 1 m³ = 1000 litres.
2. 1 metric ton (tonne) = 1000 kg by definition.
3. 1 hectare = 100 m × 100 m = 10000 m².
4. All three statements are correct.

• ✓ 1. 1 m³ = 10,00,000 cm³ and 1 litre = 1000 cm³, so 1 m³ = 1000 litres.
• ✓ 2. A metric ton (tonne) is defined as 1000 kg.
• ✓ 3. A hectare is the area of a square of side 100 m: 100 × 100 = 10000 m².

Remember · Know the core metric links: 1 litre = 1000 cm³, 1 m³ = 1000 litres, 1 hectare = 10000 m², 1 tonne = 1000 kg.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 9

Any line across the 220 cm breadth meets tile sides of 60 cm or 140 cm, which can cover at most 200 cm of it. So at most 400 × 200 = 80,000 cm² can be tiled, room for fewer than 10 tiles of 8,400 cm², and a staggered layout fits 9.

1. Floor = 400 cm × 220 cm; one tile = 140 cm × 60 cm = 8,400 cm².
2. Across the 220 cm breadth, tile sides add up in 60s and 140s: the best totals are 60 + 140 = 200 and 60 + 60 + 60 = 180. So every line across the breadth leaves at least 20 cm uncovered.
3. Covered area ≤ 400 × 200 = 80,000 cm², so tiles ≤ 80,000 ÷ 8,400 ≈ 9.5, i.e. at most 9.
4. A layout of 9 (length measured 0–400, breadth 0–220): three tiles lying lengthwise, stacked, at length 0–140 (breadth 0–180); one lengthwise tile at length 140–280, breadth 0–60; two standing tiles (60 along the length) at length 140–260, breadth 60–200; two standing tiles at length 280–400, breadth 0–140; one lengthwise tile at length 260–400, breadth 140–200.
5. 3 + 1 + 2 + 2 + 1 = 9 tiles, none overlapping — the maximum.

Remember · For tiling maxima, check a 'line' bound as well as area: the lengths that fit across a side often beat the area estimate.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 1 only

The unpainted cubes form an inner block of 5 × 3 × 1 = 15. Green (5 × 3) and blue (7 × 3) faces meet only along the four edges of length 3 cm, and each such edge gives 3 − 2 = 1 cube with exactly those two faces painted — 4 cubes, not 6.

1. No paint: remove one layer from each side: (7 − 2) × (5 − 2) × (3 − 2) = 5 × 3 × 1 = 15. Statement 1 is correct.
2. Green faces are the 5 × 3 ends (across the 7 cm length); blue faces are the 7 × 3 sides (across the 5 cm width).
3. A green face meets a blue face along an edge 3 cm long; a cuboid has 4 such edges.
4. Each 3 cm edge has 3 cubes; the 2 end ones are corners with three painted faces, leaving 1 cube per edge.
5. Blue-and-green-only cubes = 4 × 1 = 4, not 6. Statement 2 is incorrect.
6. Check: two-face cubes in all = 4(7 − 2) + 4(5 − 2) + 4(3 − 2) = 20 + 12 + 4 = 36, and the green–blue share is the 4 on the 3 cm edges.

• ✓ 1. Inner block (7 − 2)(5 − 2)(3 − 2) = 15 cubes.
• ✗ 2. Green–blue edges are the four 3 cm edges, each giving one such cube: 4 in all.

Remember · Painted cuboid a × b × c: unpainted = (a − 2)(b − 2)(c − 2); cubes on an edge with two colours = edge length − 2.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 24

Eight equidistant points form 4 diameters. The angle in a semicircle is a right angle, so each diameter with any of the other 6 points gives a right-angled triangle: 4 × 6 = 24.

1. With 8 equidistant points, each point has a point exactly opposite it, giving 8 ÷ 2 = 4 diameters.
2. The angle in a semicircle is a right angle, so a diameter with any third point forms a right-angled triangle.
3. Each diameter can pair with any of the other 6 points: 4 × 6 = 24 triangles.

Remember · Angle in a semicircle = 90°: count diameters × remaining points.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Both 1 and 2

Two 8 cm squares plus two 4 cm squares use up the whole 20 × 8 sheet, and five 4 × 8 strips cut along their diagonals give 10 equal triangles. Both statements are possible.

1. Statement 1: cut two 8 × 8 squares from the 20 × 8 sheet, leaving a 4 × 8 strip; cut that into two 4 × 4 squares. That is exactly 4 square sheets with nothing left over.
2. Statement 2: cut the sheet into five 4 × 8 rectangles and cut each along a diagonal: 10 triangles, each of area 16 cm² (160 ÷ 10).
3. Both statements are correct.

• ✓ 1. 8 × 8 + 8 × 8 + 4 × 4 + 4 × 4 = 64 + 64 + 16 + 16 = 160 cm², the whole sheet.
• ✓ 2. Five 4 × 8 rectangles halved diagonally give 10 triangles of 16 cm² each.

Remember · ‘It is possible’ needs only one construction; the squares need not be equal unless the item says so.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 2

With PQ = 3 and QR = 5, either Q lies between P and R (PR = 8) or P lies between Q and R (PR = 2); R cannot lie between P and Q because QR is longer than PQ. That gives two ratios, 3 : 8 and 3 : 2.

1. Take PQ = 3 and QR = 5 units.
2. Q between P and R: PR = 3 + 5 = 8, so PQ : PR = 3 : 8.
3. P between Q and R: QR = QP + PR, so PR = 5 − 3 = 2, giving PQ : PR = 3 : 2.
4. R between P and Q is impossible: then QR (5) would be part of PQ (3).
5. So there are 2 possible values: n = 2.

Remember · For points on a line, try each point as the middle one and drop the orders that break the given lengths.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 1000000

A cube of side 1 m holds 1 m³ = 100 × 100 × 100 cm³ = 10 lakh cm³, and each cubic centimetre is one millilitre. So it contains 1000000 mL, i.e. 1000 litres.

1. Volume = 1 m × 1 m × 1 m = 1 m³.
2. 1 m³ = 100 cm × 100 cm × 100 cm = 1000000 cm³.
3. 1 cm³ = 1 mL, so the vessel holds 1000000 mL.
4. Check: 1 m³ = 1000 litres, and 1 litre = 1000 mL.

Remember · 1 m³ = 1000 L = 10⁶ mL; 1 cm³ = 1 mL.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) u = 4v

Doubling the side multiplies a cube's volume by 2³ = 8, while the mass only doubles. So Q's mass per unit volume is 2/8 = 1/4 of P's, i.e. u = 4v.

1. Side of Q = 2 × side of P, so y = 2³ x = 8x.
2. n = 2m.
3. v = n/y = 2m/8x = (1/4)(m/x) = u/4.
4. So u = 4v.

Remember · Scaling a length by k scales volume by k³; compare ratios by writing each in terms of the smaller object.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 2 only

A circle and a square need not meet at all — a small circle can sit inside the square without touching it — so the minimum is 0, not 2. Each side is a straight segment that can cut a circle at most twice, so four sides give at most 8 points, and 8 is actually reachable.

1. A small circle inside the square, or one far away from it, meets the square in 0 points — so the minimum is 0.
2. A straight line meets a circle in at most 2 points; the square has 4 sides, so at most 4 × 2 = 8 points.
3. 8 is reached by a circle with the square's centre whose radius is more than half the side but less than half the diagonal: it crosses each side twice.

• ✗ 1. The minimum is 0 — the circle can lie wholly inside or wholly outside the square without touching it.
• ✓ 2. Four sides × at most 2 crossings each = 8, and a suitably sized concentric circle achieves it.

Remember · For 'minimum' claims, look for the case where the figures do not touch at all; for 'maximum', count lines × 2 for a circle.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 100

1 litre = 1 kg means 1 gram of water occupies 1 cm³. Since 1 cm³ = 10 × 10 × 10 = 1000 mm³, 0·1 g occupies 100 mm³.

1. 1 litre = 1000 cm³ weighs 1 kg = 1000 g, so 1 g of water = 1 cm³.
2. 1 cm³ = 10 mm × 10 mm × 10 mm = 1000 mm³.
3. 0·1 g = 0·1 cm³ = 100 mm³.

Remember · Water: 1 g = 1 cm³ = 1 mL; and 1 cm³ = 1000 mm³.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 12

Small cubes with exactly two painted faces sit along the edges of the big cube, excluding the corners. A cube has 12 edges, each with 3 − 2 = 1 such cube here, giving 12.

1. The cube splits into 3 × 3 × 3 = 27 small cubes.
2. Corner cubes have 3 painted faces; edge cubes other than corners have exactly 2.
3. Each edge has 3 small cubes, 2 of them corners, leaving 1 per edge.
4. 12 edges × 1 = 12 cubes with exactly two painted faces.
5. Check: 8 (three faces) + 12 (two) + 6 (one) + 1 (none) = 27.

Remember · Painted n-cube: 8 corners with three faces, 12(n − 2) edge cubes with two, 6(n − 2)² with one.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 8

An equilateral triangle inscribed in a circle has its vertices 120° apart — every 8th point of 24. Each point belongs to exactly one such triangle, so there are 24 ÷ 3 = 8.

1. Adjacent points are 360° ÷ 24 = 15° apart.
2. An inscribed equilateral triangle needs vertices 120° apart: 120 ÷ 15 = 8 steps.
3. Starting at points 1 to 8 gives different triangles (1-9-17, 2-10-18, …, 8-16-24); starting at 9 repeats the first.
4. Number of triangles = 24 ÷ 3 = 8.

Remember · On n equally spaced points, inscribed regular k-gons exist only if k divides n, and there are n/k of them.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 1 sq cm

With squares of side s, the rectangle measures 4s by 3s, so its diagonal is 5s by Pythagoras. 5s = 5 cm gives s = 1 cm and an area of 1 sq cm.

1. Let each square have side s; the rectangle is 4s long and 3s wide.
2. Diagonal = √((4s)² + (3s)²) = 5s.
3. 5s = 5, so s = 1 cm and each square's area is 1 sq cm.
4. Check: a 4 cm × 3 cm rectangle has diagonal √25 = 5 cm.

Remember · Spot the 3-4-5 triangle: sides in the ratio 3 : 4 make the diagonal 5 parts.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 4

Take P as the corner where the three perpendicular surfaces meet. The fly's distances from the two walls and the ceiling are its three coordinates, so 1² + 8² + h² = 9², giving h² = 16 and h = 4 m.

1. Take P as the origin, with the two walls and the ceiling as three mutually perpendicular planes.
2. The distance from each plane is one coordinate: 1 m, 8 m and h m (from the ceiling).
3. Distance from P: √(1² + 8² + h²) = 9, so 1 + 64 + h² = 81.
4. h² = 16, so h = 4 m.

Remember · Distance from a corner = √(x² + y² + z²), where x, y and z are the distances from the three faces meeting there.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

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 · Report an error

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 · Report an error

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 · Report an error

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 · Report an error

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 · Report an error

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 · Report an error