Geometry, Mensuration & Data Interpretation

This is the largest single block in the SSC CGL quantitative paper. Geometry and mensuration together supply roughly 5–7 of the 25 Tier-I questions, and Data Interpretation supplies a set of 3–5 questions on one dataset — which means one careful reading earns several marks. The three topics behave very differently: geometry rewards remembered theorems, mensuration rewards remembered formulae plus clean arithmetic, and DI rewards nothing but careful reading. All three are highly predictable, so this chapter is a formula bank you should be able to recite. Keep the −0.50 negative marking of Tier-I in mind: a mensuration formula you half-remember is a guaranteed loss, while a DI question is almost never worth skipping.

Lines, Angles and the Triangle Toolkit

Everything in plane geometry rests on four sentences. Angles on one side of a straight line add to 180°. Angles around a point add to 360°. Two lines crossing make vertically opposite angles equal. And the three angles of a triangle add to 180°. If you can apply only these four, you can already finish half the angle-chasing questions the paper sets.

Now add a transversal cutting two parallel lines. It creates eight angles, but only two distinct values — one acute and one obtuse — and they are supplementary. Corresponding angles are equal (the "F" shape), alternate interior angles are equal (the "Z" shape), and co-interior (allied) angles are supplementary (the "C" or "U" shape). So the moment one angle is given, all eight are known.

  • Worked example 1 (angle sum by ratio). The angles of a triangle are in the ratio 2 : 3 : 4. The parts add to 2 + 3 + 4 = 9, so one part is 180° ÷ 9 = 20°. The angles are 40°, 60° and 80°. Check: 40 + 60 + 80 = 180 ✓. The triangle is acute-angled, because the largest angle is under 90°.
  • Worked example 2 (exterior angle theorem). An exterior angle of a triangle is 110° and one of the two remote interior angles is 45°. The exterior angle equals the sum of the two remote interior angles, so the other one is 110° − 45° = 65°. The adjacent interior angle is 180° − 110° = 70°, and indeed 45 + 65 + 70 = 180 ✓
  • Worked example 3 (triangle inequality). Two sides of a triangle are 5 cm and 9 cm. The third side must be greater than 9 − 5 = 4 and less than 9 + 5 = 14. So the third side lies strictly between 4 and 14, and if it must be a whole number there are 9 possibilities (5, 6, 7, 8, 9, 10, 11, 12, 13). The same rule kills the classic the paper teaser "can 4, 6, 11 be the sides of a triangle?" — no, because 4 + 6 = 10 is not greater than 11. Testing the two smallest against the largest is enough.

Triangles are classified twice over. By sides: scalene (all three different), isosceles (two equal — and the angles opposite the equal sides are equal), equilateral (all three equal, so all angles 60°). By angles: acute (all under 90°), right (exactly one 90°), obtuse (one over 90°). A triangle can have at most one right angle and at most one obtuse angle, because two of them would already use up 180°. The largest side always faces the largest angle — that single sentence answers a surprising number of questions.

In a right triangle, Pythagoras gives (hypotenuse)² = (leg)² + (leg)². Memorise these six triplets, each verified here: 3-4-5 (9 + 16 = 25), 5-12-13 (25 + 144 = 169), 8-15-17 (64 + 225 = 289), 7-24-25 (49 + 576 = 625 — the standard cone, r = 7, h = 24, slant 25), 9-40-41 (81 + 1600 = 1681) and 20-21-29 (400 + 441 = 841, the one nobody recognises). Every multiple is also a triplet, so 6-8-10, 9-12-15, 30-40-50, 10-24-26 and 16-30-34 all come free. Use that instead of squaring: legs 36 and 48 divide by 12 into 3 and 4, so the hypotenuse is 5 × 12 = 60; legs 30 and 72 divide by 6 into 5 and 12, giving 13 × 6 = 78.

The Four Centres, Medians and Bisectors

A triangle has four famous centres, and the paper asks about them by name. Each is the meeting point of three special lines: medians meet at the centroid, angle bisectors at the incentre, perpendicular bisectors of the sides at the circumcentre, and altitudes at the orthocentre. Learn which line makes which centre and you will never mix them up. In an equilateral triangle all four coincide at one point, with height (√3/2)a, circumradius R = a/√3, inradius r = a/(2√3) and therefore R = 2r — for side 12 that is height ≈ 10.39, R = 4√3 ≈ 6.93 and r = 2√3 ≈ 3.46.

The four centres of a triangle — the properties the paper actually asks
CentreMade byKey propertyAngle formula (angle A at vertex A)
Centroid (G)The three medians (vertex to midpoint of opposite side)Divides every median in the ratio 2 : 1 from the vertex; always inside; each median splits the triangle into two equal areas, so the three medians give six equal areas—
Incentre (I)The three internal angle bisectorsCentre of the inscribed circle; equidistant from the three sides; inradius r = Area ÷ s, where s is the semi-perimeter; always inside∠BIC = 90° + A/2
Circumcentre (O)The three perpendicular bisectors of the sidesCentre of the circumscribed circle; equidistant from the three vertices; in a right triangle it sits exactly at the midpoint of the hypotenuse, so R = hypotenuse ÷ 2∠BOC = 2A
Orthocentre (H)The three altitudesIn a right triangle it is the right-angle vertex itself; in an obtuse triangle it lies outside; in an acute triangle inside∠BHC = 180° − A
  • Worked example 1 (the 2 : 1 ratio). A median of a triangle is 21 cm long and G is the centroid. Then the distance from the vertex to G is (2/3) × 21 = 14 cm and from G to the midpoint of the opposite side is (1/3) × 21 = 7 cm. Check 14 : 7 = 2 : 1 ✓ and 14 + 7 = 21 ✓
  • Worked example 2 (all three angle formulae at once). In triangle ABC, angle A = 70°. Then the incentre gives ∠BIC = 90° + 70/2 = 90 + 35 = 125°; the circumcentre gives ∠BOC = 2 × 70 = 140°; and the orthocentre gives ∠BHC = 180 − 70 = 110°. Three different answers from the same 70° — which is exactly why the paper asks it.
  • Worked example 3 (angle-bisector theorem). In triangle ABC, AB = 6, AC = 8, BC = 7, and AD bisects angle A meeting BC at D. The theorem says BD : DC = AB : AC = 6 : 8 = 3 : 4. Since BD + DC = 7, we get BD = (3/7) × 7 = 3 and DC = (4/7) × 7 = 4. Check 3 : 4 = 6 : 8 ✓
  • Worked example 4 (midpoint theorem). D and E are the midpoints of AB and AC in triangle ABC, and BC = 17 cm. Then DE is parallel to BC and DE = ½ × BC = 8.5 cm. The small triangle ADE is similar to ABC with ratio 1 : 2, so its area is one quarter of the whole — and the trapezium DECB holds the other three quarters.
🧠 Which centre? Match the line to the name
Use the meanings of the words. Incentre → the inscribed circle → touches the sides, so it must come from angle bisectors (a bisector is the set of points equidistant from two sides). Circumcentre → the circumscribed circle → passes through the vertices, so it must come from perpendicular bisectors (equidistant from two points). Ortho means right-angled → altitudes. Centroid is the centre of gravity → medians. Once the logic is in place there is nothing left to memorise.

Congruence, Similarity and the Thales Theorem

Congruent means identical — same shape and same size. Similar means same shape, possibly different size. Congruence is proved by SSS, SAS, ASA, AAS and (for right triangles only) RHS. Similarity needs less: AA (two equal angles force the third, so all three match), SSS (all three side ratios equal) or SAS (two side ratios equal with the included angle equal).

Once two triangles are similar with side ratio k, everything that is a length scales by k and everything that is an area scales by k². That single rule replaces a dozen separate formulae: corresponding sides, perimeters, medians, altitudes, angle bisectors, inradii and circumradii are all in the ratio k, while areas are in the ratio k². In three dimensions the same logic gives volume ratio k³, which is the whole of the mensuration scaling section later in this chapter. Similarity also gives the Basic Proportionality Theorem (BPT, or Thales): a line drawn parallel to one side of a triangle cuts the other two sides in the same ratio, AD/DB = AE/EC, and the converse is true as well — The paper uses the converse to ask whether DE is parallel to BC.

  • Worked example 1 (BPT). In triangle ABC, DE ∥ BC with D on AB and E on AC. Given AD = 4, DB = 6 and AE = 6, find EC. From AD/DB = AE/EC we get 4/6 = 6/EC, so EC = 6 × 6 ÷ 4 = 9. Cross-check the whole-side form AD/AB = AE/AC: 4/10 = 0.4 and 6/15 = 0.4 ✓
  • Worked example 2 (areas to sides). Two similar triangles have areas 81 cm² and 121 cm². Then the side ratio is √(81/121) = 9/11. If a median of the smaller is 12.6 cm, the corresponding median of the larger is 12.6 × 11 ÷ 9 = 15.4 cm. Note you take the square root going from areas to lengths, and square going the other way.
⚠️ AAA gives similarity, never congruence
Two triangles with all three angles equal need not be the same size — a 3-4-5 triangle and a 30-40-50 triangle have identical angles. So AAA is not a congruence criterion. Equally, SSA (two sides and a non-included angle) is not a congruence criterion either, except in the right-angle case where it becomes RHS. Statement questions test exactly these two false criteria, so read the letters carefully before you tick.

Quadrilaterals and Polygons

The quadrilateral family is a ladder. A trapezium has one pair of parallel sides. A parallelogram has both pairs parallel. A rhombus is a parallelogram with all sides equal. A rectangle is a parallelogram with all angles right. A square is both a rhombus and a rectangle. A kite is the odd one out — two pairs of adjacent equal sides. Every property higher on the ladder is inherited downwards, so a square owns every property listed below.

  • Parallelogram. Opposite sides equal and parallel, opposite angles equal, adjacent angles supplementary. Diagonals bisect each other but are neither equal nor perpendicular. Area = base × height.
  • Rhombus. All four sides equal. Diagonals bisect each other at right angles and each bisects the angles it joins, with d₁² + d₂² = 4a². Area = ½ d₁ d₂.
  • Rectangle. Opposite sides equal, all four angles 90°. Diagonals are equal and bisect each other, d = √(l² + b²). Area = l × b.
  • Square. All sides equal, all angles 90°. Diagonals equal, bisecting each other at 90°, d = a√2, so the area may be written as d²/2.
  • Trapezium. Exactly one pair of parallel sides; in an isosceles trapezium the non-parallel sides, the base angles and the diagonals are equal, and the segment joining the midpoints of the non-parallel sides is ½(a + b). Area = ½ (a + b) h.
  • Kite. Two pairs of adjacent equal sides and one pair of equal opposite angles. Diagonals are perpendicular and the longer one bisects the shorter. Area = ½ d₁ d₂.

For an n-sided polygon three formulae cover everything. The sum of the interior angles is (n − 2) × 180°. The sum of the exterior angles is 360°, whatever n is. And the number of diagonals is n(n − 3)/2 — each vertex joins to n − 3 others (not itself, not its two neighbours), and dividing by 2 removes the double counting. For a regular polygon divide the sums by n: each interior angle is (n − 2)180°/n and each exterior angle is 360°/n.

  • Worked example 1 (find n from the angle). Each interior angle of a regular polygon is 144°. Then each exterior angle is 180 − 144 = 36°, and since the exterior angles total 360°, n = 360 ÷ 36 = 10. Verify by the other route: (10 − 2) × 180 = 1440, and 1440 ÷ 10 = 144 ✓. Its number of diagonals is 10 × 7 ÷ 2 = 35.
  • Worked example 2 (rhombus from its diagonals). A rhombus has diagonals 16 cm and 30 cm. Its area is ½ × 16 × 30 = 240 cm². The diagonals bisect at right angles, so half-diagonals 8 and 15 form a right triangle: side = √(64 + 225) = √289 = 17 cm, and the perimeter is 4 × 17 = 68 cm. Check the identity d₁² + d₂² = 4a²: 256 + 900 = 1156 and 4 × 289 = 1156 ✓
⚠️ The exterior angle sum never changes
Candidates write "sum of exterior angles = (n − 2) × 180°" under pressure. It is always 360°, for a triangle, a decagon or a hundred-sided polygon alike — walking once around the boundary turns you through one full revolution. Only the interior sum depends on n. This one confusion accounts for a large share of the polygon questions people lose.

Circles: Chords, Tangents and Angles

Circle geometry is the most theorem-heavy part of the paper syllabus, but the theorems are few and each is one sentence long. Group them into three families — chords, angles, tangents — and add the two power-of-a-point results, and you have the entire chapter. The table below is the whole thing; read it out loud until each row is automatic.

Every circle theorem the paper asks, in one table
FamilyTheoremWorking form
ChordThe perpendicular from the centre to a chord bisects the chord (and conversely, the line from the centre to a chord midpoint is perpendicular)r² = d² + (chord/2)², where d is the distance from the centre
AngleThe angle a chord subtends at the centre is twice the angle it subtends at any point on the major arccentral = 2 × inscribed
AngleAngles in the same segment are equal; the angle in a semicircle is 90°A diameter always sees a right angle
AngleIn a cyclic quadrilateral opposite angles are supplementary, and an exterior angle equals the interior opposite angleA + C = 180°, B + D = 180°
TangentA tangent is perpendicular to the radius at the point of contactRadius–tangent–external point makes a right triangle
TangentThe two tangents from an external point are equal, and the line to the centre bisects the angle between themtangent length = √(d² − r²)
TangentAlternate segment theorem: the angle between a tangent and a chord equals the angle the chord subtends in the alternate segmentConverts a tangent angle into an inscribed angle
Power of a pointTwo chords AB and CD meeting at P satisfy PA × PB = PC × PD (true whether P is inside or outside)The intersecting-chords rule
Power of a pointFrom an external point, PT² = PA × PB where PT is the tangent and PAB a secantThe tangent–secant rule
Two circlesLength of a direct (external) common tangent, centres d apart√(d² − (r₁ − r₂)²)
Two circlesLength of a transverse (internal) common tangent√(d² − (r₁ + r₂)²)
  • Worked example 1 (chord and distance). A circle of radius 13 cm has a chord of length 24 cm. Half the chord is 12, so the distance from the centre is √(13² − 12²) = √(169 − 144) = √25 = 5 cm (the 5-12-13 triplet). Conversely, in a circle of radius 10 cm a chord lies 6 cm from the centre; half the chord is √(100 − 36) = 8, so the chord is 16 cm.
  • Worked example 2 (tangent length). A point P lies 17 cm from the centre of a circle of radius 8 cm. The tangent from P has length √(17² − 8²) = √(289 − 64) = √225 = 15 cm (the 8-15-17 triplet). Both tangents from P are 15 cm, and PT is perpendicular to the radius at T.
  • Worked example 3 (power of a point). Two chords cross inside a circle at P with PA = 4 and PB = 9. If the other chord is bisected at P, each part is √(4 × 9) = 6, since PC × PD = 36 and PC = PD. And from an external point, if a secant gives PA = 4 and PB = 9 then the tangent is PT = √36 = 6 as well.
  • Worked example 4 (two circles). Two circles of radii 8 cm and 3 cm have centres 13 cm apart. The direct common tangent is √(13² − (8 − 3)²) = √(169 − 25) = √144 = 12 cm. The transverse common tangent is √(13² − (8 + 3)²) = √(169 − 121) = √48 = 4√3 ≈ 6.93 cm. Since d = 13 is greater than r₁ + r₂ = 11, the circles lie outside each other and there are four common tangents in all.
⚠️ Read which arc, and read which segment
The commonest circle mistake is halving the central angle when the point is on the minor arc. If the central angle is 130°, the major-arc angle is 65° but the minor-arc angle is 115°. Similarly in the alternate segment theorem the equal angle is in the segment on the other side of the chord, not the one next to the tangent. When the wording is ambiguous, sketch it — a five-second rough circle prevents a −0.50.

Mensuration I — Areas and Perimeters in Two Dimensions

Two-dimensional mensuration is pure recall plus careful arithmetic. Learn the table below cold. One habit saves enormous time: use π = 22/7 whenever the radius is a multiple of 7, and π = 3.14 otherwise. Examiners choose radii like 7, 14, 21 and 35 precisely so that 22/7 cancels, and if your radius is a multiple of 7 and the arithmetic is turning ugly, you have made a slip somewhere.

Two-dimensional formulae — area and perimeter
ShapeAreaPerimeter / boundary
Triangle (base b, height h)½ × b × ha + b + c
Triangle from three sides ("Heron")√[s(s − a)(s − b)(s − c)], s = (a + b + c)/22s
Equilateral triangle (side a)(√3/4) a²3a; height = (√3/2)a
Square (side a, diagonal d)a² = d²/24a; d = a√2
Rectangle (l, b)l × b2(l + b); d = √(l² + b²)
Circle (radius r)π r²2π r
Sector (angle θ°)(θ/360) π r² = ½ × arc × rarc = (θ/360) × 2π r; boundary = arc + 2r
Segment (minor)sector area − triangle areaarc + chord
Ring / annulus (R outer, r inner)π (R² − r²) = π(R + r)(R − r)2πR + 2πr
  • Worked example 1 (Heron). Sides 13, 14, 15. Then s = (13 + 14 + 15)/2 = 21, and the area is √[21 × (21−13) × (21−14) × (21−15)] = √(21 × 8 × 7 × 6) = √7056 = 84 cm². This 13-14-15 triangle with area 84 is the favourite of every paper-setter; memorise the pair. As a bonus, its inradius is 84 ÷ 21 = 4 cm.
  • Worked example 2 (circle with π = 22/7). Radius 7 cm: area = (22/7) × 49 = 154 cm², circumference = 2 × (22/7) × 7 = 44 cm. Radius 21 cm: area = (22/7) × 441 = 1386 cm², circumference = 2 × (22/7) × 21 = 132 cm. Both are exact — no decimals appear at all.
  • Worked example 3 (sector and segment). In a circle of radius 21 cm, a sector of 60° has arc length (60/360) × 132 = 22 cm and area (60/360) × 1386 = 231 cm². Cross-check with ½ × arc × r = ½ × 22 × 21 = 231 ✓. The chord of a 60° sector makes an equilateral triangle of side 21, of area (√3/4) × 441 ≈ 190.96 cm², so the minor segment is 231 − 190.96 ≈ 40.04 cm².
  • Worked example 4 (ring). A circular path of width 7 m surrounds a pond of radius 7 m, so R = 14 and r = 7. The area of the path is (22/7)(196 − 49) = (22/7) × 147 = 462 m². Using the factored form π(R + r)(R − r) = (22/7)(21)(7) = 462 ✓ — the factored form is almost always faster.
🧠 The perimeter of a semicircle is not half the circumference
Cutting a circle in half creates a new straight edge — the diameter. So the perimeter of a semicircle is πr + 2r, not πr. The same logic applies to a sector: its boundary is the arc plus the two radii. Every "cost of fencing" question is built on this single omission, and the wrong option is always exactly the arc length alone.

Mensuration II — Solids, Scaling and Recasting

Three-dimensional mensuration has three quantities per solid: volume, curved or lateral surface area (CSA/LSA) and total surface area (TSA). Two structural facts remove most of the memory load. First, for any prism-like solid (cube, cuboid, cylinder, prism) volume = base area × height, and for any pointed solid (cone, pyramid) volume = ⅓ × base area × height — the one-third is the same one-third every time. Second, TSA = LSA + the flat faces you actually have, so a closed cylinder gets two circles, an open pipe gets none, and a cone gets one.

Three-dimensional formulae — volume and surface area
SolidVolumeCurved / lateral SATotal SA & extras
Cube (edge a)a³4a²6a²; diagonal = a√3
Cuboid (l, b, h)l b h2h(l + b)2(lb + bh + hl); diagonal = √(l² + b² + h²)
Cylinder (r, h)π r² h2π r h2π r h + 2π r² = 2π r(h + r)
Cone (r, h, slant l)⅓ π r² hπ r lπ r l + π r² = π r(l + r); l = √(r² + h²)
Sphere (r)(4/3) π r³4π r²Same as CSA — a sphere has no flat face
Hemisphere (r)(2/3) π r³2π r²3π r² (curved + the flat circle)
Frustum of a cone (R, r, h)⅓ π h (R² + r² + R r)π l (R + r), with l = √(h² + (R − r)²)π l(R + r) + π R² + π r²
Prism (any base)base area × heightbase perimeter × heightLSA + 2 × base area
Pyramid (any base)⅓ × base area × height½ × base perimeter × slant heightLSA + base area
  • Worked example 1 (cube and cuboid). A cube of edge 6 cm: volume 6³ = 216 cm³, TSA 6 × 36 = 216 cm² (numerically equal, a coincidence only at a = 6), LSA 4 × 36 = 144 cm², diagonal 6√3 ≈ 10.39 cm. A cuboid 8 × 6 × 5: volume = 240 cm³, TSA = 2(48 + 30 + 40) = 2 × 118 = 236 cm², LSA = 2 × 5 × 14 = 140 cm², diagonal = √(64 + 36 + 25) = √125 = 5√5 ≈ 11.18 cm.
  • Worked example 2 (cylinder, π = 22/7). r = 7 cm, h = 10 cm: volume = (22/7) × 49 × 10 = 1540 cm³, CSA = 2 × (22/7) × 7 × 10 = 440 cm², TSA = 440 + 2 × 154 = 748 cm². A hollow pipe with R = 8, r = 6, h = 21 holds material of volume (22/7) × 21 × (64 − 36) = 66 × 28 = 1848 cm³.
  • Worked example 3 (cone — the 7-24-25 solid). r = 7 cm, h = 24 cm, so slant l = √(49 + 576) = √625 = 25 cm. Volume = ⅓ × (22/7) × 49 × 24 = ⅓ × 154 × 24 = 1232 cm³. CSA = π r l = (22/7) × 7 × 25 = 550 cm². TSA = 550 + 154 = 704 cm². Notice how the whole question is built around one Pythagorean triplet.
  • Worked example 4 (sphere and hemisphere). r = 21 cm: sphere volume = (4/3) × (22/7) × 9261 = 38808 cm³, surface = 4 × (22/7) × 441 = 5544 cm². The hemisphere is exactly half the volume, 19404 cm³, but not half the surface: its curved part is 2πr² = 2772 cm² and its total is 3πr² = 4158 cm².
  • Worked example 5 (frustum, prism, pyramid). A frustum with R = 8, r = 4, h = 3 has slant l = √(9 + 16) = 5, volume = ⅓ π × 3 × (64 + 16 + 32) = 112π ≈ 352 cm³, CSA = π × 5 × 12 = 60π, TSA = 60π + 64π + 16π = 140π. A prism on a 3-4-5 triangular base (area 6) of length 10 has volume 60 cm³, LSA = 12 × 10 = 120 cm², TSA = 132 cm². A square pyramid of base 6 and height 4 has slant height √(3² + 4²) = 5, volume ⅓ × 36 × 4 = 48 cm³, LSA = ½ × 24 × 5 = 60 cm², TSA = 96 cm².

Now the scaling rule, which is worth more marks than any single formula. If every linear dimension of a solid is multiplied by k, then every length scales by k, every area — including all surface areas — by k², and the volume by k³. Read percentages as multipliers: a 20% increase means k = 1.2, a 10% decrease means k = 0.9. Then the answer is one power away.

  • Scaling example 1. Each edge of a cube is increased by 20%, so k = 1.2. Surface area becomes 1.2² = 1.44 times, an increase of 44%. Volume becomes 1.2³ = 1.728 times, an increase of 72.8%. Confirm with numbers: edge 10 → 12, volume 1000 → 1728, and (1728 − 1000)/1000 = 72.8% ✓
  • Scaling example 2 (mixed change). A cylinder has its radius doubled and its height halved. Volume ∝ r²h, so the factor is 2² × ½ = 2 — the volume doubles, a 100% increase. Note that halving the height does not cancel doubling the radius, because the radius is squared and the height is not.

Finally, the melting and recasting family, which is the single most predictable 3D question type. The principle is one line: melting conserves volume, never surface area. So the number of small pieces from one big lump is simply big volume ÷ small volume, and because both sides carry the same π, the π always cancels — never substitute 22/7 in these questions.

  • Recasting example 1 (sphere to cones). A solid sphere of radius 6 cm is melted into small cones of radius 2 cm and height 3 cm. Sphere volume = (4/3)π × 216 = 288π. Cone volume = (1/3)π × 4 × 3 = 4π. Number of cones = 288π ÷ 4π = 72.
  • Recasting example 2 (cylinder to cones). A cylinder of radius 6 cm and height 15 cm is recast into cones of radius 3 cm and height 5 cm. Cylinder volume = π × 36 × 15 = 540π; cone volume = (1/3)π × 9 × 5 = 15π; number = 540 ÷ 15 = 36. A useful sanity check: a cone with the same r and h as a cylinder has exactly one third its volume, so three such cones fill one cylinder.
💡 The one-third belongs to the cone, not the cylinder
Under time pressure candidates write πr²h for a cone or ⅓πr²h for a cylinder. Anchor it physically: a pointed solid holds less than the box around it, so cone and pyramid carry the ⅓; a flat-topped solid fills the box, so cylinder and prism do not. And in a cone always ask whether the number given is the height or the slant height — CSA needs l, volume needs h, and mixing them is the single commonest 3D error.

Data Interpretation I — Reading a Table

A DI set gives you one dataset and three to five questions on it, so the 40 seconds you spend understanding the table are paid back several times over. DI tests no mathematics beyond percentage, ratio and average. It tests whether you read the label, the unit and the question correctly. Every dataset in this chapter is presented as a table, because a bar chart, a line graph and a pie chart are only three different pictures of a table — learn to read the table and you can read all of them.

The dataset below is invented for practice — it is not official Staff Selection Commission data and must not be quoted as such. Read the caption and all four column heads before you look at a single number.

Dataset A (illustrative, NOT official): candidates appeared and qualified at five examination centres
CentreCandidates appearedPass rate (%)Candidates qualified
Delhi4,000251,000
Jaipur5,00018900
Patna3,00030900
Kolkata6,000201,200
Bhopal2,00035700
Total20,00023.54,700

First, check the table against itself. Each qualified figure is appeared × rate ÷ 100: 4000 × 25/100 = 1000, 5000 × 18/100 = 900, 3000 × 30/100 = 900, 6000 × 20/100 = 1200, 2000 × 35/100 = 700. The appeared column sums to 4000 + 5000 + 3000 + 6000 + 2000 = 20,000 and the qualified column to 1000 + 900 + 900 + 1200 + 700 = 4,700. So the overall pass rate is 4700 ÷ 20000 = 0.235 = 23.5%. Doing this check first means every later answer is safe.

  1. Q. Which centre had the highest pass rate, and which sent the most successful candidates? The highest rate is Bhopal at 35%, but Bhopal produced the fewest qualified candidates, only 700. The most qualified came from Kolkata, 1,200, on a modest rate of 20%. This is the rate-versus-number trap in its purest form.
  2. Q. What is the ratio of qualified candidates from Delhi to those from Kolkata? 1000 : 1200. Divide both by 200 to get 5 : 6. Always reduce before you look at the options — The paper lists 5 : 6, 6 : 5, 2 : 3 and 5 : 4 and expects you to grab the reversed one.
  3. Q. Kolkata accounted for what percentage of all qualified candidates? 1200 ÷ 4700 × 100 = 25.53% (approximately 25.5%). Delhi accounted for 1000 ÷ 4700 × 100 = 21.28%. Note the denominator is the qualified total 4,700, not the appeared total 20,000 — misreading which total is wanted is the second commonest DI error.
  4. Q. Taking Delhi and Jaipur together, what is their combined pass rate? Add the parts, then divide: (1000 + 900) ÷ (4000 + 5000) = 1900 ÷ 9000 = 21.11%. The tempting wrong answer is (25 + 18)/2 = 21.5%, the simple average of the two rates — wrong because Jaipur had more candidates and so pulls the combined figure towards its lower 18%.
⚠️ Never average two percentages — weight them
A combined percentage is (sum of the parts) ÷ (sum of the wholes) × 100, never the mean of the two percentages. The two agree only when the two bases are equal. In Dataset A the combined Delhi + Jaipur rate is 21.11% while the naive average is 21.5% — small, but the paper puts both in the options precisely so that the difference decides your mark. Whenever you see the words "taken together" or "combined", go back to the raw counts.

Data Interpretation II — Pie Shares and a Bar/Line Series

A pie chart is nothing but a set of percentage shares of one total. Two conversions make every pie question mechanical: amount = share% × total, and central angle = share% × 3.6° (because 100% must fill 360°, and 360 ÷ 100 = 3.6). Two internal checks come free: the shares must add to 100, and the angles must add to 360°. The dataset below is again invented for practice, not official data.

Dataset B (illustrative, NOT official): a household budget given as a pie chart — monthly income ₹48,000
Head of expenditureShare (%)Amount (₹)Central angle
Food3014,400108°
Rent2512,00090°
Education157,20054°
Savings125,76043.2°
Transport104,80036°
Miscellaneous83,84028.8°
Total10048,000360°

Verify the table before using it. Shares: 30 + 25 + 15 + 12 + 10 + 8 = 100 ✓. Amounts: 14,400 + 12,000 + 7,200 + 5,760 + 4,800 + 3,840 = 48,000 ✓. Angles: 108 + 90 + 54 + 43.2 + 36 + 28.8 = 360 ✓. Three sums, ten seconds, and the whole set is now trustworthy.

  1. Q. What is the central angle of the Food sector? 30 × 3.6 = 108°. Reverse the same rule when the angle is given: a sector of 54° means 54 ÷ 3.6 = 15% of the total, that is ₹7,200.
  2. Q. By what percentage does spending on Rent exceed spending on Education? Excess = 12,000 − 7,200 = ₹4,800, and the base is Education, so 4800 ÷ 7200 × 100 = 66.67%. Do not answer "10", which is only the gap in percentage points (25 − 15). The question asks by what per cent, so the base matters.
  3. Q. How much is spent on Food and Education together? Add the shares first: 30 + 15 = 45%, and 45% of 48,000 = ₹21,600. Adding shares before multiplying is always faster than adding two products.
  4. Q (the base-change trap). Next year the income rises by 25% and the Food share falls from 30% to 28%. Does the family spend less on food? No. New income = 48,000 × 1.25 = ₹60,000, and 28% of 60,000 = ₹16,800, up from ₹14,400 — an increase of ₹2,400, which is 2400 ÷ 14400 × 100 = 16.67%. The share fell by 2 percentage points (a relative fall of 2 ÷ 30 = 6.67%), yet the rupees rose, because the base grew. A smaller slice of a bigger cake can still be more cake.
⚠️ Percentage point is not per cent
A share moving from 30% to 28% has fallen by 2 percentage points but by 6.67 per cent (2 ÷ 30). Likewise Rent at 25% exceeds Education at 15% by 10 percentage points but by 66.67 per cent. The paper prints both numbers in the options. The rule is mechanical: if the question says "percentage points", subtract; if it says "by what per cent", divide by the base first. The base is always the quantity that comes after the words "than", "of" or "compared to".

Finally, a bar chart or line graph given as a table. Here the danger is the word "increase", because it can mean increase over the previous year or increase over the base year, and the two give very different numbers. Again, the figures below are invented for practice, not official data.

Dataset C (illustrative, NOT official): production of two products (in thousand units), 2020 to 2024
YearProduct AProduct BTotalGrowth in total over previous year
202012080200— (base year)
20211509024020%
202218012030025%
202321015036020%
202424021045025%
Five-year total9006501,550Average per year = 310

Check it: A sums to 120 + 150 + 180 + 210 + 240 = 900, B to 80 + 90 + 120 + 150 + 210 = 650, and 900 + 650 = 1,550, which matches the sum of the yearly totals 200 + 240 + 300 + 360 + 450 = 1,550 ✓. The average yearly total is 1550 ÷ 5 = 310, so only 2023 and 2024 are above average. The growth column is arithmetic you can redo: 240/200 = 1.20, 300/240 = 1.25, 360/300 = 1.20, 450/360 = 1.25.

  1. Q. By what percentage did total production in 2022 exceed that in 2020? Over the base year: (300 − 200)/200 × 100 = 50%. But over the previous year it is only (300 − 240)/240 × 100 = 25%. Both are correct answers to different questions, and the paper lists both. Over 2020 the series runs 20%, 50%, 80%, 125%; over the previous year it runs 20%, 25%, 20%, 25%.
  2. Q. In which year did Product B show its largest percentage increase over the previous year? B grows 80 → 90 → 120 → 150 → 210, giving 12.5%, 33.33%, 25% and 40%. The largest is in 2024. Note that the largest absolute rise, 60 thousand units, is also 2024 here — but that agreement is a coincidence, not a rule.
  3. Q. Product A rises by exactly 30 thousand units every year. Is its growth rate constant? No — and this is the finest trap in the whole topic. The increases are 30, 30, 30, 30, but the rates are 30/120 = 25%, 30/150 = 20%, 30/180 = 16.67% and 30/210 = 14.29%. A constant absolute rise gives a falling percentage rise, because the base keeps growing.
  4. Q. What is the ratio of A to B in 2024, and how did B share of the total change? In 2024, 240 : 210 = 8 : 7 (in 2020 it was 120 : 80 = 3 : 2). B share of the total was 80/200 = 40% in 2020 and 210/450 = 46.67% in 2024 — a rise of 6.67 percentage points, which is a relative rise of 6.67 ÷ 40 = 16.67%.

Key takeaways

  • Four sentences carry all angle-chasing: straight line 180°, point 360°, vertically opposite equal, triangle 180°. A transversal on parallel lines makes only θ and 180° − θ, and an exterior angle equals the sum of the two remote interior angles. Memorise the triplets 3-4-5, 5-12-13, 8-15-17, 7-24-25, 9-40-41 and 20-21-29, and scale them instead of squaring: legs 36 and 48 give hypotenuse 60 in three seconds.
  • Centroid splits every median 2 : 1 from the vertex. With angle A given, ∠BIC = 90° + A/2 (incentre), ∠BOC = 2A (circumcentre), ∠BHC = 180° − A (orthocentre). In a right triangle the circumcentre is the hypotenuse midpoint and the orthocentre is the right-angle vertex.
  • Similar triangles with side ratio k have every length in ratio k and every area in ratio k². Go from areas to lengths by square root, from lengths to areas by squaring. AAA and SSA prove similarity, never congruence.
  • For an n-gon: interior sum (n − 2)180°, exterior sum always 360° whatever n, diagonals n(n − 3)/2. Each interior angle 144° means each exterior 36°, so n = 10 and there are 35 diagonals.
  • Circles: r² = d² + (chord/2)², central angle = twice the major-arc angle, cyclic opposite angles 180°, tangent ⊥ radius, tangent length √(d² − r²), PA × PB = PC × PD and PT² = PA × PB.
  • Mensuration essentials: Heron √[s(s−a)(s−b)(s−c)], equilateral (√3/4)a², rhombus ½d₁d₂, trapezium ½(a + b)h, ring π(R + r)(R − r); prism-like volume = base × height, pointed volume = ⅓ base × height; cone slant l = √(r² + h²). Use π = 22/7 whenever the radius is a multiple of 7.
  • Scaling: multiply every linear dimension by k and lengths go × k, areas × k², volumes × k³. A 20% edge increase raises surface area 44% and volume 72.8%. Melting conserves volume, never surface area, so number of pieces = big volume ÷ small volume and π always cancels.
  • DI is arithmetic plus honesty. Verify that shares add to 100 and amounts add to the total; convert a pie share to an angle by × 3.6; combine two percentages by (sum of parts) ÷ (sum of wholes), never by averaging; and always name the denominator aloud — previous year, base year, or grand total.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The angles of a triangle are in the ratio 2 : 3 : 4. The measure of its largest angle is:

    1. 80°
    2. 90°
    3. 100°
    4. 75°
    Show answer

    Answer: A — 80°

    The three parts add to 2 + 3 + 4 = 9, and the angles of any triangle add to 180°, so one part equals 180 ÷ 9 = 20°. The angles are therefore 2 × 20 = 40°, 3 × 20 = 60° and 4 × 20 = 80°, and 40 + 60 + 80 = 180 ✓. The largest is 80°. Since the biggest angle is under 90°, the triangle is acute-angled — the trap option 90° would need the ratio to be 2 : 3 : 5.

  2. Each interior angle of a regular polygon measures 144°. How many diagonals does it have?

    1. 20
    2. 27
    3. 35
    4. 44
    Show answer

    Answer: C — 35

    Work through the exterior angle, which is always easier. Each exterior angle = 180 − 144 = 36°, and the exterior angles of any polygon total 360°, so n = 360 ÷ 36 = 10. Check by the interior route: (10 − 2) × 180 = 1440, and 1440 ÷ 10 = 144 ✓. Diagonals = n(n − 3)/2 = 10 × 7 ÷ 2 = 35. The trap answer 20 comes from using n = 8, and 44 from forgetting to divide by 2 (10 × 7 = 70 is the double count, not 44).

  3. A chord AB of a circle with centre O subtends an angle of 130° at O. If C is a point on the major arc, then angle ACB equals:

    1. 130°
    2. 65°
    3. 115°
    4. 50°
    Show answer

    Answer: B — 65°

    The angle a chord subtends at the centre is twice the angle it subtends at any point on the major arc, so angle ACB = 130 ÷ 2 = 65°. The trap option 115° is the answer for a point on the minor arc, because ACB and the major-arc angle are then opposite angles of a cyclic quadrilateral and add to 180° (180 − 65 = 115). Always read which arc the point lies on before halving.

  4. The areas of two similar triangles are 81 cm² and 121 cm². A median of the smaller triangle is 12.6 cm. The corresponding median of the larger triangle is:

    1. 13.8 cm
    2. 16.8 cm
    3. 18.9 cm
    4. 15.4 cm
    Show answer

    Answer: D — 15.4 cm

    In similar triangles areas are in the ratio k² while every length — side, median, altitude, angle bisector, inradius — is in the ratio k. Here k = √(81/121) = 9/11. A median is a length, so it scales by 9/11, giving 12.6 × 11 ÷ 9 = 138.6 ÷ 9 = 15.4 cm. The classic error is scaling by the area ratio 121/81 instead, which would give 18.8 cm; remember to take the square root when moving from areas to lengths.

  5. The sides of a triangle are 13 cm, 14 cm and 15 cm. Its area is:

    1. 84 cm²
    2. 91 cm²
    3. 105 cm²
    4. 78 cm²
    Show answer

    Answer: A — 84 cm²

    No height is given, so use the formula of Heron. The semi-perimeter is s = (13 + 14 + 15)/2 = 21, and the area is √[s(s − a)(s − b)(s − c)] = √(21 × 8 × 7 × 6) = √7056 = 84 cm². A useful check: the inradius must be area ÷ s = 84 ÷ 21 = 4 cm, a whole number, which is why examiners like this triangle. Option 105 is ½ × 14 × 15, obtained by wrongly treating two sides as base and height.

  6. A solid metal sphere of radius 6 cm is melted and recast into small cones each of radius 2 cm and height 3 cm. How many such cones are obtained?

    1. 24
    2. 36
    3. 72
    4. 108
    Show answer

    Answer: C — 72

    Melting conserves volume, so the number of pieces is big volume ÷ small volume. Sphere = (4/3)πr³ = (4/3)π × 216 = 288π. Cone = (1/3)πr²h = (1/3)π × 4 × 3 = 4π. Number = 288π ÷ 4π = 72. Notice that π cancels, so never substitute 22/7 in a recasting question — it only creates arithmetic risk. The trap answer 24 comes from forgetting the ⅓ in the cone volume (288 ÷ 12 = 24).

  7. If each edge of a cube is increased by 20%, the percentage increase in its volume is:

    1. 60%
    2. 72.8%
    3. 44%
    4. 20%
    Show answer

    Answer: B — 72.8%

    Treat the change as a multiplier: k = 1.2. Volume depends on the cube of the linear size, so it becomes 1.2³ = 1.728 times, an increase of 72.8%. Verify with numbers: edge 10 gives volume 1000, edge 12 gives 1728, and (1728 − 1000)/1000 × 100 = 72.8% ✓. Option 44% is the increase in surface area (1.2² = 1.44) and option 60% is 3 × 20, which is only a rough approximation valid for very small changes.

  8. In Dataset A of this chapter, 4,000 candidates appeared at Delhi with a 25% pass rate and 5,000 appeared at Jaipur with an 18% pass rate. Taking the two centres together, their combined pass rate is approximately:

    1. 21.5%
    2. 23.5%
    3. 20.0%
    4. 21.1%
    Show answer

    Answer: D — 21.1%

    A combined percentage is (sum of the parts) ÷ (sum of the wholes), never the average of the two percentages. Qualified: Delhi 25% of 4,000 = 1,000 and Jaipur 18% of 5,000 = 900, so 1,900 out of 4,000 + 5,000 = 9,000, giving 1900 ÷ 9000 × 100 = 21.11%, about 21.1%. Option 21.5% is the naive average (25 + 18)/2, which is wrong because Jaipur has the larger group and drags the combined figure towards its lower 18%. Option 23.5% is the overall rate for all five centres, a different question altogether.