SSC CGL Quantitative Aptitude Notes: Free Formulas and Worked Examples

Percentage

• x% of y = (x × y) ÷ 100.
• Percentage change = (new value − old value) ÷ old value × 100.
• Two successive changes of a% and b%: net change = a + b + (a × b) ÷ 100. Use a minus sign for a decrease.
• If A is r% more than B, then B is r ÷ (100 + r) × 100% less than A. If A is r% less than B, then B is r ÷ (100 − r) × 100% more than A.
• If a price rises by r%, consumption must fall by r ÷ (100 + r) × 100% to keep spending the same.

Worked example: a price goes up 20% and then down 10%. Net change = 20 − 10 + (20 × −10) ÷ 100 = 10 − 2 = 8% increase. Check: 100 becomes 120, then 108.

Worked example: sugar becomes 25% dearer. To keep the monthly bill unchanged, a family must cut consumption by 25 ÷ 125 × 100 = 20%.

Profit, loss and discount

• Profit = SP − CP and profit% = profit ÷ CP × 100. Loss% is also calculated on CP.
• SP = CP × (100 + profit%) ÷ 100, or SP = CP × (100 − loss%) ÷ 100.
• Discount is always on the marked price (MP): SP = MP × (100 − discount%) ÷ 100.
• MP ÷ CP = (100 + profit%) ÷ (100 − discount%).
• Two successive discounts of a% and b% equal one discount of a + b − (a × b) ÷ 100.
• A trader who sells at cost price but uses a false weight gains (true weight − false weight) ÷ false weight × 100%.

Worked example: goods are marked 40% above cost and sold at a 25% discount. SP = 1.40 × 0.75 = 1.05 times CP, so the profit is 5%.

Worked example: discounts of 20% and 10% equal a single discount of 20 + 10 − 2 = 28%. Check: 100 becomes 80, then 72.

Worked example: a shopkeeper sells at cost price but gives 900 g for every kilogram. Gain = 100 ÷ 900 × 100 = 11.11%.

Simple and compound interest

• Simple interest: SI = P × R × T ÷ 100, and amount = P + SI.
• Compound interest, yearly: amount A = P × (1 + R/100)^T, and CI = A − P.
• Half-yearly compounding: use R/2 as the rate and 2T as the number of periods. Quarterly: R/4 and 4T.
• Different rates in different years: A = P × (1 + R1/100) × (1 + R2/100) × and so on.
• CI − SI for 2 years = P × (R/100)². For 3 years = P × (R/100)² × (3 + R/100).
• At simple interest, a sum doubles in T years when R = 100 ÷ T.

Worked example: ₹10,000 at 10% a year for 2 years. SI = 10,000 × 10 × 2 ÷ 100 = ₹2,000. With compounding, A = 10,000 × 1.1 × 1.1 = ₹12,100, so CI = ₹2,100. The difference, ₹100, matches 10,000 × (0.1)² = ₹100.

Worked example: for 3 years at the same rate, CI − SI = 10,000 × 0.01 × 3.1 = ₹310. Check: CI = 13,310 − 10,000 = ₹3,310 and SI = ₹3,000.

Worked example: a sum doubles in 8 years at simple interest, so R = 100 ÷ 8 = 12.5% a year.

Ratio, proportion and partnership

• If a : b = c : d, then a × d = b × c.
• Mean proportional of a and b = √(ab). Third proportional to a and b = b² ÷ a. Fourth proportional to a, b and c = (b × c) ÷ a.
• To join A : B and B : C into one ratio, make the two B terms equal.
• In a partnership, profit is shared in the ratio of capital × time.

Worked example: A : B = 2 : 3 and B : C = 4 : 5. Make B equal to 12: A : B = 8 : 12 and B : C = 12 : 15, so A : B : C = 8 : 12 : 15.

Worked example: A puts in ₹30,000 for 12 months and B puts in ₹40,000 for 6 months. Ratio = 3,60,000 : 2,40,000 = 3 : 2. From a profit of ₹25,000, A gets ₹15,000 and B gets ₹10,000.

Averages

• Average = sum of values ÷ number of values.
• Average of the first n natural numbers = (n + 1) ÷ 2.
• Adding the same number k to every value raises the average by k.
• When one member of a group of n is replaced and the average rises by x, the newcomer's value = value of the person replaced + n × x.

Worked example: the average weight of 8 people rises by 2.5 kg when a new person replaces someone weighing 65 kg. New person = 65 + 8 × 2.5 = 85 kg.

Mixture and alligation

• Alligation rule: quantity of cheaper ÷ quantity of dearer = (dearer price − mean price) ÷ (mean price − cheaper price).
• Repeated replacement: if x litres are taken out of V litres of a pure liquid and replaced with water, n times over, the pure liquid left = V × (1 − x/V)^n.

Worked example: rice at ₹40 a kg is mixed with rice at ₹55 a kg to get a mix worth ₹45 a kg. Ratio = (55 − 45) : (45 − 40) = 10 : 5 = 2 : 1. Check: 2 kg × 40 + 1 kg × 55 = ₹135 for 3 kg, which is ₹45 a kg.

Worked example: a can holds 40 litres of milk. 4 litres are taken out and replaced with water, and this is done twice. Milk left = 40 × (36/40)² = 40 × 0.81 = 32.4 litres.

Time and work (with pipes and cisterns)

• If A finishes a job in a days, A does 1/a of it each day.
• A and B together finish in (a × b) ÷ (a + b) days.
• LCM method: take total work = LCM of the days; each worker's daily units = total ÷ their days.
• Men, days and hours: (M1 × D1 × H1) ÷ W1 = (M2 × D2 × H2) ÷ W2.
• Pipes: a filling pipe adds work and a leak or outlet subtracts it.

Worked example: A takes 12 days and B takes 18 days. Total work = LCM = 36 units. A does 3 units a day and B does 2, so together 5 a day: 36 ÷ 5 = 7.2 days. The formula agrees: 12 × 18 ÷ 30 = 7.2.

Worked example: pipe A fills a tank in 10 hours and pipe B empties it in 15 hours. With both open, each hour fills 1/10 − 1/15 = 1/30 of the tank, so it takes 30 hours.

Worked example: 12 men working 8 hours a day finish a job in 15 days. 10 men working 9 hours a day need 12 × 15 × 8 ÷ (10 × 9) = 1,440 ÷ 90 = 16 days.

Time, speed and distance

• Distance = speed × time. km/h to m/s: multiply by 5/18. m/s to km/h: multiply by 18/5.
• Average speed over two equal distances at x and y = 2xy ÷ (x + y).
• Relative speed: add the speeds when moving in opposite directions, subtract when moving in the same direction.
• A train passing a pole or a person covers its own length; passing a platform or another train, it covers the sum of both lengths.
• Boats: downstream speed = boat + stream; upstream speed = boat − stream. Boat speed = (down + up) ÷ 2 and stream speed = (down − up) ÷ 2.

Worked example: a 240 m train at 72 km/h crosses a 360 m platform. Speed = 72 × 5/18 = 20 m/s. Time = (240 + 360) ÷ 20 = 30 seconds.

Worked example: you travel to a town at 40 km/h and return at 60 km/h. Average speed = 2 × 40 × 60 ÷ 100 = 48 km/h, not 50.

Worked example: a boat goes 18 km/h downstream and 12 km/h upstream. Boat speed = 15 km/h and stream speed = 3 km/h.

Algebra identities

• (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².
• a² − b² = (a + b)(a − b).
• (a + b)³ = a³ + b³ + 3ab(a + b) and (a − b)³ = a³ − b³ − 3ab(a − b).
• a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).
• a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). So if a + b + c = 0, then a³ + b³ + c³ = 3abc.
• If x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k. If x − 1/x = k, then x² + 1/x² = k² + 2.
• The line ax + by = c meets the x-axis at (c/a, 0) and the y-axis at (0, c/b).

Worked example: x + 1/x = 4. Then x² + 1/x² = 16 − 2 = 14 and x³ + 1/x³ = 64 − 12 = 52.

Worked example: take a = 1, b = 2 and c = −3, so a + b + c = 0. Then a³ + b³ + c³ = 1 + 8 − 27 = −18, and 3abc = 3 × 1 × 2 × (−3) = −18, just as the identity says.

Worked example: the line 2x + 3y = 12 meets the axes at (6, 0) and (0, 4). The triangle it forms with the axes has area ½ × 6 × 4 = 12 square units.

Geometry facts

Triangles and polygons

• The angles of a triangle add up to 180°. An exterior angle equals the sum of the two opposite interior angles.
• Centroid: where the medians meet. It divides each median in the ratio 2 : 1, measured from the vertex.
• Incentre: where the angle bisectors meet. Angle BIC = 90° + A/2.
• Circumcentre: where the perpendicular bisectors of the sides meet. Angle BOC = 2A when angle A is acute. In a right triangle it is the midpoint of the hypotenuse.
• Orthocentre: where the altitudes meet. In an acute triangle, angle BHC = 180° − A.
• Congruence tests: SSS, SAS, ASA, AAS and RHS. In similar triangles, areas are in the ratio of the squares of corresponding sides.
• Pythagorean triplets to know: 3-4-5, 5-12-13, 8-15-17 and 7-24-25.
• Equilateral triangle of side a: height = (√3/2)a, area = (√3/4)a², inradius = a ÷ (2√3), circumradius = a ÷ √3.
• Right triangle with legs a, b and hypotenuse c: inradius = (a + b − c) ÷ 2 and circumradius = c ÷ 2.
• The interior angles of an n-sided polygon add up to (n − 2) × 180°. Each exterior angle of a regular polygon = 360° ÷ n.

Circles

• The angle an arc makes at the centre is twice the angle it makes at any point on the rest of the circle.
• The angle in a semicircle is 90°, and angles in the same segment are equal.
• Opposite angles of a cyclic quadrilateral add up to 180°.
• A perpendicular from the centre to a chord bisects the chord. A tangent is perpendicular to the radius at the point of contact.
• The two tangents from an outside point are equal. Tangent length from a point at distance d from the centre = √(d² − r²).
• Two chords AB and CD meeting at P (inside or outside the circle): PA × PB = PC × PD. For a tangent PT and a secant PAB: PT² = PA × PB.
• Two circles with radii r1 and r2 whose centres are d apart: direct common tangent = √(d² − (r1 − r2)²); transverse common tangent = √(d² − (r1 + r2)²).

Worked example: in triangle ABC, angle A = 70°. Angle BIC at the incentre = 90° + 35° = 125°.

Worked example: circles of radii 8 cm and 3 cm have centres 13 cm apart. Direct common tangent = √(169 − 25) = 12 cm. Transverse common tangent = √(169 − 121) = √48 = 4√3 cm.

Worked example: each interior angle of a regular hexagon = (6 − 2) × 180° ÷ 6 = 120°.

Mensuration formulas

Worked example: a cone has radius 7 cm and height 24 cm. Slant height = √(49 + 576) = 25 cm. Curved surface = (22/7) × 7 × 25 = 550 cm². Volume = ⅓ × (22/7) × 49 × 24 = 1,232 cm³.

Worked example: a triangle has sides 13, 14 and 15 cm. s = 21, so area = √(21 × 8 × 7 × 6) = √7,056 = 84 cm².

Worked example: a solid hemisphere of radius 7 cm has total surface area 3 × (22/7) × 49 = 462 cm².

Trigonometry values and identities

• sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
• Complementary angles: sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, sec(90° − θ) = cosec θ. So if A + B = 90°, then sin A = cos B and tan A × tan B = 1.
• π radians = 180°. Degrees to radians: multiply by π/180. Radians to degrees: multiply by 180/π.
• The greatest value of a sin θ + b cos θ is √(a² + b²), and the least is −√(a² + b²).
• Heights and distances: tan(angle of elevation) = height ÷ horizontal distance.

Worked example: sec θ + tan θ = 2. Since sec²θ − tan²θ = 1, sec θ − tan θ = 1/2. Adding the two equations, 2 sec θ = 5/2, so sec θ = 5/4.

Worked example: tan 1° × tan 2° × ... × tan 89° = 1, because each pair such as tan 1° × tan 89° equals 1, and the middle term tan 45° is also 1.

Worked example: from a point 30 m from the foot of a tower, the angle of elevation of the top is 60°. Height = 30 × tan 60° = 30√3, about 51.96 m.

Worked example: 5π/6 radians = 5 × 180° ÷ 6 = 150°.

Data interpretation and statistics: quick points

• In a pie chart, a sector's share = its central angle ÷ 360° × 100%. A 72° sector is 20% of the total.
• Mean = sum ÷ count. Median = the middle value after sorting (the average of the two middle values when the count is even). Mode = the most frequent value.
• For Tier 2: standard deviation measures spread around the mean, and the probability of an event = favourable outcomes ÷ total equally likely outcomes.

Worked example: for the data 2, 4, 4, 4, 5, 5, 7, 9 the mean is 40 ÷ 8 = 5. The squared deviations add up to 32, so the variance is 32 ÷ 8 = 4 and the standard deviation is 2.

Worked example: two coins are tossed. HH, HT, TH and TT are equally likely, so the chance of exactly one head is 2/4 = 1/2.

How to use these notes

1. Copy each formula into your own notebook once. Writing it out fixes it in memory far better than reading.
2. Cover the worked example, solve it yourself, then compare.
3. Straight after, do that topic's previous-year SSC questions against a timer.
4. Revise this page every week and mark the formulas you still forget.

For more free notes, topic by topic, look on Sikhami. You can also test a topic in a quiz battle there or join a study room to work through problems with other learners.

These notes follow the topic list in the official CGL 2026 notice on ssc.gov.in (paragraphs 13.10 and 13.11). Check the notice for the full syllabus and any updates.