Tier 2 mathematics for SSC CGL: what changes

A candidate who solved 22 of 25 Tier 1 maths questions often expects Tier 2 to feel similar. Then the paper opens with a geometry question that needs two theorems, an algebra question built on a surd, and a DI question where the options are within one percent of each other. Tier 2 mathematics doesn't add new chapters. It combines concepts, tightens the options and raises the penalty for errors.

That is good news in one way: you don't need a new syllabus. You need a different habit of solving.

Why Tier 2 maths decides ranks

Tier 1 is qualifying. Your rank is built in Tier 2 Paper I, where the scored sections add up to 390 marks and maths contributes 90 of them: 30 questions at 3 marks each. That is almost a quarter of the merit paper. Maths also shares Section I with Reasoning, so time you lose on one maths question comes out of both.

Where the extra difficulty shows

Algebra

Surds (x = √5 + 2), higher powers of x ± 1/x, and conditions that force values (if a² + b² + c² = ab + bc + ca, then a = b = c). The trap is usually a sign: x − 1/x and x + 1/x lead to different identities.

Geometry

Questions that combine a circle theorem with triangle properties: tangents with Pythagoras, chords with similar triangles, centres of a triangle with angle chasing.

Trigonometry

Longer identities, maximum and minimum values, and heights and distances with two angles of elevation. Minimum-value questions often hide a condition that rules out the "obvious" formula.

Arithmetic

Multi-stage stories: markup, discount and false weights in one question; compound interest paid in instalments; time and work with people joining or leaving.

DI and statistics

Longer sets with closer options, plus mean, median, mode and basic probability. See data interpretation for SSC for worked sets.

Five worked Tier 2-style questions

1. Algebra. If x = √5 + 2, find x³ − 1/x³.

1/x = 1/(√5 + 2) = (√5 − 2)/(5 − 4) = √5 − 2.
So x − 1/x = 4.
x³ − 1/x³ = (x − 1/x)³ + 3(x − 1/x) = 64 + 12 = 76.

The trap: using the x + 1/x identity, (x + 1/x)³ − 3(x + 1/x), which is for x³ + 1/x³.

2. Arithmetic. A shopkeeper marks goods 40% above cost, allows a 10% discount and uses a 900 g weight in place of 1 kg. Find the profit percentage.

Let 1 kg cost ₹100. Marked price = ₹140. Selling price after discount = ₹126.
But the customer receives only 900 g, which costs the shopkeeper ₹90.
Profit = (126 − 90)/90 × 100 = 40%.

Three concepts in one line: markup, discount and false weight. Solve each stage on a base of 100.

3. Geometry. From a point P, two tangents PA and PB are drawn to a circle with centre O and radius 5 cm. If OP = 13 cm, find the length of chord AB.

A tangent is perpendicular to the radius, so triangle OAP is right-angled at A: PA = √(13² − 5²) = √144 = 12.
AB is perpendicular to OP and is bisected by it. The half-chord is the altitude from A to the hypotenuse OP: OA × PA/OP = 5 × 12/13 = 60/13.
AB = 2 × 60/13 = 120/13 cm ≈ 9.23 cm.

4. Trigonometry. Find the minimum value of 9 sin²θ + 16 cosec²θ.

The reflex answer uses AM ≥ GM: 2√(9 × 16) = 24. But equality would need 9 sin²θ = 16 cosec²θ, that is sin²θ = 4/3, which is impossible.
Put s = sin²θ, with 0 < s ≤ 1. The expression 9s + 16/s decreases as s rises towards 4/3, so on this range it is smallest at s = 1.
Minimum = 9 + 16 = 25.

This is exactly the kind of question where 24 is among the options.

5. Compound interest. A loan is repaid in two equal annual instalments of ₹4,410 at 5% per annum compound interest. Find the loan amount.

Each instalment is discounted back to today.
First: 4,410/1.05 = 4,200. Second: 4,410/1.05² = 4,410/1.1025 = 4,000.
Loan = ₹8,200.

The arithmetic of guessing

With +3 for a right answer and −1 for a wrong one, a blind guess among four options gives, on average, (1/4 × 3) − (3/4 × 1) = 0. It neither helps nor hurts over many questions.

Eliminate one wrong option and the average becomes (1/3 × 3) − (2/3 × 1) = 1/3 mark per guess. Eliminate two and it is (1/2 × 3) − (1/2 × 1) = 1 mark.

So the rule is simple. Don't guess blind. If you can rule out one or two options with confidence, an educated guess is worth taking, but keep such guesses few, because each one adds risk.

Accuracy techniques

1. Re-read the question before choosing. Tier 2 wording often hides a twist: "not", "except", "approximately", "minimum".
2. Check units and the base for every percentage.
3. Verify by substitution in algebra and trigonometry. Put θ = 45° or x = 1 into the options when the question allows it.
4. Estimate first. If your calculated answer is far from your estimate, recheck before marking.
5. Check conditions before applying a formula, as in worked question 4.

Setting your attempt target

Use mocks to find the number of questions you can attempt with 90% or better accuracy.

• 26 attempts with 24 correct: 72 − 2 = 70 marks.
• 30 attempts with 24 correct: 72 − 6 = 66 marks.

The candidate who attempted fewer scored more. Once your target is set, the second job is raising the number of questions you can solve at that accuracy, not raising attempts for their own sake.

A four-week Tier 2 maths plan

For how maths fits with the other sections of Paper I, see the Tier 2 strategy guide, and for more surd and identity practice, advanced algebra.

Practice

1. If x + 1/x = √5, find x³ + 1/x³.
2. Successive discounts of 20%, 10% and 5% are equivalent to what single discount?
3. A sum becomes ₹13,310 in 3 years at 10% compound interest. Find the principal.
4. In triangle ABC, ∠B = 60° and ∠C = 40°. The bisectors of ∠B and ∠C meet at I. Find ∠BIC.
5. The mean of 20 observations is 15. One observation, 34, was wrongly recorded as 43. Find the correct mean.
6. A can finish a job in 12 days and B in 18 days. They work together for 4 days, then A leaves. In how many more days does B finish the job?
7. If sin θ + cos θ = √2, find tan θ + cot θ.

Answers

1. 2√5. x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 5√5 − 3√5.
2. 31.6%. 0.8 × 0.9 × 0.95 = 0.684, so the buyer pays 68.4%.
3. ₹10,000. 13,310 = P × 1.1³ = P × 1.331.
4. 130°. ∠A = 180 − 60 − 40 = 80°; ∠BIC = 90° + A/2 = 90° + 40°.
5. 14.55. Sum = 300; corrected sum = 300 − 43 + 34 = 291; 291/20 = 14.55.
6. 8 days. Together they do 1/12 + 1/18 = 5/36 a day; in 4 days, 20/36 = 5/9. Remaining 4/9 at 1/18 a day = 8 days.
7. 2. Squaring, 1 + 2 sin θ cos θ = 2, so sin θ cos θ = 1/2. tan θ + cot θ = 1/(sin θ cos θ) = 2.