Algebra is the language of unknowns. Instead of asking "what number plus 3 equals 7?", algebra lets you write x + 3 = 7 and work systematically. That's all it is at the root — using letters to hold the place of numbers you haven't found yet.
For SSC MTS, algebra shows up in two broad flavors:
Flavor 1 — Simplification using identities. The question gives you a scary-looking expression with cubes, fourth powers, or fractions. The entire point is that a known identity collapses it into something trivial. If you don't see the identity, you're in for two minutes of grinding. If you do, it takes ten seconds.
Flavor 2 — Solving for unknowns. This includes proportion problems, ratio substitution, and quadratic roots. The trick here is almost always substitution — replace variables with their simplest representative values and compute directly.
Think of algebraic identities like power tools. A carpenter could chisel a mortise by hand, but a router does it in five seconds. The "raw arithmetic" approach works — it just costs you time you don't have. Identities are the router.
Here's the inventory you genuinely need for SSC MTS:
a² - b² = (a+b)(a-b)a³ + b³ = (a+b)(a² - ab + b²)a³ - b³ = (a-b)(a² + ab + b²)a⁴ - b⁴ = (a²+b²)(a²-b²) = (a²+b²)(a+b)(a-b)(a+b)² = a² + 2ab + b²(a-b)² = a² - 2ab + b²(x + 1/x)² = x² + 2 + 1/x²For quadratics ax² + bx + c = 0: sum of roots = -b/a, product of roots = c/a. You will rarely need the full quadratic formula in MTS — Sum-Product gets the job done faster.
For proportions a:b = c:d, the cross-multiplication rule is ad = bc. For four numbers in proportion a, b, c, d, it means a/b = c/d, giving ad = bc.
The key skill is pattern recognition before calculation. When you see a fraction where the numerator has cubes or high powers and the denominator has a matching structure, immediately ask: does this match a³ ± b³ or a⁴ - b⁴?
The a³ + b³ pattern:
a³ + b³ = (a+b)(a² - ab + b²)
So (a³ + b³) / (a² - ab + b²) = a + b directly. No multiplication needed.
When you see [(2.7)³ + (1.3)³] / [(2.7)² - 2.7 × 1.3 + (1.3)²], the denominator is exactly a² - ab + b² with a = 2.7, b = 1.3. The whole thing equals a + b = 4.
The a⁴ - b⁴ pattern:
a⁴ - b⁴ = (a² + b²)(a² - b²)
So (a⁴ - b⁴) / (a² + b²) = a² - b², and (a⁴ - b⁴) / (a² - b²) = a² + b².
Further, a² - b² = (a+b)(a-b), which gives you another step of simplification if needed.
When you see [(0.5)⁴ - (0.4)⁴] / [(0.5)² + (0.4)²], the structure matches the first version: answer is (0.5)² - (0.4)² = (0.5+0.4)(0.5-0.4) = 0.9 × 0.1 = 0.09.
When you see [(0.6)⁴ - (0.1)⁴] / [(0.6)² - (0.1)²], it matches the second version: answer is (0.6)² + (0.1)² = 0.36 + 0.01 = 0.37.
This is a classic SSC pattern. If you're given x + 1/x = k, you can chain-square to get x² + 1/x² and then x⁴ + 1/x⁴.
Step 1: (x + 1/x)² = x² + 2 + 1/x², so x² + 1/x² = k² - 2
Step 2: (x² + 1/x²)² = x⁴ + 2 + 1/x⁴, so x⁴ + 1/x⁴ = (k²-2)² - 2
If k = 4: x² + 1/x² = 16 - 2 = 14, then x⁴ + 1/x⁴ = 196 - 2 = 194.
Memorize this chain. You do not need to find the actual value of x — that would be a waste of time.
When a question gives x : y : z = 1 : 2 : 3 and asks you to evaluate a ratio of expressions, don't work abstractly. Assign x = 1, y = 2, z = 3 directly (the smallest integers satisfying the ratio) and compute numerator and denominator separately.
This is faster than algebraic manipulation and eliminates errors. The ratio form means any scalar multiple gives the same answer — so use the simplest values.
For a, b, c, d in proportion: a/b = c/d, so ad = bc.
When the terms involve expressions in x, cross-multiply to get a quadratic, then factor. Look for the Sum-Product factoring route first: if the quadratic is x² + bx + c = 0, find two numbers that multiply to c and add to b.
For x² + px + q = 0 (standard form with leading coefficient 1):
α + β = -pαβ = qThis means you almost never need to actually solve for the roots. If the question asks for 1/α + 1/β, that's just (α + β)/(αβ) — computable from the coefficients alone.
For x² - 5x + 4 = 0: sum = 5, product = 4. So 1/α + 1/β = 5/4. Done.
Before touching any algebra simplification with a fraction, check if the denominator is the "partner factor" of an identity on the numerator. If numerator has a³ + b³ and denominator has a² - ab + b², they are mirror partners — cancel immediately to get a + b. Similarly, a⁴ - b⁴ over a² + b² gives a² - b², and over a² - b² gives a² + b². This pattern recognition takes 3 seconds; brute-force arithmetic takes 60+.
Given x + 1/x = k, memorize: x² + 1/x² = k² - 2 and x⁴ + 1/x⁴ = (k²-2)² - 2. Never try to find x itself — that involves surds and wastes 90 seconds. The chain-squaring method takes under 20 seconds: compute k², subtract 2, square that, subtract 2 again. For k = 4: 16 - 2 = 14, 196 - 2 = 194. Standard approach of solving for x: 90s+. Chain method: 15s.
When a question gives a ratio like x : y : z = 1 : 2 : 3 and asks for a ratio of expressions, plug in x=1, y=2, z=3 directly. Never expand algebraically. Count the steps: algebraic expansion involves collecting 6+ terms; substitution involves 6 arithmetic calculations on single-digit numbers. Substitution saves 4-5 steps and removes the risk of sign errors. Works for any ratio question where the final answer is dimensionless (a number or fraction).
For any question asking about symmetric functions of roots (like 1/a + 1/b, a² + b², a³ + b³), use Vieta's formulas rather than finding the actual roots. 1/a + 1/b = (a+b)/(ab) — readable directly from coefficients. a² + b² = (a+b)² - 2ab — again, no root-finding needed. Finding roots via factoring or formula: 30-40s. Vieta's route: 10s.
For a, b, c, d in proportion, write ad = bc immediately. If the four terms contain x, you get a quadratic. Then check the answer options — try substituting each option value of x back into ad = bc rather than solving the quadratic from scratch. With 2-3 options, this takes 20-30 seconds versus solving the full quadratic at 60+ seconds. Start with middle-value options to narrow down faster.
When an algebra question appears, run this check in order:
Step 1 — Is it a fraction with a cube or fourth-power numerator?
Yes → Check if denominator is the partner factor (a² ∓ ab + b² for cubes, a² ± b² for fourth powers). If yes, cancel and write the answer. Time: 10 seconds.
Step 2 — Does it involve x + 1/x or x - 1/x?
Yes → Use the chain-squaring method. Never solve for x.
Step 3 — Is it a ratio/proportion question with a ratio given like x:y:z?
Yes → Plug in the smallest integer values immediately and compute numerically.
Step 4 — Is it a quadratic with questions about its roots? Yes → Write down sum and product from Vieta's. Express the required quantity in terms of sum/product only.
Step 5 — Is it a proportion with an unknown x?
Yes → Cross-multiply to form a quadratic, then try substituting answer options before fully solving.
If none of the above — it's likely a straightforward linear equation. Isolate the variable in 2-3 steps.
Why this question: The most direct test of whether you know the a³ + b³ identity — without it, this takes 3 minutes of decimal arithmetic.
Solving path: Identify a = 2.7, b = 1.3. Numerator = a³ + b³. Denominator = a² - ab + b². By the identity a³ + b³ = (a+b)(a²-ab+b²), the fraction equals a + b = 2.7 + 1.3 = 4. Answer: option (c).
Why this question: Tests whether you can set up a proportion equation, cross-multiply correctly, and factor the resulting quadratic — all under time pressure.
Solving path: For proportion, (X+2)(3X-5) = (X+5)(2X-3). Expand left: 3X²-5X+6X-10 = 3X²+X-10. Expand right: 2X²-3X+10X-15 = 2X²+7X-15. Equation: 3X²+X-10 = 2X²+7X-15. Simplify: X²-6X+5 = 0. Factor: (X-1)(X-5) = 0, giving X = 1 or X = 5. Answer: option (a).
Why this question: Classic ratio substitution — designed to catch those who attempt abstract algebra when simple substitution works.
Solving path: Set x=1, y=2, z=3. Numerator: 1 + 2(4) + 9 = 1 + 8 + 9 = 18. Denominator: 3(1) - 2(4) + 4(9) = 3 - 8 + 36 = 31. Answer: 18/31, option (c).
Why this question: A 2025 paper question testing the a⁴ - b⁴ identity — two versions appeared in the same year, so you must know which denominator corresponds to which simplification.
Solving path: a⁴ - b⁴ = (a²+b²)(a²-b²). Denominator here is a²+b², so it cancels: answer is a²-b² = (0.5+0.4)(0.5-0.4) = 0.9 × 0.1 = 0.09. Answer: option (c).
Why this question: Vieta's formulas in action — the question is designed to trap those who waste time factoring x² - 5x + 4 = 0 into (x-1)(x-4) = 0 before computing the answer.
Solving path: For x² - 5x + 4 = 0, sum of roots a + b = 5, product ab = 4. Then 1/a + 1/b = (a+b)/(ab) = 5/4. Answer: option (a). You did not need to find that a=1, b=4.
Confusing which identity applies to which denominator in a⁴ - b⁴ questions. If the denominator is a² + b², the answer is a² - b². If the denominator is a² - b², the answer is a² + b². Mixing these up is the single most common error in this question type.
Expanding (x + 1/x)² as x² + 1/x² without the middle term. The correct expansion is x² + 2 + 1/x². Forgetting the +2 gives you x² + 1/x² = k² instead of k² - 2, which cascades into a wrong x⁴ + 1/x⁴.
In proportion cross-multiplication, expanding brackets with sign errors. When you expand (X+2)(3X-5), the constant term is 2 × (-5) = -10. Missing the sign on the cross-terms is the most common algebraic error in proportion problems.
Using ratio values like x=1, y=2, z=3 but then computing y² as 2² correctly yet writing 2y² as 2×2 = 4 instead of 2×4 = 8. The ratio substitution method is fast but requires careful arithmetic on the substituted values — don't rush the computation step after a fast setup.
Applying Vieta's to non-standard form. Vieta's formulas in the form sum = -b/a, product = c/a apply when the equation is ax² + bx + c = 0. If you have 2x² - 10x + 8 = 0, either divide through by 2 first (to get x² - 5x + 4 = 0) or use the general forms sum = 10/2 = 5, product = 8/2 = 4. Applying the mnemonic "sum = middle coefficient" to non-monic quadratics without dividing first leads to wrong answers.
In proportion questions, confirming only one root and missing the second. The quadratic x² - 6x + 5 = 0 has two roots. SSC MTS options often list both roots as a pair. If you find only one root and match it to a partial option, you may pick a wrong answer. Always solve for both roots of a quadratic from a proportion.