Algebra, at its core in SSC GD Maths, is about using a letter to stand in for a number you don't know yet. That letter — x, y, n — is called a चर (variable). An algebraic expression is any combination of constants, variables, and operations: 3x + 7, 2a - b, (n-4)/6.
Here is the analogy that makes this click instantly: think of a variable as a locked box. You know the box exists, you know some rules about what's inside it, and your job is to figure out the number inside. Every equation gives you a clue. Two equations together can crack two locked boxes.
In SSC GD, algebra almost never appears as a naked "solve 3x - 5 = 16" question. Instead it is disguised as a word problem — an age problem, a ratio problem, a number puzzle. The algebraic thinking is the same; only the costume changes.
Inequalities add one more twist: instead of finding the exact value of x, you find the range — all values that satisfy 3x - 2 > 7, for example. For SSC GD, deep inequality theory is rare. What does appear is the inequality hidden inside a word problem ("twice Nikita's age is more than Ankit's age by 4 years" — that "more than" is an equation, not an inequality, but read carefully because the exam sometimes rephrases to make it look like one).
What SSC GD actually tests:
5x and 4x style)If you can translate an English sentence into an algebraic equation in under 15 seconds, you will handle 80% of this chapter's questions on autopilot.
The most important skill here is translation. Every word-problem sentence maps to an algebraic statement:
| English phrase | Algebraic form |
|---|---|
| "sum of A and B is 12 more than sum of B and C" | A + B = B + C + 12 |
| "twice Nikita's age is more than Ankit's by 4" | 2N = A + 4 |
| "a number decreased by 4 and divided by 6 is 8" | (x - 4)/6 = 8 |
| "present ages in ratio 5:4" | S = 5k, A = 4k |
Look — the moment you assign the ratio variable k, you have converted a ratio into algebra. This is the single most reusable move in SSC GD algebra.
When two quantities are in ratio m:n, write them as mx and nx. Then apply the second condition and solve for x. You do not need to name two separate variables.
Example (Sameer and Anand): ratio 5:4 means ages are 5x and 4x. Three years later: (5x+3):(4x+3) = 11:9. Cross-multiply: 9(5x+3) = 11(4x+3) → 45x + 27 = 44x + 33 → x = 6. Anand's age = 4x = 24.
This pattern — ratio, apply future/past condition, solve — repeats across every age-ratio PYQ.
When a problem has two unknowns and gives you two independent conditions, you need two equations. Eliminate one variable:
Substitution: Express one variable in terms of the other from equation 1, substitute into equation 2.
Elimination: Multiply equations to make coefficients of one variable equal, then add or subtract.
For SSC GD, substitution is usually faster because one equation is often already solved for a single variable (e.g., 2N = A + 4 immediately gives A = 2N - 4).
This is a deliberate SSC trap. A question looks solvable but it has more unknowns than equations — you can get a ratio but not the actual value.
Classic form: "The ratio between Neelam's age and Shaan's age is 5:6. If the ratio of one-third of Neelam's age to half of Shaan's age is 5:9 — find Shaan's age."
Test: one-third of 5k is 5k/3; half of 6k is 3k. Their ratio = (5k/3):(3k) = 5/9. This matches the given condition 5:9 for all values of k. The second condition is not independent — it carries zero new information. No unique answer exists. Answer: Cannot be determined.
Rule: if both conditions in a problem reduce to the same relationship, the answer is "Cannot be determined." Don't force a number.
"A number decreased by 4 and divided by 6 is 8." → (x-4)/6 = 8 → x = 52. Now "2 is subtracted and divided by 5" → (52-2)/5 = 50/5 = 10.
Always extract the actual number first, then apply the new operation. Don't try to connect the two operations algebraically — it adds steps.
Let the fraction be p/q. Set up two equations from the two conditions, solve by substitution or elimination. These are straightforward simultaneous equation problems — don't let the fraction notation intimidate you.
Example: "double the numerator, increase denominator by 7 → 2/3; double the denominator, increase numerator by 2 → 3/5."
Equation 1: 2p/(q+7) = 2/3 → 6p = 2q + 14 → 3p = q + 7
Equation 2: (p+2)/(2q) = 3/5 → 5p + 10 = 6q → 5p = 6q - 10
From eq.1: q = 3p - 7. Substitute: 5p = 6(3p-7) - 10 = 18p - 42 - 10 → 5p = 18p - 52 → 13p = 52 → p = 4, q = 5. Fraction = 4/5.
"Sum of a number and its reciprocal is 13/6." Let number = x. → x + 1/x = 13/6 → x² - (13/6)x + 1 = 0 → multiply by 6: 6x² - 13x + 6 = 0. Use Sum-Product: need two numbers that multiply to 36 (= 6×6) and add to 13. Those are 9 and 4. So 6x² - 9x - 4x + 6 = 0 → 3x(2x-3) - 2(2x-3) = 0 → (3x-2)(2x-3) = 0 → x = 2/3 or x = 3/2.
When ages or quantities are in ratio m:n, write them as mx and nx — never as two separate variables x and y. This collapses a 2-variable problem into a 1-variable problem. Example: ratio 3:5 → numbers are 3x and 5x. One equation, one unknown, done in half the steps. Standard 2-variable setup: 6 steps. Single-variable ratio setup: 3 steps.
For equations like (5x+3)/(4x+3) = 11/9, cross-multiply immediately: 9(5x+3) = 11(4x+3). Expand both sides, collect x terms on one side. The x coefficient difference is always small (here: 45x - 44x = x), so x pops out in one step. This beats substitution by 2 steps in every ratio-age problem.
For ax² + bx + c = 0, find two numbers that multiply to a×c and add to b. Split the middle term and factor by grouping. Never reach for the quadratic formula — it's 6 steps minimum. Sum-Product factorisation: 3 steps. For the reciprocal question: 6x² - 13x + 6, find factors of 36 adding to 13 → 9 and 4. Done in 3 steps vs quadratic formula's 6.
Before solving, check: does condition 2 simplify to the exact same ratio as condition 1? Express both conditions in terms of the ratio variable k. If condition 2 holds true for all k, it adds no new information — answer is Cannot be determined. This check takes 10 seconds and saves you from wasting 90 seconds on an unsolvable problem.
For two-stage number puzzles ("find x from operation 1, then apply operation 2"), always extract the actual number first from operation 1, then plug it into operation 2 as a plain arithmetic problem. Trying to connect both operations algebraically adds 3 unnecessary steps. Extraction + arithmetic: 2 steps. Full algebraic path: 5 steps.
When you see an algebra word problem in the exam hall, run through this decision path:
Step 1 — Count unknowns and conditions.
Step 2 — Assign variables smartly.
mx and nx, not x and y.x for the unknown quantity, express others in terms of x.Step 3 — Translate the second condition into an equation. Cross-multiply ratios immediately. Don't leave fractions — they slow you down.
Step 4 — Solve and verify. Substitute your answer back into the original condition in 5 seconds. If it fits, move on. If not, you mis-set-up the equation — re-read the problem once.
Watch for: "more than by" (means difference, not total), "hence" (means in the future), "was" (means past — subtract years), "cannot be determined" (check before calculating).
Why this question: This is the prototype ratio-age problem. Every year SSC GD runs at least one like it.
Solving path: Let ages be 5x (Sameer) and 4x (Anand). Three years hence: (5x+3)/(4x+3) = 11/9. Cross-multiply: 45x + 27 = 44x + 33 → x = 6. Anand = 4×6 = 24.
Why this question: Two conditions, two unknowns — product given (which creates a quadratic) plus a linear relationship. Tests whether you can combine both cleanly.
Solving path: Let Ankit = A, Nikita = N. Given: AN = 240 and 2N = A + 4 → A = 2N - 4. Substitute: (2N-4)N = 240 → 2N² - 4N - 240 = 0 → N² - 2N - 120 = 0. Find factors of 120 that differ by 2: 12 and 10. So (N-12)(N+10) = 0 → N = 12 (age must be positive).
Why this question: The reciprocal equation disguised as a "rational number" puzzle. Tests factorisation of a symmetric quadratic.
Solving path: Let number = x. Then x + 1/x = 13/6. Multiply through by 6x: 6x² - 13x + 6 = 0. Sum-Product on 6×6 = 36, adding to 13: → 9 and 4. Factor: (3x-2)(2x-3) = 0 → x = 2/3 or x = 3/2.
Why this question: Classic ratio problem with subtraction condition. Tests whether you set up the post-subtraction ratio correctly.
Solving path: Numbers are 3x and 5x. After subtracting 9: (3x-9)/(5x-9) = 12/23. Cross-multiply: 23(3x-9) = 12(5x-9) → 69x - 207 = 60x - 108 → 9x = 99 → x = 11. Greater number = 5×11 = 55.
Why this question: This is the "Cannot be determined" trap in classic form. The exam uses it to separate careful readers from mechanical solvers.
Solving path: Let Neelam = 5k, Shaan = 6k. One-third of Neelam = 5k/3. Half of Shaan = 3k. Their ratio = (5k/3)/(3k) = 5/9. This equals the given ratio 5:9 for every value of k — it is not a new equation. No unique value of k (and therefore Shaan's age) can be determined.
Writing two separate variables when a ratio is given. If ages are in ratio 5:4, write 5x and 4x, not a and b. Two variables need two equations; the ratio already gives you one, and many students forget to use it as a constraint, creating an under-determined system.
Misreading "more than by" as addition. "2N is more than A by 4" means 2N - A = 4, not 2N + A = 4. The phrase "by 4" always signals a difference, never a sum.
Applying the second operation to a fraction instead of the number. In number puzzles, always solve for the actual integer first. Then apply the new operation to that integer. Mixing the two operations into one expression is a common algebraic error that produces wrong answers.
Forcing a "Cannot be determined" question to yield a number. If both conditions collapse into the same ratio relationship, there is no unique solution. Don't pick an arbitrary k = 1 and present an answer — the correct answer is "Cannot be determined."
Sign errors in cross-multiplication. When cross-multiplying (5x+3)/(4x+3) = 11/9, students sometimes expand 9(5x+3) as 45x + 3 instead of 45x + 27. Write the expansion explicitly — don't do it in your head.
Rejecting the negative root without checking. In quadratic age problems, both roots are mathematically valid until you apply the constraint that age must be positive. Always write down both roots, then reject the negative/zero one with a clear reason. Skipping this step loses you marks when the examiner checks working.