Algebraic Expression Tests — Pick a Level and Start
Tap a level to see every test in it. Start at Level 1 the day you finish the identity list, then move to Level 2 with the timer running — that is where two identities get chained inside one question, exactly as the real paper does it.
Level 1 Basic Checking… View tests
One identity applied cleanly — the condition arrives in the form the identity wants and a single substitution finishes it. Sum of cubes when the three letters total zero, a cyclic fraction sum, an equal-ratio chain with numbers supplied, a surd value put into a quadratic, and factorising by splitting the middle term. Attempt these without the clock while the identity list is still fresh.
Level 2 Tier I Standard Checking… View tests
Real SSC CGL Tier I difficulty — two identities chained, or the question run backwards so a value is given and a parameter is asked, or a simplification that has to happen before the identity becomes visible. Rationalising before a power ladder, building the pairwise product sum before the cube-sum formula, recovering a total from a fourth-power sum. This is the score that tells you the chapter is exam-ready.
Algebraic expressions and identities is the chapter that decides how the advanced maths block goes for you. A Tier I shift usually carries two or three questions that belong to it outright — a conditional identity with a + b + c = 0, a cyclic fraction sum, a polynomial evaluated at a surd — and Tier II leans on it far harder. What makes it worth practising is that the questions are not really different from one another: a short list of identities, used forwards, backwards and in pairs, covers almost everything the commission has asked in the last decade.
These practice tests put twenty-five algebra questions in front of you in one sitting, which is what turns the identity list from something you can recite into something you recognise on sight. Every question comes with a full solution and the shortcut, in English or Hindi.
How Many Algebraic Expression Questions Are Asked in SSC CGL?
Algebra is one of the two heaviest blocks in SSC CGL advanced maths, alongside geometry. In Tier I you can expect two to three questions across the algebra chapters in a typical shift, and in Tier II the count rises sharply because the paper has room for longer conditional identities. The questions repeat in shape far more than in content — the same six or seven identities carry almost every one of them, which is why a candidate who has drilled this chapter answers these in under a minute while others spend three.
| Where it appears | Typical count | What is asked |
|---|---|---|
| Tier I — Quantitative Aptitude | 2–3 | Conditional identity with a zero total, cyclic fraction sum, surd substitution, factorisation |
| Tier II — Mathematical Abilities | 3–5 | Two identities chained, inverse framings, equal-ratio chains, maximum and minimum values |
| Indirectly, inside other chapters | — | Simplification, surds and indices, quadratic equations all reuse the same identities |
The indirect row is the one candidates underrate. A simplification question that ends in a difference of squares, or a surd question that needs the cube-sum factorisation, is an algebra question wearing a different label.
What You Will Practice in the Algebraic Expression Chapter
The identities that carry the chapter
- a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). The single most useful line in the chapter.
- (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). This is how you move between the three symmetric sums.
- a² + b² + c² − ab − bc − ca = ½[(a − b)² + (b − c)² + (c − a)²]. Turns a condition into a statement about differences.
- A² − B² = (A − B)(A + B) and A³ ± B³ — the workhorses of every factorisation and every numeric shortcut.
- (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) − abc. Less famous, and it turns up more often than candidates expect.
What a zero total gives you
When a + b + c = 0, three consequences follow immediately, and most questions use two of them together:
- a³ + b³ + c³ = 3abc, because the first factor of the big identity vanishes.
- a + b = −c, b + c = −a, c + a = −b — so every pairwise sum collapses to a single letter.
- a² + b² + c² = −2(ab + bc + ca), from squaring the condition.
From these, the standard cyclic sums fall out: a²/bc + b²/ca + c²/ab is always 3, and a/(b + c) + b/(c + a) + c/(a + b) is always −3.
The condition that forces all three letters equal
If a² + b² + c² = ab + bc + ca, then a = b = c. The reason is worth understanding rather than memorising: the difference between the two sides is half the sum of the three squared differences, and a sum of real squares can only be zero when every square is zero. Once the letters are equal, any homogeneous expression reduces to the ratio of its coefficient sums, which is why these questions look frightening and finish in one line.
Cubic conditions and the divide-through move
A question that opens with something like a³ + 4a² + 16a = 1 and then asks for a³ + 4/a is not asking you to solve the cubic. Divide the condition through by the variable: every power drops by one and the constant becomes the reciprocal term you need. For this whole family the answer is 1 + k³, where k is the middle coefficient.
Equal-ratio chains
When x/(b + c) = y/(c + a) = z/(a + b), name the common value k. Every numerator becomes k times its own denominator, and because each letter sits in two of the three denominators, the numerators total 2k(a + b + c). Differences behave just as neatly: subtracting two numerators leaves k times the difference of their denominators.
Surd values and the power ladder
Never raise a surd to the fourth power directly. If x = 2 + √3, subtract the whole number and square: x² − 4x + 1 = 0, so x² = 4x − 1. Use that relation twice and x³ and x⁴ both come down to linear expressions, after which the whole polynomial collapses. When the surd is nested, as in √(7 − 4√3), unpack it as (p − q)² first.
Also worth memorising
- x⁴ + x² + 1 = (x² + x + 1)(x² − x + 1), reached by adding and subtracting x².
- The sum of the coefficients of a polynomial is its value at x = 1.
- For a cubic x³ + Ax² + Bx + C = 0: the roots sum to −A, their pairwise products to B, and their product to −C.
- For a fixed sum, a product is largest when the two numbers are equal — the basis of every maximum and minimum question here.
Solved Examples — SSC CGL Algebraic Expression
Example 1 — the zero-total identity
If p + q + r = 0 and pqr = −12, find p³ + q³ + r³.
The first factor of the big identity is p + q + r, which is zero, so p³ + q³ + r³ = 3pqr = 3 × (−12) = −36. Nothing else has to be computed — and note that the answer needs the factor 3, which is the commonest slip.
Example 2 — two symmetric sums chained
If a + b + c = 6, a² + b² + c² = 26 and a³ + b³ + c³ = 90, find abc.
First build the missing pairwise sum: 36 = 26 + 2(ab + bc + ca), so ab + bc + ca = 5. Then 90 = 6³ − 3 × 6 × 5 + 3abc = 216 − 90 + 3abc, giving 3abc = −36 and abc = −12. Two identities, in order — this is exactly the Level 2 shape.
Example 3 — the divide-through move
A non-zero a satisfies a³ + 4a² + 16a = 1. Find a³ + 4/a.
Divide the condition by a: a² + 4a + 16 = 1/a. Multiply by 4: 4/a = 4a² + 16a + 64. So a³ + 4/a = a³ + 4a² + 16a + 64, and the first three terms are the given left side, worth 1. The answer is 65.
Example 4 — a surd value in a polynomial
If x = 3 − 2√2, find x² − 6x + 7.
x² = 9 − 12√2 + 8 = 17 − 12√2, and −6x = −18 + 12√2. Adding, the surd parts cancel: 17 − 18 + 7 = 6. Note that (2√2)² is 8, not 4 — squaring only the number and leaving the root behind is the usual error.
Example 5 — a numeric identity shortcut
Evaluate (347² − 153²) ÷ (347 − 153).
The numerator factorises as (347 − 153)(347 + 153), so the bracket cancels and the answer is simply 500. Whenever a question hands you two large numbers, look for the identity before reaching for the multiplication.
How the Algebraic Expression Chapter Tests Work
- 25 questions per test, with a full question palette — answered, skipped, marked for review, not visited.
- Optional 15-minute timer — with the clock to build exam speed, without it while you are still learning the identities.
- Marking: +2 correct, −0.5 wrong, 0 unattempted — the same scheme as the real paper.
- Solutions, shortcut tricks and concept notes open after you submit, question by question.
- English and हिंदी, chosen at the start of every test.
- Free with a TrickySSC account — sign in once with Google or your mobile number, and every test is open to you.
How to Prepare Algebraic Expression for SSC CGL
- Write the identity list on one page and keep it beside you for the first two tests. Six identities cover the chapter; you should be able to recall each one without pausing before you attempt Level 2.
- Read the condition before the expression. Almost every question is decided by what the condition gives you — a zero total, an equal-letters condition, or a cubic to divide through — not by how the expression looks.
- Learn the three consequences of a zero total as a set. Cube sum becomes 3abc, pairwise sums become single letters, sum of squares becomes minus twice the pairwise product sum.
- Practise the identities backwards too. Tier I increasingly gives you the value and asks for a parameter; if you have only ever used an identity in one direction, that framing costs you a minute you do not have.
- For surds, build the quadratic instead of squaring. This one habit removes most of the arithmetic errors in the chapter.
- Do Level 1 untimed, the same day you finish the identity list. The aim is correct recall, not speed.
- Move to Level 2 with the 15-minute clock on. If a conditional identity takes more than about ninety seconds, you are still deriving it rather than recognising it.
- Study it alongside the neighbouring chapters. Do algebraic formula and algebraic result around it — the three share their identities almost entirely.
- Verify against a real paper. Once Level 2 is consistent, attempt a previous year paper and then a full mock test under locked sectional timing.
SSC CGL बीजीय व्यंजक एवं सर्वसमिकाएँ (Algebraic Expression) चैप्टर टेस्ट — हिंदी में
बीजीय व्यंजक SSC CGL गणित के एडवांस भाग का सबसे भारी अध्याय है। टियर I की एक शिफ्ट में इसके 2–3 प्रश्न सीधे आते हैं और टियर II में इससे भी अधिक — जैसे a + b + c = 0 वाली शर्त पर आधारित सर्वसमिका, चक्रीय भिन्नों का योग, या करणी वाले मान को बहुपद में रखकर हल करना। अच्छी बात यह है कि प्रश्न दिखने में अलग होते हैं पर विधि गिनी-चुनी ही है।
TrickySSC पर इस अध्याय के सभी टेस्ट हिंदी में उपलब्ध हैं — प्रश्न, विकल्प, विस्तृत हल और शॉर्टकट ट्रिक सब हिंदी में। दो स्तर हैं: लेवल 1 (आधारभूत) और लेवल 2 (टियर I स्तर)। प्रत्येक टेस्ट में 25 प्रश्न, वैकल्पिक 15 मिनट का टाइमर और +2 / −0.5 की वही मार्किंग जो असली परीक्षा में होती है।
याद रखने योग्य नियम: जब a + b + c = 0 हो तो a³ + b³ + c³ = 3abc होता है, किन्हीं दो अक्षरों का योग तीसरे का ऋणात्मक होता है, और वर्गों का योग युग्मवार गुणनफलों के योग का ऋणात्मक दुगुना होता है। तथा यदि a² + b² + c² = ab + bc + ca हो तो तीनों अक्षर बराबर होते हैं।