SSC CGL Algebra Questions with Answers

SSC CGL Algebra Questions with Answers: Quick Answer

This page solves 8 official SSC CGL algebra questions with answers, every one taken from the SSC CGL 2025 Tier-1 Quantitative Aptitude section papers held on 12, 13 and 14 September 2025. The count is 8 because that is how many questions in those five official shift papers genuinely use an algebraic identity, surd rationalisation, or algebraic manipulation of a variable. No question has been added from practice papers, memory-based papers, or any other source, and no count has been rounded up. Each question below shows the original options, the correct answer, a shortcut, and full step-by-step working, so you can see both the fast exam method and the complete reasoning. As of 2 November 2026, these are the algebra-identity questions from the five official 2025 Tier-1 section papers held on this site.

The 8 questions fall into three families that repeat in SSC CGL papers. Two questions ask for x² + 1/x² or x³ + 1/x³ from a given value of x + 1/x. Four questions need surd rationalisation, where multiplying by the conjugate removes the square root from a denominator. Two questions use cubes: one is a sum of three cubes, and one is a ratio of two sums of cubes that collapses to a single fraction. Learn the three families and you will recognise the next question of each type in seconds, even when the numbers change.

For the full shift papers these questions came from, read the SSC CGL previous year question paper note. To attempt a complete 100-question paper under time pressure, try the SSC CGL 2025 Tier 1 Question Paper 12 September Shift 1 test. For more timed algebra practice, try the SSC CGL Algebra Mock Test 01 test.

Algebra Identities and Surd Shortcuts Used in These Questions

Keep this box open while you work through the questions. Every solution below uses one or more of these lines, and nothing else fancy is needed.

The x + 1/x pair. Squaring x + 1/x gives x² + 2 + 1/x², because the middle term is 2 × x × 1/x = 2. So x² + 1/x² = (x + 1/x)² − 2. Cubing gives a similar result: x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x), because the middle terms add up to 3(x + 1/x). If you remember only two lines for this topic, remember these two. They turn a question that looks like hard algebra into one subtraction.

Rationalising a surd denominator. For a denominator of the form a − √b, multiply the top and bottom of the fraction by the conjugate a + √b. The denominator becomes a² − b, which has no square root left in it: (a − √b)(a + √b) = a² − b. The same idea works the other way round for a + √b. A useful special case is 1/(√n − √(n − 1)) = √n + √(n − 1), because the denominator becomes n − (n − 1) = 1. When several such fractions appear with alternating plus and minus signs, neighbouring terms cancel and only the first and last pieces survive. That cancelling pattern is called telescoping, and Question 5 below is a textbook example of it.

The three-cubes identity. For any three numbers a, b and c: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). It looks long, but in exam questions the numbers are chosen so that a + b + c is small, which keeps the arithmetic short. The companion fact for two cubes is a³ + b³ = (a + b)(a² − ab + b²), which is what makes Question 8 collapse: when the denominator is a scaled-up copy of the numerator, the whole bracket cancels and only the scale factor is left.

A warning that saves marks. Never square or cube only part of an expression. In x + 1/x questions, the common slip is writing x² + 1/x² = (x + 1/x)² and forgetting to subtract 2. In surd questions, the common slip is rationalising only the first fraction and then guessing the rest. Work every term, write every line, and check the last line against the options before moving on. Negative marking makes a guessed answer cost more than a skipped one.

Question 1 — x + 1/x = 5, Find x² + 1/x²

Previously asked in SSC CGL 2025 Tier-I, 13 September 2025, Shift 2.

Given, x + 1/x = 5, then determine the value of x² + 1/x².

1. 25
2. 24
3. 26
4. 23

Answer: 23 (option 4).

Shortcut. Use x² + 1/x² = (x + 1/x)² − 2 straight away. That is 5² − 2 = 25 − 2 = 23. This is the whole question once the identity is known, and it takes under ten seconds.

Full working. Square both sides of the given equation. The left side expands as (x + 1/x)² = x² + 2 × x × (1/x) + 1/x² = x² + 2 + 1/x², because x × 1/x = 1. The right side is 5² = 25. So x² + 2 + 1/x² = 25, and subtracting 2 from both sides gives x² + 1/x² = 23. The answer is 23.

Check against the traps. Option 1, 25, is what you get if you square 5 and forget to subtract 2. Option 3, 26, would come from adding 2 instead of subtracting it. The middle term in the square is always +2 for x + 1/x, so the correction is always a subtraction of 2, never an addition.

Question 2 — x + 1/x = 4, Find x³ + 1/x³

Previously asked in SSC CGL 2025 Tier-I, 14 September 2025, Shift 1.

Given, x + 1/x = 4 find the value of x³ + 1/x³.

1. 55
2. 57
3. 52
4. 50

Answer: 52 (option 3).

Shortcut. Use x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x). That is 4³ − 3 × 4 = 64 − 12 = 52. One line, no expansion needed.

Full working. Cube both sides of x + 1/x = 4. The expansion is (x + 1/x)³ = x³ + 3 × x² × (1/x) + 3 × x × (1/x²) + 1/x³ = x³ + 3x + 3/x + 1/x³ = x³ + 1/x³ + 3(x + 1/x). The right side is 4³ = 64. So x³ + 1/x³ + 3 × 4 = 64, which means x³ + 1/x³ + 12 = 64, and therefore x³ + 1/x³ = 52. The answer is 52.

Why the shortcut is worth memorising. The same pattern appears with different numbers in almost every SSC CGL cycle: x + 1/x = 5 gives x³ + 1/x³ = 125 − 15 = 110, and x + 1/x = 3 gives 27 − 9 = 18. If you can cube a single digit and subtract three times the given value, you already own this family. Do not solve for x itself; finding x from x + 1/x = 4 means solving a quadratic, which wastes minutes and invites arithmetic slips.

Question 3 — x = √7, Find x + 1/x

Previously asked in SSC CGL 2025 Tier-I, 12 September 2025, Shift 2.

If x = √7, determine the value of x + 1/x.

1. 8√7/7
2. 7√7/8
3. 7√7/9
4. 9√7/7

Answer: 8√7/7 (option 1).

Shortcut. Write 1/x with a rationalised denominator first: 1/√7 = √7/7. Then x + 1/x = √7 + √7/7 = (7√7 + √7)/7 = 8√7/7. Rationalising before adding is the whole trick.

Full working. Start with x + 1/x = √7 + 1/√7. Put both terms over the common denominator √7: √7 = 7/√7, because √7 × √7 = 7. So the sum is (7 + 1)/√7 = 8/√7. Now rationalise the denominator by multiplying top and bottom by √7: 8/√7 = 8√7/7. That matches option 1, and it is the only option with 8 in the numerator and 7 in the denominator. The answer is 8√7/7.

What to watch. Option 4, 9√7/7, comes from adding 7 + 1 + 1 instead of 7 + 1 in the numerator. Option 2 inverts the fraction. If your answer still has a square root in the denominator, you have not finished: SSC options are always written with a rationalised denominator, so a final answer of 8/√7 is a sign to take one more step.

Question 4 — Surd Inside a Square Root, Find x² + x − 9

Previously asked in SSC CGL 2025 Tier-I, 12 September 2025, Shift 2.

If x = √((2 + √3)/(2 − √3)), then determine the value of x² + x − 9?

1. 3√3
2. 5√3
3. 7√3
4. 9√3

Answer: 5√3 (option 2).

Shortcut. Rationalise inside the square root first: (2 + √3)/(2 − √3) is a perfect square, (2 + √3)². So x = 2 + √3, and everything after that is simple substitution.

Full working. Multiply the top and bottom of (2 + √3)/(2 − √3) by the conjugate (2 + √3). The denominator becomes (2 − √3)(2 + √3) = 4 − 3 = 1, and the numerator becomes (2 + √3)². So the fraction equals (2 + √3)², and x = √((2 + √3)²) = 2 + √3, taking the positive root since 2 + √3 is positive. Now square it: x² = (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3. Then x² + x − 9 = (7 + 4√3) + (2 + √3) − 9 = 5√3, because 7 + 2 − 9 = 0 and 4√3 + √3 = 5√3. The answer is 5√3.

Why this question is easier than it looks. The expression under the square root is designed to be a perfect square, and the −9 in the question is designed to cancel the whole-number part exactly. If your whole numbers do not cancel to zero at the end, recheck the squaring step: (2 + √3)² has a middle term of 2 × 2 × √3 = 4√3, which is the term most often written as 2√3 by mistake.

Question 5 — Telescoping Surd Expression

Previously asked in SSC CGL 2025 Tier-I, 13 September 2025, Shift 2.

Evaluate the expression: 1/(3 − √8) − 1/(√8 − √7) + 1/(√7 − √6) − 1/(√6 − √5) + 1/(√5 − 2) = ?

1. 5
2. 3
3. 2
4. 0

Answer: 5 (option 1).

Shortcut. Rationalise every term. Each denominator becomes 1, each fraction becomes a sum of two square roots, and the alternating signs cancel everything except 3 + 2 = 5. Recognising the telescoping pattern means you never need to handle the five fractions together.

Full working. Rationalise term by term, using 1/(√n − √(n − 1)) = √n + √(n − 1):

• 1/(3 − √8): multiply by (3 + √8); the denominator is 9 − 8 = 1, so the term equals 3 + √8.
• 1/(√8 − √7): multiply by (√8 + √7); the denominator is 8 − 7 = 1, so the term equals √8 + √7.
• 1/(√7 − √6) equals √7 + √6.
• 1/(√6 − √5) equals √6 + √5.
• 1/(√5 − 2): multiply by (√5 + 2); the denominator is 5 − 4 = 1, so the term equals √5 + 2.

Now apply the signs in the question: (3 + √8) − (√8 + √7) + (√7 + √6) − (√6 + √5) + (√5 + 2). The √8 terms cancel, the √7 terms cancel, the √6 terms cancel, and the √5 terms cancel. Only 3 + 2 = 5 remains. The answer is 5.

Exam tip. The denominators in this question step down by one each time: 3 and √8, then √8 and √7, then √7 and √6, and so on. That chain is the signal. Whenever you see a chain of surd fractions joined by alternating signs, rationalise one term, check that its denominator becomes 1, and then trust the cancelling pattern instead of expanding everything at once.

Question 6 — Two Conjugate Fractions Minus a Third

Previously asked in SSC CGL 2025 Tier-I, 12 September 2025, Shift 3.

Simplify: (1/(7 + √2)) + (1/(7 − √2)) − (14/(49 − 2))

1. 0
2. 1
3. 2
4. √2

Answer: 0 (option 1).

Shortcut. The first two fractions are conjugates, so their sum is 14/(49 − 2) = 14/47. The third term subtracts exactly that, leaving 0. Spotting the conjugate pair answers the question in one step.

Full working. Add the first two fractions over the common denominator (7 + √2)(7 − √2) = 49 − 2 = 47. The numerator is (7 − √2) + (7 + √2) = 14, because the √2 terms cancel. So the sum of the first two fractions is 14/47. The third term is 14/(49 − 2) = 14/47. Subtracting gives 14/47 − 14/47 = 0. The answer is 0.

What to watch. The third term is written as 14/(49 − 2) rather than 14/47 on purpose, to hide that it is identical to the sum of the first two fractions. Always simplify a denominator like 49 − 2 before deciding whether two terms match. Option 4, √2, is the leftover if the surd terms are added instead of cancelled in the numerator.

Question 7 — Sum of Three Cubes Plus 210

Previously asked in SSC CGL 2025 Tier-I, 12 September 2025, Shift 1.

31³ + 18³ − 37³ + 210 is equal to:

1. −36810
2. −14820
3. −45670
4. −23450

Answer: −14820 (option 2).

Shortcut. Set a = 31, b = 18 and c = −37, so the expression is a³ + b³ + c³ + 210. The three-cubes identity gives a³ + b³ + c³ = 3abc + (a + b + c)(a² + b² + c² − ab − bc − ca). Here a + b + c = 31 + 18 − 37 = 12, and the bracket works out to 3909, so a³ + b³ + c³ = 3 × 31 × 18 × (−37) + 12 × 3909 = −61938 + 46908 = −15030. Adding 210 gives −14820. The identity route avoids cubing three two-digit numbers separately, but the direct route below checks the same answer and is just as valid.

Full working. Compute each cube directly: 31³ = 29791, 18³ = 5832, and 37³ = 50653. Then 29791 + 5832 − 50653 + 210 = 35623 − 50653 + 210 = −15030 + 210 = −14820. Both routes agree, which is a strong check when the options are all large negative numbers close enough to punish one slipped digit. The answer is −14820.

Why the identity helps here. The numbers 31, 18 and 37 are chosen so that 31 + 18 − 37 = 12, a small number, and the +210 at the end is part of the same design. If you only ever cube directly, this question is three multiplications and a subtraction, which is fine but slow. If you recognise the a³ + b³ + c³ shape, the identity gives you a second, independent route to the same answer, and two agreeing routes mean the answer is safe.

Question 8 — Ratio of Two Sums of Cubes

Previously asked in SSC CGL 2025 Tier-I, 14 September 2025, Shift 1.

What is the value of (0.1 × 0.1 × 0.1 + 0.03 × 0.03 × 0.03)/(0.3 × 0.3 × 0.3 + 0.09 × 0.09 × 0.09).

1. 0.027
2. 0.009
3. 0.729
4. 0.037

Answer: 0.037 (option 4).

Shortcut. Let a = 0.1 and b = 0.03. The numerator is a³ + b³. The denominator is (3a)³ + (3b)³ = 27(a³ + b³), because 0.3 = 3 × 0.1 and 0.09 = 3 × 0.03. The bracket cancels, leaving 1/27, which is 0.037037… and rounds to the printed option 0.037. No cube needs to be worked out.

Full working. Write the numerator as a³ + b³ with a = 0.1 and b = 0.03, and factor it: a³ + b³ = (a + b)(a² − ab + b²). The denominator is 0.3³ + 0.09³. Since 0.3 = 3a and 0.09 = 3b, the denominator is (3a)³ + (3b)³ = 27a³ + 27b³ = 27(a³ + b³). Dividing gives (a³ + b³) ÷ (27(a³ + b³)) = 1/27. As a decimal, 1/27 = 0.037037…, so the option that matches is 0.037. The answer is 0.037.

Check the distractors. Option 1, 0.027, is 0.3³, the cube of the scaling number in the denominator, and option 3, 0.729, is 0.9³. Both are traps for anyone who cubes a single decimal instead of handling the ratio. The exact value is 1/27; the option shows it rounded to three decimal places, which is normal for SSC answer options in decimal questions.

How to Practise These 8 Questions

Attempt the 8 questions again without looking at the solutions, in the order given, with a 10-minute timer. That is a realistic pace for algebra questions inside a 15-minute Quantitative Aptitude section, and it leaves room for the rest of the section. Mark every question where you used a different method from the one shown, and check whether your method was actually faster or only felt faster. For most candidates, the identity shortcut beats expansion on Questions 1 and 2, rationalising first beats decimal approximation on Question 3, and the scaling argument beats cubing on Question 8.

Then attempt a full paper. The SSC CGL 2025 Tier 1 Question Paper 12 September Shift 1 test puts these algebra questions back among reasoning, awareness and English questions, which is the real test of whether the identities survive under time pressure. Review every algebra question you get wrong in the full paper by writing out the identity or rationalisation step by hand, not by re-reading the explanation. Writing the step is what moves it into memory.

SSC CGL Algebra Questions with Answers FAQs

How many algebra questions are solved on this page?

8 official questions are solved on this page. All 8 come from the five official SSC CGL 2025 Tier-1 Quantitative Aptitude section papers held on 12, 13 and 14 September 2025. The count is stated honestly: only questions whose solution genuinely uses an algebraic identity, surd rationalisation, or algebraic manipulation of a variable are included, and no practice or memory-based question has been added to make the list look longer.

What is the fastest way to solve x + 1/x questions in SSC CGL?

Use the two identities directly: x² + 1/x² = (x + 1/x)² − 2 and x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x). Do not solve for x itself. With x + 1/x = 5, the first identity gives 23 in one step, and with x + 1/x = 4, the second gives 52 in one step. Both questions on this page are solved that way.

Do I need to rationalise surds, or can I use decimal values?

Rationalise. Decimal approximations like √7 ≈ 2.646 work for a rough check, but SSC options are written with rationalised denominators, such as 8√7/7, and decimals will not match them exactly. Multiplying by the conjugate removes the square root from the denominator in one step, and in chains of surd fractions it makes neighbouring terms cancel, as in Question 5 on this page.

Are these real SSC CGL previous year questions?

Yes. Every question on this page is taken from a section test file on this site marked with paperType official, with the official answer key, and each question carries its own attribution naming the exam, the date, and the shift. Questions from the same shifts that belong to other topics, such as coordinate geometry, are not included here because this page covers algebra identities and surds only.

Where can I practise more SSC CGL algebra questions?

Attempt the full shift paper through the SSC CGL previous year question paper note, which links all five official 2025 Tier-1 shifts, or practise algebra on its own with the SSC CGL Algebra Mock Test 01 test. The full paper is the better test of speed, because it mixes algebra with the other three sections under the same 60-minute limit.

Sources

• This site's own test files, verified 2 November 2026: 5 official SSC CGL 2025 Tier-1 Quantitative Aptitude section papers (12 September Shift 1, Shift 2 and Shift 3; 13 September Shift 2; 14 September Shift 1), each 25 questions with the official answer key. The 8 questions on this page are drawn only from those files, and every numeric answer was re-solved independently before publishing.
• Staff Selection Commission portal, ssc.gov.in — examination notices and answer key information for the SSC CGL 2025 Tier-1 examination.

Last checked: 2 November 2026. This page will be updated if further official SSC CGL 2025 Tier-1 shift papers are added to this site.