SSC Algebra Questions with Solutions (Exam Level)

About This Practice Set

Algebra in SSC exams is not about long, difficult equations. It is about knowing a handful of identities so well that you can spot which one a question is hiding. Once you can do that, most algebra questions take less than 40 seconds.

This set has 40 exam-level algebra questions in four groups. Each one has the answer and a step-by-step solution, so you can see exactly which identity or rule was used. The questions follow the pattern of SSC CGL and CHSL Tier 1 papers.

How to use it: Try each group in about 15 minutes. If you get stuck, read the solution, then close it and solve the same question again on your own after a few hours.

Identities You Should Know by Heart

• (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²
• a² − b² = (a + b)(a − b)
• a³ + b³ = (a + b)³ − 3ab(a + b) and a³ − b³ = (a − b)³ + 3ab(a − b)
• If x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k.
• If a + b + c = 0, then a³ + b³ + c³ = 3abc.
• a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
• For ax² + bx + c = 0: sum of roots = −b/a and product of roots = c/a.
• Remainder theorem: the remainder when p(x) is divided by (x − a) is p(a).

Set A: Identities and Special Values (Q1–Q10)

Q1. If x + 1/x = 5, find the value of x² + 1/x².
Answer: 23
Solution: Square both sides: x² + 2 + 1/x² = 25, so x² + 1/x² = 25 − 2 = 23.

Q2. If x − 1/x = 3, find the value of x² + 1/x².
Answer: 11
Solution: Square both sides: x² − 2 + 1/x² = 9, so x² + 1/x² = 9 + 2 = 11.

Q3. If x + 1/x = 3, find the value of x³ + 1/x³.
Answer: 18
Solution: Use a³ + b³ = (a + b)³ − 3ab(a + b): 27 − 3 × 3 = 18.

Q4. If x² + 1/x² = 34 and x is positive, find x + 1/x.
Answer: 6
Solution: (x + 1/x)² = 34 + 2 = 36, so x + 1/x = 6.

Q5. If a + b = 10 and ab = 21, find a² + b².
Answer: 58
Solution: a² + b² = (a + b)² − 2ab = 100 − 42 = 58.

Q6. If a − b = 4 and ab = 12, find a² + b².
Answer: 40
Solution: a² + b² = (a − b)² + 2ab = 16 + 24 = 40.

Q7. If a + b = 7 and a² + b² = 29, find ab.
Answer: 10
Solution: 2ab = (a + b)² − (a² + b²) = 49 − 29 = 20, so ab = 10.

Q8. If a + b + c = 0, find the value of (a³ + b³ + c³) ÷ abc.
Answer: 3
Solution: When a + b + c = 0, a³ + b³ + c³ = 3abc, so the ratio is 3.

Q9. Find the value of (x + y)² − (x − y)² when x = 7 and y = 3.
Answer: 84
Solution: The expression equals 4xy = 4 × 7 × 3 = 84.

Q10. If x = 2 + √3, find x + 1/x.
Answer: 4
Solution: 1/x = 1/(2 + √3) = 2 − √3 after rationalising. So x + 1/x = (2 + √3) + (2 − √3) = 4.

Set B: Linear and Quadratic Equations (Q11–Q20)

Q11. Solve for x: 3x − 7 = 2x + 5.
Answer: x = 12
Solution: Bring like terms together: 3x − 2x = 5 + 7, so x = 12.

Q12. If 2x + 3y = 12 and x − y = 1, find y.
Answer: y = 2
Solution: Put x = y + 1 in the first equation: 2y + 2 + 3y = 12, so 5y = 10 and y = 2.

Q13. Solve for x: x/4 + x/6 = 10.
Answer: x = 24
Solution: LCM of 4 and 6 is 12. (3x + 2x)/12 = 10, so 5x = 120 and x = 24.

Q14. The sum of two numbers is 45 and their difference is 9. Find the numbers.
Answer: 27 and 18
Solution: Larger = (45 + 9)/2 = 27. Smaller = (45 − 9)/2 = 18.

Q15. Find the roots of x² − 7x + 12 = 0.
Answer: 3 and 4
Solution: Find two numbers with sum 7 and product 12. They are 3 and 4, so (x − 3)(x − 4) = 0.

Q16. Find the sum of the roots of 2x² − 8x + 3 = 0.
Answer: 4
Solution: For ax² + bx + c = 0, the sum of roots = −b/a = 8/2 = 4.

Q17. Find the product of the roots of 3x² + 5x − 12 = 0.
Answer: −4
Solution: Product of roots = c/a = −12/3 = −4.

Q18. Solve x² − 6x + 9 = 0.
Answer: x = 3
Solution: It is a perfect square: (x − 3)² = 0, so x = 3.

Q19. If a − b = 3 and a + b = 5, find a² − b².
Answer: 15
Solution: a² − b² = (a + b)(a − b) = 5 × 3 = 15.

Q20. Find the value of 101² − 99².
Answer: 400
Solution: (101 + 99)(101 − 99) = 200 × 2 = 400.

Set C: Cubes, Factors and Remainders (Q21–Q30)

Q21. If a + b = 5 and ab = 6, find a³ + b³.
Answer: 35
Solution: a³ + b³ = (a + b)³ − 3ab(a + b) = 125 − 90 = 35.

Q22. If a − b = 2 and ab = 15, find a³ − b³.
Answer: 98
Solution: a³ − b³ = (a − b)³ + 3ab(a − b) = 8 + 90 = 98.

Q23. Find the remainder when x³ − 2x² + x − 5 is divided by (x − 2).
Answer: −3
Solution: By the remainder theorem, put x = 2: 8 − 8 + 2 − 5 = −3.

Q24. Is (x − 1) a factor of x³ + 2x² − 5x + 2?
Answer: Yes
Solution: Put x = 1: 1 + 2 − 5 + 2 = 0. The remainder is zero, so (x − 1) is a factor.

Q25. If (x − 2) is a factor of x² + kx − 10, find k.
Answer: 3
Solution: Put x = 2: 4 + 2k − 10 = 0, so 2k = 6 and k = 3.

Q26. If x + y = 8 and x − y = 2, find xy.
Answer: 15
Solution: Adding gives x = 5 and subtracting gives y = 3. So xy = 15.

Q27. Find the value of (a + b)² + (a − b)² when a = 5 and b = 2.
Answer: 58
Solution: The expression equals 2(a² + b²) = 2 × 29 = 58.

Q28. Find the value of (x² − 9)/(x − 3) when x = 5.
Answer: 8
Solution: x² − 9 = (x + 3)(x − 3), so the fraction simplifies to x + 3 = 8.

Q29. If 3^(x + 1) = 81, find x.
Answer: 3
Solution: 81 = 3⁴. So x + 1 = 4 and x = 3.

Q30. If x + 1/x = 2, find x⁴ + 1/x⁴.
Answer: 2
Solution: x + 1/x = 2 means x = 1, so x⁴ + 1/x⁴ = 1 + 1 = 2.

Set D: Mixed Exam-Level Questions (Q31–Q40)

Q31. The sum of three consecutive integers is 72. Find the largest one.
Answer: 25
Solution: The middle number is 72 ÷ 3 = 24, so the numbers are 23, 24 and 25.

Q32. If 3x + 2y = 16 and 5x − 2y = 8, find y.
Answer: y = 3.5
Solution: Add the equations: 8x = 24, so x = 3. Then 9 + 2y = 16 and y = 3.5.

Q33. The sum of a positive number and its reciprocal is 10/3. Find the number.
Answer: 3
Solution: x + 1/x = 10/3 gives 3x² − 10x + 3 = 0, so (3x − 1)(x − 3) = 0. The whole-number solution is x = 3 (the other root is 1/3).

Q34. If a/b = 3/4, find (a + b)/(a − b).
Answer: −7
Solution: Take a = 3 and b = 4. (3 + 4)/(3 − 4) = 7/(−1) = −7.

Q35. If x + y = 10 and xy = 21, find 1/x + 1/y.
Answer: 10/21
Solution: 1/x + 1/y = (x + y)/xy = 10/21.

Q36. If a² + b² = 13 and ab = 6, find a + b (positive value).
Answer: 5
Solution: (a + b)² = 13 + 12 = 25, so a + b = 5.

Q37. If 5x − 3 = 2x + 9, find x².
Answer: 16
Solution: 3x = 12, so x = 4 and x² = 16.

Q38. Solve |x − 3| = 5.
Answer: x = 8 or x = −2
Solution: x − 3 = 5 gives 8, and x − 3 = −5 gives −2.

Q39. Twice a number increased by 7 equals 35. Find the number.
Answer: 14
Solution: 2n + 7 = 35, so 2n = 28 and n = 14.

Q40. If a + b + c = 9 and ab + bc + ca = 26, find a³ + b³ + c³ − 3abc.
Answer: 27
Solution: a² + b² + c² = 81 − 52 = 29. The expression = (a + b + c)(a² + b² + c² − ab − bc − ca) = 9 × (29 − 26) = 27.

Tips for Solving Algebra Faster

• Look before you expand. If you see x + 1/x or a + b together with ab, an identity is probably the shortest way.
• Substitute easy numbers when the question asks for a value at a given point. It is quicker than simplifying the expression.
• Check your answer by putting it back into the equation. It takes ten seconds and saves marks.
• Do not forget the second root. Quadratic equations often have two answers, and the options may include only one of them.

What to Do Next

If you scored below 30 out of 40, revise the identities list above and attempt Set A and Set B again after two days. If you scored above 30, try timing yourself: aim to finish all 40 questions in 30 minutes. For more practice in the same style, try the simplification and percentage sets on this site.