What Algebra Questions Test
- Algebraic identities (a³+b³, a+1/a patterns)
- Quadratic equations — roots and coefficients
- Linear equations and word problems
- Surds, indices and rationalisation
- Polynomials and remainder theorem
Question bank coverage by sub-pattern
How Algebra Appears in SSC CPO
Algebra is part of the Quantitative Aptitude section (25% of the SSC CPO syllabus).
Algebra Solved Examples for SSC CPO
12 exam-pattern algebra questions from the practice bank, each with a full solution.
If a + b = 5 and a² + b² = 13, then the value of ab is:
- A.6Correct
- B.12
- C.18
- D.24
Solution
We know that (a + b)² = a² + b² + 2ab. Substituting the given values, we have 5² = 13 + 2ab. So, 25 = 13 + 2ab, which means 2ab = 12. Therefore, ab = 6.
Translate the sentence into an equation: 'Half of a number plus 5 is equal to 11.'
- A.(1/2)x + 5 = 11Correct
- B.(1/2)(x + 5) = 11
- C.2x + 5 = 11
- D.(1/2)x - 5 = 11
Solution
'Half of a number' is (1/2)x. 'Plus 5' is + 5. 'Is equal to 11' is = 11. Therefore, the equation is (1/2)x + 5 = 11.
Find the range of x for which |x - 1| + |x - 2| ≤ 3
- A.[-1, 4]
- B.[0, 3]Correct
- C.[-2, 5]
- D.[-3, 6]
Solution
We have three cases: 1) x < 1, then -(x-1) - (x-2) ≤ 3 => -2x + 3 ≤ 3 => -2x ≤ 0 => x ≥ 0. So 0 ≤ x < 1. 2) 1 ≤ x < 2, then (x-1) - (x-2) ≤ 3 => 1 ≤ 3, which is always true. So 1 ≤ x < 2. 3) x ≥ 2, then (x-1) + (x-2) ≤ 3 => 2x - 3 ≤ 3 => 2x ≤ 6 => x ≤ 3. So 2 ≤ x ≤ 3. Combining all cases gives 0 ≤ x ≤ 3. Therefore, the solution is [0, 3].
If x + 1/x = √3, then what is the value of x³ + 1/x³?
- A.0Correct
- B.3√3
- C.2√3
- D.√3
Solution
x³ + 1/x³ = (x + 1/x)³ - 3(x + 1/x) = (√3)³ - 3√3 = 3√3 - 3√3 = 0.
If the roots of the equation x² - 7x + k = 0 are equal, then what is the value of k?
- A.49/4Correct
- B.7/2
- C.49/2
- D.7/4
Solution
For equal roots, discriminant = 0. b² - 4ac = 0. Here a = 1, b = -7, c = k. So 49 - 4k = 0, giving k = 49/4.
If x = 3 + 2√2, then what is the value of √x - 1/√x?
- A.2Correct
- B.√2
- C.1
- D.2√2
Solution
x = 3 + 2√2 = (√2 + 1)². So √x = √2 + 1. Then 1/√x = 1/(√2 + 1) = √2 - 1 (by rationalizing). Therefore √x - 1/√x = (√2 + 1) - (√2 - 1) = 2.
If 5^(x-1) = 125, then find the value of x.
- A.2
- B.3
- C.4Correct
- D.5
Solution
5^(x-1) = 125 = 5³. Therefore x - 1 = 3, so x = 4.
A father is 3 times as old as his son. After 12 years, he will be twice as old as his son. What is the present age of the son?
- A.12 yearsCorrect
- B.10 years
- C.15 years
- D.8 years
Solution
Let son's age = x. Father's age = 3x. After 12 years: 3x + 12 = 2(x + 12). So 3x + 12 = 2x + 24. Therefore x = 12.
If (a - b) = 4 and ab = 21, then what is the value of a³ - b³?
- A.316Correct
- B.268
- C.324
- D.412
Solution
a³ - b³ = (a - b)(a² + ab + b²). First, a² + b² = (a - b)² + 2ab = 16 + 42 = 58. So a² + ab + b² = 58 + 21 = 79. Therefore a³ - b³ = 4 × 79 = 316.
Simplify: (a + b + c)² - (a - b - c)²
- A.4a(b + c)Correct
- B.4bc
- C.2(ab + ac)
- D.4(ab + ac)
Solution
Using x² - y² = (x+y)(x-y) where x = (a+b+c) and y = (a-b-c). x + y = 2a, x - y = 2(b+c). Result = 2a × 2(b+c) = 4a(b+c).
What is the value of 2^(x+2) if 2^x = 32?
- A.64
- B.128Correct
- C.256
- D.512
Solution
2^x = 32 = 2^5, so x = 5. Therefore 2^(x+2) = 2^7 = 128.
If a + b + c = 0, then what is the value of (a²/bc) + (b²/ca) + (c²/ab)?
- A.0
- B.1
- C.3Correct
- D.-3
Solution
If a + b + c = 0, then a³ + b³ + c³ = 3abc. The given expression = (a³ + b³ + c³)/abc = 3abc/abc = 3.
Algebra — Frequently Asked Questions
What makes exam algebra different from school algebra?
Speed. Questions are built around identities — if you are expanding brackets for more than a minute, there is almost always a substitution (like x + 1/x = k) that finishes it in two lines.
Which identities are asked most often?
The a + 1/a family (finding a² + 1/a², a³ + 1/a³), a³ + b³ + c³ − 3abc, and the symmetric-expression patterns built on them.
How many Algebra practice questions does Parikshala have for SSC CPO?
282 exam-pattern Algebra questions for SSC CPO Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.