Mathematics · Quantitative Aptitude

Algebraic Equations and Expressions

142 Questions

Algebraic equations and expressions test mathematical skills in solving for unknown variables. Questions involve linear equations, ratios, and evaluating complex expressions. This topic forms a core component of quantitative aptitude sections in various competitive exams.

Linear equationsEvaluating expressionsRatio proportionsExponential equationsHypergeometric functions

Algebraic Equations and Expressions Questions

Multiple choice
  1. -2

  2. -1

  3. 1

  4. 2

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

2x - 12 = -4x - 24 Bring subject terms to one side and non subject terms to another side 2x + 4x = -24 + 12 6x = -12 Dividing both sides by 6 x = -2

Multiple choice
  1. X + 5

  2. 2X2 + 3

  3. 2X2 + 5X

  4. 3X2 + 2

  5. X2 + X + 2

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Assume X as 3. Then (1) = 8, (2) = 21, (3) = 33, (4) = 29 and (5) = 14. So (1) and (5) is not the answer. Let take X = 2 then (2) = 11, (3) = 18, and (4) = 14. Only choice (2) is odd in both cases.

Multiple choice
  1. 2

  2. 3

  3. 5

  4. None of the above

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Left side: (12x)_3 = 1 * 3^2 + 2 * 3^1 + x * 3^0 = 9 + 6 + x = 15 + x. Right side: (123)_x = 1 * x^2 + 2 * x^1 + 3 * x^0 = x^2 + 2x + 3. Solving: 15 + x = x^2 + 2x + 3, so x^2 + x - 12 = 0, giving x = 3 or x = -4. Since base must be > 3 (digit 3 present), x = 3.

Multiple choice
  1. 2/3

  2. -1/3

  3. 8/3

  4. -2

  5. 17/3

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given equation is 3(2x + 1) = 7. According to the distributive property of multiplication, 3 × 2x + 3 × 1 = 76x + 3 = 7 Transposing 3 to R.H.S, we get: 6x = 7 – 36x = 4 Dividing both sides of the equation by 6, we get: x = 4/6 = 2/3 The required value of x is 2/3. Thus, the required value of (4x – 3) is -1/3. This answer is correct.

Multiple choice
  1. if (x == 0) return 0;

  2. return 1;

  3. if (x >= 2) return 2;

  4. if (x == 0) return 1;

  5. if (x <= 1) return 1;

Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

A factorial function needs a base case to terminate recursion. For factorial, 0! = 1 and 1! = 1. Option E handles this correctly: when x <= 1, return 1. Otherwise, it recursively computes x * factorial(x-1). Options A and D are incomplete (they don't handle x=1). Option B would cause infinite recursion for x > 1. Option C would return incorrect values.