Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

In each of the following question, two equations are I and II given. You have to solve them and give answer. I. (5x^2 - 8x = -3) II. (4y^2 - 16y = -15)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship cannot be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rewrite equation I: 5x² - 8x + 3 = 0 factors to (5x - 3)(x - 1) = 0, giving x = 0.6 or x = 1. Rewrite equation II: 4y² - 16y + 15 = 0 factors to (2y - 3)(2y - 5) = 0, giving y = 1.5 or y = 2.5. Both x values (0.6, 1) are less than both y values (1.5, 2.5), so x < y is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option.(a) x^2 – 6x + 8 = 0 (b) y^2 – 8y + 15 = 0
  1. $x > y$
  2. $x ≤ y$
  3. $x ≥ y$
  4. $x < y$

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving (a): x²-6x+8=0 gives (x-2)(x-4)=0, so x=2 or 4. Solving (b): y²-8y+15=0 gives (y-3)(y-5)=0, so y=3 or 5. Since x can be 2 or 4 and y can be 3 or 5, we have cases where xy (when x=4). No single relationship always holds.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - 5x + 6 = 0) b) (2y^2 - 7y + 3 = 0)
  1. $x > y$
  2. $x \leq y$
  3. $x \geq y$
  4. $x < y$

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving x²-5x+6=0 gives x=2 or x=3. Solving 2y²-7y+3=0 gives y=1/2 or y=3. When x=2, x>y (both y=1/2 and y=3). When x=3, x>y when y=1/2, but x=y when y=3. Since the relationship varies (sometimes x>y, sometimes x=y), option E is correct.

Multiple choice

In each of the following question, two equations are I and II given. You have to solve them and give answer $I. (2x^2 – 23x + 65 = 0) (II. 2y^2 – 3y = 27)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 2x² - 23x + 65 = 0 gives x = (23 ± √(529-520))/4 = (23 ± 3)/4, so x = 5 or 6.5. Equation II: 2y² - 3y - 27 = 0 gives y = (3 ± √(9+216))/4 = (3 ± 15)/4, so y = 4.5 or -3. Comparing minimum values: x = 5 > y = 4.5. Therefore x > y is always true.

Multiple choice

In each of the following question, two equations are given. You have to solve them and give answer. I. (p^2 + 24 = 10p) II. (2q^2 + 18 = 12q)

  1. If p < q

  2. If p > q

  3. If p ≤ q

  4. If p ≥ q

  5. If p = q or relation between p and q cannot be established.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For p^2 + 24 = 10p, rearranging gives p^2 - 10p + 24 = 0, which factors to (p-4)(p-6) = 0, so p = 4 or p = 6. For 2q^2 + 18 = 12q, dividing by 2 gives q^2 - 6q + 9 = 0, which is (q-3)^2 = 0, so q = 3. Since both possible values of p (4 and 6) are greater than q = 3, we have p > q.

Multiple choice

In given question, two equations numbered I and II are given. You have to solve both the equations and give answer: I. (3p^2 – 10p + 3 = 0) II. (q^2 – 5q + 6 = 0)

  1. If p ≥ q

  2. If p > q

  3. If p ≤ q

  4. If p < q

  5. If p = q or relationship between p and q cannot be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For 3p^2 - 10p + 3 = 0, factoring gives (3p-1)(p-3) = 0, so p = 1/3 or p = 3. For q^2 - 5q + 6 = 0, factoring gives (q-2)(q-3) = 0, so q = 2 or q = 3. The relationships vary: when p = 1/3, p < q (for both q values); when p = 3, p can equal q = 3 or be greater than q = 2. Since the relationship varies, it cannot be established.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - 26x + 165 = 0) b) (y^2 + 8y - 153 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving x^2 - 26x + 165 = 0 gives x = 11 or 15. Solving y^2 + 8y - 153 = 0 gives y = -17 or 9. In all cases, x > y, so option A is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (2x^2 – 10x + 8 = 0) b) (y^2 – 13y – 140 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving 2x^2 - 10x + 8 = 0 gives x = 1 or 4. Solving y^2 - 13y - 140 = 0 gives y = 20 or -7. Since x < y (1<20, 4<20) but x > y (1>-7, 4>-7), no consistent relationship exists.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option ( a) x^2+19x+70 = 0 ) ( b) y^2+5y-234 = 0 )
  1. $x > y$
  2. $x ≤ y$
  3. $x ≥ y$
  4. $x < y$

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving x^2 + 19x + 70 = 0 gives x = -7 or -14. Solving y^2 + 5y - 234 = 0 gives y = -13 or 18. Since x can be less than (-7 < 18) or greater than (-7 > -13), no consistent relationship exists.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and give answer. a) $(x^{2} + 18x - 115 = 0)$ b) $(y^{2} – 24y + 95 = 0)$

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving x² + 18x - 115 = 0 gives x = -23 or x = 5. Solving y² - 24y + 95 = 0 gives y = 5 or y = 19. Comparing all possible values of x and y, the largest x value is 5 and the smallest y value is 5, so x ≤ y is always true.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and give answer. a) (x^2 - 35x + 306 = 0) b) (y^2 - 44y + 475 = 0)

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving x² - 35x + 306 = 0 gives x = 17 or 18 (since 17×18 = 306 and 17+18 = 35). Solving y² - 44y + 475 = 0 gives y = 19 or 25 (since 19×25 = 475 and 19+25 = 44). The maximum value of x is 18, while the minimum value of y is 19. Therefore, x < y for all possible values.

Multiple choice

Read the following Series and give the answers based on it : Series I: 2, 1.5, 3.5, 9.75, a Series II: 3, b, 16, 75, 364 Series III: c, 30, 136, 417, 838 If ‘b’ and ‘c’ are the roots of a quadratic equation, then find the quadratic equation by using variable ‘x’
  1. $\(x^2 - 4x + 4 = 0\)$
  2. $\(x^2 - 5x + 4 = 0\)$
  3. $\(x^2 + 5x + 1 = 0\)$
  4. $\(x^2 + 4x + 4 = 0\)$
  5. $None of these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Analyzing the number series patterns: Series II shows b=4, Series III shows c=1. These are the roots of x²-5x+4=0, which factors to (x-1)(x-4)=0.