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 questions two equations are given. Solve these equations and give answer : (i) (3x^2 – x – 70 = 0) (ii) (y^2 – 31y + 240 = 0)

  1. If x < y

  2. If x > y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or no relationship can be established between x and y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equation (i): 3x^2 - x - 70 = 0. Using quadratic formula: x = [1 ± √(1 + 840)]/6 = [1 ± √841]/6 = [1 ± 29]/6. Roots are x = 30/6 = 5 and x = -28/6 = -14/3 ≈ -4.67. For equation (ii): y^2 - 31y + 240 = 0. Roots are y = [31 ± √(961 - 960)]/2 = [31 ± 1]/2. Roots are y = 32/2 = 16 and y = 30/2 = 15. So x can be 5 or -4.67, while y can be 15 or 16. In all cases, x < y (5 < 15, 5 < 16, -4.67 < 15, -4.67 < 16). Option A is correct.

Multiple choice

The quadratic equation whose roots are 3 and – 1, is द्विघात समीकरण जिसके मूल 3 और – 1 हैं, है

  1. x2 – 4x + 3 = 0

  2. x2 – 2x – 3 = 0

  3. x2 + 2x – 3 = 0

  4. x2 + 4x + 3 = 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find a quadratic equation from given roots r1 and r2, use: x² - (r1 + r2)x + (r1 × r2) = 0. Here r1 = 3, r2 = -1. Sum = 3 + (-1) = 2. Product = 3 × (-1) = -3. Therefore the equation is x² - 2x - 3 = 0. Verify by factoring: x² - 2x - 3 = (x - 3)(x + 1), which gives roots 3 and -1. Option A would give roots 3 and 1; Option C gives -3 and 1; Option D gives -3 and -1.

Multiple choice

In the following question, two equations I and II are given. Solve both the equations Carefully & answer the questions given below: I. (x^3 = -704969) II. (y^2 = 7921)

  1. I > II

  2. I < II

  3. I ≥ II

  4. I ≤ II

  5. I = II, or relation cannot be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For equation I: x³ = -704969. Since the cube of a negative number is negative, x = -89 because -89 × -89 × -89 = -704969. For equation II: y² = 7921, so y = ±89 (since both 89² and (-89)² = 7921). Comparing x = -89 with y = ±89: x = y when y = -89, and x < y when y = 89. Therefore x ≤ y is always true. D is correct.

Multiple choice

$I. (x^2 + 12x + 27 = 0) II. (y^2 + 11y + 30 = 0)$ Solve the given equations and determine the relationship between x and y.

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving x² + 12x + 27 = 0 gives x = -3 or x = -9. Solving y² + 11y + 30 = 0 gives y = -5 or y = -6. Comparing the values: when x = -3, x > y for both y values (-3 > -5, -3 > -6). When x = -9, x < y for both values (-9 < -5, -9 < -6). Since the relationship changes depending on which roots we compare, we cannot establish a consistent relationship between x and y.

Multiple choice

If ( \alpha ) and ( \beta ) are the roots of the equation ( x^2 - 6x + 6 = 0 ), what is ( \alpha^3 + \beta^3 + \alpha^2 + \beta^2 + \alpha + \beta ) equal to?

  1. 150

  2. 138

  3. 128

  4. 124
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

द्विघात समीकरण से: α + β = 6 और αβ = 6। α^2 + β^2 = (α + β)^2 - 2αβ = 36 - 12 = 24। α^3 + β^3 = (α + β)^3 - 3αβ(α + β) = 216 - 108 = 108। कुल योग = 108 + 24 + 6 = 138। मूलों के योग और गुणनफल के सूत्रों से यह आसानी से निकाला जा सकता है। α = 3 ± √3 और β = 3 ∓ √3 हैं, जिससे αβ = 9 - 3 = 6 और α + β = 6 मिलता है।

Multiple choice

Values of k for the equation (kx(x - 2) + 6 = 0) so that it has two equal roots are-

  1. 0, 6

  2. 6

  3. 2, 3

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equal roots, discriminant must be zero. Equation: kx^2 - 2kx + 6 = 0. Discriminant: (-2k)^2 - 4(k)(6) = 4k^2 - 24k = 0. Factor: 4k(k - 6) = 0, giving k = 0 or k = 6. Option B only gives k=6 (incomplete), Option C gives incorrect values, and Option D is wrong since valid solutions exist.

Multiple choice

If ( \alpha, \beta ) are the roots of the equation ( x^2 + px + 12 = 0 ) and ( \alpha - \beta = 1 ) then the value of p is
  1. $\( \pm 3 \)$
  2. $\( \pm 5 \)$
  3. $\( \pm 7 \)$
  4. $\( \pm 4 \)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the quadratic equation x² + px + 12 = 0 with roots α and β, we have α + β = -p and αβ = 12. Given α - β = 1, squaring gives (α - β)² = 1, so α² - 2αβ + β² = 1. This means (α + β)² - 4αβ = 1. Substituting: p² - 4(12) = 1, giving p² - 48 = 1, so p² = 49 and p = ±7. Option C is correct. Using Vieta's formulas and the given difference of roots leads to the solution.

Multiple choice

In the following question, there are two equations. Solve the equations and answer accordingly. I. (x^2 = 144) II. (y^2 - 26y + 169 = 0)
  1. $X > Y$
  2. $X < Y$
  3. $X \geq Y$
  4. $X \leq Y$

  5. X=Y or No relation can be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From equation I: x² = 144, so x = ±12 (two values: 12 and -12). From equation II: y² - 26y + 169 = 0 factors to (y - 13)² = 0, so y = 13 (only one value). Comparing: when x = -12, x < y (since -12 < 13). When x = 12, x < y (since 12 < 13). In both cases, x < y, so answer is X < Y (option B).

Multiple choice

In the following question, there are two equations. Solve the equations and answer accordingly. (I. \quad 2x^2 - 7x + 5 = 0 ) (II. \quad 2y^2 + 4y - 6 = 0)

  1. X > Y

  2. X < Y

  3. X ≥ Y

  4. X ≤ Y

  5. X=Y or No relation can be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From equation I: 2x² - 7x + 5 = 0 factors to (2x - 5)(x - 1) = 0, so x = 1 or x = 5/2. From equation II: 2y² + 4y - 6 = 0 simplifies to y² + 2y - 3 = 0, which factors to (y + 3)(y - 1) = 0, so y = -3 or y = 1. Comparing all combinations: x = 1, y = 1 gives x = y. x = 5/2, y = -3 gives x > y. x = 5/2, y = 1 gives x > y. x = 1, y = -3 gives x > y. Since x is never less than y, X ≥ Y is correct.

Multiple choice

In the following question, there are two equations. Solve the equations and answer accordingly. (I. \quad x^3 = 343 ) (II. \quad y^2 - 196 = 0)
  1. $\(X > Y\)$
  2. $\(X < Y\)$
  3. $\(X \ge Y\)$
  4. $\(X \le Y\)$

  5. X=Y or No relation can be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

From equation I: x³ = 343, so x = 7 (only one value, the cube root of 343). From equation II: y² - 196 = 0, so y² = 196, giving y = ±14 (two values: 14 and -14). Comparing: when y = 14, x < y (since 7 < 14). When y = -14, x > y (since 7 > -14). Since the relationship changes depending on which y value we pick, no single relationship holds. The answer is 'X = Y or No relation can be established'.

Multiple choice

In the following question, there are two equations. Solve the equations and answer accordingly. I. (3x^2 + 18x + 24 = 0) II. (2y^2 - 11y + 15 = 0)

  1. X > Y

  2. X < Y

  3. X ≥ Y

  4. X ≤ Y

  5. X=Y or No relation can be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From equation I: 3x² + 18x + 24 = 0. Divide by 3: x² + 6x + 8 = 0. Factor: (x + 2)(x + 4) = 0, so x = -2 or x = -4. From equation II: 2y² - 11y + 15 = 0 factors to (2y - 5)(y - 3) = 0, so y = 5/2 or y = 3. Comparing all combinations: x = -2, y = 5/2 gives x < y. x = -2, y = 3 gives x < y. x = -4, y = 5/2 gives x < y. x = -4, y = 3 gives x < y. In all cases, X < Y, so answer B is correct.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. ( I. 6x^2 + 51x + 105 = 0 ) ( II. 2y^2 + 25y + 78 = 0 )

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

For equation I: 6x² + 51x + 105 = 0 gives roots x = -3.5 and x = -5. For equation II: 2y² + 25y + 78 = 0 gives roots y = -6 and y = -6.5. Comparing values: when x = -3.5 (greater root), x > -6 and x > -6.5. When x = -5 (smaller root), x > -6 but x < -6.5. Since one value of x (-3.5) is greater than both y values, x > y is correct.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. $I. (6x + 7y = 52) II. (14x + 4y = 35)$
  1. $\(x > y\)$
  2. $\(x \ge y\)$
  3. $\(x < y\)$
  4. $\(x \le y\)$

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

Solving the system: 6x + 7y = 52 and 14x + 4y = 35. Multiply first by 7: 42x + 49y = 364. Multiply second by 3: 42x + 12y = 105. Subtract: 37y = 259, so y = 7. Substituting: 6x + 49 = 52, so 6x = 3, x = 0.5. Since x = 0.5 and y = 7, we have x < y.

Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer: (I. 4x^2 - 43x + 105 = 0 ) (II. 7y^2 - 29y + 30 = 0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y or relation can’t be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 4x^2 - 43x + 105 = 0 → (4x-35)(x-3)=0 → x = 3 or 8.75. Equation II: 7y^2 - 29y + 30 = 0 → (7y-15)(y-2)=0 → y = 2 or 2.14. All x values (3, 8.75) are greater than all y values (2, 2.14). Therefore x > y.