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

The equation ax2 + bx + c = 0 has two roots, δ and θ. If δ2 + θ2 = 10 and the sum of the roots is 6, which of the following could represent (a, b, c)? A. (1, -6, 9) B. (2, -12, 26) C. (1, 4, -10)

  1. Only A

  2. Only B

  3. Only C

  4. Only A and B

  5. All A, B, and C
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given roots delta and theta, delta + theta = -b/a = 6 and delta * theta = c/a. Since delta^2 + theta^2 = (delta + theta)^2 - 2(delta * theta) = 10, we have 6^2 - 2(c/a) = 10, so 36 - 2(c/a) = 10, which means 2(c/a) = 26, or c/a = 13. Option A: c/a = 9/1 = 9 (No). Option B: c/a = 26/2 = 13 (Yes). Option C: c/a = -10/1 = -10 (No).

Multiple choice

In the given question, two equations, numbered I and II, are given. Solve both the equations to get the appropriate answer. I. x2 - 4 = 0 II 4y2 - 7y - 2 = 0

  1. If x = y or the relationship cannot be established

  2. If x > y

  3. If x ≥ y

  4. If x ≤ y

  5. If x < y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I gives x = 2 or x = -2. Equation II factors to (4y + 1)(y - 2) = 0, giving y = -1/4 or y = 2. Comparing these values, x can be less than, equal to, or greater than y, so no relationship can be established.

Multiple choice

Directions: Answer the question independently. The roots of the equations x2 + bx + c = 0 are equal to that of equation bx2 + cx + 8 = 0. The roots of the equation x3 - 9x2 + 15x - 7 = 0 are 'a' and 'd' with a > d. What is the value of a - b + c + d? (Given b, c > 0)

  1. -1

  2. 14

  3. 4

  4. 10
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the equations x^2 + bx + c = 0 and bx^2 + cx + 8 = 0 to have equal roots, the ratios of coefficients must be equal: 1/b = b/c = c/8. This implies b^2 = c and c^2 = 8b. Solving these yields b = 2 and c = 4. The roots of x^3 - 9x^2 + 15x - 7 = 0 are found by testing small integers; x=1 is a root, leading to (x-1)(x^2 - 8x + 7) = 0, which factors to (x-1)(x-1)(x-7) = 0. Thus, a=7 and d=1. The value of a - b + c + d is 7 - 2 + 4 + 1 = 10.

Multiple choice

If the roots of the equation x2 + 2ax + b = 0 are real and distinct and they differ by at most 2m, then b lies in the interval:

  1. (a2 - m2, a2)

  2. [a2 - m2, a2)

  3. [a2, a2 + m2]

  4. (a2, a2 + m2)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let roots be alpha and beta. |alpha - beta| <= 2m. (alpha-beta)^2 <= 4m^2. (alpha+beta)^2 - 4alpha*beta <= 4m^2. (-2a)^2 - 4b <= 4m^2. 4a^2 - 4b <= 4m^2. a^2 - b <= m^2, so b >= a^2 - m^2. Since roots are real and distinct, discriminant > 0, so 4a^2 - 4b > 0, b < a^2. Thus b is in [a^2 - m^2, a^2).

Multiple choice

If the roots of the equation x2 - px + q = 0 differ by unity, then

  1. p2 = 4q

  2. p2 = 4q - 1

  3. p2 = 4q + 1

  4. p2 = 2q
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If roots are r and r+1, their sum is 2r+1 = p and their product is r(r+1) = q. Substituting r = (p-1)/2 into the product equation gives ((p-1)/2)((p+1)/2) = q, which simplifies to (p^2-1)/4 = q, or p^2 = 4q + 1.

Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and then choose the correct option. I. 5x2 - 16x + 11 = 0 II. 5y2 - 3y – 2 = 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

I: 5x^2 - 16x + 11 = 0 -> (5x - 11)(x - 1) = 0 -> x = 1, 2.2. II: 5y^2 - 3y - 2 = 0 -> (5y + 2)(y - 1) = 0 -> y = 1, -0.4. Comparing: x=1, y=1 (x=y); x=2.2, y=-0.4 (x>y). Thus x >= y.

Multiple choice

Directions: In the given question, two equations numbered I and II are given. You have to solve both the equations and find out the relationship between x and y. I. 5x2 + 2x = 0 II. 6y2 - 5y + 1 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

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

I: 5x^2 + 2x = 0 => x(5x + 2) = 0, so x = 0 or x = -0.4. II: 6y^2 - 5y + 1 = 0 => (2y - 1)(3y - 1) = 0, so y = 0.5 or y = 0.33. Comparing values: 0 < 0.33, 0 < 0.5, -0.4 < 0.33, -0.4 < 0.5. Thus, x < y.