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 the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - ( (I) x^2-25 =0 )( (II) y^2- 9y +20=0 )
  1. $x \ge y$
  2. $x < y$
  3. $x \le y$
  4. $x > y$

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

Solving equation I: x^2 - 25 = 0 gives x = ±5. Solving equation II: y^2 - 9y + 20 = 0 gives (y-4)(y-5) = 0, so y = 4 or y = 5. When x = -5, y could be 4 (xy) or 5 (x=y). Since the relationship varies, no single relation can be established.

Multiple choice

In the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - ( (I) x^2 -x-20=0 ) ( (II) y^2+7y+12=0 )
  1. $x\geq y$
  2. $x<y$
  3. $x\leq y$
  4. $x>y$

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

Solving equation I: x^2 - x - 20 = 0 gives (x-5)(x+4) = 0, so x = 5 or x = -4. Solving equation II: y^2 + 7y + 12 = 0 gives (y+3)(y+4) = 0, so y = -3 or y = -4. When x = 5, x > y (both -3, -4). When x = -4, x could equal y = -4 or be less than y = -3. Since the relationship varies, it cannot be established.

Multiple choice

In the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - ( (I) x^2 -8x+15 =0 )( (II) y^2 -3y+2=0 )

  1. x≥y

  2. x<y

  3. x≤y

  4. x>y

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

Solving equation I: x^2 - 8x + 15 = 0 gives (x-3)(x-5) = 0, so x = 3 or x = 5. Solving equation II: y^2 - 3y + 2 = 0 gives (y-1)(y-2) = 0, so y = 1 or y = 2. In all cases (x=3 or 5; y=1 or 2), x is always greater than y. Therefore x > y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give answer. $I. (5x + 9y - 95 = 0) II. (2x + 3y - 35 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving the system: From the second equation, 3y = 35 - 2x, so y = (35 - 2x)/3. Substituting into first: 5x + 9((35 - 2x)/3) = 95, which gives 5x + 3(35 - 2x) = 95, so 5x + 105 - 6x = 95, giving -x = -10, so x = 10. Then y = (35 - 20)/3 = 5. Since x = 10 and y = 5, we have x > y.

Multiple choice

In each of the following questions two equations are given. You have to solve them and give answer - (I) (a^2 - 14a + 48 = 0) (II) (b^2 - 23b + 132 = 0)

  1. If a < b

  2. If a > b

  3. If a ≤ b

  4. If a ≥ b

  5. If a=b or Relationship cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the quadratic equations: Equation (I) a² - 14a + 48 = 0 factors to (a-6)(a-8) = 0, giving a = 6 or a = 8. Equation (II) b² - 23b + 132 = 0 factors to (b-11)(b-12) = 0, giving b = 11 or b = 12. Comparing the values: both values of a (6, 8) are less than both values of b (11, 12), so a < b is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and Give answer- $(I. \quad x^2 – 15x + 56 = 0)$ $(II. \quad 2y^2 – 10y + 12 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve equation I: x^2 - 15x + 56 = (x-7)(x-8) = 0, so x = 7 or 8. Solve equation II: divide by 2 first, then y^2 - 5y + 6 = (y-2)(y-3) = 0, so y = 2 or 3. Comparing all combinations (7 vs 2, 7 vs 3, 8 vs 2, 8 vs 3), in every case x is greater than y, so x > y is always true.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give the answer- $I. (x^2 - 9 = 0) II. (y^2 + 10y + 25 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solve equation I: x² - 9 = 0 gives x = ±3, so x = 3 or x = -3. Solve equation II: y² + 10y + 25 = 0 is (y+5)² = 0, giving y = -5 (repeated root). For both possible x values (3 and -3), we have x > y since 3 > -5 and -3 > -5. Therefore x > y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^{2} - 5 = 0) II. (4y^{2} - 24y + 35 = 0)$

  1. If x<y

  2. If x≤y

  3. If x=y or relationship can not be established

  4. If x>y

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

Solve equation I: x² - 5 = 0, so x = ±√5 ≈ ±2.236. Solve equation II: 4y² - 24y + 35 = 0. Using quadratic formula: y = (24 ± √(576 - 560))/8 = (24 ± 4)/8, giving y = 28/8 = 3.5 or y = 20/8 = 2.5. Comparing: all x values (±2.236) are less than all y values (2.5, 3.5). Thus x < y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and (I.) x^2 - 25 = 0 (II.) y^2 - 4 = 0 )Give answer-

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

From equation I: x² - 25 = 0, so x² = 25 and x = ±5. From equation II: y² - 4 = 0, so y² = 4 and y = ±2. Since x can be -5 or 5 and y can be -2 or 2, we have multiple possible relationships: x > y (when x=5 and y=2), x > y (when x=-5 and y=-2), x > y (when x=5 and y=-2), but x < y (when x=-5 and y=2). Because the relationship changes depending on which values we choose, it cannot be uniquely determined.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^2 - 1 = 0) II. (y^3 - 1 = 0)$ Give answer-

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

From I: x² - 1 = 0 gives x² = 1, so x = ±1. From II: y³ - 1 = 0 gives y³ = 1, so y = 1 (real cube root). Now compare: if x = -1, then x < y (since -1 < 1). If x = 1, then x = y. Combining both cases, we have x ≤ y (x is always less than or equal to y). Therefore 'x ≤ y' is correct.

Multiple choice

In each of the following questions, two equations are given you have to solve them and. (I. p(p^{-1}) = p )(II. q^{2} = 4) Give answer -

  1. If p>q

  2. If p≥q

  3. If p<q

  4. If p≤q

  5. If p=q or relationship can't be established.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

From equation I: p(p-1) = p gives p²-p = p, so p²-2p = 0, meaning p = 0 or p = 2. From equation II: q² = 4 gives q = 2 or q = -2. When p=0, p < q (for q=2). When p=2, p = q (for q=2) or p > q (for q=-2). The relationship varies, so it cannot be established. Option E is correct.

Multiple choice

In each of the following questions, two equations are given you have to solve them and. ( I. p^2 + 7p + 12 = 0 ) ( II. q^2 – 5q + 6 = 0 ) Give answer

  1. If p>q

  2. If p≥q

  3. If p<q

  4. If p≤q

  5. If p=q or relationship can't be established.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Factor equation I: p² + 7p + 12 = (p+3)(p+4) = 0, so p = -3 or p = -4. Factor equation II: q² - 5q + 6 = (q-2)(q-3) = 0, so q = 2 or q = 3. For all possible values, p is negative while q is positive, so p < q always. Option C is correct.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and(\text{(I)} \quad x^2+41x+418=0\\text{(II)} \quad y^2+36y+323=0

  1. If x>y

  2. If x≥y

  3. If x<y

  4. If x≤y

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

Solve x² + 41x + 418 = 0: factors to (x+19)(x+22) = 0, so x = -19 or x = -22. Solve y² + 36y + 323 = 0: factors to (y+17)(y+19) = 0, so y = -17 or y = -19. Comparing all values: x ranges from -22 to -19, y ranges from -19 to -17. Both values can equal -19, but x's maximum (-19) equals y's minimum (-19), and x can be less than y (when x=-22, y=-17). Therefore x ≤ y is correct.