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

\text{In each of the following questions, two equations are given. You have to solve them and state the correct relationship-} \ \text{I.} \quad x^2 + x = 56 \ \text{II.} \quad y^2 - 17y + 72 = 0
  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$

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

Solving x² + x - 56 = 0 gives x = -8 or x = 7 (factors: x+8 and x-7). Solving y² - 17y + 72 = 0 gives y = 8 or y = 9 (factors: y-8 and y-9). The maximum value of x is 7, and the minimum value of y is 8. Since even the largest x (7) is less than the smallest y (8), we always have x < y.

Multiple choice

In every question there are two equations you have to solve and give answer- I. (20x^2 - 17x + 3 = 0) II. (20y^2 - 9y + 1 = 0)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving 20x² - 17x + 3 = 0 gives x = 3/5 = 0.6 or x = 1/4 = 0.25. Solving 20y² - 9y + 1 = 0 gives y = 2/7 ≈ 0.286 or y = 3/7 ≈ 0.429. When x = 0.6, x > both y values. When x = 0.25, x < both y values. Since x can be greater than or less than y depending on which root we take, x ≥ y is the answer (it covers the case where x > y).

Multiple choice

In every question there are two equations you have to solve and give answer- I. (20x^2 + 31x + 12 = 0) II. (21y^2 - 23y + 6 = 0)

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

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

Solving 20x² + 31x + 12 = 0 gives x = -3/4 = -0.75 or x = -4/5 = -0.8. Solving 21y² - 23y + 6 = 0 gives y = 2/7 ≈ 0.286 or y = 3/7 ≈ 0.429. Both possible values of x (-0.75 and -0.8) are less than both possible values of y (0.286 and 0.429) since negative < positive. Therefore x < y is always true.

Multiple choice

In every question there are two equations you have to solve and give answer- I. (4x^2 - 8x + 3 = 0) II. (2y^2 - 13y + 15 = 0)

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

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

Solving 4x² - 8x + 3 = 0 gives x = 1/2 = 0.5 or x = 3/2 = 1.5. Solving 2y² - 13y + 15 = 0 gives y = 3 or y = 5/2 = 2.5. When x = 0.5, x < both y values. When x = 1.5, x < both y values (1.5 < 2.5 and 1.5 < 3). Therefore x < y for all cases, which means x ≤ y is the correct choice.

Multiple choice

In this question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. $(I. 0.4 p - 1.5 q = 6.5\II. 0.3 p + 0.2 q = 0.9)$

  1. If p < q

  2. If p ≥ q

  3. If p ≤ q

  4. If p > q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving equation II: 0.3p = 0.9 - 0.2q, so p = 3 - (2/3)q. Substituting in equation I: 0.4(3 - 2q/3) - 1.5q = 6.5, which gives 1.2 - (0.8/3)q - 1.5q = 6.5. Solving: 1.2 - 1.767q = 6.5, so -1.767q = 5.3, giving q ≈ -3. Then p = 3 - (2/3)(-3) = 3 + 2 = 5. Since p = 5 > q = -3, the answer is p > q.

Multiple choice

In this question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option.\newline I. (2p^2 + 11p + 12 = 0)\newline II. (6q^2 - 31q + 40 = 0)

  1. If p < q

  2. If p ≥ q

  3. If p ≤ q

  4. If p > q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equation I: 2p² + 11p + 12 = 0 factors as (2p + 3)(p + 4) = 0, giving p = -1.5 or p = -4. For equation II: 6q² - 31q + 40 = 0. Using the quadratic formula: q = (31 ± √(961 - 960))/12 = (31 ± 1)/12, giving q = 8/3 ≈ 2.67 or q = 5/2 = 2.5. Comparing the values: all p values (-1.5, -4) are less than all q values (2.67, 2.5). Therefore p < q.

Multiple choice

In this question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option.\newline I. (p^3 + 783 = 775)\newline II. (6q^2 - 19q + 15 = 0)

  1. If p < q

  2. If p ≥ q

  3. If p ≤ q

  4. If p > q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equation I: p³ + 783 = 775, so p³ = -8, giving p = -2. For equation II: 6q² - 19q + 15 = 0. Using the quadratic formula: q = (19 ± √(361 - 360))/12 = (19 ± 1)/12, giving q = 20/12 = 5/3 ≈ 1.67 or q = 18/12 = 3/2 = 1.5. Comparing p = -2 with q values (1.67, 1.5), we have p < q in both cases.

Multiple choice

In this question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option.\nI. (5p^2 + 4p = 0)\nII. (3q^2 - 243 = 0)

  1. If p < q

  2. If p ≥ q

  3. If p ≤ q

  4. If p > q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For equation I: 5p² + 4p = 0 factors as p(5p + 4) = 0, giving p = 0 or p = -0.8. For equation II: 3q² - 243 = 0 gives q² = 81, so q = ±9. Comparing all combinations: when q = -9 and p = 0 or -0.8, we have p > q; when q = 9 and p = 0 or -0.8, we have p < q. Since the relationship varies, no definite relation can be established.

Multiple choice

In each of these question two equations numbered I and II are given. You have to solve both the equations and give answer if - I. (p^2 + 13p + 42 = 0) II. (q^2 + 19q + 90 = 0)

  1. p < q

  2. p > q

  3. p ≤ q

  4. p ≥ q

  5. p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Factor p² + 13p + 42 = 0 as (p+6)(p+7) = 0, giving p = -6 or p = -7. Factor q² + 19q + 90 = 0 as (q+9)(q+10) = 0, giving q = -9 or q = -10. Since all p values are greater than all q values (-6, -7 > -9, -10), p > q is correct.

Multiple choice

In each of these question two equations numbered I and II are given. You have to solve both the equations and give answer if - I. (p^2 - 15p + 56 = 0) II. (q^2 - 23q + 132 = 0)

  1. p < q

  2. p > q

  3. p ≤ q

  4. p ≥ q

  5. p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equation I: p² - 15p + 56 = 0 factors to (p-7)(p-8) = 0, so p = 7 or 8. For equation II: q² - 23q + 132 = 0 factors to (q-11)(q-12) = 0, so q = 11 or 12. Since the maximum value of p (8) is less than the minimum value of q (11), p is always less than q regardless of which roots are selected. Therefore p < q.

Multiple choice

In each of these question two equations numbered I and II are given. You have to solve both the equations and give answer if - I. (p^{2} + 7p + 12 = 0) II. (q^{2} + 6q + 8 = 0)
  1. $\(p < q\)$
  2. $\(p > q\)$
  3. $\(p \leq q\)$
  4. $\(p \geq q\)$

  5. p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For equation I: p² + 7p + 12 = 0 factors to (p+3)(p+4) = 0, giving p = -3 or -4. For equation II: q² + 6q + 8 = 0 factors to (q+2)(q+4) = 0, giving q = -2 or -4. Comparing all combinations: when p=-3,q=-2 we get pq; when p=-4,q=-2 we get p

Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and given answer if $I. (p^2 - 15p + 56 = 0)$ $II. (q^2 + 4q - 96 = 0)$

  1. If p > q

  2. If p ≥ q

  3. If p < q

  4. If p ≤ q

  5. If p = q or no relation between p and q can be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Factor I: p²-15p+56 = (p-7)(p-8) = 0, so p = 7 or p = 8. Factor II: q²+4q-96 = (q+12)(q-8) = 0, so q = -12 or q = 8. Comparing all values: p=7 gives p-12), p=8 gives p=q when q=8, but p>q when q=-12. Since no single relation always holds, the answer is 'no relation can be established'.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and answer the given question. $I. (x^{2} - 24x + 144 = 0) (II. 2y^{2} - 52y + 338 = 0)$

  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

Solve equation I: x² - 24x + 144 = 0 can be factored as (x-12)² = 0, giving x = 12. Solve equation II: 2y² - 52y + 338 = 0, divide by 2 to get y² - 26y + 169 = 0, which factors as (y-13)² = 0, giving y = 13. Since 12 < 13, x < y is the correct relationship.

Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and answer the given question. $I. (20x^{2} - x - 12 = 0) II. (20y^{2} + 27y + 9 = 0)$

  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
E Correct answer
Explanation

Solve equation I: 20x² - x - 12 = 0 using quadratic formula gives x = [1 ± sqrt(1 + 960)]/40 = [1 ± 31]/40, so x = 0.8 or x = -0.75. Solve equation II: 20y² + 27y + 9 = 0 gives y = [-27 ± sqrt(729 - 720)]/40 = [-27 ± 3]/40, so y = -0.6 or y = -0.75. Since the equations share one common root (-0.75) but have different second roots, the relationship between x and y cannot be uniquely established.

Multiple choice

In each of the following questions, two equations I and II are given. You have to solve them and give answer. I. (p^2 + 16p + 63 = 0) II. (q^2 + 13q + 42 = 0)

  1. If p > q

  2. If p ≥ q

  3. If p < q

  4. If p ≤ q

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

Solve both equations by factorization: p²+16p+63=0 gives (p+7)(p+9)=0 so p = -7, -9. q²+13q+42=0 gives (q+6)(q+7)=0 so q = -6, -7. Testing all combinations: when p=-7, q=-7 gives p=q; when p=-7, q=-6 gives p