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 questions two equations I and II are given. You have to solve both the equations and give answer.( (I) ) (x^2 - 7x + 12 = 0) ( (II) ) (2y^2 + 18y + 36 = 0)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving equation (I): x² - 7x + 12 = 0 factors to (x-3)(x-4) = 0, so x = 3 or x = 4. Solving equation (II): 2y² + 18y + 36 = 0 simplifies to y² + 9y + 18 = 0, which factors to (y+3)(y+6) = 0, so y = -3 or y = -6. Both values of x (3 and 4) are greater than both values of y (-3 and -6), so x > y is always true.

Multiple choice

In the following question two equations I and II are given. You have to solve both the equations and give answer.(I) \ 16x^2 – 8x + 1 = 0 (II) \ y^2 + 0.9y – 0.22 = 0

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Equation I: 16x²-8x+1=0 factors as (4x-1)²=0, so x = 1/4 = 0.25 (repeated root). Equation II: y²+0.9y-0.22=0 uses quadratic formula: y = (-0.9 ± √(0.81+0.88))/2 = (-0.9 ± √1.69)/2 = (-0.9 ± 1.3)/2, giving y = 0.2 or y = -1.1. Since x=0.25 is greater than both y=0.2 and y=-1.1, x > y always. Option A is correct. Option E would be wrong because the relationship is clearly determined.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give the answer. $I. (2x = 5 + 15y) II. (6x - 5y + 1 = 0)$

  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

From equation I: 2x = 5 + 15y, rearranging gives x = (5 + 15y)/2. Substitute into equation II: 6((5 + 15y)/2) - 5y + 1 = 0, which simplifies to 15 + 45y - 5y + 1 = 0, so 40y = -16, giving y = -0.4. Substituting back: x = (5 + 15(-0.4))/2 = (5 - 6)/2 = -0.5. Since x = -0.5 and y = -0.4, we have x < y.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give the answer. $I. (x^2 + 13x + 30 = 0) II. (y^2 – 13y – 30 = 0)$

  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
B Correct answer
Explanation

Solving equation I: x² + 13x + 30 = (x+10)(x+3) = 0, so x = -10 or -3. Solving equation II: y² - 13y - 30 = 0. Using quadratic formula: y = (13 ± √(169 + 120))/2 = (13 ± √289)/2 = (13 ± 17)/2, so y = 15 or y = -2. Comparing all combinations: when x = -10, x < y for both y values (since -10 < 15 and -10 < -2). When x = -3, x < y for y = 15, but x > y for y = -2. Therefore, we can only definitively say that x ≤ y is always true.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and choose the correct option. $I. (x^2 - 2.8x + 1.92 = 0)$ $II. (y^2 - 3.6y + 3.2 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving equation I: x² - 2.8x + 1.92 = 0 gives x = 1.2 or x = 1.6. Solving equation II: y² - 3.6y + 3.2 = 0 gives y = 1.6 or y = 2. Comparing all values: x is always less than or equal to y (when x = 1.6 and y = 1.6, x = y; otherwise x < y). Therefore x ≤ y is correct.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and choose the correct option. $I. (x^2 - x - 2 = 0)$ $II. (y^2 + 3y + 2 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving equation I: x² - x - 2 = 0 gives x = 2 or x = -1. Solving equation II: y² + 3y + 2 = 0 gives y = -2 or y = -1. Comparing all values: 2 > -2, 2 > -1, -1 > -2, and -1 = -1. In all cases, x ≥ y (x is greater than or equal to y). Therefore option B is correct.

Multiple choice

In each of the following questions, two equations I and II are given. You have to solve them and- $( (I) \quad x^{3} – 2695 = 11129 \ (II) \quad y^{3} – 6329 – 5838 = 0 )$ 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 established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving (I): x³ - 2695 = 11129, so x³ = 13824. Since 24³ = 13824, x = 24. Solving (II): y³ - 6329 - 5838 = 0, so y³ = 12167. Since 23³ = 12167, y = 23. Comparing: 24 > 23, so x > y, making option A correct.

Multiple choice

In the given question two equations I & II are given, you have to solve them and answer the question. (I) (x^2 + 7x + 10 = 0) ()()()() (II) (y^2 + 11y + 30 = 0)

  1. if x > y

  2. if x < y

  3. if x = y or relation can’t be establish

  4. if x ≥ y

  5. if x ≤ y
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving equation (I): x^2 + 7x + 10 = 0 factors to (x+5)(x+2) = 0, giving x = -5 or x = -2. Solving equation (II): y^2 + 11y + 30 = 0 factors to (y+6)(y+5) = 0, giving y = -6 or y = -5. Comparing all possible combinations: when x = -2, it is greater than both y values (-5 and -6); when x = -5 and y = -6, x > y; when x = -5 and y = -5, x = y. In all cases, x is either equal to or greater than y.

Multiple choice

In the given question two equations I & II are given, you have to solve them and answer the question. (I) (x^2 - 17x + 72 = 0) ()()() (II) (3y^2 + 17y + 10 = 0)

  1. if x > y

  2. if x < y

  3. if x = y or relation can’t be establish

  4. if x ≥ y

  5. if x ≤ y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving equation (I): x^2 - 17x + 72 = 0 factors to (x-8)(x-9) = 0, giving x = 8 or x = 9. Solving equation (II): 3y^2 + 17y + 10 = 0 factors to (3y+2)(y+5) = 0, giving y = -2/3 or y = -5. Since both x values (8, 9) are positive and much larger than both y values (-2/3, -5), in all possible combinations x > y.

Multiple choice

In the given question two equations I & II are given, you have to solve them and answer the question.(\text{(I)}) (3x^2 + 8x + 5 = 0) (\text{(II)}) (10y^2 + 9y + 2 = 0)

  1. if x > y

  2. if x < y

  3. if x = y or relation can’t be establish

  4. if x ≥ y

  5. if x ≤ y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation (I): 3x^2 + 8x + 5 = 0 factors to (3x+5)(x+1) = 0, giving x = -5/3 or x = -1. Solving equation (II): 10y^2 + 9y + 2 = 0 factors to (5y+2)(2y+1) = 0, giving y = -2/5 or y = -1/2. Converting to decimals: x values are -1.67 and -1; y values are -0.4 and -0.5. In all combinations, y values (closer to zero) are greater than x values.

Multiple choice

In the given question two equations I & II are given, you have to solve them and answer the question. $( (I) 7x + 13y = 9 )( (II) 3x – 4y = 23 )$

  1. if x > y

  2. if x < y

  3. if x = y or relation can’t be establish

  4. if x ≥ y

  5. if x ≤ y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a system of two linear equations in two variables. From equation (II): 3x - 4y = 23, we get x = (23 + 4y)/3. Substituting in equation (I): 7(23 + 4y)/3 + 13y = 9. Solving: 161 + 28y + 39y = 27, so 67y = -134, giving y = -2. Substituting back: x = (23 + 4(-2))/3 = (23 - 8)/3 = 15/3 = 5. Since x = 5 and y = -2, we have x > y.

Multiple choice

In the given question two equations I & II are given, you have to solve them and answer the question. $( (I) x^2 + 14x + 49 = 0 )$ $( (II) 6y^2 + 61y + 153 = 0 )$

  1. if x > y

  2. if x < y

  3. if x = y or relation can’t be establish

  4. if x ≥ y

  5. if x ≤ y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation (I): x^2 + 14x + 49 = 0 is a perfect square - (x+7)^2 = 0, giving x = -7. Solving equation (II): 6y^2 + 61y + 153 = 0 factors to (2y+9)(3y+17) = 0, giving y = -9/2 = -4.5 or y = -17/3 ≈ -5.67. Since x = -7 and y is either -4.5 or -5.67, we have x < y in both cases (on the number line, -7 is to the left of both y values).

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- (I) (x^{2} - 3x + 2 = 0) (II) (y^{2} - 9y + 18 = 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 determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

(I) x²-3x+2=0 factors to (x-1)(x-2)=0, so x=1 or x=2. (II) y²-9y+18=0 factors to (y-3)(y-6)=0, so y=3 or y=6. Both values of x (1,2) are less than both values of y (3,6), so x < y.

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- (I) (x^{2} - 49x + 45 = 0) (II) (4y^{2} + 3y - 1 = 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 determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(I) x²-49x+45=0 gives roots x = (49±√(2401-180))/2 = (49±√2201)/2 ≈ 0.93 and 48.07. (II) 4y²+3y-1=0 factors to (4y-1)(y+1)=0, so y=1/4 or y=-1. The positive root of (I) is 48.07, which is greater than the positive root of (II) which is 0.25. So x > y.

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- ( (I) ) (6x^2 + 17x + 12 = 0 ) ( (II) ) ( 8y^2 + 10y + 3 = 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 determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

(I) 6x²+17x+12=0 factors to (2x+3)(3x+4)=0, so x=-3/2 or x=-4/3. (II) 8y²+10y+3=0 factors to (4y+3)(2y+1)=0, so y=-3/4 or y=-1/2. Comparing: x values are -1.5 and -1.33; y values are -0.75 and -0.5. Since -1.5 < -0.75 and -1.33 < -0.5, we have x < y.