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 (I) and (II) equations are given. You have to solve them and Give answer- (I) (x^2 + 3x - 28 = 0)() (II) (y^2 - 11y + 28 = 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
D Correct answer
Explanation

Solving equation (I): x² + 3x - 28 = 0 gives x = -7 or x = 4. Solving equation (II): y² - 11y + 28 = 0 gives y = 7 or y = 4. Checking all combinations: when x = -7, x < y (for y = 7 or 4); when x = 4, x ≤ y (x = y when y = 4, x < y when y = 7). In all cases, x ≤ y holds true. Option A (x > y) and C (x < y) are incorrect because they don't cover all cases. Option B (x ≥ y) fails when x = -7.

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 + 6x – 7 = 0 ) ( (II) 41y + 17 = 140 )

  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

From (I): x² + 6x - 7 = 0 factors to (x + 7)(x - 1) = 0, so x = -7 or x = 1. From (II): 41y = 140 - 17 = 123, giving y = 3. Comparing: when x = -7, x < y; when x = 1, x < y. In both cases, x < y, so option C is correct. Options A, B, D, and E are incorrect because x is consistently less than y.

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- (l) (2x^{2}– 13 x – 24 = 0) (ll) (3y^{2} + 17y + 24 = 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

Solving 2x²-13x-24=0: factor as (2x+3)(x-8)=0, giving x=8 or x=-1.5. Solving 3y²+17y+24=0: factor as (3y+8)(y+3)=0, giving y=-8/3≈-2.67 or y=-3. For all values of x (8 or -1.5) and y (-2.67 or -3), x is always greater than y. Since every possible x value exceeds every possible y value, x>y is the correct relationship.

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- (I)(3x^2 + 23x + 30 = 0) (II) (6y^2 + 13y + 5 = 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
D Correct answer
Explanation

Solving 3x²+23x+30=0: factors as (3x+20)(x+1.5)=0, giving x=-20/3≈-6.67 or x=-1.5. Solving 6y²+13y+5=0: factors as (2y+1)(3y+5)=0, giving y=-1/2=-0.5 or y=-5/3≈-1.67. Checking all combinations: -6.67<-0.5, -6.67<-1.67, -1.5<-0.5, and -1.5>-1.67. Since some x values are less than some y values and some are greater, but x is never always greater than y, the most accurate relationship is x≤y (since the maximum x value -1.5 is less than or equal to -0.5).

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- (I) (5x^{2} - 44x + 63 = 0) (II) (15y^{2} - 37y + 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
B Correct answer
Explanation

Solving 5x²-44x+63=0: factors as (5x-9)(x-7)=0, giving x=9/5=1.8 or x=7. Solving 15y²-37y+18=0: factors as (3y-2)(5y-9)=0, giving y=2/3≈0.67 or y=9/5=1.8. The possible x values are 1.8 and 7. The possible y values are 0.67 and 1.8. Comparing: minimum x=1.8 and maximum y=1.8. For all x values (1.8 or 7) and y values (0.67 or 1.8), x is always greater than or equal to y. Since x=y is possible (both can be 1.8), x≥y is correct.

Multiple choice

( \text{In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer-} )( \text{(I) } x^{2} = 1296 )( \text{(II) } y = (32768)^{1/3} )

  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=1296 gives x=±36 (two possible values: 36 or -36). Solving y=(32768)^(1/3): since 32³=32768, we get y=32. When x=36, x>y (36>32). When x=-36, x

Multiple choice

In each of the following questions (I) and (II) equations are given. You have to solve them and Give answer- (I) 12x2 - 8x - 7 = 0 (II) 10y2 + 23y + 12 = 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

Solve both quadratics. For (I): x = [8 ± √(64 + 336)]/24 = [8 ± 20]/24, giving x = 7/6 or -1/2. For (II): y = [-23 ± √(529 - 480)]/20 = [-23 ± 7]/20, giving y = -4/5 or -3/2. Comparing all four combinations: 7/6 > -4/5, 7/6 > -3/2, -1/2 > -4/5, -1/2 > -3/2. In every case x > y.

Multiple choice

$(\text{In each of the following questions two equations are given. You have to solve both the equation and find out the values of x and y and Give answer -} )$ $((I) \ (x + 2)^{2} – 25=0 \qquad (II) \ 3y^{2}+y-4=0)$

  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 established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation (I) gives x = 3 or x = -7. Equation (II) gives y = 1 or y = -4/3. When x = 3, x > y for both y values. When x = -7, x < y for both y values. Since the relationship varies depending on which root we take, the relationship cannot be established. Option E 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. $I. (x^2 + 14x = -49) II. (y^2 = 9y)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving equation I: x^2 + 14x + 49 = 0 gives (x+7)^2 = 0, so x = -7 (repeated root). Solving equation II: y^2 - 9y = 0 gives y(y-9) = 0, so y = 0 or y = 9. Comparing: -7 < 0 and -7 < 9, meaning x is always less than y.

Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer. $( (I). x^2 – 12x + 32 = 0 )$ $( (II). y^2 = 324 )$

  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
E Correct answer
Explanation

Solve x² - 12x + 32 = 0: (x - 4)(x - 8) = 0, so x = 4 or x = 8. Solve y² = 324: y = ±18. When x = 4 or 8 and y = -18, we have x > y. When y = +18, we have x < y. Since the relationship depends on which values we pick, it cannot be determined. Option E is correct.

Multiple choice

In each of the following questions, two equations are given, you have to solve them and ( (I) x^2 + 4 = 0 ) ( (II) y^2 + 9 = 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
E Correct answer
Explanation

Equation (I) x² + 4 = 0 gives x = ±2i (imaginary roots). Equation (II) y² + 9 = 0 gives y = ±3i (imaginary roots). Since both are imaginary numbers with different magnitudes (2i vs 3i), no meaningful relationship like >, <, or = can be established between x and y in the real number system.

Multiple choice

In each of the following questions, two equations are given, you have to solve them and ( (I) ) ( x^2 – 14x + 48 = 0 ) ( (II) ) ( y^2 – 23y + 132 = 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
C Correct answer
Explanation

Solving equation (I): x² - 14x + 48 = 0 gives x = 6 or x = 8. Solving equation (II): y² - 23y + 132 = 0 gives y = 11 or y = 12. Comparing the roots: maximum x value is 8 while minimum y value is 11, so x < y for all combinations of roots.

Multiple choice

In each of the following questions two equations are given. You have to solve them and give answer - ( (I) x^2 -10x+25=0 )( (II) y^2 + 26y +169 =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
B Correct answer
Explanation

Solving equation I: x² - 10x + 25 = (x-5)² = 0, so x = 5 (repeated root). Solving equation II: y² + 26y + 169 = (y+13)² = 0, so y = -13 (repeated root). Since 5 > -13, we have x > y always, making option B correct.

Multiple choice

If a and ß are the roots of the equation x2 + px +q = 0, then – a–1, –ß–1 are the roots of which one of the following equations?

  1. qx2 – px + 1 = 0

  2. qx2 + px + 1 = 0

  3. x2 + px – q = 0

  4. x2 – px + q = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a and ß are roots of x² + px + q = 0, then a + ß = -p and aß = q. For the new roots -1/a and -1/ß: sum = -(1/a + 1/ß) = -(a+ß)/(aß) = p/q and product = 1/(aß) = 1/q. The equation is x² - (p/q)x + 1/q = 0, which multiplying by q gives qx² - px + 1 = 0. Option B would give x² + (p/q)x + 1/q = 0, which is incorrect.