Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option.(a) x^2 - 32x + 247 = 0 (b) y^2 - 35y + 304 = 0
  1. $\(x > y\)$
  2. $\(x \le y\)$
  3. $\(x \ge y\)$
  4. $\(x < y\)$

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solving equation (a): x^2-32x+247=0 gives x=(32±√(1024-988))/2=(32±6)/2, so x=19 or x=13. Solving equation (b): y^2-35y+304=0 gives y=(35±√(1225-1216))/2=(35±3)/2, so y=19 or y=16. When x=13, y is always greater. When x=19, x=y when y=19 but x < y when y=19 is false. Actually, when x=19, y can be 16 (so x > y) or 19 (so x = y). Since the relationship varies, it cannot be established consistently. Option E is correct.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option a) 6x2 + 13x - 169 = 0 b) y2 + 8y - 65 = 0

  1. x > y

  2. x ≤ y

  3. x ≥ y

  4. x < y

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For equation (a): 6x²+13x-169=0 gives x=13/3≈4.33 or x=-13/2=-6.5. For equation (b): y²+8y-65=0 gives y=5 or y=-13. Since x can be 4.33 (less than y=5) or -6.5 (greater than y=-13), and y can be 5 or -13, the relationship between x and y depends on which specific roots are compared. Therefore, no consistent relationship can be established.

Multiple choice

In each of the following questions two equations (I) and (II) are given. Solve these equations and give the answer: (I) (x^2 - 324 = 0) (II) (y = \sqrt{324})
  1. $\(x < y\)$
  2. $\(x > y\)$
  3. $\(x \ge y\)$
  4. $\(x \le y\)$
  5. $\(x = y\) or no relationship can be established between \(x\) and \(y\)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For equation (I), x^2 - 324 = 0 gives x = ±18. For equation (II), y = √324 gives y = 18 (principal square root). When x = 18, x = y; when x = -18, x < y. In both cases, x ≤ y holds true.

Multiple choice

In each of these questions, two equations (a) and (b) are given. You have to solve both the equations and choose the correct option. a) (x^2 - 12x - 108 = 0) b) (y^2 + 28y + 132 = 0)
  1. $\(x > y\)$
  2. $\(x \le y\)$
  3. $\(x \ge y\)$
  4. $\(x < y\)$

  5. x = y or relationship between x and y can't be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving x^2 - 12x - 108 = 0 gives x = -6 or 18. Solving y^2 + 28y + 132 = 0 gives y = -6 or -22. Since x can be 18 (greater than y's max -6) and both equal -6, x ≥ y is correct.

Multiple choice

In each of the following questions two equations (I) and (II) are given. Solve these equations and give the answer: ( (I) x^2 - 4x - 12 = 0 ) ( (II) 3y^2 - 8y + 5 = 0 )
  1. $x < y$
  2. $x > y$
  3. $x \ge y$
  4. $x \le y$

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

Equation (I): x² - 4x - 12 = 0. Solving: x = [4 ± √(16 + 48)]/2 = [4 ± √64]/2 = [4 ± 8]/2. So x = 12/2 = 6 or x = -4/2 = -2. Equation (II): 3y² - 8y + 5 = 0. Solving: y = [8 ± √(64 - 60)]/6 = [8 ± √4]/6 = [8 ± 2]/6. So y = 10/6 = 5/3 or y = 6/6 = 1. Comparing: x = {-2, 6} and y = {1, 5/3}. We have: -2 < 1, -2 < 5/3, 6 > 1, 6 > 5/3. Since x is sometimes less than y and sometimes greater than y, no consistent relationship exists.

Multiple choice

Solve the following systems of equations using elimination method. $\dfrac{5}{x}-\dfrac{4}{y}=-2, \dfrac{2}{x}+\dfrac{3}{y}=13$
  1. $\left ( \dfrac{1}{2}, \dfrac{1}{3} \right )$
  2. $\left ( \dfrac{1}{3}, \dfrac{1}{3} \right )$
  3. $\left ( \dfrac{1}{2}, \dfrac{1}{4} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Substituting x=1/2 and y=1/3 into the first equation: 5/(1/2) - 4/(1/3) = 10 - 12 = -2. Substituting into the second: 2/(1/2) + 3/(1/3) = 4 + 9 = 13. Both equations are satisfied.

Multiple choice

Solve the following system of equations by elimination method. $\dfrac{3}{x}+\dfrac{5}{y}=\dfrac{20}{xy}, \dfrac{2}{x}+\dfrac{5}{y}=\dfrac{15}{xy}, x\neq 0, y\neq 0$
  1. $(1, 5)$
  2. $(2, 5)$
  3. $(1, 2)$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Subtracting the second equation from the first gives 1/x = 5/xy, which simplifies to y = 5. Substituting y = 5 into the second equation gives 2/x + 1 = 15/(5x), which simplifies to 2/x + 1 = 3/x, leading to x = 1.

Multiple choice

Solve the following equations by substitution method. $\dfrac{3}{x} + \dfrac{1}{y} = 7; \dfrac{5}{x} - \dfrac{4}{y} = 6. (x \neq 0, y \neq 0)$
  1. $x = \dfrac{1}{2}, y = 1$
  2. $x = \dfrac{-1}{2}, y = 1$
  3. $x = \dfrac{1}{2}, y = -1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = 1/x, v = 1/y. 3u + v = 7 and 5u - 4v = 6. Multiply first by 4: 12u + 4v = 28. Add to second: 17u = 34 => u = 2. Then x = 1/2. Substitute u=2 into 3u+v=7: 6+v=7 => v=1. Then y = 1.

Multiple choice

Solve the following system of equations by elimination method. $\dfrac{15}{x}+\dfrac{2}{y}=17, \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}, x\neq 0, y\neq 0$
  1. $\left ( 5, \dfrac{1}{7} \right )$
  2. $\left ( 2, \dfrac{1}{7} \right )$
  3. $\left ( 5, \dfrac{2}{7} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = 1/x and v = 1/y. Equations: 15u + 2v = 17 and u + v = 36/5. From second, u = 36/5 - v. Substitute: 15(36/5 - v) + 2v = 17 => 108 - 15v + 2v = 17 => 13v = 91 => v = 7. Then u = 36/5 - 7 = 1/5. So x = 5, y = 1/7.

Multiple choice

Solve the following system of equations by elimination method. $\dfrac {2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}, \dfrac{3}{x}+\dfrac{2}{y}=0, x\neq 0, y\neq 0$
  1. $(6, -4)$
  2. $(6, 4)$
  3. $(-6, -4)$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = 1/x, v = 1/y. 2u + 2v/3 = 1/6 => 12u + 4v = 1. 3u + 2v = 0 => v = -1.5u. Substitute: 12u + 4(-1.5u) = 1 => 12u - 6u = 1 => 6u = 1 => u = 1/6. x = 6. v = -1.5(1/6) = -0.25 = -1/4. y = -4. Solution (6, -4).