Quantitative Aptitude

Quadratic Equations

367 Questions
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Quadratic Equations Questions

Multiple choice

Solve the following pair of linear equation: $\dfrac{5}{2x} + \dfrac{2}{3y} = 7, \, \dfrac{3}{x} + \dfrac{2}{y} = 12, \, x \neq 0 , \, y \neq 0$
  1. $x=\dfrac {1}{2},y=\dfrac {1}{4}$
  2. $x=\dfrac {1}{4},y=\dfrac {1}{3}$
  3. $x=\dfrac {1}{2},y=\dfrac {1}{3}$
  4. $x=\dfrac {1}{4},y=\dfrac {1}{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let u=1/x, v=1/y. 5/2u + 2/3v = 7 => 15u + 4v = 42. 3u + 2v = 12 => 6u + 4v = 24. Subtracting: 9u = 18, u = 2, so x = 1/2. 6(2) + 4v = 24 => 4v = 12, v = 3, so y = 1/3.

Multiple choice

Solve the following system of equation for x and y $\displaystyle \frac{2}{x}+\frac{3}{y}=2;\, \frac{8}{x}-\frac{6}{y}=2$
  1. $\displaystyle \frac{1}{2},\frac{1}{3}$
  2. $\displaystyle \frac{1}{3},\frac{1}{2}$

  3. 2, 3

  4. 3, 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let u = 1/x and v = 1/y. The equations become 2u + 3v = 2 and 8u - 6v = 2. Multiplying the first by 2 gives 4u + 6v = 4, and adding this to the second gives 12u = 6, so u = 1/2 (x=2). Substituting back, 2(1/2) + 3v = 2 gives 3v = 1, so v = 1/3 (y=3).

Multiple choice

Solve the system of the equations for $x$ and $y$: $\displaystyle b^{\log x}= a^{\log y}$, where $\displaystyle a> 0,b> 0$ and $\displaystyle a\neq b,ab\neq 1$
  1. $x=\dfrac1a$ and $y=\dfrac1b$
  2. $x=\dfrac1{ab}$ and $y=\dfrac1b$
  3. $x=\dfrac1b$ and $y=\dfrac1a$
  4. $x=\dfrac1{a}$ and $y=\dfrac1{ab}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Take log on both sides: log(x) * log(b) = log(y) * log(a). This is log(x)/log(a) = log(y)/log(b) = k. So x = a^k, y = b^k. If we test x = 1/a, y = 1/b, then log(1/a) * log(b) = -log(a)log(b) and log(1/b) * log(a) = -log(b)log(a). They are equal.

Multiple choice

Solve the following equations: $9x + y - 8z = 0$, $4x - 8y + 7z = 0$, $yz + zx + xy = 47$.
  1. $x=\pm3; y=\pm5; z=\pm4$
  2. $x=\pm 2; y=\pm4; z=\pm1$
  3. $x=x=\pm4; y=3; z=\pm 5$

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

Solving the linear system 9x+y-8z=0 and 4x-8y+7z=0 gives x:y:z = (7-64) : (-32-63) : (-72-4) = -57 : -95 : -76 = 3 : 5 : 4. Substituting x=3k, y=5k, z=4k into yz+zx+xy=47: 20k^2 + 12k^2 + 15k^2 = 47 => 47k^2 = 47 => k=1 or -1. Thus x=3, y=5, z=4 or x=-3, y=-5, z=-4.

Multiple choice

Solve the following equations: $3x + y - 2z = 0$, $4x - y - 3z = 0$, $x^{3} + y^{3} + z^{3} = 467$
  1. $x = 5; y = -1; z = 5$
  2. $x = 2; y = -1; z = -7$
  3. $x = 3; y = -2; z = 7$
  4. $x = 5; y = -1; z = 7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving the system of linear equations 3x + y - 2z = 0 and 4x - y - 3z = 0 leads to x = z and y = -z. Substituting into x^3 + y^3 + z^3 = 467 gives z^3 + (-z)^3 + z^3 = 467, so z^3 = 467. Checking the options, x=5, y=-1, z=7 satisfies the equations: 3(5) + (-1) - 2(7) = 15 - 1 - 14 = 0 and 4(5) - (-1) - 3(7) = 20 + 1 - 21 = 0. Also 5^3 + (-1)^3 + 7^3 = 125 - 1 + 343 = 467.

Multiple choice

The graphs of the two equations $y = ax^2+bx+c$ and $y =$ $Ax^2+Bx+C$, such that $a$ and $A$ have different signs and that the quantities $b^2 - 4ac$ and $B^2-4AC$ are both negative,

  1. intersect at two points

  2.  intersect at one point

  3. do not intersect

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The condition b^2 - 4ac < 0 implies the parabola y = ax^2+bx+c never crosses the x-axis. Since a and A have different signs, one parabola opens upward and the other downward. Because they never cross the x-axis, they are entirely on opposite sides of the x-axis, meaning they cannot intersect.

Multiple choice

Solving the following equation. $\dfrac{x-a}{a^2}+\dfrac{y-b}{b^2}=\dfrac{1}{x-b}-\dfrac{1}{y-a}-\dfrac{1}{a-b}=0$. we get $x=\dfrac{a^2}{b}, \dfrac{a(2b-a)}{b}; y=\dfrac{b^2}{a}, \dfrac{b(2a-b)}{a}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given solution x = a^2/b and y = b^2/a can be verified by substitution into the original equations. The equation structure (x-a)/a^2 + (y-b)/b^2 = 0 with 1/(x-b) - 1/(y-a) - 1/(a-b) = 0 yields the stated values, making the claim true.

Multiple choice

The quantity $(x+m)^2- (x+n)^2 = (m-n)^2$, where $m$ and $n$ are unequal non zero constants, is satisfied by $x = am +bn$, where:
  1. $a = 0, b$ has a unique non-zero value
  2. $a = 0, b$ has two non-zero values
  3. $b = 0, a$ has a unique non-zero values
  4. $b = 0, a$ has two non-zero values
  5. $a$ and $b$ each have a unique non-zero value.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Expand the equation: (x^2 + 2xm + m^2) - (x^2 + 2xn + n^2) = (m-n)^2. This simplifies to 2x(m-n) + m^2 - n^2 = m^2 - 2mn + n^2. Further: 2x(m-n) = 2n^2 - 2mn = -2n(m-n). Since m is not equal to n, we can divide by 2(m-n) to get x = -n. This matches x = am + bn with a = 0 and b = -1.