MCQ test, Set 10

Answer: (C) $-2$

Put $x = -4$, $y = 3$: $k(-4) + (3) = 11$

$\Rightarrow -4k + 3 = 11$

$\Rightarrow -4k = 11 - 3 = 8$

$\Rightarrow k = -2$

Answer: (D) $(-5, -2)$

Put each point in $3x + y$; only $(-5, -2)$ gives $-17$:

$3(-5) + (-2) = -15 - 2 = -17$

Answer: (D) $-1$

$2(-1) + 5y = -7$

$\Rightarrow -2 + 5y = -7$

$\Rightarrow 5y = -7 + 2 = -5$

$\Rightarrow y = -1$

Answer: (B) $18x + 12y = -3$

With $18x + 12y = -3$: $\displaystyle \frac{a_1}{a_2} = \frac{6}{18} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{-4}{12} = -\frac{1}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{-1}{-3} = \frac{1}{3}$

$\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, so the lines intersect: a unique solution.

Answer: (D) no solution

$\displaystyle \frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2}$, $\displaystyle \frac{b_1}{b_2} = \frac{-8}{-16} = \frac{1}{2}$, $\displaystyle \frac{c_1}{c_2} = \frac{-1}{1} = -1$

$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, so the lines are parallel: no solution.

Answer: (C) $-5$

For no solution: $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

$\displaystyle \frac{-10}{k} = \frac{2}{1}$

$\Rightarrow 2k = -10$

$\Rightarrow k = -5$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{11}{10} \ne 2$, so the pair has no solution.

Answer: (A) $12$

For infinitely many solutions: $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

$\displaystyle \frac{k}{4} = \frac{-3}{-1}$

$\Rightarrow -k = -12$

$\Rightarrow k = 12$

Also $\displaystyle \frac{c_1}{c_2} = \frac{-18}{-6} = 3$, equal to the other ratios.

Answer: (B) $3x - 12y = 12$

With $3x - 12y = 12$: $\displaystyle \frac{a_1}{a_2} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{-4}{-12} = \frac{1}{3}$, $\displaystyle \frac{c_1}{c_2} = \frac{4}{12} = \frac{1}{3}$

$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, so the lines coincide: infinitely many solutions.

Answer: (A) $24 + 8y = 56$

$3(8 + 2y) + 2y = 56$

$\Rightarrow 24 + 6y + 2y = 56$

$\Rightarrow 24 + 8y = 56$

Answer: (A) $(4, 3)$

$2x + y = 11 \quad (1)$

$3x - y = 9 \quad (2)$

$(1) \Rightarrow y = 11 - 2x$

Substitute in $(2)$: $3x - (11 - 2x) = 9$

$\Rightarrow 3x - 11 + 2x = 9$

$\Rightarrow 5x - 11 = 9$

$\Rightarrow 5x = 9 + 11$

$\Rightarrow 5x = 20$

$\Rightarrow x = 4$

$(1) \Rightarrow y = 11 - 2(4) = 3$

$\therefore (x, y) = (4, 3)$

Answer: (C) $4$

$5x + 2y = -7 \quad (1)$

$4x - 5y = -32 \quad (2)$

$(1) \times 4 \Rightarrow 20x + 8y = -28$

$(2) \times 5 \Rightarrow 20x - 25y = -160$

Subtracting, $33y = 132$

$\Rightarrow y = 4$

$(2) \Rightarrow 4x - 5(4) = -32$

$\Rightarrow 4x - 20 = -32$

$\Rightarrow 4x = -32 + 20$

$\Rightarrow 4x = -12$

$\Rightarrow x = -3$

$\therefore (x, y) = (-3, 4)$

Answer: (D) (1) by $3$ and (2) by $1$

The LCM of $2$ and $6$ is $6$.

$(1) \times 3 \Rightarrow 6x - 21y = 6$

$(2) \times 1 \Rightarrow 6x - 5y = -12$

Now the $x$ terms are equal, and subtracting removes $x$.

Answer: (B) $(-2, -2)$

$\displaystyle x = \frac{-10}{5} = -2$

$\displaystyle y = \frac{-10}{5} = -2$

Answer: (D) $13$

$a_1b_2 - a_2b_1 = (1)(5) - (2)(-4) = 5 + 8 = 13$

Answer: (C) $25x - y = -15$

Multiply every term by the LCM $5$:

$\displaystyle 5x - \frac{y}{5} = -3 \Rightarrow \frac{25x - y}{5} = -3 \Rightarrow 25x - y = -15$

Answer: (A) $(4, 4)$

$\displaystyle 3x - \frac{3y}{4} = 9 \Rightarrow \frac{12x - 3y}{4} = 9 \Rightarrow 12x - 3y = 36 \quad (1)$

$3x - 2y = 4 \quad (2)$

$(1) \Rightarrow 12x - 3y = 36$

$(2) \times 4 \Rightarrow 12x - 8y = 16$

Subtracting, $5y = 20$

$\Rightarrow y = 4$

$(2) \Rightarrow 3x - 2(4) = 4$

$\Rightarrow 3x - 8 = 4$

$\Rightarrow 3x = 4 + 8$

$\Rightarrow 3x = 12$

$\Rightarrow x = 4$

$\therefore (x, y) = (4, 4)$

Answer: (D) $(2, -2)$

Let $\displaystyle \frac{1}{x} = p$ and $\displaystyle \frac{1}{y} = q$.

$p + 5q = -2 \quad (1)$

$p - q = 1 \quad (2)$

Subtracting, $6q = -3$

$\displaystyle \Rightarrow q = -\frac{1}{2}$

$\displaystyle (1) \Rightarrow p + 5(-\frac{1}{2}) = -2$

$\displaystyle \Rightarrow p - \frac{5}{2} = -2$

$\displaystyle \Rightarrow p = -2 + \frac{5}{2}$

$\displaystyle \Rightarrow p = \frac{1}{2}$

$\displaystyle \therefore (p, q) = (\frac{1}{2}, -\frac{1}{2})$

$\displaystyle \frac{1}{x} = \frac{1}{2} \Rightarrow x = 2$

$\displaystyle \frac{1}{y} = -\frac{1}{2} \Rightarrow y = -2$

$\therefore (x, y) = (2, -2)$

Answer: (C) $3p + 6q = 4$

$\displaystyle \frac{3}{x} = 3 \times \frac{1}{x} = 3p$ and $\displaystyle \frac{6}{y} = 6q$

So $3p + 6q = 4$

Answer: (A) $5p + 7y = 1$

$\displaystyle \frac{5}{x} = 5 \times \frac{1}{x} = 5p$

So $5p + 7y = 1$

Answer: (A) $(-1, 4)$

Let $\displaystyle \frac{1}{x} = p$.

$3p - 4y + 19 = 0 \quad (1)$

$p - 4y + 17 = 0 \quad (2)$

$\displaystyle \frac{p}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\displaystyle \frac{p}{-68 + 76} = \frac{y}{19 - 51} = \frac{1}{-12 + 4}$

$\displaystyle \frac{p}{8} = \frac{y}{-32} = \frac{1}{-8}$

$\displaystyle \Rightarrow p = \frac{8}{-8}, \; y = \frac{-32}{-8}$

$\therefore (p, y) = (-1, 4)$

$\displaystyle \frac{1}{x} = -1 \Rightarrow x = -1$

$\therefore (x, y) = (-1, 4)$

Answer: (C) $(1, 0)$

Let $\displaystyle \frac{1}{x - 3} = p$ and $\displaystyle \frac{1}{y - 2} = q$.

$5p + q = -3 \quad (1)$

$6p + 4q = -5 \quad (2)$

$(1) \Rightarrow q = -3 - 5p$

Substitute in $(2)$: $6p + 4(-3 - 5p) = -5$

$\Rightarrow 6p - 12 - 20p = -5$

$\Rightarrow -14p - 12 = -5$

$\Rightarrow -14p = -5 + 12$

$\Rightarrow -14p = 7$

$\displaystyle \Rightarrow p = -\frac{1}{2}$

$\displaystyle (1) \Rightarrow q = -3 - 5(-\frac{1}{2}) = -\frac{1}{2}$

$\displaystyle \therefore (p, q) = (-\frac{1}{2}, -\frac{1}{2})$

$\displaystyle \frac{1}{x - 3} = -\frac{1}{2} \Rightarrow x - 3 = -2 \Rightarrow x = 1$

$\displaystyle \frac{1}{y - 2} = -\frac{1}{2} \Rightarrow y - 2 = -2 \Rightarrow y = 0$

$\therefore (x, y) = (1, 0)$

Answer: (B) $4$

$\displaystyle \frac{1}{x - 1} = \frac{1}{3} \Rightarrow x - 1 = 3$

$\Rightarrow x = 4$

Answer: (B) $(2, -2)$

$\displaystyle \frac{3x + 5y}{xy} = 1 \Rightarrow \frac{3}{y} + \frac{5}{x} = 1$

$\displaystyle \frac{2x + 4y}{xy} = 1 \Rightarrow \frac{2}{y} + \frac{4}{x} = 1$

Let $\displaystyle \frac{1}{x} = p$ and $\displaystyle \frac{1}{y} = q$.

$5p + 3q = 1 \quad (1)$

$4p + 2q = 1 \quad (2)$

$(1) \times 4 \Rightarrow 20p + 12q = 4$

$(2) \times 5 \Rightarrow 20p + 10q = 5$

Subtracting, $2q = -1$

$\displaystyle \Rightarrow q = -\frac{1}{2}$

$\displaystyle (2) \Rightarrow 4p + 2(-\frac{1}{2}) = 1$

$\Rightarrow 4p - 1 = 1$

$\Rightarrow 4p = 1 + 1$

$\Rightarrow 4p = 2$

$\displaystyle \Rightarrow p = \frac{1}{2}$

$\displaystyle \therefore (p, q) = (\frac{1}{2}, -\frac{1}{2})$

$\displaystyle \frac{1}{x} = \frac{1}{2} \Rightarrow x = 2$

$\displaystyle \frac{1}{y} = -\frac{1}{2} \Rightarrow y = -2$

$\therefore (x, y) = (2, -2)$

Answer: (D) $\displaystyle \frac{6}{y} + \frac{7}{x}$

Divide each term by $xy$: $\displaystyle \frac{6x + 7y}{xy} = \frac{6x}{xy} + \frac{7y}{xy}$

$\displaystyle = \frac{6}{y} + \frac{7}{x}$

Answer: (B) $(25, 25)$

Let $\displaystyle \frac{1}{\sqrt{x}} = p$ and $\displaystyle \frac{1}{\sqrt{y}} = q$.

$5p + 5q = 2 \quad (1)$

$p + 4q = 1 \quad (2)$

$(1) \Rightarrow 5p + 5q = 2$

$(2) \times 5 \Rightarrow 5p + 20q = 5$

Subtracting, $-15q = -3$

$\displaystyle \Rightarrow q = \frac{1}{5}$

$\displaystyle (2) \Rightarrow p + 4(\frac{1}{5}) = 1$

$\displaystyle \Rightarrow p + \frac{4}{5} = 1$

$\displaystyle \Rightarrow p = 1 - \frac{4}{5}$

$\displaystyle \Rightarrow p = \frac{1}{5}$

$\displaystyle \therefore (p, q) = (\frac{1}{5}, \frac{1}{5})$

$\displaystyle \frac{1}{\sqrt{x}} = \frac{1}{5} \Rightarrow \sqrt{x} = 5 \Rightarrow x = 25$

$\displaystyle \frac{1}{\sqrt{y}} = \frac{1}{5} \Rightarrow \sqrt{y} = 5 \Rightarrow y = 25$

$\therefore (x, y) = (25, 25)$