MCQ test, Set 5

Answer: (B) $6$

Put $x = -5$, $y = 2$: $k(-5) + 5(2) = -20$

$\Rightarrow -5k + 10 = -20$

$\Rightarrow -5k = -20 - 10 = -30$

$\Rightarrow k = 6$

Answer: (D) infinitely many

$(4, 0)$, $(6, 1)$, $(0, -2)$, ... all satisfy $x - 2y = 4$: every value of $x$ gives a value of $y$.

So it has infinitely many solutions (every point on its line).

Answer: (C) $\displaystyle \frac{1}{7}$

Here $a_1 = 1$, $a_2 = 7$.

$\displaystyle \frac{a_1}{a_2} = \frac{1}{7}$

Answer: (C) a unique solution

$\displaystyle \frac{a_1}{a_2} = \frac{1}{3}$, $\displaystyle \frac{b_1}{b_2} = \frac{1}{7}$, $\displaystyle \frac{c_1}{c_2} = \frac{3}{1} = 3$

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

Answer: (A) $8x + 20y = 6$

With $8x + 20y = 6$: $\displaystyle \frac{a_1}{a_2} = \frac{2}{8} = \frac{1}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{5}{20} = \frac{1}{4}$, $\displaystyle \frac{c_1}{c_2} = \frac{2}{6} = \frac{1}{3}$

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

Answer: (C) $2$

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

$\displaystyle \frac{k}{1} = \frac{10}{5}$

$\Rightarrow 5k = 10$

$\Rightarrow k = 2$

Check: $\displaystyle \frac{c_1}{c_2} = \frac{-5}{2} \ne 2$, so the pair has no solution.

Answer: (D) $10$

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

$\displaystyle \frac{k}{20} = \frac{-4}{-8}$

$\Rightarrow -8k = -80$

$\Rightarrow k = 10$

Also $\displaystyle \frac{c_1}{c_2} = \frac{14}{28} = \frac{1}{2}$, equal to the other ratios.

Answer: (A) infinitely many solutions

$\displaystyle \frac{a_1}{a_2} = \frac{8}{2} = 4$, $\displaystyle \frac{b_1}{b_2} = \frac{-20}{-5} = 4$, $\displaystyle \frac{c_1}{c_2} = \frac{36}{9} = 4$

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

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

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

$5x - y = 21 \quad (2)$

$(1) \Rightarrow y = 7 - 2x$

Substitute in $(2)$: $5x - (7 - 2x) = 21$

$\Rightarrow 5x - 7 + 2x = 21$

$\Rightarrow 7x - 7 = 21$

$\Rightarrow 7x = 21 + 7$

$\Rightarrow 7x = 28$

$\Rightarrow x = 4$

$(1) \Rightarrow y = 7 - 2(4) = -1$

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

Answer: (D) $15 + 29y = -14$

$5(3 + 5y) + 4y = -14$

$\Rightarrow 15 + 25y + 4y = -14$

$\Rightarrow 15 + 29y = -14$

Answer: (C) $1$

$3x - 2y = 7 \quad (1)$

$5x + 3y = 18 \quad (2)$

$(1) \times 5 \Rightarrow 15x - 10y = 35$

$(2) \times 3 \Rightarrow 15x + 9y = 54$

Subtracting, $-19y = -19$

$\Rightarrow y = 1$

$(1) \Rightarrow 3x - 2(1) = 7$

$\Rightarrow 3x - 2 = 7$

$\Rightarrow 3x = 7 + 2$

$\Rightarrow 3x = 9$

$\Rightarrow x = 3$

$\therefore (x, y) = (3, 1)$

Answer: (B) $(2, 1)$

$3x - 5y = 1 \quad (1)$

$9x + 5y = 23 \quad (2)$

$(1) \times 3 \Rightarrow 9x - 15y = 3$

$(2) \Rightarrow 9x + 5y = 23$

Subtracting, $-20y = -20$

$\Rightarrow y = 1$

$(1) \Rightarrow 3x - 5(1) = 1$

$\Rightarrow 3x - 5 = 1$

$\Rightarrow 3x = 1 + 5$

$\Rightarrow 3x = 6$

$\Rightarrow x = 2$

$\therefore (x, y) = (2, 1)$

Answer: (A) $-6$

$a_1b_2 - a_2b_1 = (2)(3) - (4)(3) = 6 - 12 = -6$

Answer: (A) $(3, 2)$

$\displaystyle x = \frac{33}{11} = 3$

$\displaystyle y = \frac{22}{11} = 2$

Answer: (B) $(4, 4)$

$\displaystyle \frac{3x}{2} - \frac{7y}{4} = -1 \Rightarrow \frac{6x - 7y}{4} = -1 \Rightarrow 6x - 7y = -4 \quad (1)$

$3x - 5y = -8 \quad (2)$

$(1) \Rightarrow 6x - 7y = -4$

$(2) \times 2 \Rightarrow 6x - 10y = -16$

Subtracting, $3y = 12$

$\Rightarrow y = 4$

$(2) \Rightarrow 3x - 5(4) = -8$

$\Rightarrow 3x - 20 = -8$

$\Rightarrow 3x = -8 + 20$

$\Rightarrow 3x = 12$

$\Rightarrow x = 4$

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

Answer: (D) $3x + 2y = 8$

Multiply every term by the LCM $4$:

$\displaystyle \frac{3x}{4} + \frac{y}{2} = 2 \Rightarrow \frac{3x + 2y}{4} = 2 \Rightarrow 3x + 2y = 8$

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

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

$4p + 5q = -6 \quad (1)$

$4p + 3q = -4 \quad (2)$

Subtracting, $2q = -2$

$\Rightarrow q = -1$

$(1) \Rightarrow 4p + 5(-1) = -6$

$\Rightarrow 4p - 5 = -6$

$\Rightarrow 4p = -6 + 5$

$\Rightarrow 4p = -1$

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

$\displaystyle \therefore (p, q) = (-\frac{1}{4}, -1)$

$\displaystyle \frac{1}{x} = -\frac{1}{4} \Rightarrow x = -4$

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

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

Answer: (A) $3p - q = 7$

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

So $3p - q = 7$

Answer: (B) $7p + 2y = 11$

$\displaystyle \frac{7}{x} = 7 \times \frac{1}{x} = 7p$

So $7p + 2y = 11$

Answer: (C) $\displaystyle (\frac{1}{4}, -3)$

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

$5p + 3y - 11 = 0 \quad (1)$

$p - 3y - 13 = 0 \quad (2)$

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

$\displaystyle \frac{p}{-39 - 33} = \frac{y}{-11 + 65} = \frac{1}{-15 - 3}$

$\displaystyle \frac{p}{-72} = \frac{y}{54} = \frac{1}{-18}$

$\displaystyle \Rightarrow p = \frac{-72}{-18}, \; y = \frac{54}{-18}$

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

$\displaystyle \frac{1}{x} = 4 \Rightarrow x = \frac{1}{4}$

$\displaystyle \therefore (x, y) = (\frac{1}{4}, -3)$

Answer: (C) $(0, 5)$

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

$3p + q = 2 \quad (1)$

$5p + 3q = 4 \quad (2)$

$(1) \Rightarrow q = 2 - 3p$

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

$\Rightarrow 5p + 6 - 9p = 4$

$\Rightarrow -4p + 6 = 4$

$\Rightarrow -4p = 4 - 6$

$\Rightarrow -4p = -2$

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

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

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

$\displaystyle \frac{1}{x + 2} = \frac{1}{2} \Rightarrow x + 2 = 2 \Rightarrow x = 0$

$\displaystyle \frac{1}{y - 3} = \frac{1}{2} \Rightarrow y - 3 = 2 \Rightarrow y = 5$

$\therefore (x, y) = (0, 5)$

Answer: (A) $\displaystyle (-\frac{1}{2}, 2)$

$\displaystyle \frac{2x + 2y}{xy} = -3 \Rightarrow \frac{2}{y} + \frac{2}{x} = -3$

$\displaystyle \frac{2x - 2y}{xy} = 5 \Rightarrow \frac{2}{y} - \frac{2}{x} = 5$

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

$2p + 2q = -3 \quad (1)$

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

Subtracting, $4p = -8$

$\Rightarrow p = -2$

$(1) \Rightarrow 2q + 2(-2) = -3$

$\Rightarrow 2q - 4 = -3$

$\Rightarrow 2q = -3 + 4$

$\Rightarrow 2q = 1$

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

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

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

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

$\displaystyle \therefore (x, y) = (-\frac{1}{2}, 2)$

Answer: (B) $\displaystyle \frac{5}{y} - \frac{3}{x}$

Divide each term by $xy$: $\displaystyle \frac{5x - 3y}{xy} = \frac{5x}{xy} - \frac{3y}{xy}$

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

Answer: (D) $(1, 16)$

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

$2p + 4q = 3 \quad (1)$

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

$(1) \times 5 \Rightarrow 10p + 20q = 15$

$(2) \times 2 \Rightarrow 10p - 8q = 8$

Subtracting, $28q = 7$

$\displaystyle \Rightarrow q = \frac{1}{4}$

$\displaystyle (1) \Rightarrow 2p + 4(\frac{1}{4}) = 3$

$\Rightarrow 2p + 1 = 3$

$\Rightarrow 2p = 3 - 1$

$\Rightarrow 2p = 2$

$\Rightarrow p = 1$

$\displaystyle \therefore (p, q) = (1, \frac{1}{4})$

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

$\displaystyle \frac{1}{\sqrt{y}} = \frac{1}{4} \Rightarrow \sqrt{y} = 4 \Rightarrow y = 16$

$\therefore (x, y) = (1, 16)$

Answer: (B) $49$

$\sqrt{x} = 7$

$\Rightarrow x = 7^2 = 49$