Solutions: Linear Equations - Test 5

Q1. (b) intersecting at (-4, -5)

$x=-4$ is vertical, $y=-5$ is horizontal. Intersection is $(-4, -5)$.

Q2. (c) 2

Coincident: $\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16}$. $\frac{1}{2} = \frac{1}{k} \Rightarrow k=2$.

Q3. (b) 50 sq. units

Intercepts are $(10, 0)$ and $(0, 10)$. Area = $\frac{1}{2} \times 10 \times 10 = 50$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Substitute $(3,1)$: $2(3) + 1 - q^2 - 3 = 0 \Rightarrow 4 - q^2 = 0 \Rightarrow q = \pm 2$.

Q5. Add eqs: $-226x - 226y = -678 \Rightarrow x+y=3$.
Subtract eqs: $530x - 530y = 530 \Rightarrow x-y=1$.
Solving $x+y=3, x-y=1 \Rightarrow 2x=4 \Rightarrow x=2, y=1$.
x = 2, y = 1 [2]

Q6. Infinite solutions: $\frac{2}{2\alpha} = \frac{3}{\alpha+\beta} = \frac{7}{28}$.
$\frac{1}{\alpha} = \frac{1}{4} \Rightarrow \alpha = 4$.
$\frac{3}{4+\beta} = \frac{1}{4} \Rightarrow 12 = 4+\beta \Rightarrow \beta = 8$.
α = 4, β = 8 [2]

Q7. Let fraction $x/y$. $x+y = 2x+4 \Rightarrow y-x=4$.
$\frac{x+3}{y+3} = \frac{2}{3} \Rightarrow 3x+9 = 2y+6 \Rightarrow 3x-2y = -3$.
Substitute $y=x+4$: $3x - 2(x+4) = -3 \Rightarrow 3x-2x-8=-3 \Rightarrow x=5$.
$y = 9$. Fraction is $5/9$. [2]

Q8. Let 1 woman take $x$ days, 1 man take $y$ days.
$\frac{2}{x} + \frac{5}{y} = \frac{1}{4}$ and $\frac{3}{x} + \frac{6}{y} = \frac{1}{3}$.
Let $1/x=u, 1/y=v$. $2u+5v=1/4, 3u+6v=1/3$.
Solving yields $u=1/18, v=1/36$.
Woman: 18 days, Man: 36 days [3]

Q9. Let rows $x$, students per row $y$. Total $xy$.
$(y+3)(x-1) = xy \Rightarrow xy - y + 3x - 3 = xy \Rightarrow 3x - y = 3$.
$(y-3)(x+2) = xy \Rightarrow xy + 2y - 3x - 6 = xy \Rightarrow -3x + 2y = 6$.
Adding eqs: $y = 9$. Substitute: $3x - 9 = 3 \Rightarrow 3x = 12 \Rightarrow x = 4$.
Total students = $4 \times 9 = 36$. [3]

Q10. Let train speed $x$, bus speed $y$.
$\frac{60}{x} + \frac{240}{y} = 4$ ...(i)
$\frac{100}{x} + \frac{200}{y} = 4 + \frac{10}{60} = \frac{25}{6}$. ...(ii)
Let $1/x=u, 1/y=v$. $60u + 240v = 4 \Rightarrow 15u + 60v = 1$.
$100u + 200v = 25/6 \Rightarrow 24u + 48v = 1$.
Solving gives $u = 1/60, v = 1/80$.
Train: 60 km/h, Bus: 80 km/h [5]

(i) Let length $x$, breadth $y$. Area $xy$.
$(x-5)(y+3) = xy - 9 \Rightarrow 3x - 5y = 6$.
$(x+3)(y+2) = xy + 67 \Rightarrow 2x + 3y = 61$. [2]

(ii) Solving $3x - 5y = 6$ and $2x + 3y = 61$:
Multiply first by 3, second by 5: $9x - 15y = 18$ and $10x + 15y = 305$.
$19x = 323 \Rightarrow x = 17$.
$3(17) - 5y = 6 \Rightarrow 51 - 6 = 5y \Rightarrow 45 = 5y \Rightarrow y = 9$.
Length = 17 units, Breadth = 9 units. [1]

(iii) Area = $17 \times 9 = 153$ sq. units. [1]