Solutions: Linear Equations - Test 1

Q1. (b) 6

For no solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{1}{3} = \frac{2}{k} \Rightarrow k = 6$. Also $\frac{1}{3} \neq \frac{5}{15}$ is false? Wait, $3x+ky+15=0 \Rightarrow 3x+ky=-15$.
So $\frac{1}{3} = \frac{2}{k} \neq \frac{5}{-15}$. $\frac{1}{3} \neq -\frac{1}{3}$. True. So $k=6$.

Q2. (a) intersecting or coincident

Consistent means at least one solution.

Q3. (b) $x=9, y=5$

$x+y=14, x-y=4$. Adding: $2x=18 \Rightarrow x=9$. $y=14-9=5$.

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

$\frac{1}{3} = \frac{-2}{-6} = \frac{-3}{-9} = \frac{1}{3}$. Coincident lines.

Q5. $2x + 3y = 11$ ...(i), $2x - 4y = -24$ ...(ii)
Subtract (ii) from (i): $7y = 35 \Rightarrow y = 5$. [1]
Put $y=5$ in (i): $2x + 15 = 11 \Rightarrow 2x = -4 \Rightarrow x = -2$. [1]

Q6. For no solution: $\frac{k}{12} = \frac{3}{k} \neq \frac{k-3}{k}$.
$k^2 = 36 \Rightarrow k = \pm 6$. [1]
If $k=6$: $\frac{6}{12} = \frac{3}{6} = \frac{1}{2}$. RHS $\frac{3}{6} = \frac{1}{2}$. This gives coincident lines (infinite solutions).
If $k=-6$: $\frac{-6}{12} = \frac{3}{-6} = -\frac{1}{2}$. RHS $\frac{-9}{-6} = \frac{3}{2}$. $-\frac{1}{2} \neq \frac{3}{2}$.
So $k = -6$. [1]

Q7. Let pencil cost $x$, pen cost $y$.
$5x + 7y = 50$ ...(i)
$7x + 5y = 46$ ...(ii)
Add (i)+(ii): $12(x+y) = 96 \Rightarrow x+y=8$. Subtract: $-2x+2y=4 \Rightarrow -x+y=2$.
Solving $x+y=8$ and $-x+y=2$: $2y=10 \Rightarrow y=5, x=3$.
Pencil: ₹3, Pen: ₹5 [2]

Q8. $3x - 5y = 4$ ...(i), $9x - 2y = 7$ ...(ii)
From (i), $x = \frac{4+5y}{3}$. Substitute in (ii):
$9(\frac{4+5y}{3}) - 2y = 7 \Rightarrow 3(4+5y) - 2y = 7$. [1]
$12 + 15y - 2y = 7 \Rightarrow 13y = -5 \Rightarrow y = -5/13$. [1]
$x = \frac{4 + 5(-5/13)}{3} = \frac{52-25}{39} = \frac{27}{39} = \frac{9}{13}$. [1]

Q9. Let number be $10x + y$. Reverse: $10y + x$.
Sum: $11(x+y) = 66 \Rightarrow x+y=6$.
Diff: $x-y=2$ or $y-x=2$.
Case 1: $x+y=6, x-y=2 \Rightarrow x=4, y=2$. Number 42.
Case 2: $x+y=6, y-x=2 \Rightarrow y=4, x=2$. Number 24.
Numbers are 42 and 24 [3]

Q10. $x - y = -1 \Rightarrow y = x+1$. Points: $(0,1), (2,3), (-1,0)$.
$3x + 2y = 12 \Rightarrow y = \frac{12-3x}{2}$. Points: $(0,6), (2,3), (4,0)$. [2]
Plotting graph: Lines intersect at $(2,3)$. [1]
Vertices of triangle with x-axis: $(-1,0), (4,0), (2,3)$.
Base = 5 units, Height = 3 units.
Area = $\frac{1}{2} \times 5 \times 3 = 7.5$ sq units. [2]

(i) Let fixed charge = $x$, charge per km = $y$.
$x + 10y = 105$
$x + 15y = 155$ [1]

(ii) Subtracting: $5y = 50 \Rightarrow y = 10$.
$x + 100 = 105 \Rightarrow x = 5$.
Fixed charge = ₹5, Per km = ₹10. [2]

(iii) For 25 km: $x + 25y = 5 + 25(10) = 5 + 250 = 255$.
₹ 255 [1]