Solutions: Linear Equations - Test 6

Q1. (d) intersecting at (a, b)

$x=a$ is vertical, $y=b$ is horizontal. Intersection is $(a, b)$.

Q2. (c) 6 and 36

Let son $x$, father $y$. $y=6x$. $y+4 = 4(x+4) \Rightarrow 6x+4 = 4x+16 \Rightarrow 2x=12 \Rightarrow x=6, y=36$.

Q3. (b) $k \neq 10$

Unique solution: $\frac{1}{5} \neq \frac{2}{k} \Rightarrow k \neq 10$.

Q4. (b) -1

Solving $2x+3y=11$ and $2x-4y=-24$: Subtracting gives $7y=35 \Rightarrow y=5$.
$2x+15=11 \Rightarrow 2x=-4 \Rightarrow x=-2$.
$y = mx+3 \Rightarrow 5 = m(-2)+3 \Rightarrow 2 = -2m \Rightarrow m = -1$.

Q5. (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.

Q6. Let book $x$, pen $y$. $5x+7y=79$ and $7x+5y=77$.
Add: $12(x+y)=156 \Rightarrow x+y=13$. Subtract: $-2x+2y=2 \Rightarrow -x+y=1$.
Solving: $2y=14 \Rightarrow y=7, x=6$.
Cost of 1 book + 2 pens = $6 + 2(7) = 20$. ₹ 20 [2]

Q7. $\angle C = 3\angle B = 2(\angle A + \angle B)$. Also $\angle A + \angle B + \angle C = 180$.
From $3\angle B = 2(\angle A + \angle B) \Rightarrow \angle B = 2\angle A \Rightarrow \angle A = 0.5\angle B$.
Substitute in sum: $0.5\angle B + \angle B + 3\angle B = 180 \Rightarrow 4.5\angle B = 180 \Rightarrow \angle B = 40^\circ$.
$\angle A = 20^\circ, \angle C = 120^\circ$. [2]

Q8. Add: $200(x+y)=1000 \Rightarrow x+y=5$.
Subtract: $-2x+2y=-2 \Rightarrow -x+y=-1$.
Solving: $2y=4 \Rightarrow y=2, x=3$. x=3, y=2 [2]

Q9. Let speed $x$, time $y$. Distance $xy$.
$(x+10)(y-2) = xy \Rightarrow -2x + 10y = 20 \Rightarrow -x + 5y = 10$.
$(x-10)(y+3) = xy \Rightarrow 3x - 10y = 30$.
Multiply first by 2: $-2x + 10y = 20$. Add to second: $x = 50$.
$-50 + 5y = 10 \Rightarrow 5y = 60 \Rightarrow y = 12$.
Distance = $50 \times 12 = 600$ km. [3]

Q10. Incomes $9x, 7x$. Expenditures $4y, 3y$.
$9x - 4y = 2000$ and $7x - 3y = 2000$.
Multiply first by 3, second by 4: $27x - 12y = 6000$ and $28x - 12y = 8000$.
Subtract: $-x = -2000 \Rightarrow x = 2000$.
Incomes: $9(2000) = 18000$, $7(2000) = 14000$. [3]

(i) $2x + 3y = 350$ and $x + 2y = 200$. [1]

(ii) From 2nd eq, $x = 200 - 2y$. Substitute in 1st: $2(200-2y) + 3y = 350 \Rightarrow 400 - 4y + 3y = 350 \Rightarrow y = 50$.
$x = 200 - 100 = 100$. Adult: ₹100, Child: ₹50. [2]

(iii) $4(100) + 5(50) = 400 + 250 = 650$. [1]

(i) Let speeds $u, v$. Same dir: $5(u-v) = 100 \Rightarrow u-v=20$.
Opposite: $1(u+v) = 100 \Rightarrow u+v=100$. [1]

(ii) Adding eqs: $2u = 120 \Rightarrow u = 60$ km/h.
$v = 40$ km/h. [2]

(iii) Distance by A in 5 hours = $60 \times 5 = 300$ km. [1]