Pair of Linear Equations in Two Variables: CBSE Class 10 Mathematics practice questions

C. $y = \dfrac{x}{2}$ and $3x + 4y = 20$

Games are half of rides, so $y = \frac{x}{2}$. Each ride costs ₹3 and each game ₹4, so the money is $3x + 4y = 20$. Swapping the letters in the first equation ($x = \frac{y}{2}$) says rides are half the games, swapping the prices gives $4x + 3y = 20$, and $x + y = 20$ treats the number of rides plus games as the money spent. The correct pair has the solution $x = 4,\ y = 2$: $3(4) + 4(2) = 20$.

D. They are parallel

$\frac{a_1}{a_2} = \frac{9}{3} = 3$, $\frac{b_1}{b_2} = \frac{-6}{-2} = 3$ and $\frac{c_1}{c_2} = \frac{15}{7} \ne 3$. When $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$ the lines are parallel and the pair has no solution. Intersection would need $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, coincidence needs all three ratios equal, and "perpendicular" is not decided by these ratios.

A. $p = 5$

Coincident lines need $\frac{2}{p+q} = \frac{3}{2p-q} = \frac{7}{21} = \frac{1}{3}$. So $p + q = 6$ and $2p - q = 9$. Adding gives $3p = 15$, so $p = 5$ (and $q = 1$). Using $2p + q = 9$ by a sign slip gives $p = 3$, $p = 1$ is the value of $q$, and $6$ is only $p + q$.