Question

Match the two columns

Column-I (Equations)		Column-II (Solutions)
(A)	$2x+y=7$	(p)	$\left(0,9\right)$
(B)	$x-2y=4$	(q)	$\left(1,5\right)$
(C)	$x-4y=0$	(r)	$\left(0,0\right)$
(D)	$px+y=9$	(s)	$\left(4,0\right)$

Choose the correct options:

• A. (A)⇢s; (B)⇢p; (C)⇢q; (D)⇢r
• B. (A)⇢q; (B)⇢r; (C)⇢p; (D)⇢s
• C. (A)⇢q; (B)⇢s; (C)⇢r; (D)⇢p
• D. (A)⇢p; (B)⇢q; (C)⇢s; (D)⇢r

Answer 1

The correct option is C

(A)⇢q; (B)⇢s; (C)⇢r; (D)⇢p

Explanation for (a)

Step 1: Substituting the given solution in equation A

Substituting $\left(p\right)⇢$$\left(0,9\right)$ in equation $A$,

$$\begin{gathered} 2\mathrm{x}+\mathrm{y}=7 \\ \Rightarrow RHS=7 \\ \Rightarrow \mathrm{LHS}=2\left(0\right)+9 \\ \Rightarrow =9 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$

Substituting$q-\to$$\left(1,5\right)$ in equation $A$,

$$\begin{gathered} 2\mathrm{x}+\mathrm{y}=7 \\ \Rightarrow RHS=7 \\ \Rightarrow \mathrm{LHS}=2\left(1\right)+5 \\ \Rightarrow =7 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

Hence, the correct match for $A$is$\left(q\right)$

Step 2: Substituting the given solution in equation $B$

Substituting$p-\to$$\left(0,9\right)$ in equation $B$,

$$\begin{gathered} \mathrm{x}-2\mathrm{y}=4 \\ \Rightarrow RHS=4 \\ \Rightarrow \mathrm{LHS}=0-2\left(9\right) \\ \Rightarrow =-18 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$

Substituting $r-\to$$\left(0,0\right)$ in equation $B$,

$$\begin{gathered} \mathrm{x}-2\mathrm{y}=4 \\ \Rightarrow RHS=4 \\ \Rightarrow \mathrm{LHS}=0-2\left(0\right) \\ \Rightarrow =0 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$

Substituting $s\to$$\left(4,0\right)$ in equation $B$,

$$\begin{gathered} \mathrm{x}-2\mathrm{y}=4 \\ \Rightarrow RHS=4 \\ \Rightarrow \mathrm{LHS}=4-2\left(0\right) \\ \Rightarrow =4 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

Hence, the correct match for $B$ is $\left(s\right)$ .

Step 3: Substituting the given solution in equation $C$

Substituting $p-\to$$\left(0,9\right)$ in equation $C$,

$$\begin{gathered} \mathrm{x}-4\mathrm{y}=0 \\ \Rightarrow RHS=0 \\ \Rightarrow \mathrm{LHS}=0-4\left(9\right) \\ \Rightarrow =-36 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$

Substituting$r-\to$ $\left(0,0\right)$ in equation $C$ ,

$$\begin{gathered} \mathrm{x}-4\mathrm{y}=0 \\ \Rightarrow RHS=0 \\ \Rightarrow \mathrm{LHS}=0-4\left(0\right) \\ \Rightarrow =0 \\ \Rightarrow \mathrm{LHS}=\mathrm{RH}S \end{gathered}$$

Hence, the correct match for $C$ is $\left(r\right)$ .

Step 4: Substituting the given solution in equation D

Substituting $p-\to$$\left(0,9\right)$ in equation $D$,

$$\begin{gathered} \mathrm{px}+\mathrm{y}=9 \\ \Rightarrow RHS=9 \\ \Rightarrow \mathrm{LHS}=\mathrm{p}\left(0\right)+9 \\ \Rightarrow =9 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

Hence, the correct match for $D$ is $\left(p\right)$

The columns will be matched in the following pairs:

Thus, we can conclude that the correct answer is o$\left(A\right)⇢q;\left(B\right)⇢s;\left(C\right)⇢r;\left(D\right)⇢p$ption C.