Class 10 Pair of Linear Equations in Two Variables Mock Test (CBSE)

Topics in Pair of Linear Equations in Two Variables

• Graphical method of solution
• Consistent and inconsistent pairs
• Substitution method
• Elimination method

Pair of Linear Equations in Two Variables practice questions with answers

1. Q1Multiple choice·1 mark
   For what value of $k$ do the equations $kx + 3y = k - 3$ and $12x + ky = k$ represent parallel lines?

   1. (a)$6$
   2. (b)$-6$
   3. (c)$12$
   4. (d)$-12$
   Show answer and step marks

   1. Correct option: (b) $-6$. For parallel lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Here $\frac{k}{12} = \frac{3}{k} \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$. Testing $k = -6$ gives $\frac{-6-3}{-6} = \frac{3}{2}$, which is not equal to $\frac{a_1}{a_2} = -\frac{1}{2}$. Thus, $k = -6$ yields parallel lines.+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   For what value of $k$ do the pair of linear equations $3x - y = 5$ and $6x - 2y = k$ represent coincident lines?

   1. (a)$10$
   2. (b)$-10$
   3. (c)$5$
   4. (d)$0$
   Show answer and step marks

   1. Correct option: (a) $10$, since the condition for coincident lines is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, which gives $\frac{3}{6} = \frac{-1}{-2} = \frac{-5}{-k}$ and hence $k = 10$.+1

   Answer: (a)
3. Q3Multiple choice·1 mark
   The pair of linear equations $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$ has:

   1. (a)a unique solution
   2. (b)infinitely many solutions
   3. (c)no solution
   4. (d)two solutions
   Show answer and step marks

   1. Correct option: (c) no solution. Here $\frac{a_1}{a_2} = \frac{1}{-3}$, $\frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}$, and $\frac{c_1}{c_2} = \frac{5}{1} = 5$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair has no solution.+1

   Answer: (c)
4. Q4Short answer·2 marks
   Find the value of $k$ for which the given pair of linear equations has no solution: $kx + 3y = k - 3$ $12x + ky = k$
   Show answer and step marks

   1. For no solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.+0.5
   2. Substitute: $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = 6$ or $k = -6$.+0.5
   3. For $k = 6$: $\frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2}$ and $\frac{c_1}{c_2} = \frac{3}{6} = \frac{1}{2}$, which are equal, so the lines coincide (infinitely many solutions); reject $k = 6$.+0.5
   4. For $k = -6$: $\frac{a_1}{a_2} = \frac{-6}{12} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2}$ and $\frac{c_1}{c_2} = \frac{-9}{-6} = \frac{3}{2} \neq -\frac{1}{2}$, so there is no solution. Thus $k = -6$.+0.5

   Answer: $k = -6$
5. Q5Short answer·2 marks
   Solve the following pair of linear equations by the substitution method: $2x + 3y = 7$ $x - y = 1$
   Show answer and step marks

   1. From $x - y = 1$, $x = y + 1$. Substitute in $2x + 3y = 7$: $2(y + 1) + 3y = 7$.+1
   2. Simplify: $5y + 2 = 7 \implies 5y = 5 \implies y = 1$.+0.5
   3. Substitute $y = 1$ in $x = y + 1$: $x = 2$.+0.5

   Answer: $x = 2, y = 1$
6. Q6Long answer·5 marks
   A boat goes $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10\text{ hours}$. Also, it can go $40\text{ km}$ upstream and $55\text{ km}$ downstream in $13\text{ hours}$. Find the speed of the boat in still water and the speed of the stream.
   Show answer and step marks

   1. Let the speed of the boat in still water be $x\text{ km/h}$ and of the stream $y\text{ km/h}$. Upstream speed $= (x - y)\text{ km/h}$, downstream speed $= (x + y)\text{ km/h}$.+1
   2. Using time $= \frac{\text{distance}}{\text{speed}}$: $\frac{30}{x-y} + \frac{44}{x+y} = 10$ and $\frac{40}{x-y} + \frac{55}{x+y} = 13$.+1
   3. Put $p = \frac{1}{x-y}$, $q = \frac{1}{x+y}$: $30p + 44q = 10$ and $40p + 55q = 13$.+0.5
   4. Multiply the first by $4$ and the second by $3$: $120p + 176q = 40$ and $120p + 165q = 39$; subtracting, $11q = 1 \implies q = \frac{1}{11}$.+0.5
   5. Substitute: $30p + 4 = 10 \implies p = \frac{1}{5}$.+0.5
   6. So $x - y = 5$ and $x + y = 11$.+0.5
   7. Adding: $2x = 16 \implies x = 8\text{ km/h}$.+0.5
   8. Then $y = 11 - 8 = 3\text{ km/h}$.+0.5

   Answer: Speed of the boat in still water = $8\text{ km/h}$ and Speed of the stream = $3\text{ km/h}$

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