LINEAR EQUATIONS IN TWO VARIABLES - MCQ - EM worksheet | Live ... - Free Printable

Let's solve the problems step by step:

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1. If linear equation in two variables \( x + 2y = 3 \) and \( 2x + 4y = k \) coincides, then the value of \( k \) is:

Solution:
For two lines to coincide, their equations must be proportional. That is, if \( ax + by + c = 0 \) and \( a'x + b'y + c' = 0 \) are coincident, then:
\[
\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}
\]

Given:
\[
x + 2y = 3 \quad \text{(Equation 1)}
\]
\[
2x + 4y = k \quad \text{(Equation 2)}
\]

Rewriting Equation 1:
\[
x + 2y - 3 = 0
\]

Rewriting Equation 2:
\[
2x + 4y - k = 0
\]

For these lines to coincide:
\[
\frac{1}{2} = \frac{2}{4} = \frac{-3}{-k}
\]

The first two ratios are already equal:
\[
\frac{1}{2} = \frac{2}{4}
\]

Now, equate the third ratio:
\[
\frac{1}{2} = \frac{-3}{-k}
\]

Simplify:
\[
\frac{1}{2} = \frac{3}{k}
\]

Cross-multiply:
\[
k = 6
\]

Answer:
\[
\boxed{B}
\]

---

2. \( 2x + 3y - 9 = 0 \) and \( 4x + 6y - 18 = 0 \). This pair of linear equations have ---- solutions.

Solution:
To determine the nature of the solutions, we compare the coefficients of the two equations:
\[
2x + 3y - 9 = 0 \quad \text{(Equation 1)}
\]
\[
4x + 6y - 18 = 0 \quad \text{(Equation 2)}
\]

Rewriting Equation 2:
\[
4x + 6y - 18 = 0
\]

Notice that Equation 2 is simply a multiple of Equation 1:
\[
4x + 6y - 18 = 2(2x + 3y - 9)
\]

This means the two equations represent the same line. Therefore, they have infinitely many solutions.

Answer:
\[
\boxed{D}
\]

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3. In the equation \( x + y = 7 \), if \( x = 3 \), then the value of \( y \) is:

Solution:
Substitute \( x = 3 \) into the equation \( x + y = 7 \):
\[
3 + y = 7
\]

Solve for \( y \):
\[
y = 7 - 3 = 4
\]

Answer:
\[
\boxed{B}
\]

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4. If \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) are the pair of coinciding lines, then the ratios of their coefficients are:

Solution:
For two lines to coincide, their coefficients must be proportional:
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\]

Answer:
\[
\boxed{C}
\]

---

5. If the pair of equations \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, then the value of \( k \) is:

Solution:
For two lines to be parallel, their slopes must be equal. The general form of a line is \( ax + by + c = 0 \), and its slope is \( -\frac{a}{b} \).

Given:
\[
3x + 2ky = 2 \quad \Rightarrow \quad 3x + 2ky - 2 = 0
\]
\[
2x + 5y + 1 = 0
\]

The slope of the first line is:
\[
-\frac{3}{2k}
\]

The slope of the second line is:
\[
-\frac{2}{5}
\]

Set the slopes equal:
\[
-\frac{3}{2k} = -\frac{2}{5}
\]

Simplify:
\[
\frac{3}{2k} = \frac{2}{5}
\]

Cross-multiply:
\[
3 \cdot 5 = 2 \cdot 2k
\]
\[
15 = 4k
\]

Solve for \( k \):
\[
k = \frac{15}{4}
\]

Answer:
\[
\boxed{A}
\]

---

6. The pair of equations \( 2x - 5y + 4 = 0 \) and \( 2x + y - 8 = 0 \) has:

Solution:
To determine the nature of the solutions, we compare the coefficients of the two equations:
\[
2x - 5y + 4 = 0 \quad \text{(Equation 1)}
\]
\[
2x + y - 8 = 0 \quad \text{(Equation 2)}
\]

The ratios of the coefficients are:
\[
\frac{a_1}{a_2} = \frac{2}{2} = 1, \quad \frac{b_1}{b_2} = \frac{-5}{1} = -5, \quad \frac{c_1}{c_2} = \frac{4}{-8} = -\frac{1}{2}
\]

Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines are not parallel. Therefore, they intersect at exactly one point.

Answer:
\[
\boxed{C}
\]

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7. The values of \( x \) and \( y \) when a point lies on the linear equation \( 2x - 3y = 12 \) are:

Solution:
We need to find a point \((x, y)\) that satisfies the equation \( 2x - 3y = 12 \). Let's test the given options:

- Option A: \( x = 0, y = -3 \)
\[
2(0) - 3(-3) = 0 + 9 = 9 \quad (\text{Not equal to 12})
\]

- Option B: \( x = 2, y = 3 \)
\[
2(2) - 3(3) = 4 - 9 = -5 \quad (\text{Not equal to 12})
\]

- Option C: \( x = 3, y = -2 \)
\[
2(3) - 3(-2) = 6 + 6 = 12 \quad (\text{Equal to 12})
\]

- Option D: \( x = -2, y = 3 \)
\[
2(-2) - 3(3) = -4 - 9 = -13 \quad (\text{Not equal to 12})
\]

Answer:
\[
\boxed{C}
\]

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8. Identify the wrong statement with respect to a pair of linear equations:

Solution:
Let's analyze each statement:

- A: If lines are parallel, there is no solution.
- True. Parallel lines never intersect, so there is no solution.

- B: If the lines are perpendicular to each other, there is no solution.
- False. Perpendicular lines intersect at exactly one point, so there is a unique solution.

- C: Many solutions if the lines coincide each other.
- True. Coincident lines have infinitely many solutions.

- D: A unique solution if they intersect.
- True. Intersecting lines have exactly one solution.

The wrong statement is B.

Answer:
\[
\boxed{B}
\]

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9. In the equation \( 2x - y = 5 \), if \( y = 1 \), then the value of \( x \) is:

Solution:
Substitute \( y = 1 \) into the equation \( 2x - y = 5 \):
\[
2x - 1 = 5
\]

Solve for \( x \):
\[
2x = 6 \quad \Rightarrow \quad x = 3
\]

Answer:
\[
\boxed{A}
\]

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10. The cost of 5 pens and 7 pencils is Rs 50. Which of the following equation describes the above statement:

Solution:
Let the cost of one pen be \( x \) and the cost of one pencil be \( y \). The total cost is given by:
\[
5x + 7y = 50
\]

Answer:
\[
\boxed{B}
\]

---

11. The solution of the equation \( x - y = 2 \) and \( x + y = 4 \) is:

Solution:
We solve the system of equations:
\[
x - y = 2 \quad \text{(Equation 1)}
\]
\[
x + y = 4 \quad \text{(Equation 2)}
\]

Add the two equations:
\[
(x - y) + (x + y) = 2 + 4
\]
\[
2x = 6 \quad \Rightarrow \quad x = 3
\]

Substitute \( x = 3 \) into Equation 1:
\[
3 - y = 2 \quad \Rightarrow \quad y = 1
\]

Thus, the solution is \( (x, y) = (3, 1) \).

Answer:
\[
\boxed{C}
\]

---

12. The line represented by \( 6x - 8y + 8 = 0 \) and \( 6x - 6y + 16 = 0 \):

Solution:
To determine the relationship between the lines, we compare their coefficients:
\[
6x - 8y + 8 = 0 \quad \text{(Equation 1)}
\]
\[
6x - 6y + 16 = 0 \quad \text{(Equation 2)}
\]

The ratios of the coefficients are:
\[
\frac{a_1}{a_2} = \frac{6}{6} = 1, \quad \frac{b_1}{b_2} = \frac{-8}{-6} = \frac{4}{3}, \quad \frac{c_1}{c_2} = \frac{8}{16} = \frac{1}{2}
\]

Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines are not parallel. Therefore, they intersect.

Answer:
\[
\boxed{A}
\]

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13. If the equations \( 4x + Py + 8 = 0 \) and \( 2x + 2y + 2 = 0 \) have unique solutions, then the value of \( P \) is:

Solution:
For two lines to have a unique solution, their slopes must be different. The general form of a line is \( ax + by + c = 0 \), and its slope is \( -\frac{a}{b} \).

Given:
\[
4x + Py + 8 = 0 \quad \Rightarrow \quad \text{Slope} = -\frac{4}{P}
\]
\[
2x + 2y + 2 = 0 \quad \Rightarrow \quad \text{Slope} = -\frac{2}{2} = -1
\]

For unique solutions, the slopes must be different:
\[
-\frac{4}{P} \neq -1
\]

Simplify:
\[
\frac{4}{P} \neq 1 \quad \Rightarrow \quad P \neq 4
\]

Answer:
\[
\boxed{B}
\]

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14. \( x + y = 20 \) and \( x - y = 4 \) solve for \( x \) and \( y \), and substitute the values of \( x \) and \( y \) in \( y = mx + 3 \). Then the value of \( m \) is:

Solution:
We solve the system of equations:
\[
x + y = 20 \quad \text{(Equation 1)}
\]
\[
x - y = 4 \quad \text{(Equation 2)}
\]

Add the two equations:
\[
(x + y) + (x - y) = 20 + 4
\]
\[
2x = 24 \quad \Rightarrow \quad x = 12
\]

Substitute \( x = 12 \) into Equation 1:
\[
12 + y = 20 \quad \Rightarrow \quad y = 8
\]

Thus, \( x = 12 \) and \( y = 8 \).

Now substitute \( x = 12 \) and \( y = 8 \) into \( y = mx + 3 \):
\[
8 = m(12) + 3
\]

Solve for \( m \):
\[
8 = 12m + 3 \quad \Rightarrow \quad 5 = 12m \quad \Rightarrow \quad m = \frac{5}{12}
\]

Answer:
\[
\boxed{A}
\]

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Final Answers:

\[
\boxed{B, D, B, C, A, C, C, B, A, B, C, A, B, A}
\]