301 Moved Permanently - Free Printable

Let's solve each problem step by step:

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Problem 1:

Question: A pair of linear equations in two variables is called a consistent if it has:
- (A) Unique solution
- (B) No solution
- (C) Infinitely many solutions
- (D) Either (a) or (c)

Solution:
A system of linear equations is called consistent if it has at least one solution. This means the system can either have:
1. A unique solution (the lines intersect at exactly one point).
2. Infinitely many solutions (the lines are coincident).

Thus, the correct answer is:
(D) Either (a) or (c)

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Problem 2:

Question: In \( \Delta ABC \), if \( \angle A = x^\circ \), \( \angle B = 3x^\circ \), \( \angle C = y^\circ \), and \( 3y - 5x = 30 \), then \( \Delta ABC \) is:
- (A) A right-angled triangle
- (B) An isosceles triangle
- (C) An equilateral triangle
- (D) A right-angled isosceles triangle

Solution:
The sum of the angles in a triangle is \( 180^\circ \). Therefore:
\[
\angle A + \angle B + \angle C = 180^\circ
\]
Substituting the given values:
\[
x + 3x + y = 180
\]
\[
4x + y = 180 \quad \text{(Equation 1)}
\]

We are also given:
\[
3y - 5x = 30 \quad \text{(Equation 2)}
\]

Now, solve these two equations simultaneously.

From Equation 1:
\[
y = 180 - 4x \quad \text{(Equation 3)}
\]

Substitute \( y = 180 - 4x \) into Equation 2:
\[
3(180 - 4x) - 5x = 30
\]
\[
540 - 12x - 5x = 30
\]
\[
540 - 17x = 30
\]
\[
-17x = 30 - 540
\]
\[
-17x = -510
\]
\[
x = \frac{510}{17} = 30
\]

Now, substitute \( x = 30 \) back into Equation 3 to find \( y \):
\[
y = 180 - 4(30)
\]
\[
y = 180 - 120 = 60
\]

So, the angles are:
\[
\angle A = x = 30^\circ, \quad \angle B = 3x = 90^\circ, \quad \angle C = y = 60^\circ
\]

Since \( \angle B = 90^\circ \), \( \Delta ABC \) is a right-angled triangle.

Thus, the correct answer is:
(A) A right-angled triangle

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Problem 3:

Question: The value of \( k \) for which the system of equations
\[
2x + 3y = 5
\]
\[
4x + ky = 10
\]
has an infinite number of solutions is:
- (A) 1
- (B) 3
- (C) 6
- (D) 0

Solution:
For a system of linear equations to have infinitely many solutions, the equations must be dependent, meaning one equation is a scalar multiple of the other.

Given:
\[
2x + 3y = 5 \quad \text{(Equation 1)}
\]
\[
4x + ky = 10 \quad \text{(Equation 2)}
\]

Notice that Equation 2 can be written as:
\[
4x + ky = 10
\]
Divide the entire equation by 2:
\[
2x + \frac{k}{2}y = 5
\]

For the equations to be dependent, the coefficients of \( x \) and \( y \) must be proportional. Comparing with Equation 1:
\[
2x + 3y = 5
\]
we see that:
\[
\frac{k}{2} = 3
\]
\[
k = 6
\]

Thus, the correct answer is:
(C) 6

---

Problem 4:

Question: The coordinates of the points where the lines \( 3x - y = 5 \) and \( 6x - y = 10 \) meet the \( y \)-axis are:
- (A) \( (0, -5) \), \( (0, -10) \)
- (B) \( (-5, 0) \), \( (-10, 0) \)
- (C) \( (-5, 0) \), \( (0, -10) \)
- (D) \( (0, 5) \), \( (0, 10) \)

Solution:
To find the points where the lines meet the \( y \)-axis, set \( x = 0 \) in each equation.

For the line \( 3x - y = 5 \):
\[
3(0) - y = 5
\]
\[
-y = 5
\]
\[
y = -5
\]
So, the point is \( (0, -5) \).

For the line \( 6x - y = 10 \):
\[
6(0) - y = 10
\]
\[
-y = 10
\]
\[
y = -10
\]
So, the point is \( (0, -10) \).

Thus, the correct answer is:
(A) \( (0, -5) \), \( (0, -10) \)

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Problem 5:

Question: The equations \( x + 2y = 4 \) and \( 2x + y = 5 \) are:
- (A) Inconsistent
- (B) Homogeneous linear equations
- (C) Consistent and have a unique solution
- (D) Consistent and have infinitely many solutions

Solution:
To determine the nature of the system, we check if the equations are consistent and whether they have a unique solution or infinitely many solutions.

Given:
\[
x + 2y = 4 \quad \text{(Equation 1)}
\]
\[
2x + y = 5 \quad \text{(Equation 2)}
\]

Write the equations in the standard form \( ax + by + c = 0 \):
\[
x + 2y - 4 = 0
\]
\[
2x + y - 5 = 0
\]

Compare the coefficients:
\[
\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{2}{1}, \quad \frac{c_1}{c_2} = \frac{-4}{-5} = \frac{4}{5}
\]

For the system to be consistent:
\[
\frac{a_1}{a_2} \neq \frac{b_1}{b_2}
\]

Here:
\[
\frac{1}{2} \neq \frac{2}{1}
\]

Since the ratios are not equal, the system is consistent and has a unique solution.

Thus, the correct answer is:
(C) Consistent and have a unique solution

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Problem 6:

Question: If \( 8x - 9y = 6xy \) and \( 10x + 6y = 19xy \), then the value of \( xy \) is:
- (A) 1
- (B) 2
- (C) 3
- (D) 4

Solution:
Given:
\[
8x - 9y = 6xy \quad \text{(Equation 1)}
\]
\[
10x + 6y = 19xy \quad \text{(Equation 2)}
\]

Divide both equations by \( xy \) (assuming \( x \neq 0 \) and \( y \neq 0 \)):
\[
\frac{8x}{xy} - \frac{9y}{xy} = \frac{6xy}{xy}
\]
\[
\frac{8}{y} - \frac{9}{x} = 6 \quad \text{(Equation 3)}
\]

\[
\frac{10x}{xy} + \frac{6y}{xy} = \frac{19xy}{xy}
\]
\[
\frac{10}{y} + \frac{6}{x} = 19 \quad \text{(Equation 4)}
\]

Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Then:
\[
8v - 9u = 6 \quad \text{(Equation 5)}
\]
\[
10v + 6u = 19 \quad \text{(Equation 6)}
\]

Solve these equations simultaneously. Multiply Equation 5 by 2:
\[
16v - 18u = 12 \quad \text{(Equation 7)}
\]

Multiply Equation 6 by 3:
\[
30v + 18u = 57 \quad \text{(Equation 8)}
\]

Add Equations 7 and 8:
\[
(16v - 18u) + (30v + 18u) = 12 + 57
\]
\[
46v = 69
\]
\[
v = \frac{69}{46} = \frac{3}{2}
\]

Substitute \( v = \frac{3}{2} \) into Equation 5:
\[
8\left(\frac{3}{2}\right) - 9u = 6
\]
\[
12 - 9u = 6
\]
\[
-9u = 6 - 12
\]
\[
-9u = -6
\]
\[
u = \frac{6}{9} = \frac{2}{3}
\]

Recall \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \):
\[
\frac{1}{x} = \frac{2}{3} \implies x = \frac{3}{2}
\]
\[
\frac{1}{y} = \frac{3}{2} \implies y = \frac{2}{3}
\]

Thus:
\[
xy = \left(\frac{3}{2}\right)\left(\frac{2}{3}\right) = 1
\]

Thus, the correct answer is:
(A) 1

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Problem 7:

Question: If \( 2a - 3 = 5 \) and \( 3b + 1 = 2 \), then \( 3b - 2a \) is:
- (A) \(-3\)
- (B) \(-7\)
- (C) \(7\)
- (D) \(-9\)

Solution:
Solve for \( a \) and \( b \) from the given equations.

From \( 2a - 3 = 5 \):
\[
2a = 5 + 3
\]
\[
2a = 8
\]
\[
a = 4
\]

From \( 3b + 1 = 2 \):
\[
3b = 2 - 1
\]
\[
3b = 1
\]
\[
b = \frac{1}{3}
\]

Now, calculate \( 3b - 2a \):
\[
3b - 2a = 3\left(\frac{1}{3}\right) - 2(4)
\]
\[
= 1 - 8
\]
\[
= -7
\]

Thus, the correct answer is:
(B) \(-7\)

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Final Answers:

1. (D) Either (a) or (c)
2. (A) A right-angled triangle
3. (C) 6
4. (A) \( (0, -5) \), \( (0, -10) \)
5. (C) Consistent and have a unique solution
6. (A) 1
7. (B) \(-7\)

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