Class 10 Maths: Pair of Linear Equations in two Variables – Extra Questions with Answers

Ans: C

Solution:
The second equation can be rewritten as $x + 3y = 4$. The first equation is $x + 3y = 4$. Since the two equations are identical, they represent the same line. Therefore, there are infinitely many solutions.
Answer: Infinitely many

Ans: D

Solution:
For parallel lines, the ratio of coefficients of x and y must be equal.
So, 4/3 = 3k/6
=> 24 = 9k
=> k = 8/3
Answer: 8/3

Ans: D

Solution:
The second equation is just the first equation multiplied by 2. Therefore, the two equations represent the same line. Two lines that overlap have infinitely many solutions.
Answer: Infinitely many solutions

Ans: A

Solution:
Let the cost of a spoon be ‘s’ and the cost of a fork be ‘f’.
We have two equations:
15s + 20f = 1425 (1)
20s + 15f = 1575 (2)

Multiply (1) by 4 and (2) by 3:
60s + 80f = 5700 (3)
60s + 45f = 4725 (4)

Subtract (4) from (3):
35f = 975
f = 975/35 = 27.857

Substitute f in (1):
15s + 20(27.857) = 1425
15s + 557.14 = 1425
15s = 867.86
s = 867.86 / 15= 57.857

Alternatively:
Multiply (1) by 3: 45s + 60f = 4275
Multiply (2) by 4: 80s + 60f = 6300
Subtract the first from the second: 35s = 2025
s = 2025/35 = 57.857
Substitute s in (1):
15(57.857) + 20f = 1425
867.855 + 20f = 1425
20f = 557.145
f = 27.857

Multiply (1) by 20 and (2) by 15
300s + 400f = 28500 (5)
300s + 225f = 23625 (6)
Subtracting (6) from (5): 175f = 4875
f = 27.857
Substituting in (1): 15s + 20(27.857) = 1425
15s + 557.14 = 1425
15s = 867.86
s = 57.857

Let’s use simpler approach.
From (1): 15s + 20f = 1425 => 3s + 4f = 285 (7)
From (2): 20s + 15f = 1575 => 4s + 3f = 315 (8)
Multiply (7) by 4 and (8) by 3:
12s + 16f = 1140
12s + 9f = 945
Subtract: 7f = 195
f = 195/7 = 27.85
3s + 4(195/7) = 285
3s + 780/7 = 285
3s = 285 – 780/7
3s = (1995-780)/7
3s = 1215/7
s = 1215/21 = 57.85

From (7) *3 and (8) *4
9s + 12f = 855
16s + 12f = 1260
7s = 405
s = 405/7 = 57.857
s=57.86
f=27.86

15s + 20f = 1425, dividing by 5: 3s + 4f = 285
20s + 15f = 1575, dividing by 5: 4s + 3f = 315
3s + 4f = 285 *4 => 12s + 16f = 1140
4s + 3f = 315 *3 => 12s + 9f = 945
7f = 195 => f = 27.86
4s + 3*27.86 = 315
4s = 315 – 83.58
4s = 231.42
s = 57.86

Answer: Cost of a spoon: ₹57.86, Cost of a fork: ₹27.86

Ans: D

Solution:
Let the numerator be $n$ and the denominator be $d$.
We are given that $n + d = 2n + 6$, so $d = n + 6$.
Also, we are given that $(n-1)/(d-1) = 3/4$.
Substituting $d = n+6$, we have $(n-1)/(n+6-1) = 3/4$, so $(n-1)/(n+5) = 3/4$.
Cross-multiplying gives $4(n-1) = 3(n+5)$, so $4n-4 = 3n+15$, which gives $n=19$.
Then $d = n+6 = 19+6 = 25$.
Therefore the fraction is $19/25$.

Answer: 19/25

Ans: D

Solution:
For a unique solution, the lines must not be parallel. The slopes of the lines are $-q/4$ and $-1$. The lines are parallel when $-q/4 = -1$, so $q=4$. Therefore, for a unique solution, $q$ must not be 4.

Answer: $q \ne 4$

Ans: C

Solution: For the system to have no solution, the lines must be parallel and distinct. This occurs when the slopes are equal but the y-intercepts are different. The slope of the first equation is $-2$. The slope of the second equation is $-\frac{t+1}{t}$. Setting the slopes equal gives $-\frac{t+1}{t} = -2$, so $t+1 = 2t$, and $t=1$.
If $t=1$, the equations become $2x+y=3$ and $2x+y=5$, which are parallel and distinct.
Answer: 1

Ans: C

Solution: Let the original fraction be $\frac{x}{y}$.
Given: $\frac{x+2}{y+2} = \frac{2}{3}$ and $\frac{x-1}{y-1} = \frac{1}{2}$.
From the first equation, $3(x+2) = 2(y+2) \Rightarrow 3x+6 = 2y+4 \Rightarrow 3x-2y = -2$.
From the second equation, $2(x-1) = y-1 \Rightarrow 2x-2 = y-1 \Rightarrow 2x-y = 1$.
Multiply the second equation by 2: $4x-2y = 2$.
Subtract the first equation from the modified second equation: $(4x-2y) – (3x-2y) = 2-(-2) \Rightarrow x = 4$.
Substitute x=4 into $2x-y = 1 \Rightarrow 2(4)-y=1 \Rightarrow 8-y=1 \Rightarrow y = 7$.
The original fraction is $\frac{4}{7}$.
Answer: $\frac{4}{7}$

Ans: B

Solution:
Multiply the first equation by $a$ and the second by $b$:
$7a^2x + 7aby = 4a^2 + 3ab$
$7b^2x – 7aby = 4b^2 – 3ab$
Adding the two equations:
$7(a^2 + b^2)x = 4a^2 + 4b^2$
$x = \frac{4(a^2 + b^2)}{7(a^2 + b^2)} = \frac{4}{7}$

Multiply the first equation by $b$ and the second by $a$:
$7abx + 7b^2y = 4ab + 3b^2$
$7abx – 7a^2y = 4ab – 3a^2$
Subtracting the second equation from the first:
$7(b^2 + a^2)y = 3b^2 + 3a^2$
$y = \frac{3(a^2 + b^2)}{7(a^2 + b^2)} = \frac{3}{7}$

Answer: $x = \frac{4}{7}, y = \frac{3}{7}$

Ans: A

Solution:
The given equations are:
$$\frac{3x+4y}{xy} = 5 \quad \text{and} \quad \frac{6x-4y}{xy} + 2 = 0$$
We can rewrite the equations as:
$$\frac{3x}{xy} + \frac{4y}{xy} = 5 \Rightarrow \frac{3}{y} + \frac{4}{x} = 5 \quad (1)$$
$$\frac{6x}{xy} – \frac{4y}{xy} + 2 = 0 \Rightarrow \frac{6}{y} – \frac{4}{x} + 2 = 0 \Rightarrow \frac{6}{y} – \frac{4}{x} = -2 \quad (2)$$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. Then the equations become:
$$4u + 3v = 5 \quad (3)$$
$$-4u + 6v = -2 \quad (4)$$
Adding equations (3) and (4), we get:
$$(4u+3v) + (-4u+6v) = 5 + (-2)$$
$$9v = 3$$
$$v = \frac{3}{9} = \frac{1}{3}$$
Since $v = \frac{1}{y}$, we have $\frac{1}{y} = \frac{1}{3}$, which gives $y = 3$.
Substitute $v = \frac{1}{3}$ into equation (3):
$$4u + 3(\frac{1}{3}) = 5$$
$$4u + 1 = 5$$
$$4u = 4$$
$$u = 1$$
Since $u = \frac{1}{x}$, we have $\frac{1}{x} = 1$, which gives $x = 1$.
Thus, we have $x = 1$ and $y = 3$.
Check the solution in the original equations:
$$\frac{3(1)+4(3)}{(1)(3)} = \frac{3+12}{3} = \frac{15}{3} = 5$$
$$\frac{6(1)-4(3)}{(1)(3)} + 2 = \frac{6-12}{3} + 2 = \frac{-6}{3} + 2 = -2 + 2 = 0$$
The solution is $x = 1$ and $y = 3$.

Answer: $x=1, y=3$