Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 28

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Question 1
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$$A$$ is $$25$$ years older than $$B$$. In $$15$$ years, $$A$$ will be twice of $$B$$. Find the present ages of $$A$$ and $$B$$.

A
Present age of $$A$$ is $$40$$ years and

Present age of $$B$$ is $$15$$ years

B
Present age of $$A$$ is $$37$$ years and

Present age of B is $$12$$ years

C
Present age of $$A$$ is $$35$$ years and

Present age of $$B$$ is $$10$$ years

D
Present age of $$A$$ is $$45$$ years and

Present age of $$B$$ is $$20$$ years

Solution

Let the present age of $$A$$ be $$x$$ and the present age of $$B$$ be $$y$$.

It is given that $$A$$ is $$25$$ years older than $$B$$ that is:

$$x=y+25$$
$$x-y=25..........(1)$$

Also, In $$15$$ years, $$A$$ will be twice of $$B$$ that is:

$$x+15=2(y+15)$$
$$x+15=2y+30$$
$$x-2y=30-15$$
$$x-2y=15..........(2)$$

Now subtract equation 1 from equation 2 to eliminate $$x$$, because the coefficients of $$x$$ are same. So, we get

$$(x-x)+(-2y+y)=15-25$$

i.e. $$-y=-10$$

i.e. $$y=10$$

Substituting this value of $$y$$ in (1), we get

$$x-10=25$$

i.e. $$x=25+10=35$$

Hence, the present age of $$A$$ is $$35$$ years and the present age of $$B$$ is $$10$$ years.
Question 2
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A man is 24 years older than his son. 12 years ago, he was five times as old as his son. Find the present ages of both.

A
Present age of father is $$44$$ years and Present age of son is $$20$$ years

B
Present age of father is $$42$$ years and Present age of son is $$18$$ years

C
Present age of father is $$60$$ years and Present age of son is $$36$$ years

D
Present age of father is $$48$$ years and Present age of son is $$24$$ years

Solution

Let the present age of the father be $$x$$ years and the present age of the son be $$y$$ years.

It is given that the man is $$24$$ years older than his son that is:

$$x=y+24$$
$$\Rightarrow x-y=24\quad ..........(1)$$

Also, $$12$$ years ago, he was five times as old as his son that is:

$$(x-12)=5(y-12)$$
$$\Rightarrow x-12=5y-60$$
$$\Rightarrow x-5y=-60+12$$
$$\Rightarrow x-5y=-48\quad .......(2)$$

On subtracting $$(1)$$ from $$(2)$$, we get:

$$(x-x)+(-5y+y)=-24-48$$

$$\Rightarrow -4y=-72$$

$$\Rightarrow y=18$$

Substituting this value of $$y$$ in $$(1)$$, we get:

$$x-18=24$$

$$\Rightarrow x=24+18=42$$

Hence, the present age of the father is $$42$$ years and the present age of the son is $$18$$ years.
Question 3
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Find the fraction such that it becomes $$\displaystyle \frac{1}{2}$$ if 1 is added to the numerator, and $$\displaystyle \frac{1}{3}$$ if 1 is added to the denominator.

A
$$\dfrac{3}{8}$$

B
$$\dfrac{1}{6}$$

C
$$\dfrac{8}{7}$$

D
$$\dfrac{7}{3}$$

Solution

Let the fraction be $$\dfrac { x }{ y }$$
$$\therefore \dfrac { x+1 }{ y } = \dfrac { 1 }{ 2 } $$ ...(1)

$$\dfrac { x }{ y+1 } = \dfrac { 1 }{ 3 }$$ ...(2)
So from $$1$$ we get, $$2x+2 = y $$ ...(3)
From $$2$$ we get, $$3x = y+1 $$ ...(4)
Substitute $$eq(3)$$ in $$eq(4)$$, we get $$3x = 2x+2+1$$
$$ x = 3\\ y = 2x+2 = 8$$

The fraction is $$\dfrac { 3 }{ 8 }$$
Question 4
1 / -0

Solve the following pair of equations:
$$7x+6y= 71$$
$$5x-8y= -23$$

A
$$x= -5;y= 4$$

B
$$x= 3,y= -2$$

C
$$x= 5,y= 6$$

D
$$x= 1,y= 3$$

Solution

$$\\ 7x+6y=71;\\ y=\dfrac{71-7x}{6};\\$$
Substituting:
$$5x-8\left(\dfrac{71-7x}{6}\right)=-23$$
$$\Rightarrow 30x-568+56x=-138$$
$$\Rightarrow 86x=430$$
$$\Rightarrow x=5$$
Resubstituting the value of $$x$$, we get
$$y=\dfrac{71-35}{6}=6$$
Therefore
$$x=5, y=6$$
Question 5
1 / -0

Solve the following pair of equations using substitution method:
$$\displaystyle \frac{5y}{2}-\displaystyle \frac{x}{3}= 8$$
$$\displaystyle \frac{y}{2}+\displaystyle \frac{5x}{3}= 12$$

A
$$x= 2;y= 7$$

B
$$x= 6;y= 4$$

C
$$x= 7;y= 1$$

D
$$x= 8;y= 3$$

Solution

$$\dfrac { 5y }{ 2 } -\dfrac { x }{ 3 } =8\quad ...(1)\\ \dfrac { y }{ 2 } +\dfrac { 5x }{ 3 } =12\quad ...(2)$$

Let $$\dfrac{y}{2} = a$$ and $$\dfrac{x}{3} = b$$
So, the original equations reduce to
$$5a - b = 8$$ ...$$(1)$$
$$a + 5b = 12$$ ...$$(2)$$
From $$(1), b = 5a - 8$$

Substituting $$b = 5a - 8$$ in $$(2)$$, we get
$$a + 5(5a - 8) = 12$$
$$a + 25a - 40 = 12$$
$$26a = 52$$
$$a = 2$$

So, $$b = 5a - 8$$
$$b = 10 - 8 = 2$$

But $$\dfrac{y}{2} = a$$ or $$y = 2a = 4$$
$$\dfrac{x}{3} = b$$ or $$x = 3b = 6$$

$$x = 6, y = 4$$
Option $$B$$.
Question 6
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Solve the following pairs of equations, graphically:
$$3x - 4y = 1$$ and $$x - 2y + 1 = 0$$

A
$$(-3, -2)$$

B
$$(-3, 2)$$

C
$$(3, -2)$$

D
$$(3, 2)$$

Solution

The solution set of equation $$3x-4y=1$$ is as follows:

If $$x=-1$$ then

$$3\times (-1)-4y=1\\ \Rightarrow -3-4y=1\\ \Rightarrow -4y=1+3\\ \Rightarrow -4y=4\\ \Rightarrow y=-1$$

If $$x=3$$ then

$$3\times (3)-4y=1\\ \Rightarrow 9-4y=1\\ \Rightarrow -4y=1-9\\ \Rightarrow -4y=-8\\ \Rightarrow y=2$$

The required points are $$(-1,-1),(3,2)$$.

Also, the solution set of equation $$x-2y+1=0$$ or $$x-2y=-1$$ is as follows:

If $$x=1$$ then

$$1-2y=-1\\ \Rightarrow -2y=-1-1\\ \Rightarrow -2y=-2\\ \Rightarrow y=1$$

If $$x=3$$ then

$$3-2y=-1\\ \Rightarrow -2y=-1-3\\ \Rightarrow -2y=--4\\ \Rightarrow y=2$$

The required points are $$(1,1),(3,2)$$.

Plot all the above points in the graph. Since the point of intersection in the graph is $$(3,2)$$.

Hence the solution is $$x=3,y=2$$.
Question 7
1 / -0

Solve the following pairs of equations, graphically:
$$3x - 2y = 0$$ and $$y + 3 = 0$$

A
$$(-2, -3)$$

B
$$(2, -3)$$

C
$$(-2, 3)$$

D
$$(-3, -3)$$

Solution

The solution set of equation $$3x-2y=0$$ is as follows:

If $$x=0$$ then

$$3\times 0-2y=0\\ \Rightarrow 0-2y=0\\ \Rightarrow -2y=0\\ \Rightarrow y=0$$

If $$x=-2$$ then

$$3\times (-2)-2y=0\\ \Rightarrow -6-2y=0\\ \Rightarrow -2y=6\\ \Rightarrow y=-3$$

The required points are $$(0,0),(-2,-3)$$.

Another equation of line given is $$y+3=0$$ or $$y=-3$$.

Plot all the above points in the graph. Since the point of intersection in the graph is $$(-2,-3)$$.

Hence the solution is $$x=-2,y=-3$$.
Question 8
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Solve the following pairs of equations, graphically:
$$x + y = 0$$ and $$y = 5$$

A
$$(5, -5)$$

B
$$(-5, -5)$$

C
$$(-5, 5)$$

D
$$(5, 5)$$

Solution

The solution set of equation $$x+y=0$$ is as follows:
If $$x=0$$ then,
$$0+y=0\\ \Rightarrow y=0$$

If $$x=-5$$ then
$$-5+y=0\\ \Rightarrow y=5$$

The required points to plot are $$(0,0),(-5,5)$$.

Another equation of a line given is $$y=5$$.

Plot all the above points of the line $$x+y=0$$ and the line parallel to $$x$$ axis that is $$y=5$$ in the graph. Since the point of intersection in the graph is $$(-5,5)$$.

Hence the solution is $$x=-5,y=5$$.
Question 9
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Solve the following pair of linear (simultaneous) equations by the method of elimination :
$$x+y= 7$$
$$5x+12y= 7$$

A
$$x= 11$$ and $$y=-4$$

B
$$x= 1$$ and $$y=7$$

C
$$x= 13$$ and $$y=6$$

D
$$x= 1$$ and $$y=-3$$

Solution

$$x+y=7$$ .... (1)
$$5x+12y=7$$ ....(2)

Multiply eq. (1) by 5, we get
$$5x+5y=35$$ ....(3)

Subtracting eq(2) from eq.(3)
$$5x+5y=35$$
$$5x+12y=7$$
$$-$$ $$-$$ $$-$$
---------------------
$$-7y=28$$
$$y=-4$$

Substituting value of $$y$$ in eq,(1), we get,
$$x-4=7$$
$$x=11$$
Question 10
1 / -0

Solve the following pairs of equations, graphically :
$$4x + 3y = 1$$ and $$2x - y = 3$$

A
$$(1, 0)$$

B
$$(1, -1)$$

C
$$(2, -1)$$

D
$$(2, -3)$$

Solution

Given,
$$ 4x + 3y = 1$$
$$\Rightarrow y = \dfrac {1-4x}{3} $$

Substituting different values of $$ x $$ to get corresponding values of $$ y $$, we get

$$ ( 0, \dfrac {1}{3}), (1, -1), (-1, \dfrac {5}{3}), (2, -\dfrac {7}{3}), (-2,3) $$

Plotting these points on the graph, we get the graph of the equation, $$ 4x + 3y = 1 $$

Similarly,
Given second eqn,
$$ 2x-y= 3 $$
$$\Rightarrow y = 2x - 3$$

Substituting different values of $$ x $$ to get corresponding values of $$ y $$, we get

$$ ( 1, -1), (0,-3), (2,1), (-1,-5), (3,3) $$

Plotting these points on the graph, we get the graph of the equation, $$ 2x-y= 3 $$

The point of intersection of the graphs of the lines is the solution of the equations which is $$ (1,-1) $$

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Pair of Linear Equations in Two Variables Test - 28

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Pair of Linear Equations in Two Variables Test - 28

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Question 1

Present age of $$A$$ is $$40$$ years and Present age of $$B$$ is $$15$$ years

Present age of $$A$$ is $$37$$ years and Present age of B is $$12$$ years

Present age of $$A$$ is $$35$$ years and Present age of $$B$$ is $$10$$ years

Present age of $$A$$ is $$45$$ years and Present age of $$B$$ is $$20$$ years

Solution

Question 2

Present age of father is $$44$$ years and Present age of son is $$20$$ years

Present age of father is $$42$$ years and Present age of son is $$18$$ years

Present age of father is $$60$$ years and Present age of son is $$36$$ years

Present age of father is $$48$$ years and Present age of son is $$24$$ years

Solution

Question 3

$$\dfrac{3}{8}$$

$$\dfrac{1}{6}$$

$$\dfrac{8}{7}$$

$$\dfrac{7}{3}$$

Solution

Question 4

$$x= -5;y= 4$$

$$x= 3,y= -2$$

$$x= 5,y= 6$$

$$x= 1,y= 3$$

Solution

Question 5

$$x= 2;y= 7$$

$$x= 6;y= 4$$

$$x= 7;y= 1$$

$$x= 8;y= 3$$

Solution

Question 6

$$(-3, -2)$$

$$(-3, 2)$$

$$(3, -2)$$

$$(3, 2)$$

Solution

Question 7

$$(-2, -3)$$

$$(2, -3)$$

$$(-2, 3)$$

$$(-3, -3)$$

Solution

Question 8

$$(5, -5)$$

$$(-5, -5)$$

$$(-5, 5)$$

$$(5, 5)$$

Solution

Question 9

$$x= 11$$ and $$y=-4$$

$$x= 1$$ and $$y=7$$

$$x= 13$$ and $$y=6$$

$$x= 1$$ and $$y=-3$$

Solution

Question 10

$$(1, 0)$$

$$(1, -1)$$

$$(2, -1)$$

$$(2, -3)$$

Solution

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Present age of $$A$$ is $$40$$ years and

Present age of $$B$$ is $$15$$ years

Present age of $$A$$ is $$37$$ years and

Present age of B is $$12$$ years

Present age of $$A$$ is $$35$$ years and

Present age of $$B$$ is $$10$$ years

Present age of $$A$$ is $$45$$ years and

Present age of $$B$$ is $$20$$ years