Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 21

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Question 1
1 / -0

The pair of linear equations $$ 4x +6y= 9\ and\ 2x+ 3y= 6$$ has

A
No solution

B
Many solutions

C
Two solutions

D
One solution.

Solution

Given system of linear equations
$$4x+6y- 9=0\Rightarrow a_1=4,\,b_1=6,\,c_1=-9$$
$$2x+3y-6=0\Rightarrow a_2=2,\,b_2=3,\,c_2=-6$$.

As we know a pair of linear equations is inconsistent (no solution) if
$$\cfrac{{a}_{1}}{{a}_{2}} = \cfrac{{b}_{1}}{{b}_{2}} \ne \cfrac{{c}_{1}}{{c}_{2}}$$
We have $$\dfrac{4}{2}=\dfrac{6}{3} \neq\dfrac{-9}{-6}$$
Hence, no solution.
Question 2
1 / -0

The graph of the linear equation $$2x + 3y = 6$$ intersects the $$y-$$axis at the point

A
$$(2, 0)$$

B
$$(0, 3)$$

C
$$(3, 0)$$

D
$$(0, 2)$$

Solution

$$2x + 3y =6$$
Since , the line intersects at $$y-$$axis.
Put $$x =0$$ in the given equation
$$\Rightarrow y =2$$
Hence, the line intersects $$y-$$axis at $$(0,2)$$
Question 3
1 / -0

Determine the vertices of the triangle formed by the lines
$$y = x, 3y = x $$ and $$ x + y = 8$$.

A
$$(0, 0), (0, 4), (6, 2)$$

B
$$(0, 0), (4, 4), (0, 2)$$

C
$$(0, 0), (7, 7), (5, 2)$$

D
$$(0, 0), (4, 4), (6, 2)$$

Solution

The vertices of the triangle will be the point of intersection of given lines taken two at a time. So, we need to find the points of intersection of the lines:

A. Taking
$$ y = x $$ $$ \dots (1)$$
$$ 3y = x$$   $$ \dots (2)$$
Let, the point of intersection of above two lines is $$A$$.
Subtract $$(1)$$ from $$(2)$$:
$$ 2y = 0$$
$$ y = 0$$
Put this value of $$y$$ in equation $$(1)$$:
$$ y=x=0 $$
Hence, the coordinates of A are $$(0,0) $$.

B. Taking
$$ y = x $$ $$ \dots (2)$$
$$ x+y = 8$$   $$ \dots (3)$$
Let, the point of intersection of above two lines is $$B$$.
Subtract $$(2)$$ from $$(3)$$:
$$ x+y-y = 8-x$$
$$ 2x = 8$$
$$ x= 4$$
Put this value of $$x$$ in equation $$(3)$$:
$$ y=x=4 $$
Hence, the coordinates of B are $$(4,4) $$.

C. Taking
$$ x+y = 8 $$ $$ \dots (3)$$
$$ x = 3y$$   $$ \dots (1)$$
Let, the point of intersection of above two lines is $$C$$.
Subtract equation $$(1)$$ from $$(3)$$:
$$ x+y-x = 8-3y$$
$$ y=8-3y$$
$$4y = 8$$
$$y=2$$
Put this value of $$y$$ in equation $$(1)$$:
$$ x=3(2) = 6 $$
Hence, the coordinates of C are $$(6,2) $$.

The coordinates of the triangle are $$(0,0), (4,4)$$ and $$(6,2). $$
Question 4
1 / -0

The cost of $$4$$ pens and $$4$$ pencil boxes is Rs $$100$$. Three times the cost of a pen is Rs $$15$$ more than the cost of a pencil box. The cost of a pen and a pencil box respectively are

A
Rs $$15$$ and Rs $$8$$

B
Rs $$10$$ and Rs $$15$$

C
Rs $$12$$ and Rs $$10$$

D
Rs $$16$$ and Rs $$12$$

Solution

Let the cost of a pen and a pencil be $$x$$ and $$y$$ respectively.
$$4x+4y=100$$
$$x+y=25$$ ...(1)
$$3x=y+15$$
$$3x-y=15$$ ...(2)
Adding (1) and (2), we get
$$4x=40$$
$$x=10$$
Substitute $$x=10$$ in equation (1) to get $$y=15$$.
Question 5
1 / -0

Solve the following pair of simultaneous equations:
$$2x+3y = 12; \, \, 5x-3y=9$$

A
$$x=1;\, \, y=0$$

B
$$x=3;\, \, y=2$$

C
$$x=7;\, \, y=3$$

D
$$x=2;\, \, y=5$$

Solution

Given equations are
$$2x+3y=12\quad\quad\dots(i)$$
$$5x-3y=9\quad\quad\dots(ii)$$

Add equations $$(i)$$ and $$(ii) $$ to eliminate $$y$$,
$$\begin{array}{l}\underline {\begin{array}{ccccccccccccccc}{2x}& + &{3y}& = &{12}\\{5x}& - &{3y}& = &9\end{array}} \\\begin{array}{ccccccccccccccc}{7x}&{}&{\;\;\;\;\;\;\;}& = &{\;21}\end{array}\\\quad\;\quad\quad\quad \quad x = 3\end{array}$$

Substitute this value in equation $$(i)$$,
$$2(3)+3y=12$$
$$\Rightarrow 6+3y=12$$
$$\Rightarrow 3y=6$$
$$\Rightarrow y=2$$

Therefore, the solution is $$x=3, y=2$$.
Question 6
1 / -0

The graph of the linear equation $$2x + 3y = 6$$ is a line which meets the $$x-$$axis at the point

A
$$(0, 2)$$

B
$$(2, 0)$$

C
$$(3, 0)$$

D
$$(0, 3)$$

Solution

$$2x + 3y =6 $$ will meet the $$x-$$axis at $$y=0$$.
For $$y=0$$, $$x =3$$.
$$\therefore$$ The point is $$(3,0)$$
Question 7
1 / -0

Are the following pair of linear equations consistent?
$$-3x-4y=12$$,
$$ 4y+3x=12$$

A
No

B
Yes

C
Data Insufficient

D
None of these

Solution

The given pair of linear equations can be written as
$$-3x-4y-12=0 $$ and $$ 3x+4y-12=0$$
Here $$a_1=-3,b_1=-4,c_1=-12$$
and $$a_2=3,b_2=4,c_2=-12$$
$$\therefore \frac{a_1}{a_2}=\frac{-3}{3}=-1$$
$$\frac{b_1}{b_2}=\frac{-4}{4}=-1$$
$$\frac{c_1}{c_2}=\frac{-12}{-12}=1$$
$$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$$
Hence the given pair of linear equations is inconsistent.
Question 8
1 / -0

If $$p+q=k,\ p-q=n$$ and $$k > n$$, then $$q$$ is ________ .

A
positive

B
negative

C
$$0$$

D
none of the above

Solution

$$p+q=k$$ ------- (1)
$$p-q=n$$ ------- (2)
Subtracting (2) from (1),

$$2q=(k-n)$$
Since $$k>n$$,

$$(k-n)$$ is positive, therefore, $$q=\dfrac{k-n}{2}$$, which is positive
Question 9
1 / -0

Directions For Questions

If we have two simultaneous equations
$$ax+by=c$$ -------(1)
and $$bx+ay=d$$, then in order to solve -------(2)
we find (1) $$+$$ (2) and then (1) $$-$$ (2), we shall get
$$(a+b)x+(a+b)y=c+d$$
i.e. $$\displaystyle x+y =\frac{c+d}{a+b}$$
and $$(a-b)x-(a-b)y=c-d$$
i.e. $$x-y=\dfrac{c-d}{a-b}$$
To find $$x$$, (3)+(4) gives, $$2x=\dfrac{c+d}{a+b}+\dfrac{c-d}{a-b}$$
$$\Rightarrow \displaystyle x=\frac{1}{2} \left ( \frac{c+d}{a+b}+\frac{c-d}{a-b} \right )$$
To find $$y$$, (3)-(4) gives, $$y=\displaystyle \frac{1}{2} \left ( \frac{c+d}{a+b}-\frac{c-d}{a-b} \right )$$
Read the above passage carefully and mark the correct choice.

...view full instructions

The solution of
$$217x+131y=913$$
$$131x+217y=827$$ is

A
$$x=2, y=3$$

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

C
$$x=2, y=2$$

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

Solution

Let $$ 217x+131y=913$$ .....(1)
and $$ 131x+217y=827$$ ......(2)
Here, according to the given passage
$$ a=217, b=131, c=913$$ and $$d=827$$
$$ \therefore 2x= \left[ \dfrac { c+d }{ a+b } +\dfrac { c-d }{ a-b } \right] $$
$$ =\dfrac { 1740 }{ 348 } +\dfrac { 86 }{ 86 } =\dfrac { 2088 }{ 348 } $$
$$ \therefore x=\dfrac { 2088 }{ 348\times 2 } =3$$
and $$2y= \left[ \dfrac { c+d }{ a+b } -\dfrac { c-d }{ a-b } \right] $$
$$ =\dfrac { 1740 }{ 348 } -1=\dfrac { 1392 }{ 348 } $$
$$ \therefore y=\dfrac { 1392 }{ 348\times 2 } =2$$
$$ \therefore x=3, y=2$$
Question 10
1 / -0

The graphical representation of the pair of equations $$x+2y-4=0$$ and $$2x+4y-12=0$$ is:

A
intersecting lines

B
parallel lines

C
coincident lines

D
all the above

Solution

For two lines to be parallel,
$$ \dfrac { a_1 }{ a_2 } =\dfrac { b_1 }{ b_2 } \neq \dfrac { c_1 }{ c_2 }$$ -----equation$$(1)$$
where, $$a$$ and $$b$$ are coefficients of $$x$$ and $$y$$ and $$c$$ is constant
Here, the lines are $$x+2y-4=0$$ and $$2x+4y-12=0$$
$$a_1 =1 , b_1 =2 , c_1 = -4 $$ and $$a_2=2, b_2=4, c_2 =-12$$
As we can see that,
$$ \dfrac { 1 }{ 2 } =\dfrac { 2 }{ 4 } \neq \dfrac { -4 }{ -12 } $$
We can say that the given lines are parallel.

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Pair of Linear Equations in Two Variables Test - 21

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Pair of Linear Equations in Two Variables Test - 21

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Question 1

No solution

Many solutions

Two solutions

One solution.

Solution

Question 2

$$(2, 0)$$

$$(0, 3)$$

$$(3, 0)$$

$$(0, 2)$$

Solution

Question 3

$$(0, 0), (0, 4), (6, 2)$$

$$(0, 0), (4, 4), (0, 2)$$

$$(0, 0), (7, 7), (5, 2)$$

$$(0, 0), (4, 4), (6, 2)$$

Solution

Question 4

Rs $$15$$ and Rs $$8$$

Rs $$10$$ and Rs $$15$$

Rs $$12$$ and Rs $$10$$

Rs $$16$$ and Rs $$12$$

Solution

Question 5

$$x=1;\, \, y=0$$

$$x=3;\, \, y=2$$

$$x=7;\, \, y=3$$

$$x=2;\, \, y=5$$

Solution

Question 6

$$(0, 2)$$

$$(2, 0)$$

$$(3, 0)$$

$$(0, 3)$$

Solution

Question 7

No

Yes

Data Insufficient

None of these

Solution

Question 8

positive

negative

$$0$$

none of the above

Solution

Question 9

$$x=2, y=3$$

$$x=3, y=2$$

$$x=2, y=2$$

$$x=3, y=3$$

Solution

Question 10

intersecting lines

parallel lines

coincident lines

all the above

Solution

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