Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 19

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Question 1
1 / -0

The value of $$k$$ for which the following system of equations has infinitely many solutions
$$5x+2y=k$$
$$10x+4y=3$$

A
$$\cfrac{1}{2}$$

B
$$3$$

C
$$\cfrac{3}{2}$$

D
$$1$$

Solution

For infinitely many solutions, both equations need to be identical in the ratios
i.e.
$$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$
where, $$a_1,b_1,c_1$$ are the coefficients of a equation
and $$a_2,b_2,c_2 $$ are the coefficients of another equation

Thus from
$$5x+2y=k$$
$$10x+4y=3$$

To have infinitely many solutions
$$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$

$$\dfrac{5}{10}=\dfrac{2}{4}=\dfrac{k}{3}$$

Taking
$$\dfrac{2}{4}=\dfrac{k}{3}$$

$$\Rightarrow \dfrac{1}{2}=\dfrac{k}{3}$$

Thus,$$2k=3$$
$$\Rightarrow k=\cfrac{3}{2}$$ is the required answer.
Question 2
1 / -0

Which of the following condition is true if the system of equations below is shown to be consistent and dependent?
$$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$$

A
$$\displaystyle \frac{a_1}{a_2}= \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$

B
$$\displaystyle \frac{a_1}{a_2}= \frac{b_1}{b_2}= \frac{c_1}{c_2}$$

C
$$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$

D
$$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}= \frac{c_1}{c_2}$$

Solution

If the system of equations are consistent, then there can be exactly one solution or infinite solutions.
The solutions are infinite when they are dependent.

We know that, for two linear equations
$$a_{1}x+b_{1}=c_{1}$$ and $$a_{2}x+b_{2}y=c_{2}$$
If $$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$$, the system is consistent and has infinitely many solutions.

Hence, a system of equations will be consistent and dependent if $$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$$.
Question 3
1 / -0

The value of $$k$$ for which the system $$kx+3y=7$$ and $$2x-5y=3$$ has no solution is:

A
$$7$$ and $$k=-\cfrac{3}{14}$$

B
$$4$$ and $$k=\cfrac{3}{14}$$

C
$$\cfrac{6}{5}$$ and $$k\ne \cfrac{14}{3}$$

D
$$-\cfrac{6}{5}$$ and $$k\ne \cfrac{14}{3}$$

Solution

$$kx+3y=7$$
$$2x-5y=3$$
Here, $${a}_{1}={k}_{1}, {b}_{1}=3, {c}_{1}=-7$$
$${a}_{2}=2, {b}_{2}=-5,{c}_{2}=-3$$
For no solution, $$\cfrac{k}{2}=\cfrac{3}{-5}\ne \cfrac{-7}{-3}$$
$$\Rightarrow k=\cfrac{-6}{5}$$ and $$k\ne \cfrac{14}{3}$$
Question 4
1 / -0

The ratio of income of two persons is $$9 : 7$$ and the ratio of their expenditure is $$4 : 3$$. If each of them manages to save Rs. $$2000$$ per month, find their monthly income.

A
$$Rs. 18,000 , Rs. 14,000$$

B
$$Rs. 1,000 , Rs. 14,000$$

C
$$Rs. 18,000 , Rs. 1,000$$

D
None of these

Solution

Let their salaries be $$9x$$ and $$7x$$.
Let their expenditure be $$4y$$ and $$3y$$.

According to the question,
$$9x-4y=2000$$ $$—(1)$$
$$7x-3y=2000$$ $$—(2)$$

From $$(1)$$
$$x=\dfrac{2000+4y}{9}$$ $$—(3)$$

On putting $$x$$ in $$(2)$$, we get
$$7\times \dfrac{(2000+4y)}{9}-3y=2000$$

$$\dfrac{(14000+28y)}{9}-3y=2000$$

$$\dfrac{14000+28y-27y}{9}=2000$$

$$14000+y=18000$$

$$y=4000$$

Now, put $$y$$ in (3)
$$x=\dfrac{2000+4\times 4000}{9}$$

$$x=\dfrac{2000+16000}{9}$$

$$x=\dfrac{18000}{9}=2000$$

So,
Salary of first person $$=9\times 2000=Rs.\ 18000$$
Salary of second person $$=7\times 2000=Rs.\ 14000$$

Hence, this is the answer.
Question 5
1 / -0

Given that $$2y=-6$$ and $$x-3y=13$$, the value of $$x=$$________

A
$$\dfrac { 13 }{ 9 } $$

B
$$4$$

C
$$19$$

D
$$22$$

Solution

Given $$2y=-6$$
$$\Rightarrow y=\dfrac{-6}{2}$$
$$\Rightarrow y=-3$$ ............ $$(i)$$
$$x-3y=13$$ ............. $$(ii)$$ (given )
Put the value of $$y$$ in $$(ii)$$, we get
$$x-3(-3)=13$$
$$\Rightarrow x+9=13$$
$$\Rightarrow x=13-9=4$$
Then $$x=4$$
Question 6
1 / -0

Solve the following system of equations using elimination method.
$$3x+4y=24, 20x-11y=47$$

A
$$(4 , 3)$$

B
$$(2 , 3)$$

C
$$(4 , 2)$$

D
None of these

Solution

Consider the given equations.
$$3x+4y=24$$ $$-------( 1 )$$

$$20x-11y=47$$ $$-------(2)$$

From $$(1)$$ and $$(2)$$
On multiplying by $$20$$ in equation $$(1)$$ and multiplying by $$3$$ in equation $$(2)$$ and subtract, we get

$$60x + 80y = 480$$
$$60x -33y =141$$
—————————
$$113y = 339$$
$$y=3$$ $$[$$ put in $$( 2 )\ ]$$

$$20x-11\times 3=47$$

$$20x-33=47$$

$$20x=80$$

$$x=4$$

Hence, the value of $$x$$ is $$4$$ and the value of $$y$$ is $$3$$.
Question 7
1 / -0

Fill in the blank: _______ system of linear equation has no solution.

A
Inconsistent

B
Consistent

C
Dependent

D
Empty

Solution

Inconsistent system of linear equation has no solution.
Question 8
1 / -0

If $$\bar{x}-Z=3,$$ $$\bar{x}+Z =45$$ and $$2\bar{x}+Z=3M$$ then find $$M$$.

A
$$26$$

B
$$22$$

C
$$24$$

D
$$23$$

Solution

Given equations are
$$\bar{x}-Z=3$$ ...... $$(1)$$
$$\bar{x}+Z=45$$ ...... $$(2)$$
$$3M=2\bar{x}+Z$$ ....... $$(3)$$
Adding 1 and 2 equations, we get
$$2\bar{x}=48$$
$$\therefore \bar{x}=24$$
Now, $$\bar{x}+Z=45$$
$$\therefore Z=45-\bar{x}$$ $$=45-24=21$$
$$3M=2\bar{x}+Z$$
$$=2(24)+Z$$
$$=48+21$$
$$=69$$
$$\therefore M=23$$
Question 9
1 / -0

Determine graphically the vertices of the triangle, the equations of whose side are give below:
$$y=x,y =0$$ and $$3x+3y=10$$

A
$$(0,0) \, \left(\dfrac{10}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$$

B
$$(0,0) \, \left(\dfrac{17}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$$

C
$$(0,0) \, \left(\dfrac{14}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$$

D
$$(0,0) \, \left(\dfrac{10}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{8}{3}\right)$$

Solution

As shown in the graph , to find vertices of required triangle,
We can find intersection of lines
1. $$3x+3y=10$$ and $$y=x$$ which is $$(\dfrac{5}{3},\frac{5}{3})$$
2. $$y=0$$ and $$3x+3y =10$$ which is $$(\dfrac{10}{3},0)$$
3. $$y=x$$ and $$y=0$$ which is origin $$(0,0)$$
Question 10
1 / -0

The solution of $$8x+5y=9$$, $$3x+2y=4$$ lie on $$x+y=k$$, then $$k=$$

A
$$3$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

$$ 8x+5y=9 ---(1) \times 2$$
$$3x+2y=4 ---(2) \times 5$$

subtracting both
$$(16x+10y) - (15x+10y)=18-20$$
$$x=-2$$

Putting $$x=-2$$ in $$(2)$$
$$3x+2y=4$$
$$\Rightarrow 3(-2)+2y=4$$
$$\Rightarrow 2y=10$$
$$\Rightarrow y=5$$

Since point $$(-2,5)$$ lies on $$x+y=k$$
So, $$-2+5=k$$
$$\Rightarrow k=3$$

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 19

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Pair of Linear Equations in Two Variables Test - 19

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SHARING IS CARING

Question 1

$$\cfrac{1}{2}$$

$$3$$

$$\cfrac{3}{2}$$

$$1$$

Solution

Question 2

$$\displaystyle \frac{a_1}{a_2}= \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$

$$\displaystyle \frac{a_1}{a_2}= \frac{b_1}{b_2}= \frac{c_1}{c_2}$$

$$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$

$$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}= \frac{c_1}{c_2}$$

Solution

Question 3

$$7$$ and $$k=-\cfrac{3}{14}$$

$$4$$ and $$k=\cfrac{3}{14}$$

$$\cfrac{6}{5}$$ and $$k\ne \cfrac{14}{3}$$

$$-\cfrac{6}{5}$$ and $$k\ne \cfrac{14}{3}$$

Solution

Question 4

$$Rs. 18,000 , Rs. 14,000$$

$$Rs. 1,000 , Rs. 14,000$$

$$Rs. 18,000 , Rs. 1,000$$

None of these

Solution

Question 5

$$\dfrac { 13 }{ 9 } $$

$$4$$

$$19$$

$$22$$

Solution

Question 6

$$(4 , 3)$$

$$(2 , 3)$$

$$(4 , 2)$$

None of these

Solution

Question 7

Inconsistent

Consistent

Dependent

Empty

Solution

Question 8

$$26$$

$$22$$

$$24$$

$$23$$

Solution

Question 9

$$(0,0) \, \left(\dfrac{10}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$$

$$(0,0) \, \left(\dfrac{17}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$$

$$(0,0) \, \left(\dfrac{14}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$$

$$(0,0) \, \left(\dfrac{10}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{8}{3}\right)$$

Solution

Question 10

$$3$$

$$4$$

$$5$$

$$2$$

Solution

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