Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Solve the following system of linear equations graphically: $$2x + y = 6, x - 2y + 2 = 0$$. Find the vertices of the triangle formed by the above two lines and the $$x$$-axis. Also find the area of the triangle.

A
Vertices of the triangle are $$A(1,2), 8(5, 0)$$ and $$C(-2,0)$$ and Area: $$10$$ square units

B
Vertices of the triangle are $$A(4,2), 8(2, 0)$$ and $$C(-2,0)$$ and Area: $$15$$ square units

C
Vertices of the triangle are $$A(0,2), 8(1, 0)$$ and $$C(-2,0)$$ and Area: $$7$$ square units

D
Vertices of the triangle are $$A(2,2), 8(3, 0)$$ and $$C(-2,0)$$ and Area: $$5$$ square units

Solution

$$2x+y=6$$
$$\implies y=6-2x$$
Let, $$x=1$$, then $$y=6-2(1)=4$$
Similarly, when $$x=2, y=2$$
$$x-2y+2=0$$
$$\implies x=2y-2$$
Let, $$y=0, $$ then $$x=2(0)-2=-2$$
Similarly, when $$y=2, x=2$$
Plotting the above points on the graph, we get, the vertices of the traingle, i.e. $$A(2,3), B(3,0), C(-2,0)$$
$$\therefore$$ Area of the triangle $$=\dfrac{1}{2}\times 5 \text{ units} \times 2 \text{ units}$$
$$=5$$ sq. units
Question 2
1 / -0

Rs. 58 is divided among 150 children such that each girl gets 25p and each boy get 50p. How many boys are there ?

A
$$52$$

B
$$54$$

C
$$68$$

D
$$62$$

Solution

Let number of boys be $$x$$ and number of girls be $$y$$
Since, Total number of children's are $$150$$
$$\therefore$$ $$x+y=150$$ ....(1)
Rs. 58 is divided among 150 children such that each girl gets 25p and each boy get 50p
$$\therefore$$ Out of $$Rs. 58$$ Boys will get Rs. $$0.25 x$$ and girls will get Rs. $$0.5y$$.
$$\therefore$$ $$\dfrac{25}{100}x+\dfrac{50}{100}y=58$$

$$\therefore$$$$x+2y=232$$ ...(2)
Subtract equation (1) from equation (2), we get $$y = 82$$
And put $$y = 82$$ in equation (1)
$$x=68$$ and $$y=82$$
Number of boys are $$68$$
Question 3
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Solve the following pair of equations by reducing them to a pair of linear equations:

$$\dfrac {1}{(3x+y)}+\dfrac {1}{(3x-y)}=\dfrac {3}{4},\ \dfrac {1}{2(3x+y)}-\dfrac {1}{2(3x-y)}=\dfrac {-1}{8}$$

A
$$x=2,\,\,y=7$$

B
$$x=5,\,\,y=0$$

C
$$x=1,\,\,y=1$$

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

Solution

Given:
$$\dfrac {1}{(3x+y)}+\dfrac {1}{(3x-y)}=\dfrac {3}{4}, \dfrac {1}{2(3x+y)}+\dfrac {1}{2(3x-y)}=\dfrac {-1}{8}$$

$$let \dfrac{1}{3x+y}=a and \dfrac{1}{3x-y}=b$$

Then$$a+b=\dfrac{3}{4}\Rightarrow 4a+4b=3........eq1$$

$$\dfrac{a}{2}-\dfrac{b}{2}=\dfrac{-1}{8}$$

$$\dfrac{a-b}{2}=\dfrac{-1}{8}$$

$$a-b=\dfrac{-2}{8}\Rightarrow a-b=\dfrac{-1}{4}$$

$$4a-4b=-1..............eq2$$

Adding eq1 and eq2 we get

$$8a=2$$

$$a=\dfrac{2}{8}\Rightarrow \dfrac{1}{4}$$

Substituting the value of a in eq1 we get

$$4\times \dfrac{1}{4}+4b=3$$

$$4b=2$$

$$b=\dfrac{2}{4}\Rightarrow \dfrac{1}{2}$$

Then $$\dfrac{1}{3x+y}=\dfrac{1}{4}$$

$$3x+y=4................eq3$$

$$\dfrac{1}{3x-y}=\dfrac{1}{2}$$

$$3x-y=2...............eq4$$

Adding eq3 and eq4 we get
$$6x=6$$
$$x=1$$

Substituting $$x=1$$ in eq3 we get
$$3\times 1 +y=4$$
$$y=1$$

Thus $$x=1$$ and $$y=1$$
Question 4
1 / -0

Two years ago, a father was five times as old as his son. Two years later, his age will be $$8$$ more than $$3$$ times the age of the son. Find the present age of father.

A
$$22$$ years

B
$$60$$ years

C
$$39$$ years

D
$$42$$ years

Solution

Let the present age of the father be $$x$$ and that of the son be $$y$$
Two years ago their ages were $$(x-2)$$ and $$(y-2)$$ respectively

Then, according to question, we have
$$(x-2)=5(y-2)$$
$$\Rightarrow x-2=5y-10$$
$$\Rightarrow x-5y=-8......(1)$$

Two years later their ages will be $$(x+2)$$ and $$(y+2)$$ respectively
Then again,
$$(x+2)=3(y+2)+8$$
$$\Rightarrow x+2=3y+6+8$$
$$\Rightarrow x-3y=12.....(2)$$

Subtracting equation (1) from (2), we get
$$\Rightarrow (x-3y)-( x-5y)=12-(-8) $$
$$\Rightarrow x-3y- x+5y=12+8 $$
$$\Rightarrow 2y=20 $$
$$\Rightarrow y=10$$

Putting $$y = 10$$ in equation (2), we get
$$x-3\times 10=12$$
$$\Rightarrow x=12+30$$
$$\Rightarrow x=42$$
The present ages of father and son are $$42$$ years and $$10$$ years respectively.
So, $$Op-D$$
Question 5
1 / -0

Solve the following pair of equations by reducing them to a pair of linear equations:

$$\dfrac {10}{(x+y)}+\dfrac {2}{(x-y)}=4, \dfrac {15}{(x+y)}-\dfrac {5}{(x-y)}=-2$$

A
$$x=1,\,\,y=3$$

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

C
$$x=8,\,\,y=1$$

D
$$x=0,\,\,y=3$$

Solution

Given:
$$\dfrac{10}{x+y}+\dfrac{2}{x-y}=4$$

$$\dfrac{15}{x+y}+\dfrac{5}{x-y}=-2$$

let $$\dfrac{1}{x+y}=a \ and \ \dfrac{1}{x-y}=b$$

$$10a+2b=4.......eq1$$
$$15a-5b=-2........eq2$$

Multiply eq1 with 3 and eq2 with 2 we get
$$30a+6b=12......eq3$$
$$30a-10b=-4......eq4$$

subtracting eq3 and eq4 we get
$$16b=16$$
$$b=1$$

Substituting $$b=1$$ in eq1 we get
$$10a+2\times 1=4$$
$$10a=2$$
$$a=\dfrac{1}{5}$$

As $$\dfrac{1}{x+y}=a=\dfrac{1}{5}$$

$$x+y=5............eq5$$
and $$\dfrac{1}{x-y}=b=1$$

$$x-y=1...........eq6$$

subtracting eq5 and eq6 we get
$$2x=6$$
$$x=3$$

putting $$x=3$$ in eq 5 we get
$$3+y=5$$
$$y=2$$

hence $$x=3$$ and $$y=2$$
Question 6
1 / -0

Solve graphically the following system of linear equations:
$$3x + y+ 1 = 0$$
$$2x - 3y + 8 = 0$$

A
$$x = 0,\, y = 6$$

B
$$x = - 1,\, y = 2$$

C
$$x = 1,\, y = 1$$

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

Solution

$$3x+y+1=0$$
$$\implies y=-1-3x$$
Let, $$x=-1$$, then $$y=-1-3(-1)=2$$
Similarly, when $$x=1, y=-4$$
And,
$$2x-3y+8=0$$
$$\implies x=\dfrac{3y-8}{2}$$
Let, $$y=2,$$ then $$x=\dfrac{3(2)-8}{2}=-1$$
Similarly, when $$y=0, x=-4$$
Plotting the above points on the graph, we get, the intersecting point at$$(-1,2)$$
$$\therefore x=-1, y=2$$
Question 7
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Albert buys $$4$$ horses and $$9$$ cows for Rs. $$13400$$. If he sells the horses at $$10\%$$ profit and the cows at $$20\%$$ profit, then he earns a total profit of Rs. $$1880$$. The cost of a horse is

A
Rs. $$1000$$

B
Rs. $$2000$$

C
Rs. $$2500$$

D
Rs. $$3000$$

Solution

Let the C.P. of $$1$$ horse be Rs. $$x$$ and C.P. of $$1$$ cow $$=$$ Rs. $$y$$
Then $$4x+9y=13400$$ ........(i)
Also $$10\%$$ of $$4x + 20\%$$ of $$9y = 1880$$
$$\displaystyle \frac{10}{100}4x+\frac{20}{100}9y=1880$$
$$\displaystyle \frac{2}{5}x+\frac{9}{5}y=1880\Rightarrow 2x+9y=9400$$ ...........(ii)
Subtracting eqn (ii) from eqn (i), we get
$$(4x + 9y) - (2x + 9y) = 13400 - 9400$$
$$\displaystyle \Rightarrow$$ $$2x = 4000$$
$$\displaystyle \Rightarrow$$ $$x = 2000$$
Therefore, the cost of a horse is Rs. $$2000$$.
Question 8
1 / -0

Mr. Joshi has $$430$$ cabbage-plants which he wants to plant out. Some $$25$$ to a row and the rest $$20$$ to a row. If there are to be $$18$$ rows in all how many rows of $$25$$ will there be?

A
$$10$$

B
$$14$$

C
$$8$$

D
$$12$$

Solution

Let '$$x$$' rows of $$25$$ and '$$y$$' rows of $$20$$.
Given, $$25x + 20y = 430$$
$$5x+4y=86$$ ....(1)
Also $$x + y = 18$$ ....(2)
Multiply equation (1) by $$4$$, we get
$$4x+4y=72$$ ....(3)
Subtract equations (1) and (3), we get
$$x=14$$
Hence, answer is $$14$$.
Question 9
1 / -0

The difference between two numbers is $$ 7$$ and their sum is $$35$$. What will be their product?

A
$$324$$

B
$$294$$

C
$$79$$

D
$$245$$

Solution

$$x + y = 35$$
$$x - y = 7$$
Add both the equations
$$\displaystyle \Rightarrow $$$$ 2x = 42$$ $$\displaystyle \Rightarrow $$ $$x = 21$$
$$\displaystyle \therefore $$ $$21 + y = 35 $$ $$\displaystyle \Rightarrow $$ $$y = 14$$

$$\displaystyle \therefore $$ $$x$$ $$\displaystyle \times $$ $$ y = 21 $$ $$\displaystyle \times $$ $$ 14 = 294$$
Question 10
1 / -0

For what value of $$k$$ does the system of equations $$\displaystyle 2x+ky=11\:and\:5x-7y=5$$ has no solution?

A
$$\displaystyle \frac{-14}{5}$$

B
$$\displaystyle \frac{-11}{5}$$

C
$$\displaystyle \frac{-14}{9}$$

D
$$\displaystyle \frac{14}{5}$$

Solution

The equations are $$ 2x+ky=11$$ and $$5x-7y=5$$
or $$2x+ky-11=0$$ and $$5x-7y-5=0$$
They will have no solution if $$ \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 } }. .......(1) $$
Here $$ { a }_{ 1 }=2,\quad { a }_{ 2 }=5,\quad { b }_{ 1 }=k,\quad { b }_{ 2 }=-7,\quad { c }_{ 1 }=-11,\quad { c }_{ 2 }=-5$$
$$\therefore \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { 2 }{ 5 } , \dfrac { { b }_{ 1 } }{ { b }_{ 2 } } =\dfrac { k }{ -7 } ,\quad \dfrac { { c }_{ 1 } }{ { c }_{ 2 } } =\dfrac { -11 }{ -5 } =\dfrac { 11 }{ 5 }$$

Now from $$(1)$$
$$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$$
$$\therefore \quad \dfrac { 2 }{ 5 } =\dfrac { k }{ -7 } \\ \Rightarrow k=\dfrac { -14 }{ 5 }$$
Also the first two ratios $$ \neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 } } $$
$$ \therefore k=\dfrac { -14 }{ 5 }$$

Hence, the value of $$k$$ is $$\dfrac { -14 }{ 5 } $$

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Pair of Linear Equations in Two Variables Test - 46

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Pair of Linear Equations in Two Variables Test - 46

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Question 1

Vertices of the triangle are $$A(1,2), 8(5, 0)$$ and $$C(-2,0)$$ and Area: $$10$$ square units

Vertices of the triangle are $$A(4,2), 8(2, 0)$$ and $$C(-2,0)$$ and Area: $$15$$ square units

Vertices of the triangle are $$A(0,2), 8(1, 0)$$ and $$C(-2,0)$$ and Area: $$7$$ square units

Vertices of the triangle are $$A(2,2), 8(3, 0)$$ and $$C(-2,0)$$ and Area: $$5$$ square units

Solution

Question 2

$$52$$

$$54$$

$$68$$

$$62$$

Solution

Question 3

$$x=2,\,\,y=7$$

$$x=5,\,\,y=0$$

$$x=1,\,\,y=1$$

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

Solution

Question 4

$$22$$ years

$$60$$ years

$$39$$ years

$$42$$ years

Solution

Question 5

$$x=1,\,\,y=3$$

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

$$x=8,\,\,y=1$$

$$x=0,\,\,y=3$$

Solution

Question 6

$$x = 0,\, y = 6$$

$$x = - 1,\, y = 2$$

$$x = 1,\, y = 1$$

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

Solution

Question 7

Rs. $$1000$$

Rs. $$2000$$

Rs. $$2500$$

Rs. $$3000$$

Solution

Question 8

$$10$$

$$14$$

$$8$$

$$12$$

Solution

Question 9

$$324$$

$$294$$

$$79$$

$$245$$

Solution

Question 10

$$\displaystyle \frac{-14}{5}$$

$$\displaystyle \frac{-11}{5}$$

$$\displaystyle \frac{-14}{9}$$

$$\displaystyle \frac{14}{5}$$

Solution

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