Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Solve the system of equations $$65x-33y=97, $$ and $$33x-65y=1$$ by substitution method.

A
$$(2, 1)$$

B
$$(1, 1)$$

C
$$(2, 2)$$

D
None of these

Solution
Question 2
1 / -0

The sum of two numbers is $$25$$ and their difference is $$7$$, then the numbers are.

A
$$20$$ and $$5$$

B
$$18$$ and $$7$$

C
$$15$$ and $$10$$

D
$$9$$ and $$16$$

Solution

let the two numbers be$$ A$$ and $$B$$

so it is given that

$$ A + B = 25 $$...(1)

$$ A - B = 7 $$ ...(2)

Adding the equations we get

$$ A + B + A - B = 25 +7 $$

$$ 2A = 32 $$

$$A = 16$$

Using the above value of A in (1)

So $$ 16 + B = 25 $$

$$=>B = 9$$

Thus $$ A =16$$ and $$B =9 $$
Question 3
1 / -0

Two lines $$x+ 2y + 7 = 0$$ and $$2x + ky + 18 = 0$$ do not intersect each other. Find the value of $$k.$$

A
$$3$$

B
$$2$$

C
$$1$$

D
$$4$$

Solution

$$x + 2y + 7 = 0$$ and $$2x + ky + 18 = 0$$ do not intersect each other, this means that the pair of equations is inconsistent.

On comparing $$x + 2y + 7 = 0$$ with $$a_1x+b_1y+c_1=0$$ and $$2x+ky+18=0$$ with $$a_2x+b_2y+c_2=0$$ we get,
$$a_1=1, b_1=2, c_1=7, a_2=2,b_2=k, c_2=18$$

Thus, we have
$$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$$

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

$$\Rightarrow k = 4$$

Hence, if $$k=4$$ then the pair of equations will be inconsistent.
Question 4
1 / -0

Solve the following system of equations by substitution method.

$$\dfrac{15}{x}+\dfrac{2}{y}=17, \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}, x\neq 0, y\neq 0$$

A
$$\left ( 5, \dfrac{1}{7} \right )$$

B
$$\left ( 2, \dfrac{1}{7} \right )$$

C
$$\left ( 5, \dfrac{2}{7} \right )$$

D
None of these

Solution

Given equations are

$$\dfrac{15}{x}+\dfrac{2}{y}=17,$$

$$ \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5} $$

$$Let \ \dfrac{1}{x}=a, \ \ \dfrac{1}{y}=b$$

$$\therefore \ 15a+2b=17$$......(i)

and $$a+b=\dfrac{36}{5}$$

$$\Rightarrow a=\dfrac{36}{5}-b$$....(ii)

Putting equation (ii) in (i) we get,

$$15\left[\dfrac{36}{5}-b\right]+2b=17$$

$$\Rightarrow 108-13b=17$$

$$\Rightarrow b=7$$

$$\therefore \ y=\dfrac{1}{7}$$

Putting value of $$b$$ in equation (i), we get,

$$15a+(2\times7)=17$$

$$\Rightarrow a=\dfrac{1}{5}$$

$$\therefore \ x=5$$

Hence, $$\left(5,\dfrac17\right)$$ is the solution of the given system of equations.
Question 5
1 / -0

Solve the following systems of equations using elimination method.
$$0.5x+0.8y=0.44, 0.8x+0.6y=0.5$$

A
$$(0.4, 0.3)$$

B
$$(0.2, 0.3)$$

C
$$(0.4, 0.1)$$

D
None of these

Solution

Consider the given equations.
$$0.5x+0.8y=0.44$$ $$-------( 1 )$$

$$0.8x+0.6y=0.5$$ $$-------(2)$$

From $$(1)$$ and $$(2)$$
On multiplying by $$0.8$$ in equation $$(1)$$ and multiplying by $$0.5$$ in equation $$(2)$$, and subtracting we get

$$0.4x + 0.64y = 0.352$$
$$0.4x +0.3y =0.25$$
—————————
$$0.34y = 0.102$$
$$y=0.3$$    $$[$$ put in $$( 2 )\ ]$$

$$0.8x+0.6\times 0.3=0.5$$

$$0.8x+0.18=0.5$$

$$0.8x=0.32$$

$$x=0.4$$

Hence, the value of $$x$$ is $$0.4$$ and the value of $$y$$ is $$0.3$$.
Question 6
1 / -0

Sahil has Rs.12 less than three times what Sohan has. If both have a total of Rs.88, how many rupees each must have got?

A
Sohan Rs.25 , Sahil-Rs.63

B
Sohan Rs.25 , Sahil-Rs.33

C
Sohan Rs.55 , Sahil-Rs.63

D
Sohan Rs.25 , Sahil-Rs.55

Solution

Let Sahil have rupees $$x$$ and Sohan has rupees $$y$$.

Sahil has Rs.12 less than 3 times Sohan
$$\implies x=3y-12$$

$$\implies x-3y=-12 ---eqn(1)$$

Total both have Rs.88
$$x+y=88 --- eqn(2) $$

Substracting equation 1 from 2 we get $$y=25$$

Then substituting in any one of the equation we get $$x=63$$
Question 7
1 / -0

Solve the following puzzles using the equations:
The present age of Beena is 2 years more than Reeta.The present age of Teena is 3 years more than Beena.If the sum of the ages of Reeta , Beena , and Teena is 79 years, find the present age of all the three of them.

A
Reeta 24 years , Beena 26 years , Teena 29 years.

B
Reeta 24 years , Beena 21 years , Teena 29 years.

C
Reeta 24 years , Beena 29 years , Teena 29 years.

D
Reeta 24 years , Beena 26 years , Teena 30 years.

Solution

Let Beena's present age be $$b$$, Reeta's present age be $$r$$ and Teena's present age be $$t$$

Present age of Beena is 2 years more than of Reeta
$$\implies b=r+2$$
$$\implies b-r=2 --eqn(1)$$

Present age of Teena is 3years more than Beena
$$\implies t=b+3$$
$$\implies t-b=3 --eqn(2)$$

Sum of all the present ages is 79
$$\implies b+t+r=79--eqn(3)$$

Adding Equation 1 and 2 we get $$t-r=5 --eqn(4)$$
Adding equation 3 and 4 we get $$b+2t=84--eqn(5)$$
Adding equation 2 and 5 we get

$$3t=87$$

$$t=29$$

Thus substituting $$t$$ in equation 2 and equation 4 we will get $$b=26$$
and $$r=24$$
Question 8
1 / -0

The solution of the system of equations $$\displaystyle \frac{2x+5y}{xy} = 6$$ and $$\displaystyle \frac{4x-5y}{xy} + 3 = 0$$ (where $$x \neq 0, y \neq 0$$), respectively is ___________.

A
$$1, 2$$

B
$$0, 0$$

C
$$-1, 2$$

D
$$1, -2$$

Solution

Equation $$1 : \dfrac{2x+5y}{xy} = 6$$
$$\Rightarrow 2x+5y = 6xy$$ ........ $$(1)$$

Equation $$2 : \dfrac{4x-5y}{xy} + 3 = 0$$
$$\Rightarrow 4x-5y = -3xy$$ ........ $$(2)$$

Adding both the equations, we get $$6x = 3xy$$
$$\therefore y = 2$$

Substituting value of $$y$$ in $$(1)$$ we get
$$2x + 10 = 12x$$
$$\therefore x = 1$$
Question 9
1 / -0

Solve the following puzzles using the equations:
In a village , the number of women is 89 more than the number of men.The number of children is 400 more than the number of men.If the total population of the village is 4989. Find the number of men , women and children.

A
Men-1500 , Women 1589 , Children 1900.

B
Men-1500 , Women 1600 , Children 1900.

C
Men-1500 , Women 1589 , Children 1200.

D
Men-1400 , Women 1589 , Children 1900.

Solution
Question 10
1 / -0

Solve for $$x$$ and $$y$$ in the following question:
$$\displaystyle \frac{2}{x + 2y} + \frac{1}{2x - y} + \frac{5}{9} = 0$$, $$\displaystyle \frac{9}{x + 2y} + \frac{6}{2x - y} + 4 = 0$$

A
$$x = 1, y = 2$$

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

C
$$\displaystyle x = 2, y = \frac{1}{2}$$

D
$$\displaystyle x = \frac{1}{2}, y=2$$

Solution

$$\dfrac{2}{x+2y} + \dfrac{1}{2x-y} + \dfrac{5}{9} = 0......(1)$$

$$\dfrac{9}{x+2y} + \dfrac{6}{2x-y} + 4 = 0......(2)$$

Let $$\dfrac{1}{x+2y} = a$$ and $$\dfrac{1}{2x-y} = b$$

Substituting the assumed values in equations $$1$$ and $$2$$, we get:
$$2a+b=\dfrac{-5}{9}$$ or
$$18a+9b=-5......(3)$$
and
$$9a+6b=-4......(4)$$

Solving equations $$(3)$$ and $$(4)$$ using elimination method:
Multiplying equation $$(4)$$ by $$2$$ and subtracting $$(3)$$ from it, we get
$$3b=-3$$ or
$$b=-1$$
$$\implies 2x-y = -1 ......(5)$$

Substituting $$b=-1$$ in $$(4)$$
$$9a-6 = -4$$
$$a = \dfrac{2}{9}$$

$$\Rightarrow x+2y = \dfrac{9}{2}......(6)$$

Solving equations $$(5)$$ and $$(6)$$ using elimination method:
Multiplying equation $$(6)$$ by $$2$$ and subtracting $$(5)$$ from it, we get
$$2x+4y-2x+y=9+1$$
$$\therefore y = 2$$
Substituting in $$(5)$$, we get
$$x = \dfrac{1}{2}$$
$$y=2x+1=2.\dfrac{1}{2}+1=2$$

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Pair of Linear Equations in Two Variables Test - 57

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Pair of Linear Equations in Two Variables Test - 57

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Question 1

$$(2, 1)$$

$$(1, 1)$$

$$(2, 2)$$

None of these

Solution

Question 2

$$20$$ and $$5$$

$$18$$ and $$7$$

$$15$$ and $$10$$

$$9$$ and $$16$$

Solution

Question 3

$$3$$

$$2$$

$$1$$

$$4$$

Solution

Question 4

$$\left ( 5, \dfrac{1}{7} \right )$$

$$\left ( 2, \dfrac{1}{7} \right )$$

$$\left ( 5, \dfrac{2}{7} \right )$$

None of these

Solution

Question 5

$$(0.4, 0.3)$$

$$(0.2, 0.3)$$

$$(0.4, 0.1)$$

None of these

Solution

Question 6

Sohan Rs.25 , Sahil-Rs.63

Sohan Rs.25 , Sahil-Rs.33

Sohan Rs.55 , Sahil-Rs.63

Sohan Rs.25 , Sahil-Rs.55

Solution

Question 7

Reeta 24 years , Beena 26 years , Teena 29 years.

Reeta 24 years , Beena 21 years , Teena 29 years.

Reeta 24 years , Beena 29 years , Teena 29 years.

Reeta 24 years , Beena 26 years , Teena 30 years.

Solution

Question 8

$$1, 2$$

$$0, 0$$

$$-1, 2$$

$$1, -2$$

Solution

Question 9

Men-1500 , Women 1589 , Children 1900.

Men-1500 , Women 1600 , Children 1900.

Men-1500 , Women 1589 , Children 1200.

Men-1400 , Women 1589 , Children 1900.

Solution

Question 10

$$x = 1, y = 2$$

$$x = 2, y = 1$$

$$\displaystyle x = 2, y = \frac{1}{2}$$

$$\displaystyle x = \frac{1}{2}, y=2$$

Solution

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