Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the numerator and the denominator of a fraction is equal to $$7$$. Four times the numerator is $$8$$ less than $$5$$ times the denominator. Find the fraction.

A
$$\displaystyle \frac{1}{6}$$

B
$$\displaystyle \frac{3}{4}$$

C
$$\displaystyle \frac{4}{7}$$

D
$$\displaystyle \frac{7}{11}$$

Solution

Let the fraction be $$ \dfrac {x}{y} $$

Given, the sum of the numerator and the denominator of a fraction is equal to $$ 7 $$

$$\Rightarrow x + y = 7 $$ $$...(1)$$

Also, four times the numerator is $$ 8 $$ less than $$ 5 $$ times the denominator

$$\Rightarrow 4x = 5y - 8 $$

or $$ 4x -5y = -8 $$ $$...(2)$$

Multiplying equation $$ (1) $$ with $$ 5 $$

We get $$ 5x + 5y = 35 $$ $$...(3)$$

Adding equations $$ (2) $$ and $$ (3), $$

$$4x-5y=-8$$

$$5x+5y=35$$
______________

$$9x\ \ \ \ \ \ \ \ =27$$

$$ \Rightarrow 9x = 27 $$

$$\Rightarrow x = 3 $$

Substituting $$ x = 3 $$ in the equation $$ (1) $$

We get $$ 3 + y = 7 $$

$$\Rightarrow y = 4 $$

Hence, the fraction is $$\dfrac34$$
Question 2
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$$1250$$ persons went to see a circus-show. Each adult paid Rs. $$75$$ and each child paid Rs. $$25$$ for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to Rs. $$61,250$$.

A
Adults $$=600$$ and children $$=650$$

B
Adults $$=300$$ and children $$=450$$

C
Adults $$=800$$ and children $$=700$$

D
Adults $$=640$$ and children $$=800$$

Solution

Let $$ x $$ be the number of adults and $$ y $$ children.

Given, $$ x + y = 1250 $$ --- $$(1)$$

And $$ 75x + 25y = 61250 $$
$$\Rightarrow 3x + y = 2450 $$ --- $$(2)$$

Subtracting equation $$ (1) $$from $$ (2) $$, we get $$ 2x = 1200 \Rightarrow x = 600 $$
Substituting $$ x = 600 $$ in the equation $$ (1) $$, we get $$ 600 + y = 1250 \Rightarrow y = 650 $$
Hence, $$ 600 $$ adults and $$ 650 $$ children visited.
Question 3
1 / -0

Rohit says to Ajay "Give me a hundred, I shall then become twice as rich as you." Ajay replies "If you give me ten, I shall be six times as rich as you." How much does each have originally?

A
Rohit has Rs. $$20$$ and Ajay has Rs. $$ 90$$

B
Rohit has Rs. $$50$$ and Ajay has Rs. $$ 200$$

C
Rohit has Rs. $$ 40 $$ and Ajay has Rs. $$170$$

D
Rohit has Rs. $$ 56$$ and Ajay has Rs. $$190$$

Solution

$$\\ Let\quad the\quad numbers\quad be\quad x,\quad w,\quad z\quad and\quad y.\\ 2(x-100)=y+100;\\ x+10=6(y-10);\\ x=6y-70;\\ Substituting\quad the\quad value\quad in\quad the\quad 1st\quad equation:\\ 2(6y-70-100)=y+100;\\ Therefore:\\ y=40\\ x=170\\ $$
Question 4
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The sum of two-digit numbers and the number obtained by reversing the order of the digit is $$121$$. Find the number, if the digits differ by $$3$$.

A
$$47$$ or $$74$$

B
$$36$$ or $$63$$

C
$$67$$ or $$76$$

D
$$94$$ or $$49$$

Solution

Let the two digit number be $$ 10x + y $$

On reversing its digits, we get the number $$ 10y + x $$

Given, $$ (10x + y) + (10y + x) = 121 $$
$$ 11x + 11y = 121 $$
=> $$ x + y = 11 $$ --- $$(1)$$

Given, $$ x - y = 3 $$ -- $$(2)$$
Adding $$(1)$$ and $$(2),$$ we get
$$ 2x = 14 $$ => $$ x = 7 $$

Substituting $$ x = 7 $$ in the equation $$ (2) $$, we get $$ 7 -y = 3 \Rightarrow y = 4 $$

Hence, the number is $$ 74$$ or $$ 47 $$
Question 5
1 / -0

The pair of linear equations $$2x+5y=k, kx+15y=18$$ has infinitely many solutions if

A
$$k=3$$

B
$$k=6$$

C
$$k=9$$

D
$$k=18$$

Solution

The pair of linear equations
$$2x+5y=k$$
and $$ kx+15y=18$$
Here,$$ a_{1}=2,b_{1}=5,c_{1}=-k$$
and $$ a_{2}=k,b_{2}=15,c_{2}=-18$$

The system of equation has infinitely many solutions if
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$

$$\Rightarrow \displaystyle \frac{2}{k}=\frac{5}{15}=\frac{-k}{-18}$$

$$\Rightarrow \displaystyle \frac{2}{k}=\frac{1}{3} =\frac{k}{18} $$

Taking,
$$\displaystyle \frac{2}{k}=\frac{1}{3} \Rightarrow k=6$$

Taking,
$$\displaystyle \frac{1}{3} =\frac{k}{18} \Rightarrow k=6$$

So, answer is $$k=6$$
Question 6
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The pair of equations $$3x+4y=k, 9x+12y=6$$ has infinitely many solutions if

A
$$k=2$$

B
$$k=6$$

C
$$k\neq 6$$

D
$$k=3$$

Solution

The equations are
$$ 3x+4y-k=0$$
$$ 9x+12y-6=0$$
Here, $$ a_{1}=3,b_{1}=4,c_{1}=-k$$
and $$ a_{2}=9,b_{2}=12,c_{2}=-6$$
For the system to have infinite solutions condition is
$$ \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{3}{9}=\frac{4}{12}=\frac{-k}{-6}$$

$$\dfrac{4}{12}=\dfrac{-k}{-6}$$

$$\Rightarrow \dfrac{1}{3}=\dfrac{-k}{-6}$$

$$\Rightarrow 3k=6$$

$$\Rightarrow k=2$$
Question 7
1 / -0

The pair of linear equations $$3x+7y=k, 12x+2ky=4k+1$$ do not have any solution if

A
$$k=7$$

B
$$k=14$$

C
$$k=21$$

D
$$k=28$$

Solution

The pair of linear equations
$$ 3x+7y-k=0$$
$$ 12x+2ky-(4k+1)=0$$
Here,$$ a_{1}=3,b_{1}=7,c_{1}=-k$$
and $$a_{2}=12,b_{2}=2k,c_{2}=-(4k+1)$$
The system has no solution if
$$\displaystyle\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{3}{12}=\frac{7}{2k}\neq \frac{-k}{-(4k+1)}$$
$$\Rightarrow \frac{1}{4}=\frac{7}{2k} \neq \frac{k}{4k+1}$$
$$\Rightarrow 2k=28 $$
$$\Rightarrow k=14 $$
Question 8
1 / -0

If the system of equations
$$3x + y =1$$
$$\left(2k - 1\right)x + \left(k - 1\right)y = 2k + 1$$ is inconsistent, then $$k =$$

A
$$1$$

B
$$-1$$

C
$$-2$$

D
$$2$$

Solution

The system of equations is inconsistent if
$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$
$$\therefore$$ $$\displaystyle\frac{3}{2k - 1} = \displaystyle\frac{1}{k - 1} \neq \displaystyle\frac{1}{2k + 1}$$
$$\Rightarrow 3k - 3 = 2k - 1$$
$$\Rightarrow k = 2$$
Question 9
1 / -0

The difference between two whole numbers is $$26$$ and one number is three times the other. Find the numbers.

A
$$45$$ and $$15$$

B
$$48$$ and $$12$$

C
$$60$$ and $$20$$

D
$$39$$ and $$13$$

Solution

Let the numbers be $$ x $$ and $$ y $$

Given, $$ x - y = 26 $$ --- $$(1)$$
Also, $$ x = 3y $$ --- $$(2)$$

From equations $$ (1) $$ and $$ (2) $$, we get $$ 3y - y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13 $$
So, $$ x = 3y = 3 \times 13 = 39 $$

Numbers are $$ 39, 13 $$
Question 10
1 / -0

The value of k for which the system of linear equations: $$kx+4y=k-4$$, $$16x+ky=k$$, has infinitely many solutions, is

A
$$0$$

B
$$8$$

C
$$-8$$

D
$$4$$

Solution

$$kx+4y=k-4$$
$$16x+ky=k$$

$${a}_{1}=k,{b}_{1}=4,{c}_{1}=-(k-4)$$
$${a}_{2}=16,{b}_{2}=k,{c}_{2}=-k$$

Here condition is $$\displaystyle\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$$

$$\displaystyle\frac{k}{16}=\frac{4}{k}=\frac{k-4}{k}$$

$$\displaystyle\frac{k}{16}=\frac{4}{k}\Longrightarrow {k}^{2}=64\Longrightarrow k=\pm 8$$

Also, $$\frac{4}{k}=\frac{k-4}{k}\Longrightarrow 4k={k}^{2}-4k$$

$$\Longrightarrow {k}^{2}-8k=0\Longrightarrow k(k-8)=0$$

$$k=0$$ or $$k=8$$ but $$k=0$$ is not possible.
Therefore, $$k$$ is $$8$$ is correct value for infinitely many solutions.

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Pair of Linear Equations in Two Variables Test - 36

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Pair of Linear Equations in Two Variables Test - 36

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Question 1

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{3}{4}$$

$$\displaystyle \frac{4}{7}$$

$$\displaystyle \frac{7}{11}$$

Solution

Question 2

Adults $$=600$$ and children $$=650$$

Adults $$=300$$ and children $$=450$$

Adults $$=800$$ and children $$=700$$

Adults $$=640$$ and children $$=800$$

Solution

Question 3

Rohit has Rs. $$20$$ and Ajay has Rs. $$ 90$$

Rohit has Rs. $$50$$ and Ajay has Rs. $$ 200$$

Rohit has Rs. $$ 40 $$ and Ajay has Rs. $$170$$

Rohit has Rs. $$ 56$$ and Ajay has Rs. $$190$$

Solution

Question 4

$$47$$ or $$74$$

$$36$$ or $$63$$

$$67$$ or $$76$$

$$94$$ or $$49$$

Solution

Question 5

$$k=3$$

$$k=6$$

$$k=9$$

$$k=18$$

Solution

Question 6

$$k=2$$

$$k=6$$

$$k\neq 6$$

$$k=3$$

Solution

Question 7

$$k=7$$

$$k=14$$

$$k=21$$

$$k=28$$

Solution

Question 8

$$1$$

$$-1$$

$$-2$$

$$2$$

Solution

Question 9

$$45$$ and $$15$$

$$48$$ and $$12$$

$$60$$ and $$20$$

$$39$$ and $$13$$

Solution

Question 10

$$0$$

$$8$$

$$-8$$

$$4$$

Solution

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