Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

A certain two digits number is equal to five times the sum of its digits. If $$9$$ were added to the number, its digits would be reversed. The sum of the digits of the number is:

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

Let the ones and tens digit of the number be $$y$$ and $$x.$$

According to question, we have
$$10x+y=5\left( x+y \right) $$
$$\Rightarrow 5x-4y=0\quad\quad\quad\dots(i)$$

And,
$$10x+y+9=10y+x$$
$$\Rightarrow 9x-9y=-9$$
$$\Rightarrow x-y=-1\quad\quad\quad\dots(ii)$$

Multiply $$(ii)$$ by $$-4$$ we get,
$$\Rightarrow -4x+4y=4\quad\quad\quad\dots(iii)$$

Add equations $$(i)$$ and $$(iii),$$
$$\left( {5x - 4y} \right) + \left( { - 4x + 4y} \right) = 0 + 4$$
$$\Rightarrow 5x - 4x - 4y + 4y = 4$$
$$\Rightarrow x = 4$$

Substitute $$x=4$$ in equation $$(ii),$$
$$4 - y = - 1$$
$$\Rightarrow - y = - 5$$
$$\Rightarrow y=5$$

Thus, the sum of the digits of the number is equal to $$x+y=9.$$.
Question 2
1 / -0

Solve for $$x$$ and $$y $$
$$\displaystyle \frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5};\, \, \frac{5}{3x+2y}+\frac{1}{3x-2y}=2$$

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

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

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

D
$$\displaystyle x=\frac{1}{5},y=\frac{1}{5}$$

Solution
Question 3
1 / -0

The expression $$ax + b$$ is equal to $$13$$ when $$x$$ is $$5$$ and $$ax + b$$ is equal to $$29$$ when $$x$$ is $$13$$. The value of expression when $$x$$ is $$0.5$$

A
$$2.9$$

B
$$\dfrac {29}{5}\times 13$$

C
$$4$$

D
$$\dfrac {29\times 5}{13}$$

Solution

$$5a+ b= 13$$ .......eq1 & $$13a+ b= 29$$ ......eq2
Subtract eq1 from eq2, we get
$$a=2 $$
$$b = 3$$
when $$x=0.5\times 2+3=1+3=4$$.
Question 4
1 / -0

When one is added to each of two given numbers, their ratio becomes $$3 : 4$$ and when $$5$$ is subtracted from each, the ratio becomes $$7:10$$. The numbers are

A
$$8, 11$$

B
$$11, 15$$

C
$$26, 35$$

D
$$27, 36$$

Solution

Let the two numbers be $$x$$ and $$y$$.
According to the given condition,
$$\dfrac {x+1}{y+1}=\dfrac {3}{4}$$
$$\Rightarrow 4x+4=3y+3$$
$$4x-3y=-1$$ ........$$(i)$$
Also,
$$\dfrac {x-5}{y-5}=\dfrac {7}{10}$$
$$\Rightarrow 10x-50=7y-35$$
$$\Rightarrow 10x-7y=15$$ .......$$(ii)$$
Multiply equation $$(1)$$ and $$(2)$$ by $$7$$ and $$3$$ respectively.
$$[4x -3y = -1] \times 7$$
$$[10x -7y = 15] \times 3$$
$$28x -21y =-7$$.............$$(4)$$
$$30x-21y=45$$...............$$(5)$$
Subtract equation $$(4)$$ from $$(5)$$
$$2x = 52$$
$$\Rightarrow x=26$$
Putting the value of $$x$$ in $$(i)$$
$$4(26) -3y =-1$$
$$-3y = -105$$
$$y = 35$$
$$\therefore$$ Numbers are $$26, 35$$.
Question 5
1 / -0

Solve graphically: $$5x-6y+30=0$$; $$5x+4y-20=0$$. Also, find the vertices of the triangle formed by the above two lines and $$x$$-axis.

A
$$(3,5),(4,2),(6,9)$$

B
$$(0,5),(-6,0),(4,0)$$

C
$$(0,2),(7,2),(0,4)$$

D
$$(2,3),(1,7),(3,9)$$

Solution

For $$5x - 6y + 30 = 0$$:
x 0 -6 6
y 5 0 10
For $$x - y - 1 = 0$$:
x 0 4 -4
y 5 0 10
As the graph shows, vertices of triangle are $$(0, 5), (-6, 0) \& (4, 0)$$.
Question 6
1 / -0

In a zoo there are some pigeons and some rabbits. If the total number of heads is $$300$$ and the total number of legs is $$750$$, then how many pigeons are there?

A
$$150$$

B
$$175$$

C
$$200$$

D
$$225$$

Solution

Let, the total number of pigeons be $$p$$ and the total number of rabbits be $$r$$
Both the pigeons and rabbits has one head each
so, $$p+r=300$$ ...........................................(1)

Each pigeon has two legs and each rabbit has four legs
so, $$2p+4r=750$$................................................(2)

Multiply by (1) by $$4$$
Then $$4p+ 4r=1200$$................................................(3)

By subtraction (2) from (3)
$$(4p+ 4r)-(2p+4r)=1200-750$$
$$\Rightarrow 4p+ 4r-2p-4r=450$$
$$\Rightarrow 2p=450$$
$$\Rightarrow p=225$$

Hence, the number of pigeons in the zoo is $$225$$.
Question 7
1 / -0

The total salary of $$15$$ men and $$8$$ women is $$Rs.\ 3050$$. The difference of salaries of $$5$$ women and $$3$$ men is $$Rs.\ 50$$. Find the sum of the salaries of $$3$$ men and $$3$$ women

A
$$Rs.\ 900$$

B
$$Rs.\ 850$$

C
$$Rs.\ 750$$

D
$$Rs.\ 1000$$

Solution

Let the salary of $$1$$ men be $$x$$ and the salary of $$1$$ woman be $$y.$$

As per the question,
$$15x + 8y =3050\quad\quad...(1)$$
$$3x-5y=-50\quad\quad...(2)$$

Multiply equation $$(2)$$ by $$5$$,
$$15x -25y = -250\quad\quad...(3)$$

Subtract equation $$(3)$$ from equation $$(1),$$
$$\left( {15x + 8y} \right) - \left( {15x - 25y} \right) = 3050 - \left( { - 250} \right)$$
$$\Rightarrow 15x + 8y - 15x + 25y = 3050 + 250$$
$$\Rightarrow 33y = 3300$$
$$\Rightarrow y= 100$$

Substitute the value of $$y$$ in equation $$(2),$$
$$3x-500 =-50$$
$$\Rightarrow 3x= 450$$
$$\Rightarrow x= 150$$

Now, total salary of $$3$$ men and $$3$$ women will be,
$$3x + 3y= 3 \times 150 + 3 \times 100$$
$$=450 + 300$$
$$ =Rs.\ 750$$
Question 8
1 / -0

A students walks from his house at less 4 km per hour and reaches his school late by 5 minutes. If his speed has been increase by 5 km per hour then he would have reached 10 minutes early. The distance of the school from his house is

A
$$ \displaystyle \frac{3}{5}km $$

B
$$5$$km

C
$$6$$km

D
$$4$$km

Solution

Let the actual time be 't'
and distance be x
Now according to first condition, we have
$$\dfrac{x}{4}=t+\dfrac{5}{60}$$
And according to Second condition, we have
$$\dfrac{x}{5}=t-\dfrac{10}{60}$$
$$60x=4t+20$$
$$60x=5t-50$$
Subtracting equation 1 from 2, we have
$$0=t-70$$
$$\therefore t=70$$
Putting the value of t we get $$x=5km$$
Question 9
1 / -0

Solve for $$x$$ and $$y$$, if $$2x+ 3y= 8$$ and $$x+2y=5$$.

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

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

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

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

Solution

$$2x+3y=8$$ .............(1)
and $$x+2y=5$$ .......(2)
From (1), we get $$x=\frac {8-3y}{2}$$ .......... (3)
From (2), we get $$x=5-2y$$ .......... (4)
$$\therefore \frac {8x-3y}{2}=5-2y$$
or $$8x -3y= 10 -4y$$ or $$3y= 10 -4y$$ or $$y= 2$$
Using $$y = 2$$ in (4), we get $$x = 5 -4 = 1$$. Therefore,
$$x=1,y=2$$
Question 10
1 / -0

The sum of the digits of a two-digit number is $$5$$. The digit obtained by increasing the digit in tens' place by unity is one-eighth of the number. Then the number is

A
less than $$30$$

B
lies between $$30$$ and $$40$$

C
more than $$37$$

D
lies between $$40$$ and $$50$$

Solution

$$x+y=5$$
and $$x+1=\dfrac {1}{8}(10x+y)$$ .........(1)
or $$8x+8=10x+y$$
or $$2x+y=8$$ .........(2)
Subtract equation (1) from equation (2)
$$2x+y=8$$
$$+ x+y=5$$
$$\underline {(-) (-) (-)}$$
$$x=3$$
$$\therefore y=2$$
Number is $$32$$.

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Pair of Linear Equations in Two Variables Test - 48

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Pair of Linear Equations in Two Variables Test - 48

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Question 1

$$6$$

$$7$$

$$8$$

$$9$$

Solution

Question 2

$$x = 1, y = 3$$

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

$$x = 1, y = 1$$

$$\displaystyle x=\frac{1}{5},y=\frac{1}{5}$$

Solution

Question 3

$$2.9$$

$$\dfrac {29}{5}\times 13$$

$$4$$

$$\dfrac {29\times 5}{13}$$

Solution

Question 4

$$8, 11$$

$$11, 15$$

$$26, 35$$

$$27, 36$$

Solution

Question 5

$$(3,5),(4,2),(6,9)$$

$$(0,5),(-6,0),(4,0)$$

$$(0,2),(7,2),(0,4)$$

$$(2,3),(1,7),(3,9)$$

Solution

Question 6

$$150$$

$$175$$

$$200$$

$$225$$

Solution

Question 7

$$Rs.\ 900$$

$$Rs.\ 850$$

$$Rs.\ 750$$

$$Rs.\ 1000$$

Solution

Question 8

$$ \displaystyle \frac{3}{5}km $$

$$5$$km

$$6$$km

$$4$$km

Solution

Question 9

$$x=1, y=2$$

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

$$x=-2, y=2$$

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

Solution

Question 10

less than $$30$$

lies between $$30$$ and $$40$$

more than $$37$$

lies between $$40$$ and $$50$$

Solution

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