Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle \frac{a}{4}\, -\, \frac{b}{3}\, =\, 0\,;\, \frac{3a\, +\, 8}{5}\, =\, \frac{2b\, -\, 1}{2}$$

A
$$a = -4.5, b = 12$$

B
$$a = 4, b = -5$$

C
$$a = 14, b = 10.5$$

D
$$a = 12, b = 11.5$$

Solution

$$\dfrac { a }{ 4 } -\dfrac { b }{ 3 } =0\\ \Rightarrow a =\dfrac { 4b }{ 3 } $$
Substituting for '$$a$$' in the other equation.
$$\dfrac { 3a+8 }{ 5 } =\dfrac { 2b-1 }{ 2 } \\ 6a+16 =10b - 5\\ 6*\dfrac { 4b }{ 3 } + 16 = 10b - 5\\ 8b +16 = 10b - 5\\ 21 = 2b\\ b = 10.5$$

So, $$a = \dfrac { 4\times10.5 }{ 3 } = 14$$

$$\therefore a = 14, b = 10.5$$

Option $$C$$
Question 2
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle \frac{3}{a}\, +\, \frac{4}{b}\, =\, 2\,;\, \frac{9}{a}\, -\, \frac{4}{b}\, =\, 2$$

A
$$a= 3, b = 4$$

B
$$a = 1, b = -2$$

C
$$a = 5, b = -3$$

D
$$a = 0, b = 6$$

Solution

$$\displaystyle \frac{3}{a}\, +\, \frac{4}{b}\, =\, 2\,;\, \frac{9}{a}\, -\, \frac{4}{b}\, =\, 2$$
Let $$\dfrac{1}{a} = x$$ and $$\dfrac{1}{b} = y$$
So, $$3x + 4y = 2$$ and $$9x - 4y = 2$$
Adding the above equations, we get
$$12x = 4$$
$$x = \dfrac{1}{3}$$

Substituting $$x = \dfrac{1}{3}$$ in either equation, we get $$y =\dfrac{1}{4}$$
So, $$a = \dfrac{1}{x} = 3$$ and $$b = \dfrac{1}{y} = 4$$

$$a = 3, b = 4$$
Option $$A$$.
Question 3
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle\, 4x\, +\, \frac{3}{y}\, =\, 1\,; 3x\, -\, \frac{2}{y}\, =\, 5$$

A
$$x=-6,y=7$$

B
$$x=-3,y=1$$

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

D
$$x=2,y=0$$

Solution

On putting $$\dfrac1y=Y$$, we get

$$4x+3Y=1$$ ...(i)
$$3x-2Y=5$$ ...(ii)
On multiplying (i) by 3 and (ii) by 4 and subtracting, we get
$$12x+9Y=3$$
$$\underline {\underset {-}12x\underset {+}{-}8Y=\underset{-}20}$$
$$17Y=-17$$
$$\therefore Y=-1$$
On putting $$Y=-1$$ in (i), we get
$$4x+3(-1)=1\Rightarrow 4x=4\Rightarrow x=1$$
$$\therefore x=1,y=\dfrac1Y=\dfrac1{-1}=-1$$
Question 4
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle\, y\, -\, \frac{3}{x}\, =\, 8\, ;\, 2y\, +\, \frac{7}{x}\, =\, 3$$

A
$$x = - 7$$ and $$y = 3$$

B
$$x = 0$$ and $$y = 3$$

C
$$x = - 5$$ and $$y = 9$$

D
$$x = - 1$$ and $$y = 5$$

Solution

The equation $$y-\frac { 3 }{ x } =8$$ can be solved as:

$$y-\frac { 3 }{ x } =8\\ \Rightarrow \frac { xy-3 }{ x } =8\\ \Rightarrow xy-3=8x\quad .........(1)$$

The equation $$2y+\frac { 7 }{ x } =3$$ can be solved as:

$$2y+\frac { 7 }{ x }=3\\ \Rightarrow \frac { 2xy+7 }{ x } =3\\ \Rightarrow 2xy+7=3x\quad .........(2)$$

Multiply the equation 1 by $$2$$. Then we get:

$$2xy-6=16x.........(3)$$

Subtract equation 3 from equation 1 to eliminate $$xy$$, because the coefficients of $$xy$$ are same. So, we get

$$(2xy-2xy)+(7+6)=3x-16x$$

i.e. $$-13x=13$$

i.e. $$x=-1$$

Substituting this value of $$x$$ in the equation 1, we get

$$-y-3=-8$$

i.e. $$-y=-5$$

i.e. $$y=5$$

Hence, the solution of the equations is $$x=-1,y=5$$.
Question 5
1 / -0

Solve the following pairs of equations, graphically:
$$x = 0$$ and $$x + 3y = 6$$

A
$$(0,2)$$

B
$$(2,0)$$

C
$$(0,-2)$$

D
$$(-2,0)$$

Solution

The solution set of equation $$x+3y=6$$ is as follows:

If $$x=0$$ then

$$0+3y=6\\ \Rightarrow 3y=6\\ \Rightarrow y=2$$

If $$x=3$$ then

$$3+3y=6\\ \Rightarrow 3y=6-3=3\\ \Rightarrow y=1$$

The required points are $$(0,2),(3,1)$$.

Another equation of line given is $$x=0$$.

Plot all the above points in the graph. Since the point of intersection in the graph is $$(0,2)$$.

Hence the solution is $$x=0,y=2$$.
Question 6
1 / -0

Find two numbers such that twice of the first added to the second gives $$21$$, and twice the second added to the first gives $$27$$.

A
$$5$$ and $$11$$

B
$$9$$ and $$16$$

C
$$3$$ and $$18$$

D
$$3$$ and $$17$$

Solution

Let the first number is $$x$$ and second is $$y$$
Given twice of first added with second gives $$21$$
$$\therefore 2x+y=21$$ ...$$(1)$$
And given twice of second added with first gives $$27$$
$$\therefore x+2y=27$$ ...$$(2)$$
Multiply $$(1)$$ by $$2$$
Then $$4x+2y=42$$ ...$$(3)$$
Subtract $$(3)$$ with $$(2)$$
Then $$3x=15$$
Or $$x=5$$
Put the value of $$x=5$$ in $$(1)$$
Then $$10+y=21$$
Or $$y=21-10$$
Or $$y=11$$
Then first number is $$5$$ and second is $$11$$.
Question 7
1 / -0

A man's age is three times that of his son and in twelve years he will be twice as old as his son would be. What are their present ages?

A
Present age of man is $$60$$ years and

Present age of son is $$20$$ years

B
Present age of man is $$45$$ years and

Present age of son is $$15$$ years

C
Present age of man is $$54$$ years and

Present age of son is $$18$$ years

D
Present age of man is $$36$$ years and

Present age of son is $$12$$ years

Solution

Let the present age of son be $$x$$ years
As per question the present age of father is three times
then present age of father is $$3x$$
As per question after $$12$$ years the age of father will be twice the age of son
$$3x+12=2(x+12)$$
$$\Rightarrow 3x+12=2x+24$$
$$\Rightarrow 3x-2x=24-12$$
$$\Rightarrow x=12$$
Then present age of son is $$12$$ years
So present age of father $$=3\times12=36$$ years
Question 8
1 / -0

Find two numbers whose sum is $$15$$ and difference is $$3$$.

A
$$13$$ and $$2$$

B
$$7$$ and $$8$$

C
$$9$$ and $$6$$

D
$$3$$ and $$12$$

Solution

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

It is given that the sum of the numbers is $$15$$ that is

$$x+y=15..........(1)$$

Also the difference of the numbers is $$3$$ that is

$$x-y=3..........(2)$$.

Add equations 1 and 2 to eliminate $$y$$, because the coefficients of $$y$$ are opposite. So, we get

$$(x+x)+(y-y)=15+3$$

i.e. $$2x=18$$

$$\implies x=9$$

Substituting this value of $$x$$ in (1), we get

$$9+y=15$$

i.e. $$y=15-9=6$$

Hence, the two numbers are $$9$$ and $$6$$.
Question 9
1 / -0

Find a fraction which reduces to $$\displaystyle \frac{2}{3}$$ if the numerator and the denominator are each increased by $$1$$, and reduces to $$\displaystyle \frac{3}{5}$$ if the numerator and the denominator are each decreased by $$2$$.

A
$$\displaystyle \frac{13}{12}$$

B
$$\displaystyle \frac{11}{17}$$

C
$$\displaystyle \frac{9}{4}$$

D
$$\displaystyle \frac{13}{3}$$

Solution

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

Given, if the numerator and the denominator are each increased by $$ 1 $$, then fraction $$ = \dfrac {x + 1}{y + 1} = \dfrac {2}{3} $$
$$ \Longrightarrow 3x + 3 = 2y + 2 $$
$$ \Longrightarrow 3x - 2y = -1 $$ ....................... $$(1)$$

Also, if the numerator and the denominator are each decreased by $$ 2 $$, then fraction $$ = \dfrac {x - 2}{y - 2} = \dfrac {3}{5} $$
$$ \Longrightarrow 5x - 10 = 3y - 6 $$
$$ \Longrightarrow 5x - 3y = 4 $$ ....................... $$(2)$$

Multiplying equation $$ (1) $$ with $$ 5 $$ we get, $$ 15x - 10y = -5 $$ ...$$(3)$$

Multiplying equation $$ (2) $$ with $$ 3 $$ we get, $$ 15x - 9y = 12 $$ ....$$(4)$$
Subtracting equations $$ (3) $$ from $$ (4) $$, we get,
$$ -9y-(-10y) = 12-(-5) $$
$$y=17$$
Substituting $$ y = 17 $$ in the equation $$ (2) $$, we get,
$$ 5x - 3(17) = 4$$
$$ 5x = 55$$
$$\Longrightarrow x = 11 $$

Hence, the fraction is $$ \dfrac {11}{17}. $$
Question 10
1 / -0

Find two numbers, which differ by $$7$$, such that twice the greater added to five times the smaller gives $$42$$.

A
$$19$$ and $$12$$

B
$$12$$ and $$5$$

C
$$11$$ and $$4$$

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

Solution

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

It is given that two numbers differ by $$7$$

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

Also, twice the greater number added to five times the smaller gives $$42$$

$$\Rightarrow 2x+5y=42$$ $$...(2)$$

Multiply equation $$(1)$$ by $$2$$

$$2x-2y=14$$ $$...(3)$$

Let us subtract equation $$(3)$$ from equation $$(2)$$ to eliminate $$x$$, because the coefficients of $$x$$ are the same.

$$(2x+5y)-(2x-2y)=42-14$$

$$\Rightarrow (2x-2x)+(5y+2y)=28$$

$$\Rightarrow 7y=28$$

$$\Rightarrow y=4$$

Substituting this value of $$y$$ in $$(1),$$

we get $$x-4=7$$

i.e. $$x=11$$

Hence, the two numbers are $$11$$ and $$4$$

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Pair of Linear Equations in Two Variables Test - 27

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Pair of Linear Equations in Two Variables Test - 27

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Question 1

$$a = -4.5, b = 12$$

$$a = 4, b = -5$$

$$a = 14, b = 10.5$$

$$a = 12, b = 11.5$$

Solution

Question 2

$$a= 3, b = 4$$

$$a = 1, b = -2$$

$$a = 5, b = -3$$

$$a = 0, b = 6$$

Solution

Question 3

$$x=-6,y=7$$

$$x=-3,y=1$$

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

$$x=2,y=0$$

Solution

Question 4

$$x = - 7$$ and $$y = 3$$

$$x = 0$$ and $$y = 3$$

$$x = - 5$$ and $$y = 9$$

$$x = - 1$$ and $$y = 5$$

Solution

Question 5

$$(0,2)$$

$$(2,0)$$

$$(0,-2)$$

$$(-2,0)$$

Solution

Question 6

$$5$$ and $$11$$

$$9$$ and $$16$$

$$3$$ and $$18$$

$$3$$ and $$17$$

Solution

Question 7

Present age of man is $$60$$ years and Present age of son is $$20$$ years

Present age of man is $$45$$ years and Present age of son is $$15$$ years

Present age of man is $$54$$ years and Present age of son is $$18$$ years

Present age of man is $$36$$ years and Present age of son is $$12$$ years

Solution

Question 8

$$13$$ and $$2$$

$$7$$ and $$8$$

$$9$$ and $$6$$

$$3$$ and $$12$$

Solution

Question 9

$$\displaystyle \frac{13}{12}$$

$$\displaystyle \frac{11}{17}$$

$$\displaystyle \frac{9}{4}$$

$$\displaystyle \frac{13}{3}$$

Solution

Question 10

$$19$$ and $$12$$

$$12$$ and $$5$$

$$11$$ and $$4$$

$$16$$ and $$9$$

Solution

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Present age of man is $$60$$ years and

Present age of son is $$20$$ years

Present age of man is $$45$$ years and

Present age of son is $$15$$ years

Present age of man is $$54$$ years and

Present age of son is $$18$$ years

Present age of man is $$36$$ years and

Present age of son is $$12$$ years