Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

10 years ago the age of the father was 5 times that of the son. 20 years hence the age of the father will be twice that of the son. The present age of the father (in years) is

A
$$40$$

B
$$45$$

C
$$60$$

D
$$70$$

Solution

Let the present age of father be $$x$$ years. and that of son be $$y$$ years
10 years ago the age of the father was 5 times that of the son.
$$\therefore$$ $$x-10=5(y-10)$$
$$\therefore$$ $$x-5y=-40$$ ...(1)
20 years hence, the age of the father will be twice that of the son.
$$\therefore$$ $$x+20=2(y+20)$$
$$\therefore$$ $$x-2y=20$$ ...(2)
Subtracting (1) and (2) we get ,
$$y=20$$ and $$x=60$$
$$\therefore$$ The present age of father is $$60$$ years
Question 2
1 / -0

The graphs of $$2x+3y-6=0, 4x-3y-6=0, x=2$$ and $$ \displaystyle y=\dfrac{2}{3} $$ intersect in

A
four points

B
one point

C
in no points

D
infinite number of points

Solution

Solving $$2x + 3y - 6 = 0$$ and $$4x - 3y - 6 = 0$$ we get
$$x = 2$$ and $$\displaystyle y=\dfrac{2}{3}$$
Hence from all the four given equations the
value of $$x$$ is $$2$$ and $$y$$ is $$\displaystyle \frac{2}{3}$$
Therefore the graphs of all the four lines intersect
in one point
Question 3
1 / -0

What values of x and y satisfies the equations $$\displaystyle \frac{x}{6}+\frac{y}{15}=4$$ and $$\displaystyle \frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$$

A
$$(6,15)$$

B
$$(18,15)$$

C
$$(15,18)$$

D
$$(12,30)$$

Solution

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $$\frac { x }{ 3 } -\frac { y }{ 12 } =4\frac { 3 }{ 4 }$$ or $$4x-y=57$$ and write it as

$$y=4x-57...........(1)$$

Substitute the value of $$y$$ in Equation $$\frac { x }{ 6 } +\frac { y }{ 15 } =4$$ or $$15x+6y=360$$. We get

$$15x+6(4x-57)=360$$

i.e. $$15x+24x-342=360$$

i.e. $$39x=702$$

i.e. $$x=18$$

Therefore, $$x=18$$

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

$$y=72-57$$

i.e. $$y=15$$

Hence, the solution is $$x = 18, y =15$$.
Question 4
1 / -0

The ratio of the ages of A and B ten years before was $$3 : 5$$. The ratio of their present ages is $$2 : 3$$ . Their ages are respectively

A
$$30,50$$

B
$$20,30$$

C
$$40,60$$

D
$$16,24$$

Solution

Let the present ages of $$A$$ and $$B$$ are $$x$$ and $$y$$ years respectively
$$\displaystyle \therefore \dfrac{x-10}{y-10}=\dfrac{3}{5}$$ .....(i)
or $$5x -50=3y-30$$ ......(ii)
and $$\displaystyle \dfrac{x}{y}=\dfrac{2}{3} $$or $$3x=2y$$ or $$y=\dfrac{3x}{2}$$
putting the value of y in equation (i) we get
$$5x-50=$$$$\displaystyle \dfrac{9x}{2}-30$$
$$10x-100=9x-60$$
or $$x=40$$ $$\displaystyle \therefore y=\dfrac{3\times 40}{2}=60$$
Question 5
1 / -0

In the system of equations $$\dfrac {12}{x+y}+\dfrac {8}{x-y}=8$$ and $$\dfrac {27}{x+y}-\dfrac {12}{x-y}=3$$, the values of $$x$$ and $$y$$ will be

A
$$\dfrac {5}{3}$$ and $$\dfrac {1}{3}$$

B
$$\dfrac {5}{2}$$ and $$\dfrac {1}{2}$$

C
2 and $$\dfrac {1}{3}$$

D
$$\dfrac {5}{4}$$ and $$\dfrac {1}{4}$$

Solution

If $$\dfrac {1}{x+y}=a$$ and $$\dfrac {1}{x-y}=b$$, then the equations will be:
$$12a + 8b = 8$$........ (i)
and $$27a -12b = 3$$ ........... (ii)
On multiplying equation (i) by 3 and equation (ii) by 2 and adding the two equations,
$$3(12a + 8b) + 2(27a -12b) = 8\times 3+3 \times 2$$
$$\therefore 36a + 24b + 54a -24b = 24 + 6$$
$$\therefore 90a=30\Rightarrow a=\dfrac {30}{90}=\dfrac {1}{33}$$
On substituting this value of a in equation (i),
$$12\times \dfrac {1}{3}+8b=8\therefore 4+8b=8$$
$$\Rightarrow 8b=4\Rightarrow b=\dfrac {1}{2}$$
$$\because \dfrac {1}{x+y}=a=\dfrac {1}{3}\Rightarrow x+y=3$$ ......(iii)
and $$\dfrac {1}{x+y}=b=\dfrac {1}{2}\Rightarrow x-y=2$$ ......(iv)
On adding equation (iii) and (iv),
$$2x=5\Rightarrow x=\dfrac {5}{2}$$
On substituting this value of x in equation (iii),
$$\dfrac {5}{2}+y=3\Rightarrow y=3-\dfrac {5}{2}$$
$$\Rightarrow y=\dfrac {1}{2}$$
$$\therefore x=\dfrac {5}{2}$$ and $$y=\dfrac {1}{2}$$
Question 6
1 / -0

Solve equations using substitution method: $$x-y = 3$$ and $$x + y = 0$$

A
$$\dfrac{3}{2}$$ and $$1$$

B
$$\dfrac{-3}{2}$$ and $$-1$$

C
$$\dfrac{3}{2}$$ and $$-\dfrac{3}{2}$$

D
$$0$$ and $$\dfrac{3}{2}$$

Solution

$$x- y = 3\dots (1)$$
$$x + y = 0\dots (2)$$
From equation $$(1)$$ :
$$y = x - 3$$
Substitute the value of $$y$$ in equation $$(2)$$ :
$$x + y = 0$$
$$x + x- 3 = 0$$
$$2x = 3$$
$$x=\dfrac{3}{2}$$

Now, Substitute $$x = \dfrac{3}{2}$$ in equation $$(1)$$:
$$x- y = 3$$
$$\dfrac{3}{2}- y = 3$$
$$y = \dfrac{3}{2}- 3$$
$$y = -\dfrac{3}{2}$$
Therefore the solution is:
$$x =\dfrac{3}{2}$$ and $$y = -\dfrac{3}{2}$$
Question 7
1 / -0

Find the value of x and y using substitution method:
$$5x - 6y = 2$$ and $$6x - 5y = 9$$

A
$$4$$ and $$-3$$

B
$$-3$$ and $$-4$$

C
$$-4$$ and $$-3$$

D
$$4$$ and $$3$$

Solution

$$5x -6y = 2$$ ------- $$(1)$$
$$6x - 5y = 9$$ ------- $$(2)$$

From equation $$(1)$$:

$$\Rightarrow 6y = 5x- 2$$

$$\Rightarrow y =\dfrac{5x-2}{6}$$

Substitute the value of $$y$$ in equation $$(2)$$ :

$$\Rightarrow 6x - 5y = 9$$

$$\Rightarrow6x- 5(\dfrac{5x-2}{6}) = 9$$

$$\Rightarrow36x -25x + 10 = 54$$

$$\Rightarrow11x = 54- 10$$

$$\Rightarrow x = \dfrac{44}{11} = 4$$

Now, Substitute $$x = 4$$ in equation $$(1)$$

$$\Rightarrow5x - 6y = 2$$

$$\Rightarrow5(4)-6y = 2$$

$$\Rightarrow 20 - 2 = 6y$$

$$\Rightarrow18 = 6y$$

$$\Rightarrow y =\dfrac{18}{6}=3$$

$$\therefore$$ the solution is $$x = 4$$ and $$y = 3$$
Question 8
1 / -0

Solve equations using substitution method:
$$2x -y = 3$$ and $$4x + y = 3$$

A
$$-1$$ and $$1$$

B
$$-2$$ and $$1$$

C
$$1$$ and $$-1$$

D
$$-1$$ and $$-1$$

Solution

$$2x- y = 3$$ ------- (1)
$$4x + y = 3$$ ------ (2)
From equation 1 :
$$y = 2x - 3$$
Substitute the value of $$y$$ in equation 2 :
$$4x + y = 3$$
$$4x + 2x -3 = 3$$
$$6x = 3 + 3$$
$$x = \dfrac{6}{6} = 1$$
Now, Substitute $$x = 1$$ in equation 1 :
$$2x -y = 3$$
$$2 (1)- y = 3$$
$$2- 3 = y$$
$$y = -1$$
Therefore the solution is: $$x = 1$$ and $$y = -1$$
Question 9
1 / -0

Solve equations using substitution method:
$$2x + y = 5$$ and $$2x- 3y = 3$$

A
$$\frac{1}{4}$$ and $$\frac{-1}{2}$$

B
$$\frac{1}{4}$$ and $$\frac{1}{2}$$

C
$$\frac{-9}{4}$$ and $$\frac{-1}{2}$$

D
$$\frac{9}{4}$$ and $$\frac{1}{2}$$

Solution

$$2x + y = 5$$ ------- (1)
$$2x- 3y = 3$$------ (2)
From equation 1 :
$$y = 5 -2x$$
Substitute the value of y in equation 2 :
$$2x -3(5- 2x) = 3$$
$$2x -15 + 6x = 3$$
$$8x = 3 + 15$$
$$x = \dfrac{18}{8} = \dfrac{9}{4}$$
Now, Substitute $$x = \dfrac{9}{4}$$ in equation 1 :
$$2(\dfrac{9}{4}) + y = 5$$
$$18 + 4y = 20$$
$$4y = 20 -18$$
$$y = \dfrac{2}{4}$$
$$y = \dfrac{1}{2}$$
Therefore the solution is: $$x = \dfrac{9}{4}$$ and $$y = \dfrac{1}{2}$$
Question 10
1 / -0

If $$2$$ tables and $$3$$ chairs cost $$Rs. 4500$$ and $$3$$ tables and $$2$$ chairs cost $$Rs. 5000$$, then how much does a table cost?

A
Rs. $$1100$$

B
Rs. $$1200$$

C
Rs. $$1150$$

D
Rs. $$1300$$

Solution

Let the cost of table and chair is $$x$$ and $$y$$.
Then, $$2x+3y=4500$$ ....(1)
and $$3x+2y=5000$$ ...(2)
Add eq(1) and eq(2)
$$5x+5y=9500$$
$$x+y = 1900$$ ....(3)
Subtract eq(1) from eq(2)
$$x-y= 500 $$ ....(4)
Solving eq(3) and eq(4) we get
$$2x=2400$$
$$x =1200$$

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 49

Result

Pair of Linear Equations in Two Variables Test - 49

Score

-

Rank

-

SHARING IS CARING

Question 1

$$40$$

$$45$$

$$60$$

$$70$$

Solution

Question 2

four points

one point

in no points

infinite number of points

Solution

Question 3

$$(6,15)$$

$$(18,15)$$

$$(15,18)$$

$$(12,30)$$

Solution

Question 4

$$30,50$$

$$20,30$$

$$40,60$$

$$16,24$$

Solution

Question 5

$$\dfrac {5}{3}$$ and $$\dfrac {1}{3}$$

$$\dfrac {5}{2}$$ and $$\dfrac {1}{2}$$

2 and $$\dfrac {1}{3}$$

$$\dfrac {5}{4}$$ and $$\dfrac {1}{4}$$

Solution

Question 6

$$\dfrac{3}{2}$$ and $$1$$

$$\dfrac{-3}{2}$$ and $$-1$$

$$\dfrac{3}{2}$$ and $$-\dfrac{3}{2}$$

$$0$$ and $$\dfrac{3}{2}$$

Solution

Question 7

$$4$$ and $$-3$$

$$-3$$ and $$-4$$

$$-4$$ and $$-3$$

$$4$$ and $$3$$

Solution

Question 8

$$-1$$ and $$1$$

$$-2$$ and $$1$$

$$1$$ and $$-1$$

$$-1$$ and $$-1$$

Solution

Question 9

$$\frac{1}{4}$$ and $$\frac{-1}{2}$$

$$\frac{1}{4}$$ and $$\frac{1}{2}$$

$$\frac{-9}{4}$$ and $$\frac{-1}{2}$$

$$\frac{9}{4}$$ and $$\frac{1}{2}$$

Solution

Question 10

Rs. $$1100$$

Rs. $$1200$$

Rs. $$1150$$

Rs. $$1300$$

Solution

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