Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 22

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Question 1
1 / -0

If $$x=a, y=b$$ is the solution of the equations $$x-y=2$$ and $$x+y=4$$, then the values of $$a$$ and $$b$$ are, respectively:

A
$$3$$ and $$5$$

B
$$5$$ and $$3$$

C
$$3$$ and $$1$$

D
$$-1$$ and $$-3$$

Solution

Given, $$x-y=2$$ and $$x+y=4$$.

Adding both the equations, we get $$2x=6$$
$$\Rightarrow x=3$$

Putting the value of $$x$$ in the first equation, we get
$$3-y=2$$ $$\Rightarrow y=1$$

As stated in the question, $$x=a$$ and $$y=b$$.
Thus, $$a=3$$ and $$b=1$$
Question 2
1 / -0

The ages of Hari and Harry are in the ratio $$5:7$$. If four years from now, the ratio of their ages will be $$3:4$$, then the present age of

A
Hari is $$20$$ years and Harry is $$28$$ years.

B
Hari is $$28$$ years and Harry is $$20$$ years.

C
Hari is $$25$$ years and Harry is $$35$$ years.

D
Hari is $$35$$ years and Harry is $$25$$ years.

Solution

Suppose Hari's present age and Harry's present age are $$x$$ and $$y$$ years respectively.
Therefore, $$\dfrac{x}{y}=\dfrac{5}{7}$$
$$7x-5y=0$$
$$\Rightarrow x=\dfrac{5y}{7}$$ ...(1)

After four years, their ages will be $$x+4$$ and $$y+4$$ years respectively.
Therefore, $$\dfrac{x+4}{y+4}=\dfrac{3}{4}$$
$$4x+16=3y+12$$
$$4x-3y=-4$$ ...(2)

Putting the value of $$x$$ from (1) in equation (2), we get
$$4\left(\dfrac{5y}{7}\right)-3y=-4$$
$$20y-21y=-28$$
$$y=28$$ years

Putting the above value of $$y$$ in equation (1), we get
$$x=\dfrac{140}{7} = 20$$ years.

Hence, option A.
Question 3
1 / -0

Solve for $$x$$ and $$y$$:
$$4x+\dfrac{6}{y}=15 $$ ; $$3x-\dfrac{4}{y}=7$$

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

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

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

D
none of the above

Solution

Given equations are
$$4x+\cfrac { 6 }{ y } =15$$ ------equation$$(1)$$
$$ 3x-\cfrac { 4 }{ y } =7$$ --------equation$$(2) $$
From equation $$(1)$$ and equation $$(2)$$ can be written as
$$4xy+6=15y$$ -----equation$$(3)$$
$$3xy-4=7y$$ ------equation$$(4)$$
Multiplying equation$$(3)$$ by $$3$$ and equation$$(4)$$ by $$4$$ and subtracting
$$12xy+18=45y$$
$$12xy-16=28y$$
After solving, we get $$y=2$$
From equation$$(1)$$,
$$4x+\cfrac { 6 }{ y } =15$$ --------equation$$(1)$$
$$4x=12$$
$$x=3,y=2$$
Question 4
1 / -0

Suresh is half his father's Age. After $$20$$ years, his father's age will be one and a half times the Suresh's age. What is his father's age now?

A
$$40$$

B
$$20$$

C
$$26$$

D
$$30$$

Solution

Let Suresh's present age be $$x$$
Let Father's present age be $$y$$
Therefore applying condition no. $$1$$
$$x=\dfrac{y}{2}$$
$$y-2x=0$$ ...............................................................(i)
Also applying condition no. 2
$$\dfrac{3(x+20)}{2}=y+20$$
$$2y-3x=20$$ ......................................................(ii)
Multiplying equation (i) by $$2$$
$$2y-4x=0$$ ......................................................(iii)
Subtracting equation (iii) from equation (ii), we get
$$(2y-3x-2y+4x)=20$$
Therefore $$x=20 $$ years.
Substituting in equation (i), we get
$$y=40 $$ years
Hence, the age of the father is $$40$$ years.
Question 5
1 / -0

Solve the systems of equations:

$$\displaystyle \frac{x-2}{4} + \frac{y+1}{3} = 2 ; \ \frac{x+1}{7}+ \frac{y-3}{2} = \frac{1}{2}$$

A
$$(6, 2)$$

B
$$(2, 2)$$

C
$$(2, 3)$$

D
$$(3, 4)$$

Solution

$$\displaystyle \frac{x-2}{4} + \frac{y+1}{3} = 2$$ ......... (1)

$$\displaystyle \frac{x+1}{7} + \frac{y-3}{2} = \frac{1}{2} $$ ..... (2)

First, we need to simplify the two equations. Let's multiply both sides of equation (1) by 12 and simplify.

$$\displaystyle 12 \left ( \frac{x-2}{4} + \frac{y+1}{3} \right ) = 12 (2)$$

$$3(x-2) + 4(y+1) = 24$$
$$3x - 6 + 4y + 4 =24$$
$$3x + 4y - 2 =24$$
$$3x + 4y = 26$$

Let's multiply both sides of eq. (2) by 14

$$\displaystyle 14 \left ( \frac{x+1}{7} + \frac{y-3}{2} \right ) = 14 \left ( \frac{1}{2} \right )$$

$$2(x+1)+7(y-3)=7$$
$$ 2x + 2 + 7y - 21 = 7$$
$$2x + 7y - 19 = 7$$
$$2x+7y = 26$$

Now we have the following system to solve
$$3x+4y = 26 $$ .............. (3)
$$2x+7y = 26$$ ............... (4)

Because changing the form of their eq (3) or eq. (4) in preparation for the substitution method would produce a fractional form, let's use the elimination-by-addition method. We can start by multiplying equation (4) by -3
$$3x+4y=26$$ ............. (5)
$$-6x - 21y =-78$$ .......... (6)

No, we can replace eq. (6) with an eq. we form by multiplying eq. (5) by 2 then adding that result to eq. (6)
$$3x+4y=26$$ ............ (7)
$$-13y = -26$$ ............ (8)

From eq. (8) we can find the value of y.
$$-13 y = -26 \Rightarrow y=2$$

Now we can substitute 2 for y in eq. (7)
$$3x+4y = 26$$
$$3x + 4(2) = 26$$
$$3x= 18 \Rightarrow x=6$$

The solution set is $$(6, 2).$$
Question 6
1 / -0

The values of x and y satisfying the two equations $$32x + 33y = 31$$, $$33x + 32y = 34$$ respectively will be

A
$$- 1, 2$$

B
$$2, - 1$$

C
$$0, 0$$

D
$$2, 3$$

Solution

Multiplying $$ 32x+33y=31 $$ with $$ 33 $$, we get $$ 1056x + 1089y = 1023 $$ ---- (1)

Multiplying $$ 33x+32y=34 $$ with $$ 32 $$, we get $$ 1056x + 1024y = 1088 $$ ---- (2)

Subtracting (2) from (1),
$$1056x + 1089y - 1056x - 1024y = 1023 - 1088 $$

$$ 65y = - 65 $$
$$ y = -1 $$

Substituting $$

y = - 1 $$ in $$ 32x+33y=31 $$ , we get $$ 32x - 33 = 31
$$32x = 64$$
$$ x = 2 $$
Question 7
1 / -0

Solution of the equations $$\cfrac{x + 3}{4} + \cfrac{2y + 9}{3} = 3$$ and $$\cfrac{2x - 1}{2} - \cfrac{y + 3}{4} = 4 \cfrac{1}{2}$$ is

A
$$x = - 5, y = - 3$$

B
$$x = - 5, y = 3$$

C
$$x = 5, y = 3$$

D
$$x = 5, y = -3$$

Solution

$$\dfrac{x+3}{4}+\dfrac{2y+9}{3}=3$$

Multiplying both sides by $$12$$ we get,

$$\Rightarrow$$ $$3x+9+8y+36=36$$

$$\Rightarrow$$ $$3x+8y=-9$$ ----- ( 1 )

$$\dfrac{2x-1}{2}-\dfrac{y+3}{4}=4\dfrac{1}{2}$$

$$\dfrac{2x-1}{2}-\dfrac{y+3}{4}=\dfrac{9}{2}$$

Multiplying both sides by $$4$$, we get

$$\Rightarrow$$ $$4x-2-y-3=18$$

$$\Rightarrow$$ $$4x-y=23$$ ---- ( 2 )

Multiplying equation ( 1 ) by $$4$$,

$$12x+32y=-36$$ ----- ( 3 )

Multiplying equation ( 2 ) by $$3$$,

$$12x-3y=69$$ ---- ( 4 )

Subtracting equation ( 4 ) from ( 3 ) we get,

$$\Rightarrow$$ $$35y=-105$$

$$\therefore$$ $$y=-3$$

Substituting $$y=-3$$ in equation ( 1 )

$$3x+8(-3)=-9$$

$$3x-24=-9$$

$$3x=15$$

$$\therefore$$ $$x=5$$

$$\therefore$$ $$x=5$$ and $$y=-3$$
Question 8
1 / -0

If $$(3)^{x + y} = 81$$ and $$(81)^{x - y} = 3$$, then the values of $$x$$ and $$y$$ are

A
$$\frac{17}{8}$$,$$\frac{9}{8}$$

B
$$\frac{17}{8}$$, $$\frac{11}{8}$$

C
$$\frac{17}{8}$$, $$\frac{15}{8}$$

D
$$\frac{11}{8}$$, $$\frac{15}{8}$$

Solution

$$3^{x+y} = 81$$
$$\Rightarrow 3^{x+y} = 3^{4}$$
or $$x+y = 4$$ ...(i)
Also $$(81)^{x-y} = 3$$
$$\Rightarrow 3^{4(x-y)} = 3^{1}$$,
$$4x-4y = 1$$ ...(ii)
Solving equations (i) and (ii), we get
$$x=\frac{17}{8}$$ and $$y = \frac{15}{8}$$
Question 9
1 / -0

If $$\left (x+y,1 \right )$$ $$=$$ $$\left (3,y-x \right )$$, then $$x$$ $$=$$ , $$y$$ $$=$$

A
2, 1

B
-1, -2

C
1, 2

D
2, -1

Solution

We have, $${(x+y,1)}={(3,y-x)}$$
$$\implies x+y=3$$ and $$y-x=1$$
$$\Rightarrow x+y=3$$ and $$-x+y=1$$
Adding both equations, we get
$$2y=4$$
$$\Rightarrow y=2$$
Put the value of $$y$$ in $$x+y=3$$, we get
$$x+2=3$$
$$\Rightarrow x=1$$
So, $$\text{C}$$ is the correct option.
Question 10
1 / -0

If $$\displaystyle \frac{3}{x}- \frac{2}{y} =5$$ and $$\displaystyle \frac{4}{x} - \frac{5}{y} = 2$$, then $$\displaystyle \frac{1}{x} - \frac{1}{y} =?$$

Where $$ (x, y \neq 0)$$

A
$$1$$

B
$$-1$$

C
$$5$$

D
None of these

Solution

$$\displaystyle \frac{3}{x} - \frac{2}{y} = 5$$ ........ eq(1)

$$\displaystyle \frac{4}{x} - \frac{5}{y} = 2$$ ........ eq(2)

Multiply (1) by $$-4$$ and (2) by $$3$$ and add them

$$-\displaystyle \frac{12}{x} + \frac{8}{y} = -20$$ ...........eq(3)

$$\displaystyle \frac{12}{x} - \frac{15}{y}=6$$ .................eq(4)
adding eq(3) and eq(4)
$$\displaystyle -\frac{7}{y}= -14 \Rightarrow y = \frac{1}{2}$$

Put $$y = \displaystyle \frac{1}{2} $$ in eq. (1)

$$\displaystyle \frac{3}{x} - 4 = 5 \Rightarrow \frac{3}{x} = 9 \Rightarrow x = \frac{1}{3}$$

$$\displaystyle \frac{1}{x} = 3$$ and $$\displaystyle \frac{1}{y} = 2$$

$$\therefore$$ $$\displaystyle \frac{1}{x} - \frac{1}{y} = 3 -2 =1$$

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Pair of Linear Equations in Two Variables Test - 22

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Pair of Linear Equations in Two Variables Test - 22

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Question 1

$$3$$ and $$5$$

$$5$$ and $$3$$

$$3$$ and $$1$$

$$-1$$ and $$-3$$

Solution

Question 2

Hari is $$20$$ years and Harry is $$28$$ years.

Hari is $$28$$ years and Harry is $$20$$ years.

Hari is $$25$$ years and Harry is $$35$$ years.

Hari is $$35$$ years and Harry is $$25$$ years.

Solution

Question 3

$$x=1, y=2$$

$$x=1, y=1$$

$$x=2, y=2$$

none of the above

Solution

Question 4

$$40$$

$$20$$

$$26$$

$$30$$

Solution

Question 5

$$(6, 2)$$

$$(2, 2)$$

$$(2, 3)$$

$$(3, 4)$$

Solution

Question 6

$$- 1, 2$$

$$2, - 1$$

$$0, 0$$

$$2, 3$$

Solution

Question 7

$$x = - 5, y = - 3$$

$$x = - 5, y = 3$$

$$x = 5, y = 3$$

$$x = 5, y = -3$$

Solution

Question 8

$$\frac{17}{8}$$,$$\frac{9}{8}$$

$$\frac{17}{8}$$, $$\frac{11}{8}$$

$$\frac{17}{8}$$, $$\frac{15}{8}$$

$$\frac{11}{8}$$, $$\frac{15}{8}$$

Solution

Question 9

2, 1

-1, -2

1, 2

2, -1

Solution

Question 10

$$1$$

$$-1$$

$$5$$

None of these

Solution

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