Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 58

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Question 1
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When a bucket is half full, the weight of the bucket and the water is $$10\text{ kg}$$. When the bucket is two-thirds full, the total weight is $$11\text{ kg}$$. What is the total weight, in kg, when the bucket is completely full?

A
$$12 \text{ kg}$$

B
$$12\displaystyle\frac{1}{2} \text{ kg}$$

C
$$12\displaystyle\frac{2}{3}\text{ kg}$$

D
$$13\text{ kg}$$

Solution

Let the weight of the empty bucket be $$B$$ kg and total weight of water it can accommodate be $$W$$ kg.

$$\therefore B + \cfrac{W}{2} = 10 \ \ ...(1)$$

Also, $$B + \cfrac{2W}{3} = 11 \ \ ...(2)$$

Subtracting (1) from (2), we have

$$\cfrac{2W}{3} - \cfrac{W}{2} = 1$$

$$\therefore \cfrac{4W - 3W}{6} = 1$$ or $$W = 6$$
Substituting $$W=6 $$ in (1)

$$\implies B+\dfrac{6}{2}=10$$
$$\therefore B = 7$$

Total weight when completely filled $$= B + W = 13$$ kg
Question 2
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If $$2x + 3y = 24$$ and $$2x - 3y = 12$$, then the value of $$xy$$ is ___________.

A
$$10$$

B
$$12$$

C
$$18$$

D
$$14$$

Solution

The given equations are,
$$ 2x + 3y = 24$$ ....(1)
$$ 2x - 3y = 12$$ ....(2)

Adding Equations (1) and (2), we get
$$ 2x + 3y+2x-3y = 24+12$$
$$\Rightarrow 4x = 36$$
$$\Rightarrow x = 9$$
Putting $$ x = 9$$ in Equation (1), we get
$$ 2\times9 + 3y = 24$$
$$\Rightarrow 18 + 3y = 24$$
$$\Rightarrow 3y = 6$$
$$\Rightarrow y =2$$
Thus $$ xy = 9\times2 = 18$$
Question 3
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Choose the correct answer from the alternatives given.
If $$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{x \, + \, 2y \, - \, 14}{3} \, = \, \frac{3x \, + \, y \, - \, 12}{11}$$ then find the values of x and y, respectively.

A
$$2,6$$

B
$$4,8$$

C
$$3,5$$

D
$$4,5$$

Solution

Given that
$$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{x \, + \, 2y \, - \, 14}{3}$$
$$3x + 3y - 24 = 2x + 4y - 28$$
$$x - y = -4$$ ....... (1)
and $$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{3x \, + \, y \, - \, 12}{11}$$
$$11x + 11y - 88 = 6x + 2y - 24$$
$$5x + 9y = 64$$ ....... (2)
Multiply (1)by 9
$$9x-9y=-36$$..................(3)
adding (2) and (3)
$$14x=28$$
$$x=\dfrac{28}{14}=2$$
put value of x in (1)
$$2-y=-4$$
$$-y=-4-2$$
$$-y=-6$$
$$y=6$$
we get $$x = 2, y = 6$$
Question 4
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$$ax + 2y = 5$$
$$3x - 6y = 20$$
In the system of equations above, $$a$$ is a constant. if the system has one solution, which of the following CANNOT be the value of $$a$$?

A
$$-1$$

B
$$4$$

C
$$1$$

D
$$3$$

Solution

The given equations are :

$$ax + 2y +(-5) = 0\quad\quad\quad\dots(i)$$

$$3x+(-6)y+(-20)=0\quad\quad\quad\dots(ii)$$

As we can see they are of the form $$ax + by +c =0$$

Also the condition for some given system of equations to have solution is,

$$\Rightarrow \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} $$

Here $$a_1, a_2, b_1, b_2$$ all are constants, and their values are respectively:
$$a_1 = a$$
$$a_2 = 3$$
$$b_1 = 2$$
$$b_2 = -6$$

So if the given system of equations has exactly one solution then,
$$\Rightarrow \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} $$

$$\Rightarrow \dfrac{a}{3} \neq \dfrac{2}{-6} $$

$$\Rightarrow a \neq -1$$

Hence out of all the values given in the options $$a$$ can not be equal to $$-1.$$

Hence option $$A$$ is correct.
Question 5
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Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically and find their present ages.

A
$$x=12,y=49$$

B
$$x=12,y=42$$

C
$$x=10,y=49$$

D
$$x=10,y=42$$

Solution

Present Age of Ravish $$= y$$ years
Present Age of Aarushi $$= x$$ years

According to Question

$$7$$ Years ago,
$$y-7=7(x-7)\Rightarrow 7x-y-42=0.............(1)$$

and $$3$$ Years from now
$$y+3=3(x+3)\Rightarrow 3x-y+6=0............(2)$$

From (1) and (2)
$$(1)-(2)\Rightarrow 7x-3x-y+y-42-6=0\Rightarrow 4x=48\Rightarrow x=12$$

putting $$x=12$$ in Equation (2)
we get,
$$3 \times 12-y+6=0\Rightarrow y=42$$
Question 6
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$$\dfrac{22}{x+y}+\dfrac{15}{x-y}=5$$
$$\dfrac{55}{x+y}+\dfrac{45}{x-y}=14$$

A
$$x=-8$$, $$y=3$$

B
$$x=-3$$, $$y=8$$

C
$$x=8$$, $$y=3$$

D
$$x=8$$, $$y=-3$$

Solution

$$let\>(\frac{1}{x+y})=u\>and\>(\frac{1}{x-y})=v\\then\>22u+165v=5\rightarrow\>(1.)\\and\>55+45v=14\rightarrow\>(2.)\\multiply\>equation\>(1.)by\>3,we\>get\\66u+45v=15\rightarrow\>(1.)\\+55u+45v=14\rightarrow\>(2.)\\upon\>subtraction\>of\>(1.)with(2.),we\>get\>\\\>11u+0=1\\\therefore\>u=(\frac{1}{11})or\>x+y\>=11\rightarrow\>(3.)\\and\>then\>\\v=\>(\frac{5-22u}{15})=(\frac{5-2}{15})=(\frac{1}{5})\\\therefore\>x-y=5\rightarrow\>(4.)\\add\>(3.)and\>(4.)equation\>we\>get\>\\2x=16\\\therefore\>x=8\\and\>y\>=11-x=11-8=3$$
Question 7
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Find the solution of pair of equations $$\dfrac{x}{10}+\dfrac{y}{5}-1=0$$ and $$\dfrac{x}{8}+\dfrac{y}{6}=15.$$ Hence, find $$\lambda$$, if $$y=\lambda x+5$$.

A
$$\dfrac{-165}{17}$$

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

C
$$\dfrac{361}{696}$$

D
$$\dfrac{340}{17}$$

Solution

$$\dfrac{x}{10}+\dfrac{y}{5}-1=0$$
$$\dfrac{x}{10}+\dfrac{y}{5}=1$$
$$\dfrac{x+2y}{10}=1$$
$$x+2y=10$$.....................(1)
now consider,$$\dfrac{x}{8}+\dfrac{y}{6}=15$$
$$\dfrac{6x+8y}{48}=15$$
$$3x+4y=24\times{15}=360$$.................................(2)
now multiply (1)by 3
$$3x+6y=30$$ ........................(3)
subtracting (2) and (3)
$$2y=-330$$
$$y=-165$$
put in(1)
$$x+2(-165)=10$$
$$x-330=10$$
$$x=340$$
now, $$y=\lambda{x}+5$$
$$\implies{-165=\lambda{(340)}+5}$$
$$-170=\lambda{(340)}$$
$$\lambda{=\dfrac{-170}{340}=\dfrac{-1}{2}}$$
Question 8
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Solve by using substitution method. $$3x +4y =10$$ & $$2x - 2y =2$$

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

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

C
$$x=1, y=0$$

D
$$x=-2, y=3$$

Solution

$$3x+4y= 10 \ .....(1)$$
$$2x-2y=2\ \ .....(2)$$
equation $$(2)$$ can be written as
$$\Rightarrow x-y=1$$
$$\Rightarrow x=y+1\ .....(3)$$

substitute $$x$$ value in equation $$(2)$$
$$\Rightarrow 3x+4y= 10 $$
$$\Rightarrow 3(y+1)+4y=10$$
$$\Rightarrow 3y+3+4y=10$$
$$\Rightarrow 7y=7$$
$$\Rightarrow y=1$$

then, substitute $$y$$ value in equation $$(3)$$
$$\Rightarrow x=y+1=1+1=2$$

$$\therefore x=2, y=1$$
Question 9
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Solve the following pair of equations graphically; $$2x -3y =1, $$ & $$4x -3y +1=0$$

A
$$x=1$$ $$y=-1$$

B
$$x=-1$$ $$y=-1$$

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

D
$$x=1$$ $$y=1$$

Solution

Plot each line by taking two points on them.
$$2x-3y=1$$
$$x=0\implies y =\frac{-1}{3}$$ and $$y=0\implies x=\frac{1}{2}$$
Join the two points $$(0,\frac{-1}{3})$$ and $$(\frac{1}{2},0)$$ by drawing line.
$$4x-3y+1=0$$
$$x=0\implies y =\frac{-1}{3}$$and $$y=0\implies x=\frac{1}{4}$$
Join the two points $$(0,\frac{-1}{3})$$ and $$(\frac{1}{4},0)$$ by drawing line.
From the graph, the point of intersection of both lines is $$ (-1,-1)$$ and hence solution of given system of linear equation is $$x=-1, y=-1$$
Question 10
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Solve each pair of equation by using the substitution method.
$$x+\dfrac{6}{y}=6$$
$$3x-\dfrac{8}{y}=5$$

A
$$x=3$$ and $$y=-2$$

B
$$x=30$$ and $$y=2$$

C
$$x=3$$ and $$y=2$$

D
None of these

Solution

$$x+\dfrac {6}{y}=6......(i)$$
$$3x-\dfrac {8}{y}=5.....(ii)$$
From equation $$(i)$$, we get
$$\Rightarrow \ x=6-\dfrac {6}{y}...(iii)$$
Substituting value of $$x$$ in equation $$(ii)$$, we get
$$3\left (6-\dfrac {6}{y}\right)-\dfrac {8}{y}=5$$
$$\Rightarrow \ 18-\dfrac {18}{y}-\dfrac {8}{y}=5$$
$$\Rightarrow \ \dfrac {-26}{y}=-13$$
$$\Rightarrow \ y=2$$
Substituting the value of $$y$$ in (iii)
$$\ x=6-\dfrac {6}{2}=3$$
$$\therefore x = 3, y = 2$$

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 58

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Pair of Linear Equations in Two Variables Test - 58

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SHARING IS CARING

Question 1

$$12 \text{ kg}$$

$$12\displaystyle\frac{1}{2} \text{ kg}$$

$$12\displaystyle\frac{2}{3}\text{ kg}$$

$$13\text{ kg}$$

Solution

Question 2

$$10$$

$$12$$

$$18$$

$$14$$

Solution

Question 3

$$2,6$$

$$4,8$$

$$3,5$$

$$4,5$$

Solution

Question 4

$$-1$$

$$4$$

$$1$$

$$3$$

Solution

Question 5

$$x=12,y=49$$

$$x=12,y=42$$

$$x=10,y=49$$

$$x=10,y=42$$

Solution

Question 6

$$x=-8$$, $$y=3$$

$$x=-3$$, $$y=8$$

$$x=8$$, $$y=3$$

$$x=8$$, $$y=-3$$

Solution

Question 7

$$\dfrac{-165}{17}$$

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

$$\dfrac{361}{696}$$

$$\dfrac{340}{17}$$

Solution

Question 8

$$x=2, y=1$$

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

$$x=1, y=0$$

$$x=-2, y=3$$

Solution

Question 9

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

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

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

$$x=1$$ $$y=1$$

Solution

Question 10

$$x=3$$ and $$y=-2$$

$$x=30$$ and $$y=2$$

$$x=3$$ and $$y=2$$

None of these

Solution

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