Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 47

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Question 1
1 / -0

In covering a distance of $$30$$ km, Abhay takes $$2$$ hours more than Sameer. If Abhay doubles his speed, then he would take $$1$$ hour less than Sameer. What is Abhay's speed? (in km/hr)

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Let Abhay's speed be $$x$$ km/hr and Sameer's speed by $$y$$ km/hr
Then $$\displaystyle \frac{30}{x}-\frac{30}{y}=2$$ ..... (i)
and $$\displaystyle \frac{30}{y}-\frac{30}{2x}=1$$ ......(ii)
Adding equation (i) and (ii), we get
$$\displaystyle \frac{30}{x}-\frac{30}{2x}=3$$
$$\Rightarrow \displaystyle \frac{30}{2x}=3$$
$$\Rightarrow 2x=10$$
$$\Rightarrow x=5$$ km/hr
Question 2
1 / -0

Solve the following pair of equations:

$$\displaystyle \frac{25}{x+y}-\frac{3}{x-y}=1$$ and $$\dfrac{40}{x+y}+\dfrac{2}{x-y}=5$$,

A
$$x = 8 , y = 6$$

B
$$x = 4 , y = 6$$

C
$$x = 6, y = 4$$

D
None of these

Solution

$$\displaystyle \frac{25}{x+y}-\frac{3}{x-y}=1,\frac{40}{x+y}+\frac{2}{x-y}=5$$

Let, $$\dfrac{1}{x +y} = m$$ and $$\dfrac{1}{x -y} = n$$

$$25 m - 3n = 1$$ ...(1)
$$40 m + 2n = 5$$ ...(2)
Multiply eq(1) by 2 and eq(2) by 3 and add
$$50 m + 120 m= 17$$
$$m = \dfrac{1}{10}$$
Hence, $$n = \dfrac{1}{2}$$
Thus, $$x + y = 10$$
an $$x - y = 2$$
or, $$x = 6, y = 4$$
Question 3
1 / -0

If Rs. $$50$$ is distributed amount $$150$$ children giving $$50$$ p to each boy and $$25$$ p to each girl. Then the number of boys is

A
$$25$$

B
$$40$$

C
$$36$$

D
$$50$$

Solution

Let number of boys $$= x$$
Number of girls $$= y$$
Then, $$x + y = 150$$ .....(1)
and $$0.5x + 0.25 y = 50$$
$$\Rightarrow 2x + y = 200$$ ......(2)
Subtracting (1) from (2), we get
$$x = 50$$
Thus, number of boys are $$50$$.
Question 4
1 / -0

The electricity bill of a certain establishment is partly fixed and partly varies as the number of units of electricity consumed. When in a certain month 540 units are consumed, the bill is Rs. 1800. In another month 620 units are consumed and the bill is Rs. 2040. In yet another month 500 units are consumed then, the bill for that month would be

A
Rs. 1560

B
Rs. 1680

C
Rs. 1840

D
Rs. 1950

Solution

Let the fixed amount be Rs. $$x$$ and the cost of each unit be Rs. $$y$$ . Then,
$$540y + x = 1800$$ ...(i) and
$$620y + x = 2040$$ ...(ii)
On subtracting (i) from (ii), we get $$80y = 240 \Rightarrow y = 3$$
Putting y = 3 in (i), we get :
$$540 \times 3 + x = 1800 \Rightarrow x = \left(1800 - 1620\right) = 180$$
$$\therefore$$ Fixed charges $$= Rs. 180,$$ Charge per unit $$= Rs. 3$$
Total charges for consuming $$500$$ units $$= Rs.\left (180 + 500 \times 3\right) = Rs. 1680$$
Question 5
1 / -0

The equations $$2x - 3y + 5 = 0$$ and $$6y - 4x = 10$$, when solved simultaneously has

A
Only one solution

B
No solution

C
Only two solutions

D
Infinite number of solutions

Solution

Given equations: $$2x-3y+5=0$$ and $$6y-4x=10$$

Rewriting the equations, we get
$$2x - 3y + 5=0$$ ...$$(i)$$
$$-4x+6y-10=0$$ ...$$(ii)$$

Comparing the coefficients from $$(i)$$ and $$(ii)$$ gives

$$\dfrac{a_1}{a_2} =\dfrac2{-4},\, \dfrac{b_1}{b_2} =\dfrac{-3}6,\, \dfrac{c_1}{c_2}=\dfrac5{-10}$$

i.e, $$\dfrac{a_1}{a_2}=-\dfrac12,\, \dfrac{b_1}{b_2}=-\dfrac12,\, \dfrac{c_1}{c_2}=-\dfrac12$$

$$\Rightarrow \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$

Therefore, equations have infinite numbers of solutions.

Hence, Option $$D$$ is correct.
Question 6
1 / -0

The real numbers $$x$$ and $$y$$ are such that $$\displaystyle x+\frac{2}{y}=\frac{8}{3}$$ and $$\displaystyle y+\frac{2}{x}=3$$ .
The value of $$xy$$, is

A
$$\displaystyle \frac{4}{3}$$

B
$$\displaystyle \frac{16}{9}$$

C
$$2$$

D
$$4$$

Solution

$$\displaystyle x+\frac{2}{y}=\frac{8}{3}$$ ...(i)

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

From (i)
$$\displaystyle x+\frac{2x}{3x-2}=\frac{8}{3}$$

$$\Rightarrow \displaystyle 3(3x^2-2x+2x)=8(3x-2)$$

$$\Rightarrow \displaystyle 9x^2-24x+16=0$$

$$\Rightarrow \displaystyle (3x-4)^2=0$$

$$\Rightarrow \displaystyle x=\frac43$$

From (ii), we get
$$y=\dfrac{3\times\frac43-2}{\frac43}=\dfrac32$$
Now, $$xy=\dfrac43\times\dfrac32=2$$
Hence, option C is correct.
Question 7
1 / -0

Given $$3x - 4y = 7 $$ and $$ x + cy = 13$$, for what value of $$c$$ will the two equations not have a solution ?

A
$$\displaystyle \frac{3}{4}$$

B
$$\displaystyle \frac{4}{3}$$

C
$$-4$$

D
$$\displaystyle \frac{-4}{3}$$

Solution

$$3x - 4y = 7 $$ $$\Rightarrow a_1=3,\,b_1=-4,\,c_1=7$$
$$ x + cy = 13\Rightarrow a_2=1,\,b_2=c,\,c_2=13$$
As we know the system of linear equation has no solution if
$$ \dfrac{a_{1}}{a_{2}}= \dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}} $$
$$\therefore$$ $$ \dfrac{3}{1}= \dfrac{-4}{c} \ne\dfrac{7}{13}$$
$$c = \dfrac{-4}{3}$$
Hence, option D is correct.
Question 8
1 / -0

If $$\displaystyle \frac{x+y}{x-y}=\frac{5}{3}\: and\: \frac{x}{\left ( y+2 \right )}=2$$ the value of (x , y) is

A
(4, 1)

B
(2, 8)

C
(1, 4)

D
(8, 2)

Solution

$$\dfrac{x}{\left ( y+2 \right )}=2$$
$$x=2y+4$$ ..... $$(1)$$
$$\displaystyle \dfrac{x+y}{x-y}=\frac{5}{3}$$ ..... $$(2)$$
From eqs $$(1)$$ and $$(2)$$
$$\dfrac { 2y+4+y }{ 2y+4-y } =\dfrac { 5 }{ 3 } $$
$$\dfrac { 3y+4 }{ y+4 } =\dfrac { 5 }{ 3 } $$
$$9y+12=5y+20$$
$$4y=8$$
$$y=2$$
and $$x=2y+4=2\times 2+4=8$$
Hence, $$(x , y)=(8,2)$$
Question 9
1 / -0

If $$\displaystyle r+s=t$$ and $$\displaystyle x+t=y-2s$$, then which of the following must be true ?

A
$$\displaystyle r+2s-x=y-t$$

B
$$\displaystyle r+2s-t=x+y$$

C
$$\displaystyle x+r=y+s$$

D
$$\displaystyle x+r=y-3s$$

Solution

$$\displaystyle r+s=t$$---eq1 and
$$\displaystyle x+t=y-2s$$----eq2
putting the value of t in eq 2
$$\displaystyle x+r+s=y-2s$$
$$\displaystyle x+r=y-3s$$
Answer (D) $$\displaystyle x+r=y-3s$$
Question 10
1 / -0

The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father are respectively,

A
$$4$$ and $$24$$

B
$$5$$ and $$30$$

C
$$6$$ and $$36$$

D
$$3$$ and $$24$$

Solution

Let the present age of father's be $$x$$ years and present age of son's be $$y$$ years.
According to the problem
$$x=6y$$
After $$4$$ years
$$x+4 = 4(y+4)$$
Hence we get two equations
$$x=6y$$ ......... $$(1)$$
$$x+4 = 4(y+4)$$ ..... $$(2)$$
Simplifying eq $$(2)$$
$$x+4 = 4y+16$$
$$x-4y = 12$$
Put $$x=6y$$ in eq $$(2)$$, we get
$$6y-4y = 12$$
$$2y=12$$
$$y=6$$ years
and $$x=6y=36$$ years
Present age of son $$=6$$ years
and present age of father $$=36$$ years

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Pair of Linear Equations in Two Variables Test - 47

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Pair of Linear Equations in Two Variables Test - 47

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 2

$$x = 8 , y = 6$$

$$x = 4 , y = 6$$

$$x = 6, y = 4$$

None of these

Solution

Question 3

$$25$$

$$40$$

$$36$$

$$50$$

Solution

Question 4

Rs. 1560

Rs. 1680

Rs. 1840

Rs. 1950

Solution

Question 5

Only one solution

No solution

Only two solutions

Infinite number of solutions

Solution

Question 6

$$\displaystyle \frac{4}{3}$$

$$\displaystyle \frac{16}{9}$$

$$2$$

$$4$$

Solution

Question 7

$$\displaystyle \frac{3}{4}$$

$$\displaystyle \frac{4}{3}$$

$$-4$$

$$\displaystyle \frac{-4}{3}$$

Solution

Question 8

(4, 1)

(2, 8)

(1, 4)

(8, 2)

Solution

Question 9

$$\displaystyle r+2s-x=y-t$$

$$\displaystyle r+2s-t=x+y$$

$$\displaystyle x+r=y+s$$

$$\displaystyle x+r=y-3s$$

Solution

Question 10

$$4$$ and $$24$$

$$5$$ and $$30$$

$$6$$ and $$36$$

$$3$$ and $$24$$

Solution

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