Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Form the pair of linear equations for the following problem and find their solution by substitution method.
The difference between the two numbers is $$26$$ and one number is three times the other.

A
$$x - y = 26$$ and $$x = 3y$$, $$x=33$$ and $$,y=11$$

B
$$x - y = 26$$ and $$x = 3y$$, $$x=39$$ and $$y=13$$

C
$$x - y = 13$$ and $$x = 3y$$, $$x=75$$ and $$y=25$$

D
None of these

Solution

Let the two numbers be $$x$$ and $$y\ (x> y)$$.

In question given that the difference between two number is $$26$$.
$$\Rightarrow x- y = 26$$ .......(i)

And the one number is $$3$$ times of the other number
$$\Rightarrow x = 3y$$ .......(ii)

Now putting $$x = 3y$$ in equation (1)
$$\Rightarrow 3y - y = 26$$
$$\Rightarrow 2y = 26$$
$$ \Rightarrow y = 13$$

Now substituting $$y$$ value in (2),
$$\Rightarrow x = 3 \times 13$$
$$\Rightarrow x = 39$$

Hence, $$x = 39$$ , $$y = 13$$
Question 2
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Solve the following pair of linear equations by the substitution method.
$$\dfrac {3x}{2}-\dfrac {5y}{3}=-2,\,\, \dfrac {x}{3}+\dfrac {y}{2}=\dfrac {13}{6}$$

A
$$x=3, y=3$$

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

C
$$x=2, y=3$$

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

Solution

Given Pair of equations are:
$$\frac{3x}{2}-\frac{5y}{3} = -2\Rightarrow 9x-10y=-12$$ ...eq(i)

$$\frac{x}{3}+\frac{y}{2} = \frac{13}{6}\Rightarrow 2x+3y=13$$ ...eq(ii)
Now from eq2 will get
$$2x+3y = 13$$
Now using substitution method
$$ y = \frac{13-2x}{3}$$
Now substituting y in eq (i) will get
$$\Rightarrow 9x-\frac{10\left ( 13-2x \right )}{3} = -12$$
$$\Rightarrow 27x-130+20x= -36$$
$$\Rightarrow 47x = -36+130$$
$$\Rightarrow 47x = 94$$
$$\Rightarrow x = 2$$
Now substituting x in eq (ii) will get
$$\Rightarrow 2\times 2 +3y = 13$$
$$\Rightarrow 4 +3y = 13$$
$$\Rightarrow 3y = 13-4$$
$$\Rightarrow 3y = 9$$
$$\Rightarrow y = 3$$
$$\Rightarrow x = 2 , y = 3$$
Question 3
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Solve the following pair of linear equations by the substitution method
$$s-t=3, \dfrac {s}{3}+\dfrac {t}{2}=6$$

A
$$s = 2, t = 7$$

B
$$s = 9, t = 6$$

C
$$s = 3, t = 1$$

D
$$s = 6, t = -9$$

Solution

$$s-t=3$$.....(i)
$$\frac {s}{3}+\frac {t}{2}=6$$.....(ii)
From (i) $$s = t + 3$$ ...(iii)
Substituting s from (iii) in (ii), we get
$$\frac {t+3}{3}+\frac {t}{2}=6$$
$$\Rightarrow 2(t + 3) + 3t = 36$$
$$\Rightarrow 5t + 6 = 36$$
$$\Rightarrow t = 6$$
From (iii), $$s =6 +3 = 9$$,
Hence, $$s = 9, t = 6$$
Question 4
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Solve the following pair of linear equations by the substitution method
$$x + y = 14, x - y = 4$$

A
$$x = 8, y = 5$$

B
$$x = 6, y = 9$$

C
$$x = 7, y = 10$$

D
None of these

Solution

$$x + y = 14$$ ....(i)
$$x - y = 4$$ .....(ii)
From (ii) $$y=x - 4$$ ...(iii)
Substituting y from (iii) in (i), we get
$$x + x - 4 = 14 $$
$$\Rightarrow 2x = 18$$
$$\Rightarrow x =9$$
Substituting $$x = 9$$ in (iii), we get
$$y = 9 - 4 = 5$$,
i.e, $$y = 5$$
$$x = 9, y = 5$$
Question 5
1 / -0

Solve the following pair of linear equations by the substitution method.
$$0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3$$

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

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

C
$$x=5, y=1$$

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

Solution

The given equations are
$$\Rightarrow .2x+.3y=1.3........eq1$$
$$\Rightarrow x=\frac{1.3-.3y}{.2}.......eq2$$

and
$$\Rightarrow .4x+.5y=2.3.......eq3$$
Substitute the value of x in eq3
$$\Rightarrow .4\times\left (\frac{1.3-.3y}{.2} \right)+.5y=2.3$$
$$\Rightarrow 2.6-.6y+.5y=2.3$$
$$\Rightarrow -.1y=-.3 \Rightarrow y=3$$
Substitute $$y=3$$ in eq 2
$$\Rightarrow x=\frac{1.3-.9}{.2}\Rightarrow x=2$$
Hence x=2 and y=3
Question 6
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A fraction becomes $$\dfrac {9}{11}$$, if $$2$$ is added to both the numerator and the denominator. If $$3$$ is added to both the numerator and the denominator it becomes $$\dfrac {5}{6}$$. Find the fraction.

A
$$\dfrac{7}{9}$$

B
$$\dfrac{5}{7}$$

C
$$\dfrac{9}{11}$$

D
None of these

Solution

Let $$\dfrac {x}{y}$$ be the fraction, where $$x$$ and $$y$$ are positive integers.

$$\Rightarrow \dfrac {x+2}{y+2}=\dfrac {9}{11}$$ ..(1)
$$\Rightarrow \dfrac {x+3}{y+3}=\dfrac {5}{6}$$ ..(2)

From (1) we get
$$\Rightarrow 11x+22 = 9y+18$$
$$\Rightarrow 11x-9y = -4$$ ...(3)

From (2) we get
$$\Rightarrow 6x +18 = 5y +15$$
$$\Rightarrow 6x-5y = -3$$ ...(4)

From (4) will get
$$y = \dfrac{3+6x}{5}$$

Now substituting $$y$$ in (3),

$$\Rightarrow 11x-\dfrac{9\left ( 3+6x \right )}{5} = -4$$

$$\Rightarrow 55x-27-54x = -20$$

$$\Rightarrow 55x-54x = -20+27$$

$$\Rightarrow x = 7$$

Now substituting $$x$$ in (4)

$$\Rightarrow 6\times 7 -5y = -3$$

$$\Rightarrow 42-5y = -3$$

$$\Rightarrow -5y = -3 - 42$$

$$\Rightarrow -5y = -45$$

$$\Rightarrow y = 9$$

Hence, $$ x = 7$$ and $$ y = 9$$.

And fraction $$\dfrac{x}{y}=\dfrac{7}{9}$$
Question 7
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Five years hence, the age of jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

A
Present age of Jacob: $$50$$ years and Present age of Jacob's son: $$15$$ years

B
Present age of Jacob: $$50$$. years and Present age of Jacob's son: $$10$$ years

C
Present age of Jacob: $$40$$ years and Present age of Jacob's son: $$10$$ years

D
Present age of Jacob: $$35$$ years and Present age of Jacob's son: $$15$$ years

Solution

Let $$x$$ be the present age of Jacob's son and $$y$$ be the present age of Jacob. Then as per the question

$$(y + 5) =3(x + 5)\ ......(1)$$
$$(y- 5) = 7 (x - 5)\ ......(2)$$

From equation $$(1)$$,
$$\Rightarrow y +5 = 3x +15$$
$$\Rightarrow 3x-y = -10\ ......(3)$$

From equation $$(2)$$,
$$\Rightarrow y-5 = 7x-35$$
$$\Rightarrow 7x-y = 30\ .......(4)$$

from equation $$(3)$$ we get
$$\Rightarrow -y = -10 -3x$$
$$\Rightarrow y = 10 +3x$$

Now Substitute $$y$$ in $$(4)$$
$$\Rightarrow 7x-\left ( 10+3x \right ) = 30$$
$$\Rightarrow 7x -10-3x = 30$$
$$\Rightarrow 4x = 30+10$$
$$\Rightarrow 4x = 40$$
$$\Rightarrow x = 10$$

Now substituting $$x$$ in (1) will give
$$\Rightarrow 3\times 30-y = -10$$
$$\Rightarrow 30 - y = -10$$
$$\Rightarrow -y = -40$$
$$\Rightarrow y = 40$$

Hence, Jacob's son's current age $$=$$ $$10$$ and Jacob's Current age $$=$$ $$40$$.
Question 8
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Form the pair of linear equations for the following problem and find their solution by substitution method.
The larger of two supplementary angles exceeds the smaller by $$18$$ degrees.

A
$$x + y = 100$$ and $$x - y = 18$$, $$x=90$$ and $$\,y=16$$

B
$$x + y = 60$$ and $$x - y = 18$$, $$x=95$$ and $$\,\,y=70$$

C
$$x + y = 180$$ and $$x - y = 18$$, $$x=99$$ and $$\,\,y=81$$

D
None of these

Solution

Let the supplementary angles be $$x$$ and $$y$$, $$(x> y) $$

$$\therefore$$ Sum of both angles is $$180^0$$

$$\Rightarrow x + y = 180$$ ...$$(1)$$

The larger angle exceeds the smaller by $$18^0$$

$$\Rightarrow x - y = 18$$ ...$$(2)$$

$$\Rightarrow y=x -18$$

Put value of $$y$$ in (1), we get

$$2x=198$$

$$\Rightarrow x = 99$$

Now substituting the value of $$x$$ in equation $$(1)$$

$$\Rightarrow 99+y = 180$$

$$ \Rightarrow y = 180-99$$

$$\Rightarrow y = 81$$

Hence, $$ x = 99^0 $$ and $$ y = 81^0$$

The pair of linear equations for the given problem are
$$ x + y = 180$$ and $$ x - y = 18$$ .
Question 9
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Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
Five years ago Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

A
Nuri is $$40$$ years old and Sonu is$$ 10 $$years old.

B
Nuri is $$50$$ years old and Sonu is$$ 20$$ years old.

C
Nuri is $$70$$ years old and Sonu is $$30$$ years old.

D
Nuri is $$60$$ years old and Sonu is $$10$$ years old.

Solution

Let present age of Nuri= $$x$$ years
Let present age of sonu=$$y$$ years
Five years ago
Nuri was $$\left(x-5 \right) years$$
Sonu   was $$ \left(y-5 \right) years$$
According to the first condition
$$\Rightarrow \left(x-5 \right)=3\left(y-5 \right)$$
$$\Rightarrow x-5=3y-15 \Rightarrow x-3y=-10.........eq1$$
Ten years later
Nuri will be $$ \left(x+10    \right) years$$
Sonu will be $$\left(y+10   \right)$$
$$\Rightarrow \left(x+10   \right)=2\left(y+10 \right)$$
$$\Rightarrow x+10=2y+20 \Rightarrow x+2y=10.....eq2$$
Subtracting $$eq 1 and eq2$$
$$\Rightarrow -y=-2- \Rightarrow y=20$$
put y=20 in $$eq1$$
$$x-3y=10 \Rightarrow x=10-60 \Rightarrow x=50$$
Hence, present age of Nuri= 50 years
Let present age of sonu=20 years
Question 10
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Solve the following pair of equations by reducing them to a pair of linear equations:

$$\dfrac {1}{(x-1)}+\dfrac {1}{(y-2)}=2, \ \dfrac {6}{(x-1)}-\dfrac {2}{(y-2)}=1$$

A
$$x=\dfrac{11}{5},\,\,y=\dfrac{13}{11}$$

B
$$x=\dfrac{13}{5},\,\,y=\dfrac{30}{11}$$

C
$$x=\dfrac{4}{5},\,\,y=\dfrac{20}{11}$$

D
None of these

Solution

The given equations are
$$ \dfrac { 1 }{ x-1 } +\dfrac { 1 }{ y-2 } =2$$ and
$$ \dfrac { 6 }{ x-1 } -\dfrac { 2 }{ y-2 } =1 $$

Let $$ \dfrac { 1 }{ x-1 } =u$$ and $$ \dfrac { 1 }{ y-2 } =v$$
$$ u+v=2$$ .......(i) and
$$ 6u-2v=1$$ ......(ii)

Multiply (i) by $$6$$
$$\Rightarrow 6u+6v=12$$ ......(iii)
Subtract (ii) from (iii)
$$\Rightarrow 8v=11$$
$$\Rightarrow v=\dfrac { 11 }{ 8 } $$
Therefore $$ \dfrac { 1 }{ y-2 } =\dfrac { 11 }{ 8 } $$
$$ \Rightarrow y-2=\dfrac { 8 }{ 11 } $$
$$ \Rightarrow y=\dfrac { 30 }{ 11 } $$

Putting $$ v=\dfrac { 11 }{ 8 } $$ in (i), we get

$$ u+\dfrac { 11 }{ 8 } =2$$
$$ \Rightarrow u=\dfrac { 5 }{ 8 } $$
$$ \Rightarrow \dfrac { 1 }{ x-1 } =\dfrac { 5 }{ 8 }$$
$$ \Rightarrow x-1=\dfrac { 8 }{ 5 } $$
$$ \Rightarrow x=\dfrac { 13 }{ 5 }$$
Therefore, $$ \left( x,y \right) =\left( \dfrac { 13 }{ 5 } ,\dfrac { 30 }{ 11 } \right) $$

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Pair of Linear Equations in Two Variables Test - 44

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Pair of Linear Equations in Two Variables Test - 44

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Question 1

$$x - y = 26$$ and $$x = 3y$$, $$x=33$$ and $$,y=11$$

$$x - y = 26$$ and $$x = 3y$$, $$x=39$$ and $$y=13$$

$$x - y = 13$$ and $$x = 3y$$, $$x=75$$ and $$y=25$$

None of these

Solution

Question 2

$$x=3, y=3$$

$$x=1, y=3$$

$$x=2, y=3$$

$$x=0, y=0$$

Solution

Question 3

$$s = 2, t = 7$$

$$s = 9, t = 6$$

$$s = 3, t = 1$$

$$s = 6, t = -9$$

Solution

Question 4

$$x = 8, y = 5$$

$$x = 6, y = 9$$

$$x = 7, y = 10$$

None of these

Solution

Question 5

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

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

$$x=5, y=1$$

$$x=2, y=3$$

Solution

Question 6

$$\dfrac{7}{9}$$

$$\dfrac{5}{7}$$

$$\dfrac{9}{11}$$

None of these

Solution

Question 7

Present age of Jacob: $$50$$ years and Present age of Jacob's son: $$15$$ years

Present age of Jacob: $$50$$. years and Present age of Jacob's son: $$10$$ years

Present age of Jacob: $$40$$ years and Present age of Jacob's son: $$10$$ years

Present age of Jacob: $$35$$ years and Present age of Jacob's son: $$15$$ years

Solution

Question 8

$$x + y = 100$$ and $$x - y = 18$$, $$x=90$$ and $$\,y=16$$

$$x + y = 60$$ and $$x - y = 18$$, $$x=95$$ and $$\,\,y=70$$

$$x + y = 180$$ and $$x - y = 18$$, $$x=99$$ and $$\,\,y=81$$

None of these

Solution

Question 9

Nuri is $$40$$ years old and Sonu is$$ 10 $$years old.

Nuri is $$50$$ years old and Sonu is$$ 20$$ years old.

Nuri is $$70$$ years old and Sonu is $$30$$ years old.

Nuri is $$60$$ years old and Sonu is $$10$$ years old.

Solution

Question 10

$$x=\dfrac{11}{5},\,\,y=\dfrac{13}{11}$$

$$x=\dfrac{13}{5},\,\,y=\dfrac{30}{11}$$

$$x=\dfrac{4}{5},\,\,y=\dfrac{20}{11}$$

None of these

Solution

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