2 Marks Questions

Question 12 Marks

Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
5x - 8y + 1 = 0; ...(i)
3x - $\frac{24}{5}$y + $\frac{3}{5}$ = 0 ...(ii)

Answer

$a_1 = 5,\ b_1 = -8,\ c_1 = 1$ and $a_2 = 3,\ b_2 =$ $\frac{-24}{5}$, $c_2 =$ $\frac{3}{5}$
$\frac { a _ { 1 } } { a _ { 2 } }$ = $\frac{5}{3}$ ...(i)
$\frac { b _ { 1 } } { b _ { 2 } }$ =$\frac { - 8 } { - 24 / 5 }$ = $\frac{5}{3}$ ...(ii)
and $\frac { c _ { 1 } } { c _ { 2 } }$ = $\frac { 1 } { 3 / 5 }$ = $\frac{5}{3}$ ...(iii)
Form (i), (ii) and (iii)
$\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }$
$\therefore$The pair of equations has infinitely many solutions.

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Question 22 Marks

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Solve the pair of the linear equation obtained by the elimination method.

Answer

Suppose the fixed charge be Rs. x and the extra charge per day be Rs y.
According to the question, Mona paid Rs 27 for a book kept for 7 days,
$\Rightarrow$ $x + 4y = 27$........(i)
Tanvy paid Rs.21 for a book kept for 5 days,
$\Rightarrow$ $x+ 2y = 21$.........(ii)
Subtracting (ii) from (i),
$\Rightarrow 2y = 6$
$\Rightarrow$ $y = 3$
Substituting $y = 3$ in (ii), we get $x = 15$
The fixed charge is Rs. 15 and the charge per day is Rs 3.

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Question 32 Marks

Solve the pair of linear equations by substitution method:
$\sqrt { 2 } x - \sqrt { 3 } y = 0$
$\sqrt { 3 } x - \sqrt { 8 } y = 0$

Answer

The given equations are
$\sqrt 2 x - \sqrt 3 y = 0$ ............(i)
$\sqrt 3 x - \sqrt 8 y = 0$ .............(ii)
From equation (i), we obtain:
$x = \frac{{ \sqrt 3 y}}{{\sqrt 2 }}$ ...(iii)
Substituting this value in equation (ii), we obtain:
$\sqrt 3 \left( { \frac{{\sqrt 3 y}}{{\sqrt 2 }}} \right) - \sqrt 8 y = 0$
$ \frac{{3y}}{{\sqrt 2 }} - 2\sqrt 2 y = 0$
$y\left( { \frac{3}{{\sqrt 2 }} - 2\sqrt 2 } \right) = 0$
y = 0
Substituting the value of y in equation (iii), we obtain:
x = 0
$\therefore $ x = 0, y = 0
Hence the solution of given equation is (0,0).

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Question 42 Marks

Solve the pair of linear equations by substitution method: 3x – y = 3; 9x – 3y = 9

Answer

3x - y = 3
9x - 3y = 9
The given pair of linear equations is
3x - y = 3..............(1)
9x - 3y = 9.............(2)
From equation(1),
y = 3x - 3...................(3)
9x - 3(3x - 3) = 9
$\Rightarrow$ 9x - 9x + 9 = 9
$\Rightarrow$ 9 = 9
which is true. Therefore, equation (1) and (2) have infinitely many solutions.

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Question 52 Marks

Is the pair of linear equation consistent/inconsistent? If consistent, obtain the solution graphically: 2x – 2y – 2 = 0; 4x – 4y – 5 = 0

Answer

2 x - 2 x - 2 = 0................(1)
4 x - 4 y - 5 = 0..................(2)
Here, $a _ { 1 } = 2 , \quad b = - 2 , c _ { 1 } = - 2$
$a _ { 2 } = 4 , b _ { 2 } = - 4 , c _ { 2 } = - 5$
We see that $\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$
Hence, the lines represented by the equations(1) and ( 2 ) are parallel.
Therefore, equations ( 1) and (2) have no solution, i.e., the given pair of a linear equation is inconsistent.

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Question 62 Marks

Is the pair of linear equation consistent/inconsistent? If consistent, obtain the solution graphically: x – y = 8; 3x – 3y = 16

Answer

x - y = 8.................(1)
3 x - 3 y = 16.............(2)
Here, $a _ { 1 } = 1 , b _ { 1 } = - 1 , c _ { 1 } = - 8$
$a _ { 2 } = 3 , b _ { 2 } = - 3 , c _ { 2 } = - 16$
We see that $\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$
Hence, the lines represented by the equations(1) and (2) are parallel.
Therefore, equations (1) and ( 2 ) have no solution, i.e., the given pair of linear equation is inconsistent.

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Question 72 Marks

On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } $ and $\frac {c_1}{c_2}$, find out whether the pair of linear equations are consistent, or inconsistent: $ \frac { 3 } { 2 } x + \frac { 5 } { 3 } y = 7,$ 9x − 10y = 14

Answer

Given equations are:
$ \frac { 3 } { 2 } x + \frac { 5 } { 3 } y = 7$&

9x - 10y = 14
Comparing equation $ \frac { 3 } { 2 } x + \frac { 5 } { 3 } y = 7$ with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$
and 9x − 10y = 14 with
$ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $ a _ { 1 } = \frac { 3 } { 2 }$, $ b _ { 1 } = \frac { 5 } { 3 }$, $ c _ { 1 } = - 7 , a _ { 2 } = 9 , b _ { 2 } = - 10 , c _ { 2 } = - 14$
$ \frac { a _ { 1 } } { a _ { 2 } } = \frac { \frac { 3 } { 2 } } { 9 } = \frac { 1 } { 6 }$ and $ \frac { b _ { 1 } } { b _ { 2 } } = \frac { \frac { 5 } { 3 } } { - 10 } = \frac { - 1 } { 6 }$
Here $ \frac { a _ { 1 } } { a _ { 2 } } \neq \frac { b _ { 1 } } { b _ { 2 } }$
Therefore, equations have unique solution.Hence, they are consistent.

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Question 82 Marks

On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } $ and $\frac { c _ { 1 } } { c _ { 2 } }$, find out whether the pair of linear equations are consistent, or inconsistent: 2x − 3y = 8, 4x − 6y = 9.

Answer

Given equations are as:2x − 3y = 8
4x − 6y = 9
Comparing equation 2x − 3y = 8 with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$
and 4x − 6y = 9 with
$ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $a_1=2, b_1=-3, c_1=-8, a_2=4, b_2=-6, c_2=-9$
Here $ \frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }$because $ \frac { 2 } { 4 } = \frac { - 3 } { - 6 } \neq \frac { - 8 } { - 9 } \Rightarrow \frac { 1 } { 2 } = \frac { 1 } { 2 } \neq \frac { - 8 } { - 9 }$
Therefore, equations have no solution because they are parallel.
Hence, they are inconsistent.

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Question 92 Marks

Answer

Given equations are $3 x+2 y=5$
$2 x-3 y=7$
Comparing equation $3 x +2 y =5$ with $a_1 x+b_2 y+c_1=0$ and $2 x-3 y=7$
with $a_2 x+b_2 y+c_2=0$,
We get, $a_1=3, b_1=2, c_1=-5, a_2=2, b_2=-3, c_2=-7$ $\frac{a_1}{a_2}=\frac{3}{2}$ and $\frac{b_1}{b_2}=\frac{2}{-3}$
Here $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ which means equations have unique solution.
Hence they are consistent.

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Question 102 Marks

On comparing the ratios $ \frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } } \text { and } \frac { c _ { 1 } } { c _ { 2 } }$, find out whether the lines representing the pair of linear equations intersect at a point, are parallel or coincide: 6x − 3y + 10 = 0; 2x – y + 9 = 0.

Answer

Given equations are $6 x-3 y+10=0$
$2 x-y+9=0$
Comparing equation $6 x -3 y +10=0$ with $a_1 x+b_1 y+c_1=0$ and $2 x-y+9=0$ with
$a_2 x+b_2 y+c_2=0$
We get, $a_1=6, b_1=-3, c_1=10, a_2=2, b_2=-1, c_2=9$
We have $\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ because $\frac{6}{2}=\frac{-3}{-1} \neq \frac{10}{9} \Rightarrow \frac{3}{1}=\frac{3}{1} \neq \frac{10}{9}$ Hence, lines are parallel to each other.

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Question 112 Marks

On comparing the ratios $\frac { a _ { 1 } } { a _ { 2 } } , \frac { b _ { 1 } } { b _ { 2 } }$ and $\frac {c_1}{c_2}$, find out whether the lines representing the pair of linear equations intersect at a point, are parallel or coincident: 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

Answer

Given equations are9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing equation 9x + 3y + 12 = 0 with $ a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0$
and 18x + 6y + 24 = 0 with
$ a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$,
We get, $a_1=9, b_1=3, c_1=12, a_2=18, b_2=\mid 6, c_2=24$
We have $ \frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }$ because$ \frac { 9 } { 18 } = \frac { 3 } { 6 } = \frac { 12 } { 24 } \Rightarrow \frac { 1 } { 2 } = \frac { 1 } { 2 } = \frac { 1 } { 2 }$
Hence, lines are coincident.

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Question 122 Marks

Answer

Formulation: Let the number of girls be x and the number of boys be y.
It is given that total ten students took part in the quiz.
$\therefore$ Number of girls+ Number of boys = 10
i.e. x + y =10
It is also given that the number of girls is 4 more than the number of boys.
$\therefore$ Number of girls= Number of boys + 4
i.e. x = y+4
or, x-y = 4
Thus, the algebraic representation of the given situation is
x + y=10 ........(i)
x - y =4 ..........(ii)
Add (i) and (ii) we get
x + y + x - y = 10 + 4
2x = 14
x = 7
Put x = 7 in (i)
x + y = 10
7 + y = 10
y = 10 -7
y = 3
So, value of x = 7 and y = 3
Graphical Representation: Now putting y = 0 in x + y = 10, we get
x = 10. Similarly, by putting x = 0 in x + y = 10, we get y = 10.
Thus, two solution of equation (i) are:

x	10	0
y	0	10

Similarly, two solutions of equation (ii) are:
putting y = 0 in x - y = 4, we get
x = 4. Similarly, by putting x = 0 in x + y = 10, we get y = -4.

x	4	0
y	0	-4

Now, we plot the points A (10, 0), B (0, 10), P (4, 0) and Q (0, -4) corresponding to these solutions on the graph paper and draw the lines AB and PQ representing the equations x + y = 10 and x - y - 4 as shown in Fig.

We observe that the two lines representing the two equations are intersecting at the point (7, 3).

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Question 132 Marks

Two rails are represented by the equations: x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other?

Answer

The pair of linear equations are given as:
x + 2y – 4 = 0 ...(i)
2x + 4y – 12 = 0 ...(ii)
We express x in terms of y from equation (i), to get
x = 4 – 2y
Now, we substitute this value of x in equation (ii), to get
2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0
i.e., – 4 = 0
Which is a false statement. Therefore, the equations do not have a common solution. So, the two rails will not cross each other.

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