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#### Question 1:

Choose the correct answer from the given four options:
Graphically, the pair of equations
6x – 3y + 10 = 0
2xy + 9 = 0
represents two lines which are
(A) intersecting at exactly one point.
(B) intersecting at exactly two points.
(C) coincident.
(D) parallel.

Comparing the coefficients of two equations, with general equation of the form $ax+by+c=0$
we get

$\begin{array}{ccc}⇒\frac{6}{2}& =& \frac{-3}{-1}\ne \frac{10}{9}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}⇒\frac{3}{1}& =& \frac{3}{1}\ne \frac{10}{9}\end{array}$

As , this means that lines are parallel, so there will be no solution. Hence, the correct answer is option D.

#### Question 2:

Choose the correct answer from the given four options:
The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have
(A) a unique solution
(B) exactly two solutions
(C) infinitely many solutions
(D) no solution

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

As    which means that lines are parallel, so there will be no solution.

Hence, the correct answer is option D.

#### Question 3:

Choose the correct answer from the given four options:
If a pair of linear equations is consistent, then the lines will be
(A) parallel
(B) always coincident
(C) intersecting or coincident
(D) always intersecting

Consistent equations means the equations must have solution which is obtained by either of conditions:
1.  lines coincide (with infinitely many solutions) or
2.  lines intersect (with unique solution)
Hence, the correct answer is option C.

#### Question 4:

Choose the correct answer from the given four options:
The pair of equations y = 0 and y = –7 has
(A) one solution
(B) two solutions
(C) infinitely many solutions
(D) no solution

The give pair of equations are y = 0 and y = $-$7.
The equation y = 0 represents x-axis and
Also y = $-$7 is line parallel to x-axis (no x-intercept). We can see this graphically, both lines are parallel and thus will have no solution.
Hence, the correct answer is option D.

#### Question 5:

Choose the correct answer from the given four options:
The pair of equations x = a and y = b graphically represents lines which are
(A) parallel
(B) intersecting at (b, a)
(C) coincident
(D) intersecting at (a, b)

It is very clear that the line x = a will be parallel to y axis and y = b will be parallel to x axis.
So they will be intersecting at a point having x coordinate as a and y coordinate as b
Hence, the correct answer is option D.

#### Question 6:

Choose the correct answer from the given four options:
For what value of k, do the equations 3xy + 8 = 0 and 6xky = –16 represent coincident lines?
(A) $\frac{1}{2}$
(B) $-\frac{1}{2}$
(C) 2
(D) –2

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

As , which means that lines are coincident

$\begin{array}{ccc}⇒\frac{3}{6}& =& \frac{-1}{-k}=\frac{8}{16}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}⇒\frac{1}{2}& =& \frac{1}{k}=\frac{1}{2}\end{array}$
Solving we get k = 2
Hence, the correct answer is option C.

#### Question 7:

Choose the correct answer from the given four options:
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(A) $\frac{-5}{4}$

(B) $\frac{2}{5}$

(C) $\frac{15}{4}$

(D) $\frac{3}{2}$

Comparing the coefficients of two equations, with general equation of the form ax +by + c = 0, we get:

As lines are parallel which means that
Substituting the values, we get $\begin{array}{rcl}& & \begin{array}{ccc}\frac{3}{2}& =& \frac{2k}{5}\ne \frac{-2}{1}\end{array}\\ & & \end{array}$
$⇒$15 = 4k
$⇒k=\frac{15}{4}$
Hence, the correct answer is option C.

#### Question 8:

Choose the correct answer from the given four options:
The value of c for which the pair of equations cxy = 2 and 6x – 2y = 3 will have infinitely many solutions is
(A) 3
(B) – 3
(C) –12
(D) no value

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

As there are infinitely many solutions which means
Substituting the values, we get $\begin{array}{rcl}& & \begin{array}{ccc}\frac{c}{6}& =& \frac{-1}{-2}=\frac{-2}{-3}\end{array}\\ & & \end{array}$
$⇒$

Since, c has different values, so for no value of c the pair of equations will have infinitely many solutions
Hence, the correct answer is option D.

#### Question 9:

Choose the correct answer from the given four options:
One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be
(A) 10x + 14y + 4 = 0
(B) –10x – 14y + 4 = 0
(C) –10x + 14y + 4 = 0
(D) 10x – 14y = –4

The condition for pair of equation to be dependent linear equations is
Comparing the coefficients of the first equation, with general equation of the form ax + by + c = 0, we get

Let the second equation be  ${a}_{2}x+{b}_{2}y+{c}_{2}=0$
, where k is any arbitrary constant.
Putting $k=\frac{-1}{2}$, we get

The required equation becomes $10x\mathit{-}14y+4=0$
Hence, the correct answer is option D.

#### Question 10:

Choose the correct answer from the given four options:
A pair of linear equations which has a unique solution x = 2, y = –3 is
(A) x + y = –1

2x – 3y = –5
(B) 2x + 5y = –11
4 x + 10y = –22
(C) 2xy = 1
3x + 2y = 0
(D) x – 4y –14 = 0
5x y – 13 = 0

As x = 2 and y$-$3 is a unique solution of pair of equation, then these values must satisfy that pair of equations among the given options

From option (A), we see that
$2+\left(-3\right)=1$  and $2×\left(2\right)-3×\left(-3\right)=13\ne 5$

From option (B),we see that
$2×2+5×\left(-3\right)=-11$  and $4×\left(2\right)+10×\left(-3\right)=-22$

Since, both the equations are satisfied by the unique solution  x = 2 and y = $-$3.
Hence, the correct answer is option B.

#### Question 11:

Choose the correct answer from the given four options:
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

Given:
.....(1)
$x+y=4$                           .....(2)
Adding (1) and (2), we get
$2x=6$
$⇒x=3$
Substituting the value of x in (2), we get
$3+y=4$
$⇒y=1$
Hence, the correct answer is option C.

#### Question 12:

Choose the correct answer from the given four options:
Aruna has only Re 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Re 1 and Rs 2 coins are, respectively
(A) 35 and 15
(B) 35 and 20
(C) 15 and 35
(D) 25 and 25

Let the number of Re 1 coins = x
and the number of Rs 2 coins = y
Now, by given conditions:
Total number of coins = $x+y=50$  .....(1)
Amount of money = $1x+2y=75$     .....(2)
Subtracting (1) from (2), we get $y=25$
Substituting the value of in (1), we get $x=25$.
Hence, the correct answer is option D.

#### Question 13:

Choose the correct answer from the given four options:
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

Let the present age of father be x years and the present age of son be y years.
As per information given,
$x=6y$                          .....(1)
After four years, the relation between their ages is given by,
$\left(x+4\right)=4\left(y+4\right)$          .....(2)
Substituting the value of x from (1) in (2), we get
$\left(6y+4\right)=4y+16$
$⇒2y=12$
$⇒y=6$
Substituting the value of y in (1), we get $x=36$.
Hence, the correct answer is option C.

#### Question 1:

Do the following pair of linear equations have no solution? Justify your answer.
(i) 2x + 4y = 3

12y + 6x = 6

(ii) x = 2y
y = 2x

(iii) 3x + y – 3 = 0
$2x+\frac{2}{3}y=2$

We know that for two equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0$ and  ${a}_{2}x+{b}_{2}y+{c}_{2}=0$ , the condition for no solution is given by  ,

(i)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$\begin{array}{ccc}⇒\frac{2}{6}& =& \frac{4}{12}\ne \frac{-3}{-6}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}⇒\frac{1}{3}& =& \frac{1}{3}\ne \frac{1}{2}\end{array}$
Hence, the given pair of linear equations has no solution.

(ii)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$\begin{array}{c}⇒\frac{1}{2}\ne \frac{-2}{-1}\end{array}\phantom{\rule{0ex}{0ex}}$
As

​Hence, the given pair of linear equations has unique solution.

(iii)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$\begin{array}{ccc}⇒\frac{3}{2}& =& \frac{1}{\frac{2}{3}}=\frac{3}{2}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}⇒\frac{3}{2}& =& \frac{3}{2}=\frac{3}{2}\end{array}$
Hence, the given pair of linear equations is coincident.

#### Question 2:

Do the following equations represent a pair of coincident lines? Justify your answer.
(i)
$3x+\frac{1}{7}y=3\phantom{\rule{0ex}{0ex}}7x+3y=7$

(ii)
–2x – 3y = 1
6y + 4x = – 2

(iii)
$\frac{x}{2}+y+\frac{2}{5}=0\phantom{\rule{0ex}{0ex}}4x+8y+\frac{5}{16}=0$

We know that for two equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0$ and  ${a}_{2}x+{b}_{2}y+{c}_{2}=0$ , the condition for co incident lines is ,

(i)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$\begin{array}{ccc}⇒\frac{3}{7}& \ne & \frac{\frac{1}{7}}{3}\end{array}\phantom{\rule{0ex}{0ex}}$
Hence, the given pair of linear equations has unique solution and does not represent a pair of coincident lines.

(ii)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$\begin{array}{c}⇒\frac{1}{2}=\frac{-3}{6}\end{array}=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}$
$\begin{array}{c}⇒\frac{1}{2}=\frac{1}{2}\end{array}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$
Since ,

​Hence, the given pair of linear equations represent a pair of coincident lines.

(iii)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$\begin{array}{ccc}⇒\frac{\frac{1}{2}}{4}& =& \frac{1}{8}\ne \frac{\frac{2}{5}}{\frac{5}{16}}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}⇒\frac{1}{8}& =& \frac{1}{8}\ne \frac{32}{25}\end{array}$
Since ,
Hence, the given pair of linear equations are parallel and does not represent a pair of coincident lines.

#### Question 3:

(i)
–3x– 4y = 12
4y + 3x = 12

(ii)
$\frac{3}{5}x-y=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{1}{5}x-3y=\frac{1}{6}$

(iii)
2ax + by = a
4ax + 2by – 2a = 0; a, b  ≠ 0

(iv)
x + 3y = 11
2 (2x + 6y) = 22

The condition for linear equations to be consistent  is either    for infinite many solutions or   for unique solutions.

(i)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

$⇒\frac{-3}{3}=\frac{-4}{4}\ne \frac{-12}{-12}\phantom{\rule{0ex}{0ex}}⇒\frac{-1}{1}=\frac{-1}{1}\ne \frac{1}{1}$

Hence, the given pair of linear equations has no solution  i.e. inconsistent.

(ii)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

$⇒\frac{3}{5}}{1}{5}}\ne \frac{-1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{1}\ne \frac{-1}{3}\phantom{\rule{0ex}{0ex}}$

Hence, the given pair of linear equations has unique solution i.e. consistent.

(iii)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

$⇒\frac{2a}{4a}=\frac{b}{2b}=\frac{-a}{-2a}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$

Hence, the given pair of linear equations has infinite solution i.e. consistent.

(iv)

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$⇒\frac{1}{4}=\frac{3}{12}\ne \frac{-11}{-22}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{4}=\frac{1}{4}\ne \frac{1}{2}$

Hence, the given pair of linear equations has no solution  i.e. inconsistent.

#### Question 4:

For the pair of equations
λ x + 3y = –7
2x + 6y = 14
to have infinitely many solutions, the value  of λ should be 1. Is the statement true? Give reasons.

In order to have infinite solutions

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

$⇒\frac{\lambda }{2}=\frac{3}{6}=\frac{7}{-14}\phantom{\rule{0ex}{0ex}}⇒\frac{\lambda }{2}=\frac{1}{2}=-\frac{1}{2}$
Solving above we get λ = 1 and λ$-$1.
So, for no value of λ, the given pair of linear equations has infinitely many solutions.
Hence, the given statement is false.

#### Question 5:

For all real values of c, the pair of equations
x – 2y = 8
5x – 10y = c
have a unique solution. Justify whether it is true or false.

In order to have unique solution

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

Since, $\frac{1}{5}=\frac{-2}{-10}\phantom{\rule{0ex}{0ex}}$ , we can say that given pair of equations will not have a unique solution for any real value of c.

#### Question 6:

The line represented by x = 7 is parallel to the x-axis. Justify whether the statement is true or not. From the figure it is clear that , the line x = 7 is parallel to the y-axis and not the x-axis.
Hence, the given statement is false.

#### Question 1:

For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have
(i) no solution?
(ii) infinitely many solutions?
(iii) a unique solution?

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

(i) no solution
In order to have no solution the condition is

On solving it we get ${\lambda }^{2}=1$  .....(1) and
${\lambda }^{2}\ne \lambda$  .....(2)
and ${\lambda }^{}\ne 1$
Hence the value of λ for which given pair of linear equations have no solution is$-$1.

(ii) infinitely many solutions
In order to have infinitely many solution the condition is

Solving it we get ${\lambda }^{2}=1$  .....(1) and
${\lambda }^{2}=\lambda$  .....(2)
and
Hence the value of λ  for which given pair of linear equations has infinitely many solutions is 1.

(iii) unique solution
In order to have unique solution the condition is

Solving it we get ${\lambda }^{2}\ne 1$
So for all real values of λ except 1 and $-$1 , the given pair of linear equations has unique solution.

#### Question 2:

For which value(s) of k will the pair of equations
kx + 3y = k – 3
12x + ky = k
have no solution?

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

In order to have no solution, the condition is

Hence, required value of k for which the given pair of linear equations has no solution is k$-$ 6.

#### Question 3:

For which values of a and b, will the following pair of linear equations have infinitely many solutions?
x + 2y = 1
(a b)x + (a + b)y = a + b – 2

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

In order to have infinite many solution, the condition is

On taking first and second terms, we get

On taking first and third terms, we get

On substituting b = 1 in (1), we get a = 3
Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

#### Question 4:

Find the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of equations:
(i) 3xy – 5 = 0 and 6x – 2yp = 0,

if the lines represented by these equations are parallel.

(ii) – x + py = 1 and px y = 1,
if the pair of equations has no solution.

(iii) – 3x + 5y = 7 and 2px – 3y = 1,
if the lines represented by these equations are intersecting at a unique point.

(iv) 2x + 3y – 5 = 0 and px – 6y – 8 = 0,
if the pair of equations has a unique solution.

(v) 2x + 3y = 7 and 2px + py = 28 – qy,
if the pair of equations have infinitely many solutions.

(i) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

If the lines are parallel then

Hence, the given pair of linear equations are parallel for all real values of p except 10.

(ii) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

In order to have no solution, the condition is

Hence, required value of p for which the given pair of linear equations has no solution is p = 1.

(iii) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

In order to have unique solution, the condition is

Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except $\frac{9}{10}$.

(iv) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

In order to have unique solution, the condition is

Hence, the pair of linear equations has a unique solution for all values of p except - 4.

(v) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

In order to have infinitely many solution, the condition is

From first and third terms, we get  $p=4$
From second and third terms, we get $12=p+q$
$⇒q=8$
Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

#### Question 5:

Two straight paths are represented by the equations x – 3y = 2 and –2x + 6y = 5.
Check whether the paths cross each other or not.

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get

As we can see that

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

#### Question 6:

Write a pair of linear equations which has the unique solution x = – 1, y = 3. How many such pairs can you write?

In order to have unique solution, the condition is
Let the equations be:
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$

Since, x $-$ 1 and y = 3 is the unique solution of these two equations, then it must satisfy the two above equations

Since different values of and  satisfy (1) and (2) respectively
Hence, infinitely many pairs of linear equations are possible.

#### Question 7:

If 2x + y = 23 and 4xy = 19, find the values of 5y – 2x and $\frac{y}{x}-2$.

Given equations are
2x + y  =  23    .....(1)
4x $-$ y = 19     .....(2)
On adding both equations, we get
6x = 42
So, x = 7
Put the value of x in (1), we get
2(7) + y = 23
y = 23 - 14
so, y = 9
Hence, the value of 5y $-$ 2x = 5(9) $-$ 2(7) = 45 $-$ 14 = 31

Also $\frac{y}{x}-2$ = $\frac{9}{7}-2$ =$\frac{-5}{7}$
Hence, the value of 5y $-$ 2x = 31 and the value of $\frac{y}{x}-2$ =$\frac{-5}{7}$.

#### Question 8:

Find the values of x and y in the following rectangle [see in Figure]. By property of rectangle, we know that opposite sides are equal i.e. CD = AB
So, x + 3y = 13 .....(1)
So, 3x + y = 7 .....(2)
On multiplying (2) by 3 and then subtracting from (1), we get
3(3x + y) = 3(7)
$⇒$x + 3y = 13
8x = 8
$⇒$x = 1
On substituting x = 1 in (1), we get
y = 4
Hence, the required values of x and y are 1 and 4 respectively.

#### Question 9:

Solve the following pairs of equations:
(i)

(ii)
$\frac{x}{3}+\frac{y}{4}=4\phantom{\rule{0ex}{0ex}}\frac{5x}{6}-\frac{y}{8}=4$

(iii)

(iv)

(v)
43x + 67y = – 24
67x + 43y = 24

(vi)

(vii)

(i)
Given pair of linear equations is

Now, multiplying (1) by 2 and then adding with (2), we get

$⇒5x=6\phantom{\rule{0ex}{0ex}}⇒x=\frac{6}{5}$
Now, put the value of x in (1), we get
$1.2+y=3.3$
$⇒y=2.1$
Hence, the required values of x and y are 1.2 and 2.1, respectively.

(ii)
Given pair of linear equations is

On multiplying both sides of (1) by LCM (3, 4) = 12 and multiplying both sides of (2) by LCM (6, 8) = 24, we get

Now, adding (3) and (4), we get
$24x=144\phantom{\rule{0ex}{0ex}}⇒x=6$
Now, substituting the value of x in (3), we get
$4×6+3y=48\phantom{\rule{0ex}{0ex}}⇒3y=24\phantom{\rule{0ex}{0ex}}⇒y=8$
Hence, the required values of x and y are 6 and 8 respectively.

(iii)
Given pair of linear equations is

Let $u=\frac{1}{y}$ then above equation becomes

On multiplying (3) by 8 and (4) by 6 and then adding both of them, we get
$32x+48u=120\phantom{\rule{0ex}{0ex}}36x-48u=84\phantom{\rule{0ex}{0ex}}$
$68x=204$
$⇒x=3$
Now, substituting the value of x in (3), we get
$4×3+6u=15\phantom{\rule{0ex}{0ex}}⇒6u=3\phantom{\rule{0ex}{0ex}}⇒u=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$
Since, $y=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$
$⇒y=2$
Hence, the required values of x and y are 3 and 2 respectively.

(iv) Given pair of linear equations is

Let $u=\frac{1}{x}$ and $v=\frac{1}{y}$then the equation becomes

On, multiplying (4) by 2 and then adding with (3) we get
$\left(u-2v\right)+\left(4u+2v\right)=-2+32\phantom{\rule{0ex}{0ex}}⇒5u=30\phantom{\rule{0ex}{0ex}}⇒u=6$
Now, put the value of u in (4), we get
$2×6+v=16\phantom{\rule{0ex}{0ex}}⇒v=4$
Since,
$u=\frac{1}{x}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{6}$
As
$v=\frac{1}{y}\phantom{\rule{0ex}{0ex}}⇒y=\frac{1}{4}$
Hence, the required values of x and y are $\frac{1}{6}$ and $\frac{1}{4}$ respectively.

(v) Given pair of linear equations is
43x + 67= – 24          .....(1)
67+ 43= 24             .....(2)
On multiplying (1) by 43 and (2) by 67 and then subtracting both of them, we get
$\left[{43}^{2}x+67\left(43\right)y\right)\right]-\left[{67}^{2}x+43\left(67\right)y\right]=\left(-24×43\right)-\left(24×67\right)\phantom{\rule{0ex}{0ex}}⇒{43}^{2}x-{67}^{2}x=-24\left(43+67\right)\phantom{\rule{0ex}{0ex}}⇒{67}^{2}x-{43}^{2}x=24×110\phantom{\rule{0ex}{0ex}}⇒110×24x=24×110\phantom{\rule{0ex}{0ex}}⇒x=1$
Now, put the value of x in (1), we get
$43×1+67×y=-24\phantom{\rule{0ex}{0ex}}⇒67×y=-67\phantom{\rule{0ex}{0ex}}⇒y=-1$
Hence, the required values of x and y are 1 and $-$1 respectively.

(vi) Given pair of linear equations is

On multiplying (1) by $\frac{1}{a}$ and then subtracting from (2), we get

Now, putting the value of y in (2), we get
$\frac{x}{{a}^{2}}+\frac{{b}^{2}}{{b}^{2}}=2\phantom{\rule{0ex}{0ex}}⇒\frac{x}{{a}^{2}}+1=2\phantom{\rule{0ex}{0ex}}⇒\frac{x}{{a}^{2}}=1\phantom{\rule{0ex}{0ex}}⇒x={a}^{2}$
Hence, the required values of x and y are  ${a}^{2}$ and ${b}^{2}$ respectively.

(vii) Given pair of linear equations is

Now put and $\frac{1}{y}=v$ then the pair of equation becomes

Now substracting (4) from (3) we get
$3v=\frac{-6}{3}\phantom{\rule{0ex}{0ex}}⇒v=\frac{-2}{3}$
Now putting value of v in (3), we get
$u=\frac{6}{3}=2$

$x=\frac{1}{u}=\frac{1}{2}$

$y=\frac{1}{v}=-\frac{3}{2}$

Hence, the required values of x and y are $\frac{1}{2}$ and $-\frac{3}{2}$ respectively.

#### Question 10:

Find the solution of the pair of equations .
Hence, find λ, if y = λx + 5.

Given pair of equations is

Now, multiplying both sides of (1) by LCM (10, 5) = 10, we get

Again, multiplying both sides of (2) by LCM (8, 6) = 24, we get

On, multiplying (3) by 2 and then subtracting from (4), we get
$\left(3x-2x\right)+\left(4y-4y\right)=340\phantom{\rule{0ex}{0ex}}⇒x=340$
Put the value of x in (3), we get
$340+2y=10\phantom{\rule{0ex}{0ex}}⇒y=-165$
Given, y = λx + 5
$⇒-165=\lambda ×340+5\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{-1}{2}$
Hence, the value of λ is $\frac{-1}{2}.$

#### Question 11:

By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.
(i) 3x + y + 4 = 0

6x – 2y + 4 = 0

(ii) x – 2y = 6
3x – 6y = 0

(iii) x + y = 3
3x + 3y = 9

(i) Given equations are
3x + y + 4 = 0      .....(1)
When x = 0, then y = $-$ 4
When x$-$ 1, then y = $-$ 1

6x $-$ 2y + 4 = 0      .....(2)
When x = 0, then y = 2
When x$-$ 1, then y$-$1 Plotting the points B(0, $-$ 4) and A( $-$ 2, 2), we get the straight line AB. Plotting the points Q(0, 2) and P(1, 5) we get the straight line PQ. The lines AB and PQ intersect at C ( $-$ 1, $-$ 1).

(ii) Given equations are
x – 2y = 6   .....(1)
When x = 0, then y$-$ 3
When x = 6, then y = 0

3x $-$ 6= 0   .....(2)
When x = 0, then y = 0
When x = 2, then y = 1 On plotting we find the two lines to be parallel
$\therefore$ The following pair of equations are not consistent.

(iii)  Given equations are
x + y = 3     ..…(1)
When x = 0, then y = $-$ 3
When x = 3, then y = 0

3x + 3y = 9   .....(2)
When x = 0, then y = $-$ 3
When x = 3, then y = 0 $\therefore$ The given pair of linear equations is coincident and having infinitely many solutions.

#### Question 12:

Draw the graph of the pair of equations 2x + y = 4 and 2xy = 4. Write the vertices of the triangle formed by these lines and the y-axis. Also find the area of this triangle.

The given pair of linear equations

Table for (1)

 x 0 2 y 4 0 Points A B

Table for (2)
 x 0 2 y $-$ 4 0 Points C B

Graphical representation of both lines. Here, both lines and y-axis form triangle.
Hence, the vertices of triangle ABC are A (0, 4), B(2, 0) and C(0, $-$ 4).
∴ Required area of $△$ABC= 2 $△$Area of AOB

ABC = $2×\left(\frac{1}{2}×4×2\right)$ = 8 square units.
Hence, the required area of $△$ABC is 8 square units.

#### Question 13:

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x y = 1. How many such lines can we find?

Given pair of linear equations is

Comparing with ax + by + c = 0, we get

$⇒\frac{{a}_{1}}{{a}_{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{b}_{1}}{{b}_{2}}=\frac{1}{-1}$
Since $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$
So, both lines intersect at a point. Therefore, the pair of equations has a unique solution. Hence, these equations are consistent.

Now, x + y = 2 or y = 2 $-$ x

If x = 0 then y = 2 and if x = 2 then y = 0.
If x = 0 then y = $-$ 1 if x =$\frac{1}{2}$ then y = 0 and if x = 1 then y = 1.
Plotting the points A (2,0) and B(0,2), we get the straight line AB. Plotting the points C(0, $-$ 1) and D$\left(\frac{1}{2},0\right)$, we get the straight line CD.
The lines AB and CD intersect at E (1, 1). Hence, infinite lines can pass through the intersection point of linear equations, like as y = x, x + 2y = 3, x + y = 2 and so on.

#### Question 14:

If x + 1 is a factor of 2x3 + ax2 + 2bx + 1, then find the values of a and b given that 2a – 3b = 4.

Given that (x + 1) is a factor of 2x3 + ax2 + 2bx + 1.
Let f(x) = 2x3 + ax2 + 2bx + 1,
then f($-$1) = 0
[We know that if $x+\alpha$ is a factor of f(x)= $a{x}^{2}+bx+c=0$ then f( $-$α) = 0]
f( $-$ 1) = 0
$⇒$$-$ 2 + a $-$ 2b + 1 = 0
$⇒$a $-$ 2b $-$ 1 = 0       …..(1)
Also, 2a $-$ 3b = 4        .....(2)
Solving (1) and (2), we get
$⇒$3b = 2a $-$ 4
$b=\frac{2a-4}{3}$
Now, put the value of b in (1), we get
$a-2\left(\frac{2a-4}{3}\right)-1=0$
$⇒$3a $-$ 2(2a $-$ 4) $-$ 3 = 0
$⇒$3a $-$ 4a + 5 = 0
$⇒$a = 5
Now, put the value of a in (1), we get
$-$ 2b $-$ 1 = 0
$⇒$4 = 2b
$⇒$b = 2
Hence, the required values of a and b are 5 and 2 respectively.

#### Question 15:

The angles of a triangle are x, y and 40°. The difference between the two angles x and y is 30°. Find x and y.

Given the angles of a triangle are x, y and 40°
So, x + y + 40 = 180    [since, the sum of all the angles of a triangle is 180$°$]
And hence x + y = 140    ..... (1)
Also, difference of angles = x $-$ y = 30     .....(2)
Adding (1) and (2), we get
2x = 170
So, x = 85
On putting, the value of x in (1), we get
y = 55
Hence, the required values of x and y are 85$°$ and 55$°$, respectively.

#### Question 16:

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

Let Salim and his daughter’s age be x and y years respectively.
Now, by first condition
Two years ago, Salim was thrice as old as his daughter.
i.e. $-$ 2 = 3(y $-$ 2)
so, x $-$ 3y$-$ 4     .....(1)
and by second condition, six years later. Salim will be four years older than twice her age.
(x + 6) = 2(y + 6) + 4
x + 6 = 2y + 16
x – 2y = 10    .....(2)
On subtracting (1) from (2), we get
x – 2y = 10
$-$ (– 3y$-$4)
We get, y = 14
Put the value of y in (2), we get
x $-$ 2(14) = 10
so, x = 38
Hence, Salim and his daughter’s age are 38 year and 14 year respectively.

#### Question 17:

The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Let the present age (in year) of father and his two children be x, y and z year respectively.
Now, by given condition, x = 2(y + z)     .....(1)
And after 20 year, (x + 20) = (y + 20) + (z + 20)
So, x = y + z + 20
Or y + z = x  $-$ 20     .....(2)
Substituting the value of (y + z) in (1), we get the present age of father
x = 2($-$ 20)
x = 40
Hence, the father’s age is 40 year.

#### Question 18:

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Let the two numbers be x and y.
Then, by first condition ratio of these two numbers = 5 : 6

And by second condition, then 8 is subtracted from each of the numbers, then ratio becomes 4 : 5
$\frac{x-8}{y-5}=\frac{4}{5}$
$⇒$5x – 4y = 8      .....(2)
Now, put the value of y in (2), we get
$5x-4\left(\frac{6}{5}\right)x=8$
$⇒\frac{x}{5}=8$
$⇒x=40$
Substituting the value of x in (1), we get
$y=\frac{6}{5}×40=48$
Hence, the required numbers are 40 and 48.

#### Question 19:

There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.

Let the number of students in halls A and B are x and y, respectively.
Now, by given condition, x – 10 = y + 10
x $-$ y = 20   .....(1)
and (x + 20) = 2(y $-$ 20)
– 2y = $-$60     .....(2)
On subtracting (2) from (1), we get
(x $-$ y) $-$ (x $-$ 2y) = 80
$⇒$y = 80
On putting y = 80 in (1), we get
x – 80 = 20
so, x = 100
Hence, 100 students are in hall A and 80 students are in hall B.

#### Question 20:

A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 16 for the book kept for four days. Find the fixed charges and the charge for each extra day.

Let the fixed charge be Rs x and additional charge for each day thereafter be Rs y.
Now by first condition,
Latika paid Rs 22 for a book kept for six days i.e. for first two days as Rs x and remaining Rs 4y.
x + 4y = 22     .....(1)
and by second condition,
Anand paid Rs 16 for a book kept for four days i.e. for first two days as Rs x and remaining Rs 2y.
x + 2y = 16     .....(2)
Now, subtracting (2) from (1), we get
2y = 6
y = 3
On substituting the value of y in (2), we get
x + 2(3) = 16
x = 10
Hence, the fixed charge is Rs 10 and the charge for each extra day Rs 3.

#### Question 21:

In a competitive examination, one mark is awarded for each correct answer while $\frac{1}{2}$ mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

Let x be the number of correct answers of the questions in a competitive examination, which results in (120 - x) as the number of wrong answers of the questions.
$⇒x×1-\left(120-x\right)×\frac{1}{2}=90\phantom{\rule{0ex}{0ex}}⇒x-\frac{\left(120-x\right)}{2}=90\phantom{\rule{0ex}{0ex}}⇒x+\frac{x}{2}-60=90\phantom{\rule{0ex}{0ex}}⇒\frac{3x}{2}=150\phantom{\rule{0ex}{0ex}}⇒x=100$
Hence, Jayanti answered 100 questions correctly .

#### Question 22:

The angles of a cyclic quadrilateral ABCD are
∠A = (6x + 10)°, ∠B = (5x
∠C = (x + y)°,     ∠D = (3y – 10)°
Find x and y, and hence the values of the four angles.

We know that, by property of cyclic quadrilateral, sum of opposite angles = 180$°$
∠A + ∠C = 180$°$
$⇒$(6x + 10)° + (x + y)° = 180$°$
So, 7x + y = 170          .....(1)
Similarly,
∠B + ∠D = 180$°$
$⇒$(5x)° + (3y – 10)° = 180$°$
So, 5x + 3y = 190       .....(2)
On multiplying (1) by 3 and then subtracting (2) from it, we get
3(7x + y) – (5x + 3y) = 3(170) – 190
16x = 320
$⇒x=20°$
Putting  $x=20°$ in (1), we get
7(20) + y = 170
$⇒y=30°$
Substituting the values of and y, we get
∠A = (6$×20$ + 10)° = 130° , ∠B = (5$×20$)° = 100°
∠C = ($20$ + 30)° = 50°, ∠D = (3$×30$ – 10)° = 80°
Hence, the values of the four angles i.e. ∠A, ∠B, ∠C, ∠D are 130°, 100°, 50° and 80° respectively.

#### Question 1:

Graphically, solve the following pair of equations:
2x + y = 6
2x y + 2 = 0
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Given equations are 2x + y = 6 and 2$-$ y + 2 = 0
Finding points for equation 2x + y $-$ 6 = 0,
We get for x = 0, y = 6 and for y = 0, x = 3.
Similarly, for equation 2x $-$ y + 2 = 0,
We get for x = 0, y = 2 and for y = 0, x = $-$1
Let ${A}_{1}$ and ${A}_{2}$ represent the areas of respectively where E is the intersection of lines. ${A}_{1}$= Area of $\frac{1}{2}×\mathrm{AC}×\mathrm{PE}$ =$\frac{1}{2}×4×4$ = 8
${A}_{2}$= Area of $\frac{1}{2}×\mathrm{BD}×\mathrm{QE}$ =$\frac{1}{2}×4×1$= 2
${A}_{1}:{A}_{2}=8:2=4:1$
Hence, the pair of equations intersect graphically at point E(1,4) i.e. x = 1 and y = 4. Also, the ratio of the areas of two triangle is 4 : 1.

#### Question 2:

Determine, graphically, the vertices of the triangle formed by the lines
y = x,     3y = x,     x + y = 8

Given linear equations are

y = x       .....(1)
3y = x     .....(2)
x + y = 8 .....(3)

For equation y = x,
If x = 1, then y = 1
If x = 0, then y = 0
If x = 2, then y = 2

For equation x = 3y,
If x = 0, then y = 0
If x = 3, then y = 1
If x = 6, then y = 2

For equation x + y = 8 or x = 8 $-$ y,
If x = 0, then y = 8
If x = 8, then y = 0
If x = 4, then y = 4

Plotting the points A (1, 1) and B (2, 2), we get the straight line AB.
Plotting the points C (3, 1) and D (6, 2), we get the straight line CD.
Plotting the point P (0, 8), Q (4, 4) and R (8, 0), we get the straight line PQR.
We see that lines AB and CD intersecting the line PR on Q and D respectively. So $∆\mathrm{OQD}$ is formed by these lines. Hence, the vertices of the $∆\mathrm{OQD}$ formed by the given lines are O(0, 0), Q(4, 4) and D(6, 2).

#### Question 3:

Draw the graphs of the equations x = 3, x = 5 and 2x y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x-axis.

Given equation of lines x = 3, x = 5 and 2x $-$ y $-$ 4 = 0.
Now, for line 2x $-$ y $-$ 4 = 0, we get
For x = 0, y$-$ 4 and for y = 0, x = 2
Draw the points P(0, $-$ 4) and Q(2, 0) and join these points and form a line PQ also draw the lines x = 3 and x = 5. Area of Quadrileral ABCD = $\frac{1}{2}$ × (distance between parallel lines)                 [since, quadrilateral ABCD is a trapezium]
$=\frac{1}{2}×\mathrm{AB}×\left(\mathrm{AD}+\mathrm{BC}\right)$
= 8 sq units                                                                                                 [since AB = OB $-$ OA = 5 $-$ 3 = 2, AD = 2 and BC = 6]
Hence, the area of the quadrilateral formed by the lines and the x-axis is 8 sq units.

#### Question 4:

The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Let the cost of a pen be Rs x and the cost of a pencil box be Rs y.
4x + 4y = 100
Or x + y = 25 .....(1)
and 3x = y + 15
3x $-$ y = 15 .....(2)
Adding (1) and (2), we get
4x = 40
So, x = 10
By substituting x = 10 in (1), we get
y = 25 $-$ 10 = 15
Hence, the cost of a pen and a pencil box are Rs 10 and Rs 15 respectively.

#### Question 5:

Determine, algebraically, the vertices of the triangle formed by the lines
3xy = 3
2x – 3y = 2
x + 2y = 8

Given equation of lines are

3x $-$ y = 3               .....(1)
2x $-$ 3y = 2             .....(2)
x + 2y = 8              .....(3)

Let lines (1), (2) and (3) represent the side of a ∆ABC i.e. AB, BC and CA respectively.
Solving lines (1) and (2), we will get the intersecting point B.
Multiplying  (1) by 3 and then subtracting (2), we get
(9$-$ 3y$-$ (2$-$ 3y) = 9 $-$ 2
$⇒$7x = 7
$⇒$x = 1
Substituting the value of x in (1), we get
3×1 $-$ y = 3
$⇒$y = 0
So, the coordinate of point or vertex B is (1, 0)

Solving lines (2) and (3), we will get the intersecting point C.
Multiplying (3) by 2 and then subtracting (2), we get
(2x + 4y$-$ (2x $-$ 3y) = 16 $-$ 2
7y = 14
$⇒$y = 2
Substituting the value of y in (3), we get
$⇒$x = 4
Hence, the coordinate of point or vertex C is (4, 2).

Solving lines (3) and (1), we will get the intersecting point A.
Multiplying (1) by 2 and then adding (3), we get
(6$-$ 2y) + (x + 2y) = 6 + 8
$⇒$7x = 14
$⇒$x = 2
Substituting the value of x in (1), we get
3×2 $-$ y = 3
$⇒$y = 3
So, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the ∆ABC formed by the given lines are A (2, 3), B(1, 0) and C (4, 2).

#### Question 6:

Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus. On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus.

Let the speed of the rickshaw and the bus are x and y km/h, respectively.
Now, she has taken time to travel 2 km by rickshaw,  ${t}_{1}=\frac{2}{x}$ hr ,
Remaining distance travelled by bus = 12 km
Time taken by bus to cover remaining distance, ${t}_{2}=\frac{12}{y}$ hr
By first condition,
.....(1)
By second condition,
Time taken by her to travel 4 km by rickshaw,  ${t}_{3}=\frac{4}{x}$ hr
Remaining distance travelled by bus = 10 km
So, Time taken to travel remaining distance by bus =  ${t}_{4}=\frac{10}{y}$ hr
Now,
${t}_{3}+{t}_{4}=\frac{1}{2}+\frac{9}{60}=\frac{13}{20}\phantom{\rule{0ex}{0ex}}$

.....(2)

Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$ then (1) and (2) becomes
.....(3)
.....(4)
Multiplying (3) by 2 and then subtract (4) from (3), we get
$\left(4u+24v\right)-\left(4u+10v\right)=1-\frac{13}{20}\phantom{\rule{0ex}{0ex}}⇒v=\frac{1}{40}$
Now, put the value of v in (3), we get
$2u+12\left(\frac{1}{40}\right)=\frac{1}{2}$
$2u=\frac{2}{10}$
$⇒u=\frac{1}{10}$
$⇒x=\frac{1}{u}=10$ km/h
km/h
Hence, the speed of rickshaw and the bus are 10 km/h and 40 km/h, respectively.

#### Question 7:

A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.

Let the speed of the stream be v km/h.
Given that, speed of person rowing in still water = 5 km/h
The speed of a person rowing in downstream = (5 + v) km/h
and the speed of a person has rowing in upstream = (5 $-$ v) km/h
Now, the time taken by person to cover 40 km downstream,
Time taken by person to cover 40 km upstream,
By condition, ${t}_{2}=3×{t}_{1}$
$⇒\frac{40}{5-v}=3×\frac{40}{5+v}$
$⇒\frac{1}{5-v}=\frac{3}{5+v}$
$⇒5+v=15-3v$
$⇒4v=10$

Hence, the speed of the stream is 2.5 km/h.

#### Question 8:

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Let the speed of the motorboat in still water and the speed of the stream are u km/h and v km/h respectively.
Then, a motorboat speed in downstream = (u + v) km/h
and a motorboat speed in upstream = (u $-$ v) km/h
Motorboat has taken time to travel 30 km upstream,
${t}_{1}=\frac{30}{u-v}\mathrm{hr}$
And motorboat has taken time to travel 28 km downstream,
${t}_{2}=\frac{28}{u+v}\mathrm{hr}$
By first condition, a motorboat can travel 30 km upstream and 38 km downstream in 7 h i.e. ${t}_{1}+{t}_{2}=7$

Now, motorboat has taken time to travel 21 km upstream and return i.e. ${t}_{3}=\frac{21}{u-v}\mathrm{hr}$ [for upstream] and ${t}_{4}=\frac{21}{u+v}\mathrm{hr}$ [for downstream]
By second condition, ${t}_{4}+{t}_{3}=5$

Let
Now (1) and (2) becomes

Multiplying (4) by 28 and substracting from (3), we get
$\left(28x-28x\right)+\left(30y-28y\right)=7-\frac{5}{21}×28$
$⇒2y=7-\frac{20}{3}$
$⇒y=\frac{1}{6}$
Substituting value of y in (4), we get
$x+\frac{1}{6}=\frac{5}{21}$
$⇒x=\frac{5}{21}-\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{14}$
As

Hence, the speed of the motorboat in still water is 10 km/h and the speed of the stream 4 km/h.

#### Question 9:

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Let the digits be x and y, then two-digit number = 10x + y

Case I:  Multiplying the sum of the digits by 8 and then subtracting 5, we get two digit number
$⇒8×\left(x+y\right)-5=10x+y\phantom{\rule{0ex}{0ex}}⇒8x+8y-5=10x+y\phantom{\rule{0ex}{0ex}}$
.....(1)

Case II: Multiplying the difference of the digits by 16 and then adding 3, we get two digit number
$⇒16×\left(x-y\right)+3=10x+y\phantom{\rule{0ex}{0ex}}⇒16x-16y+3=10x+y\phantom{\rule{0ex}{0ex}}$
.....(2)

Now, multiplying (1) by 3 and then subtracting from (2), we get
$⇒\left(6x-17y\right)-\left(6x-21y\right)=-3-\left(-15\right)\phantom{\rule{0ex}{0ex}}⇒4y=12\phantom{\rule{0ex}{0ex}}⇒y=3$
Now, put the value of y in (1), we get
$2x-21=-5\phantom{\rule{0ex}{0ex}}⇒2x=16\phantom{\rule{0ex}{0ex}}⇒x=8$

The required two-digit number is
=$10×8+3$
=$83$
Hence, the number is 83.

#### Question 10:

A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the station A to B costs Rs 2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B costs Rs 3810. Find the full first class fare from station A to B, and also the reservation charges for a ticket.

Let the cost of full first class fare be Rs and reservation charges be Rs y per ticket.

Case I : The cost of one reserved first class ticket from the stations A to B  is = Rs 2530
.....(1)
Case II : The cost of one reserved first class ticket and one reserved first class half ticket from stations A to B = Rs 3810

.....(2)
Now, multiplying (1) by 4 and then subtracting from (2), we get
$\left(3x+4y\right)-\left(4x+4y\right)=7620-10120\phantom{\rule{0ex}{0ex}}⇒-x=-2500\phantom{\rule{0ex}{0ex}}⇒x=2500$
On putting the value of x in (1), we get
$⇒2500+y=2530\phantom{\rule{0ex}{0ex}}⇒y=30$
Hence, full first class fare from stations A to B is Rs 2500 and the reservation charges for a ticket is Rs 30.

#### Question 11:

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby, getting a sum Rs 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs 1028. Find the cost price of the saree and the list price (price before discount) of the sweater.

Let the cost price of the saree and the list price of the sweater be Rs x and Rs y, respectively.
Case I: Selling price of saree at 8% profit + Selling price of sweater at 10% discount = Rs 1008

.....(1)

Case II: Selling price of saree at 10% profit + Selling price of sweater at 8% discount = Rs 1028

.....(2)

Substituting the value of y from (1) into (2), we get

$⇒$1.1 × 0.9x + 0.92 × (1008 $-$ 1.08x) = 1028 × 0.9
$⇒$0.99x $-$ 0.9936x = 925.2 $-$ 927.36
$⇒$$-$0.0036x = $-$2.16
$⇒x=\frac{2.16}{0.0036}=600$

Substituting the value of x in (1), we get
$1.08×600+0.9y=1008\phantom{\rule{0ex}{0ex}}⇒0.9y=1008-648\phantom{\rule{0ex}{0ex}}⇒y=\frac{360}{0.9}=400$
Hence, the cost price of the saree and the list price (price before discount) of the sweater are Rs 600 and Rs 400 respectively.

#### Question 12:

Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme?

Let the amount of investments in schemes A and B be Rs x and Rs y respectively.

Case I: Interest at the rate of 8% per annum on scheme A + Interest at the rate of 9% per annum on scheme B = Total amount received
$⇒\frac{\left(x×8×1\right)}{100}+\frac{\left(y×9×1\right)}{100}=1860$

So, 8x + 9y = 186000     .....(1)

Case II: Interest at the rate of 9% per annum on scheme A + Interest at the rate of 8% per annum on scheme B = ₹ 20 more as annual Interest
$⇒\frac{\left(x×9×1\right)}{100}+\frac{\left(y×8×1\right)}{100}=20+1860$
So, 9x + 8y = 188000      .....(2)
Multiplying (1) by 9 and (2) by 8 and then subtracting them, we get
$⇒\left(72x+81y\right)-\left(72x+64y\right)=186000×9-188000×8$
$⇒$17y = 170000
$⇒y=10000$
Substituting the value of y in (1), we get
8x + 9(10000) = 186000
So x = 12000
Hence, she invested Rs 12000 and Rs 10000 in two schemes A and B respectively.

#### Question 13:

Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had.

Let the number of bananas in lot A and B be x and y, respectively
Case I :
Cost of the first lot at the rate of Rs 2 for 3 bananas + Cost of the second lot at the rate of Rs 1 per banana = Amount received

Case II :
Cost of the first lot at the rate of Rs 1 per banana + Cost of the second lot at the rate of Rs 4 for 5 bananas = Amount received

On multiplying (1) by 4 and (2) by 3 and then subtracting them, we get
(8x + 12y) – (15x + 12y) = 4800 – 6900
$⇒$$-$ 7x = $-$ 2100
$⇒$x = 300
Substituting the value of x in (1), we get
2(300) + 3y = 1200
3y = 600
$⇒$y = 200
∴ Total number of bananas = Number of bananas in lot A + Number of bananas in lot B = x + y
x + y = 300 + 200 = 500