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#### Page No 18:

#### Question 1:

Choose the correct answer from the given four options:

Graphically, the pair of equations

6*x* – 3*y* + 10 = 0

2*x* – *y* + 9 = 0

represents two lines which are

(A) intersecting at exactly one point.

(B) intersecting at exactly two points.

(C) coincident.

(D) parallel.

#### Answer:

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

we get

${a}_{1}=6,{b}_{1}=-3,{c}_{1}=10\phantom{\rule{0ex}{0ex}}{a}_{2}=2,{b}_{2}=-1,{c}_{2}=9$

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

As $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$, this means that lines are parallel, so there will be no solution.

Hence, the correct answer is option (D).

#### Page No 18:

#### Question 2:

Choose the correct answer from the given four options:

The pair of equations *x* + 2*y* + 5 = 0 and –3*x* – 6*y* + 1 = 0 have

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

#### Answer:

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

${a}_{1}=1,{b}_{1}=2,{c}_{1}=5\phantom{\rule{0ex}{0ex}}{a}_{2}=-3,{b}_{2}=-6,{c}_{2}=1$

$\begin{array}{rcl}\Rightarrow \frac{1}{-3}& =& \frac{2}{-6}\ne \frac{5}{1}\\ \Rightarrow -\frac{1}{3}& =& -\frac{1}{3}\ne \frac{5}{1}\end{array}$

As $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ which means that lines are parallel, so there will be no solution.

Hence, the correct answer is option (D).

#### Page No 18:

#### 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

#### Answer:

Consistent equations means the equations must have a solution which is obtained by either of conditions:

1. lines coincide (with infinitely many solutions)

2. lines intersect (with unique solution)

Hence, the correct answer is option (C).

#### Page No 18:

#### 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

#### Answer:

The give pair of equations are *y* = 0 and *y* = $-$7.

The equation *y* = 0 represents *x*-axis and *y* = $-$7 is a 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).

#### Page No 18:

#### 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*)

#### Answer:

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).

#### Page No 18:

#### Question 6:

Choose the correct answer from the given four options:

For what value of *k*, do the equations 3*x* – *y* + 8 = 0 and 6*x* – *ky* = –16 represent coincident lines?

(A) $\frac{1}{2}$

(B) $-\frac{1}{2}$

(C) 2

(D) –2

#### Answer:

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

${a}_{1}=3,{b}_{1}=-1,{c}_{1}=8\phantom{\rule{0ex}{0ex}}{a}_{2}=6,{b}_{2}=-k,{c}_{2}=16$

As $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$, which means that lines are coincident

$\begin{array}{ccc}\Rightarrow \frac{3}{6}& =& \frac{-1}{-k}=\frac{8}{16}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}\Rightarrow \frac{1}{2}& =& \frac{1}{k}=\frac{1}{2}\end{array}$

Solving we get *k* = 2.

Hence, the correct answer is option (C).

#### Page No 19:

#### Question 7:

Choose the correct answer from the given four options:

If the lines given by 3*x* + 2*ky* = 2 and 2*x* + 5*y* + 1 = 0 are parallel, then the value of *k* is

(A) $\frac{-5}{4}$

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

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

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

#### Answer:

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

${a}_{1}=3,{b}_{1}=2k,{c}_{1}=-2\phantom{\rule{0ex}{0ex}}{a}_{2}=2,{b}_{2}=5,{c}_{2}=1$

As lines are parallel which means that $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.

Substituting the values, we get $\begin{array}{rcl}& & \begin{array}{ccc}\frac{3}{2}& =& \frac{2k}{5}\ne \frac{-2}{1}\end{array}\\ & & \end{array}$.

$\Rightarrow $15 = 4k

$\Rightarrow k=\frac{15}{4}$

Hence, the correct answer is option (C).

#### Page No 19:

#### Question 8:

Choose the correct answer from the given four options:

The value of *c* for which the pair of equations *cx* – *y* = 2 and 6*x* – 2*y* = 3 will have infinitely many solutions is

(A) 3

(B) – 3

(C) –12

(D) no value

#### Answer:

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

${a}_{1}=c,{b}_{1}=-1,{c}_{1}=-2\phantom{\rule{0ex}{0ex}}{a}_{2}=6,{b}_{2}=-2,{c}_{2}=-3$

As there are infinitely many solutions which means $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.

Substituting the values, we get $\begin{array}{rcl}& & \begin{array}{ccc}\frac{c}{6}& =& \frac{-1}{-2}=\frac{-2}{-3}\end{array}\\ & & \end{array}$.

$\Rightarrow $$\frac{c}{6}=\frac{1}{2}\mathrm{and}\frac{c}{6}=\frac{2}{3}$

$\Rightarrow c=3\mathrm{and}c=4$

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).

#### Page No 19:

#### Question 9:

Choose the correct answer from the given four options:

One equation of a pair of dependent linear equations is –5*x* + 7*y* = 2. The second equation can be

(A) 10*x* + 14*y* + 4 = 0

(B) –10*x* – 14*y* + 4 = 0

(C) –10*x* + 14*y* + 4 = 0

(D) 10*x *– 14*y *= –4

#### Answer:

The condition for pair of equation to be dependent linear equations is $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.

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

${a}_{1}=-5,{b}_{1}=7,{c}_{1}=-2\phantom{\rule{0ex}{0ex}}$

Let the second equation be ${a}_{2}x+{b}_{2}y+{c}_{2}=0$.

$\Rightarrow \frac{-5}{{a}_{2}}=\frac{7}{{b}_{2}}=\frac{-2}{{c}_{2}}=k$, where *k* is any arbitrary constant.

Putting $k=\frac{-1}{2}$, we get

${a}_{2}=10,{b}_{2}=-14,{c}_{2}=4$

$\therefore $The required equation becomes $10x\mathit{-}14y+4=0$

Hence, the correct answer is option (D).

#### Page No 19:

#### 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

*x*– 3

*y*= –5

*x*+ 5

*y*= –11

*x*+ 10

*y*= –22

*x*–

*y*= 1

*x*+ 2

*y*= 0

*x*– 4

*y*–14 = 0

*x*–

*y*– 13 = 0

#### Answer:

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

$2+(-3)=1$ and $2\times \left(2\right)-3\times (-3)=13\ne 5$

From option (B),we see that

$2\times 2+5\times (-3)=-11$ and $4\times \left(2\right)+10\times (-3)=-22$

Since, both the equations are satisfied by the unique solution *x* = 2 and *y* = $-$3.

Hence, the correct answer is option (B).

#### Page No 19:

#### 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

#### Answer:

Given:

$x-y=2$ .....(1)

$x+y=4$ .....(2)

Adding (1) and (2), we get

$2x=6$

$\Rightarrow x=3$

Substituting the value of *x *in (2), we get

$3+y=4$

$\Rightarrow y=1$

Hence, the correct answer is option (C).

#### Page No 19:

#### 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

#### Answer:

Let the number of ₹1 coins = *x*

and the number of ₹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 *y *in (1), we get $x=25$.

Hence, the correct answer is option (D).

#### Page No 19:

#### 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

#### Answer:

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

$(6y+4)=4y+16$

$\Rightarrow 2y=12$

$\Rightarrow y=6$

Substituting the value of *y* in (1), we get *$x=36$.*

Hence, the correct answer is option (C).

#### Page No 21:

#### Question 1:

Do the following pair of linear equations have no solution? Justify your answer.

(i) 2*x* + 4*y* = 3

12*y* + 6*x* = 6

(ii)* x *= 2*y*

*y*= 2

*x*

(iii) 3

*x*+

*y*– 3 = 0

#### Answer:

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 $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.

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

${a}_{1}=2,{b}_{1}=4,{c}_{1}=-3\phantom{\rule{0ex}{0ex}}{a}_{2}=6,{b}_{2}=12,{c}_{2}=-6$

$\begin{array}{ccc}\Rightarrow \frac{2}{6}& =& \frac{4}{12}\ne \frac{-3}{-6}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}\Rightarrow \frac{1}{3}& =& \frac{1}{3}\ne \frac{1}{2}\end{array}$

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:

${a}_{1}=1,{b}_{1}=-2,{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}=2,{b}_{2}=-1,{c}_{2}=0$

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

As $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$, 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:

${a}_{1}=3,{b}_{1}=1,{c}_{1}=-3\phantom{\rule{0ex}{0ex}}{a}_{2}=2,{b}_{2}=\frac{2}{3},{c}_{2}=-2$

$\begin{array}{ccc}\Rightarrow \frac{3}{2}& =& \frac{1}{{\displaystyle \frac{2}{3}}}=\frac{3}{2}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}\Rightarrow \frac{3}{2}& =& \frac{3}{2}=\frac{3}{2}\end{array}$

Hence, the given pair of linear equations is coincident.

#### Page No 21:

#### 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)

–2*x* – 3*y* = 1

6*y* + 4*x* = – 2

(iii)

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

#### Answer:

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 $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.

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

${a}_{1}=3,{b}_{1}=\frac{1}{7},{c}_{1}=-3\phantom{\rule{0ex}{0ex}}{a}_{2}=7,{b}_{2}=3,{c}_{2}=-7$

$\begin{array}{ccc}\Rightarrow \frac{3}{7}& \ne & \frac{{\displaystyle \frac{1}{7}}}{3}\end{array}\phantom{\rule{0ex}{0ex}}$

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:

$\begin{array}{rcl}{a}_{1}& =& -2,{b}_{1}=-3,{c}_{1}=-1\\ {a}_{2}& =& 4,{b}_{2}=6,{c}_{2}=2\end{array}$

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

$\begin{array}{c}\Rightarrow \frac{1}{2}=\frac{1}{2}\end{array}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$

Since $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ 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:

${a}_{1}=\frac{1}{2},{b}_{1}=1,{c}_{1}=\frac{2}{5}\phantom{\rule{0ex}{0ex}}{a}_{2}=4,{b}_{2}=8,{c}_{2}=\frac{5}{16}$

$\begin{array}{ccc}\Rightarrow \frac{{\displaystyle \frac{1}{2}}}{4}& =& \frac{1}{{\displaystyle 8}}\ne \frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{5}{16}}}\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}\Rightarrow \frac{1}{8}& =& \frac{1}{8}\ne \frac{32}{25}\end{array}$

Since $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ hence, the given pair of linear equations are parallel and does not represent a pair of coincident lines.

#### Page No 21:

#### Question 3:

Are the following pair of linear equations consistent? Justify your answer.

(i)

–3*x*– 4*y* = 12

4*y* + 3*x* = 12

(ii)

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

(iii)

2*ax* + *by *= *a*

4*ax *+ 2*by* – 2*a* = 0; *a, b * ≠ 0

(iv)

*x *+ 3*y *= 11

2 (2*x* + 6*y*) = 22

#### Answer:

The condition for linear equations to be consistent is either $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ for infinitely many solutions or $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$ for unique solutions.

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

${a}_{1}=-3,{b}_{1}=-4,{c}_{1}=-12\phantom{\rule{0ex}{0ex}}{a}_{2}=3,{b}_{2}=4,{c}_{2}=-12$

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

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

${a}_{1}=\frac{3}{5},{b}_{1}=-1,{c}_{1}=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=\frac{1}{5},{b}_{2}=-3,{c}_{2}=-\frac{1}{6}$

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

$\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

${a}_{1}=2a,{b}_{1}=b,{c}_{1}=-a\phantom{\rule{0ex}{0ex}}{a}_{2}=4a,{b}_{2}=2b,{c}_{2}=-2a$

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

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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 a*x + by + c* = 0, we get:

${a}_{1}=1,{b}_{1}=3,{c}_{1}=-11\phantom{\rule{0ex}{0ex}}{a}_{2}=4,{b}_{2}=12,{c}_{2}=-22$

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

$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

#### Page No 21:

#### Question 4:

For the pair of equations

λ *x* + 3*y* = –7

2*x* + 6*y* = 14

to have infinitely many solutions, the value of λ should be 1. Is the statement true? Give reasons.

#### Answer:

In order to have infinite solutions $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.

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

${a}_{1}=\lambda ,{b}_{1}=3,{c}_{1}=7\phantom{\rule{0ex}{0ex}}{a}_{2}=2,{b}_{2}=6,{c}_{2}=-14$

$\Rightarrow \frac{\lambda}{2}=\frac{3}{6}=\frac{7}{-14}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\lambda}{2}=\frac{1}{2}=-\frac{1}{2}$

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.

#### Page No 22:

#### Question 5:

For all real values of *c*, the pair of equations

*x* – 2*y* = 8

5*x* – 10*y* =* c*

have a unique solution. Justify whether it is true or false.

#### Answer:

False

In order to have unique solution $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.

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

${a}_{1}=1,{b}_{1}=-2,{c}_{1}=-8\phantom{\rule{0ex}{0ex}}{a}_{2}=5,{b}_{2}=-10,{c}_{2}=-c$

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*.

#### Page No 22:

#### Question 6:

The line represented by *x* = 7 is parallel to the *x*-axis. Justify whether the statement is true or not.

#### Answer:

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.

#### Page No 25:

#### Question 1:

For which value(s) of *λ*, do the pair of linear equations *λx* + *y* = *λ*^{2} and *x* + *λy* = 1 have

(i) no solution?

(ii) infinitely many solutions?

(iii) a unique solution?

#### Answer:

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

${a}_{1}=\lambda ,{b}_{1}=1,{c}_{1}=-{\lambda}^{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=1,{b}_{2}=\lambda ,{c}_{2}=-1$

(i) no solution

In order to have no solution the condition is $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.

$\Rightarrow \frac{\lambda}{1}=\frac{1}{\lambda}\ne \frac{-{\lambda}^{2}}{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

On solving it we get,

${\lambda}^{2}=1$ .....(1)

${\lambda}^{2}\ne \lambda $ .....(2)

$\Rightarrow \lambda =1,-1$ 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 $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.

$\frac{\lambda}{1}=\frac{1}{\lambda}=\frac{-{\lambda}^{2}}{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Solving it we get,

${\lambda}^{2}=1$ .....(1)

${\lambda}^{2}=\lambda $ .....(2)

$\Rightarrow \lambda =1,-1$ and $\lambda =0,1$

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 $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.

$\frac{\lambda}{1}\ne \frac{1}{\lambda}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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.

#### Page No 25:

#### Question 2:

For which value(s) of *k* will the pair of equations

*kx* + 3*y* = *k* – 3

12*x* + *ky* =* k*

have no solution?

#### Answer:

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

${a}_{1}=k,{b}_{1}=3,{c}_{1}=3-k\phantom{\rule{0ex}{0ex}}{a}_{2}=12,{b}_{2}=k,{c}_{2}=-k$

In order to have no solution, the condition is $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.

$\frac{k}{12}=\frac{3}{k}\ne \frac{3-k}{-k}$

$\Rightarrow {k}^{2}=36\mathrm{and}3k\ne {k}^{2}-3k\phantom{\rule{0ex}{0ex}}\Rightarrow k=6,-6\mathrm{and}k\ne 6,0\phantom{\rule{0ex}{0ex}}\Rightarrow k=-6$

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

#### Page No 25:

#### Question 3:

For which values of *a* and *b*, will the following pair of linear equations have infinitely many solutions?

*x* + 2*y *= 1

(*a* –* b*)*x *+ (*a* + *b*)*y* = *a* + *b* – 2

#### Answer:

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

$\begin{array}{rcl}{a}_{1}& =& 1,{b}_{1}=2,{c}_{1}=-1\\ {a}_{2}& =& a-b,{b}_{2}=a+b,{c}_{2}=-a-b+2\end{array}$

In order to have infinitely many solution, the condition is $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.

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

On taking first and second terms, we get

$a+b=2(a-b)\phantom{\rule{0ex}{0ex}}\Rightarrow a=3b.....\left(1\right)$

On taking first and third terms, we get

$a+b-2=a-b$

$\Rightarrow b=1.....\left(2\right)$

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.

#### Page No 25:

#### 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) 3*x* – *y* – 5 = 0 and 6*x* – 2*y* – *p* = 0,

(ii) –

*x*+

*py*= 1 and

*px*–

*y*= 1,

(iii) – 3

*x*+ 5

*y*= 7 and 2

*px*– 3

*y*= 1,

(iv) 2

*x*+ 3

*y*– 5 = 0 and

*px*– 6

*y*– 8 = 0,

(v) 2

*x*+ 3

*y*= 7 and 2

*px*+

*py*= 28 –

*qy*,

#### Answer:

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

${a}_{1}=3,{b}_{1}=-1,{c}_{1}=-5\phantom{\rule{0ex}{0ex}}{a}_{2}=6,{b}_{2}=-2,{c}_{2}=-p$

If the lines are parallel then $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$. So,

$\frac{3}{6}=\frac{-1}{-2}\ne \frac{-5}{-p}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow p\ne 10\phantom{\rule{0ex}{0ex}}$

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

${a}_{1}=-1,{b}_{1}=p,{c}_{1}=-1\phantom{\rule{0ex}{0ex}}{a}_{2}=p,{b}_{2}=-1,{c}_{2}=-1$

In order to have no solution, the condition is $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$. So,

$\frac{-1}{p}=\frac{p}{-1}\ne \frac{-1}{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {p}^{2}=1\mathrm{and}p\ne -1\phantom{\rule{0ex}{0ex}}\Rightarrow p=1,-1\mathrm{and}p\ne -1\phantom{\rule{0ex}{0ex}}\Rightarrow p=1$

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

${a}_{1}=-3,{b}_{1}=5,{c}_{1}=-7\phantom{\rule{0ex}{0ex}}{a}_{2}=2p,{b}_{2}=-3,{c}_{2}=-1$

In order to have unique solution, the condition is $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$. So,

$\Rightarrow \frac{-3}{2p}\ne \frac{5}{-3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow p\ne \frac{9}{10}$

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

${a}_{1}=2,{b}_{1}=3,{c}_{1}=-5\phantom{\rule{0ex}{0ex}}{a}_{2}=p,{b}_{2}=-6,{c}_{2}=-8$

In order to have unique solution, the condition is $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.

$\Rightarrow \frac{2}{p}\ne \frac{3}{-6}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow p\ne -4$

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

${a}_{1}=2,{b}_{1}=3,{c}_{1}=-7\phantom{\rule{0ex}{0ex}}{a}_{2}=2p,{b}_{2}=p+q,{c}_{2}=-28$

In order to have infinitely many solution, the condition is $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$. So,

$\Rightarrow \frac{2}{2p}=\frac{3}{p+q}=\frac{-7}{-28}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{1}{p}=\frac{3}{p+q}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

From first and third terms, we get $p=4$.

From second and third terms, we get $12=p+q$

$\Rightarrow q=8$

Hence, the pair of equations has infinitely many solutions for all values of *p *= 4 and *q *= 8.

#### Page No 26:

#### Question 5:

Two straight paths are represented by the equations *x* – 3*y *= 2 and –2*x* + 6*y* = 5.

Check whether the paths cross each other or not.

#### Answer:

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

${a}_{1}=1,{b}_{1}=-3,{c}_{1}=-2\phantom{\rule{0ex}{0ex}}{a}_{2}=-2,{b}_{2}=6,{c}_{2}=-5$

As we can see that $\frac{1}{-2}=\frac{-3}{6}\ne \frac{-2}{-5}$

$\Rightarrow \frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$

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

#### Page No 26:

#### Question 6:

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

#### Answer:

In order to have unique solution, the condition is $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.

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.

$\phantom{\rule{0ex}{0ex}}\Rightarrow -{a}_{1}+3{b}_{1}+{c}_{1}=0.....\left(1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -{a}_{2}+3{b}_{2}+{c}_{2}=0.....\left(2\right)$

Since different values of ${a}_{1},{b}_{1},{c}_{1}$and ${a}_{2},{b}_{2},{c}_{2}$ satisfy (1) and (2) respectively, hence, infinitely many pairs of linear equations are possible.

#### Page No 26:

#### Question 7:

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

#### Answer:

Given equations are

2*x* + *y *= 23 .....(1)

4*x* $-$ *y* = 19 .....(2)

On adding both equations, we get

6*x* = 42

So, *x *= 7

Putting the value of *x* in (1), we get

2(7) + *y* = 23

*y *= 23 − 14

So, *y* = 9

Hence, the value of 5*y* $-$ 2*x* = 5(9) $-$ 2(7) = 45 $-$ 14 = 31

Also $\frac{y}{x}-2$ = $\frac{9}{7}-2$ =$\frac{-5}{7}$.

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

#### Page No 26:

#### Question 8:

Find the values of *x* and *y* in the following rectangle [see in Figure].

#### Answer:

By property of rectangle, we know that opposite sides are equal i.e. CD = AB

So, *x* + 3*y* = 13 .....(1)

Similarly, AD = BC

So, 3*x* + *y* = 7 .....(2)

On multiplying (2) by 3 and then subtracting from (1), we get

8*x* = 8

$\Rightarrow $*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.

#### Page No 26:

#### Question 9:

Solve the following pairs of equations:

(i)

$x+y=3.3\phantom{\rule{0ex}{0ex}}\frac{0.6}{3x-2y}=-1,3x-2y\ne 0$

(ii)

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

(iii)

$4x+\frac{6}{y}=15\phantom{\rule{0ex}{0ex}}6x-\frac{8}{y}=14,y\ne 0$

(iv)

$\frac{1}{2x}-\frac{1}{y}=-1\phantom{\rule{0ex}{0ex}}\frac{1}{x}+\frac{1}{2y}=8,x,y\ne 0$

(v)

43*x* + 67*y *= – 24

67*x *+ 43*y *= 24

(vi)

$\frac{x}{a}+\frac{y}{b}=a+b\phantom{\rule{0ex}{0ex}}\frac{x}{{a}^{2}}+\frac{y}{{b}^{2}}=2,a,b\ne 0$

(vii)

$\frac{2xy}{x+y}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\frac{xy}{2x-y}=\frac{-3}{10},x+y\ne 0,2x-y\ne 0$

#### Answer:

(i) Given pair of linear equations is

$x+y=3.3.....\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{0.6}{3x-2y}=-1,3x-2y\ne 0$

$\Rightarrow 0.6=-3x+2y\phantom{\rule{0ex}{0ex}}\Rightarrow 3x-2y=-0.6.....\left(2\right)\phantom{\rule{0ex}{0ex}}$

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

$5x=6\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{6}{5}$

Now, putting the value of *x* in (1), we get

$1.2+y=3.3$

$\Rightarrow y=2.1$

Hence, the required values of *x* and *y* are 1.2 and 2.1, respectively.

(ii) Given pair of linear equations is

$\frac{x}{3}+\frac{y}{4}=4.....\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{5x}{6}-\frac{y}{8}=4.....\left(2\right)$

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

$4x+3y=48.....\left(3\right)\phantom{\rule{0ex}{0ex}}20x-3y=96.....\left(4\right)$

Now, adding (3) and (4), we get

$24x=144\phantom{\rule{0ex}{0ex}}\Rightarrow x=6$

Now, substituting the value of *x* in (3), we get

$4\times 6+3y=48\phantom{\rule{0ex}{0ex}}\Rightarrow 3y=24\phantom{\rule{0ex}{0ex}}\Rightarrow y=8$

Hence, the required values of *x* and *y* are 6 and 8 respectively.

(iii) Given pair of linear equations is

$4x+\frac{6}{y}=15.....\left(1\right)\phantom{\rule{0ex}{0ex}}6x-\frac{8}{y}=14,y\ne 0.....\left(2\right)$

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

$4x+6u=15.....\left(3\right)\phantom{\rule{0ex}{0ex}}6x-8u=14.....\left(4\right)$

On multiplying (3) by 8 and (4) by 6 and then adding both of them, we get

$32x+48u=120\phantom{\rule{0ex}{0ex}}36x-48u=84\phantom{\rule{0ex}{0ex}}$

$68x=204$

$\Rightarrow x=3$

Now, substituting the value of *x* in (3), we get

$4\times 3+6u=15\phantom{\rule{0ex}{0ex}}\Rightarrow 6u=3\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$

Since, $y=\frac{1}{u}$

$\Rightarrow y=2$

Hence, the required values of *x* and *y* are 3 and 2 respectively.

(iv) Given pair of linear equations is

$\frac{1}{2x}-\frac{1}{y}=-1.....\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{1}{x}+\frac{1}{2y}=8,x,y\ne 0.....\left(2\right)$

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

$\frac{u}{2}-v=-1\phantom{\rule{0ex}{0ex}}\Rightarrow u-2v=-2.....\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}u+\frac{v}{2}=8\phantom{\rule{0ex}{0ex}}\Rightarrow 2u+v=16.....\left(4\right)$

On, multiplying (4) by 2 and then adding with (3) we get

$(u-2v)+(4u+2v)=-2+32\phantom{\rule{0ex}{0ex}}\Rightarrow 5u=30\phantom{\rule{0ex}{0ex}}\Rightarrow u=6$

Now, putting the value of *u* in (4), we get

$2\times 6+v=16\phantom{\rule{0ex}{0ex}}\Rightarrow v=4$

Since,

$u=\frac{1}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{1}{6}$

As

$v=\frac{1}{y}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{1}{4}$

Hence, the required values of *x* and *y* are $\frac{1}{6}$ and $\frac{1}{4}$ respectively.

(v) Given pair of linear equations is

43*x* + 67*y *= – 24 .....(1)

67*x *+ 43*y *= 24 .....(2)

On multiplying (1) by 43 and (2) by 67 and then subtracting both of them, we get

$\left[{43}^{2}x+67\left(43\right)y)\right]-\left[{67}^{2}x+43\left(67\right)y\right]=(-24\times 43)-(24\times 67)\phantom{\rule{0ex}{0ex}}\Rightarrow {43}^{2}x-{67}^{2}x=-24(43+67)\phantom{\rule{0ex}{0ex}}\Rightarrow {67}^{2}x-{43}^{2}x=24\times 110\phantom{\rule{0ex}{0ex}}\Rightarrow 110\times 24x=24\times 110\phantom{\rule{0ex}{0ex}}\Rightarrow x=1$

Now, putting the value of *x* in (1), we get

$43\times 1+67\times y=-24\phantom{\rule{0ex}{0ex}}\Rightarrow 67\times y=-67\phantom{\rule{0ex}{0ex}}\Rightarrow y=-1$

Hence, the required values of *x* and *y* are 1 and $-$1 respectively.

(vi) Given pair of linear equations is

$\frac{x}{a}+\frac{y}{b}=a+b.....\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{x}{{a}^{2}}+\frac{y}{{b}^{2}}=2,a,b\ne 0.....\left(2\right)$

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

$\frac{x}{{a}^{2}}-\frac{x}{{a}^{2}}+\frac{y}{{b}^{2}}-\frac{y}{ab}=2-1-\frac{b}{a}\phantom{\rule{0ex}{0ex}}\Rightarrow y\left(\frac{1}{{b}^{2}}-\frac{1}{ab}\right)=\left(1-\frac{b}{a}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y={b}^{2}\phantom{\rule{0ex}{0ex}}$

Now, putting the value of *y* in (2), we get

$\frac{x}{{a}^{2}}+\frac{{b}^{2}}{{b}^{2}}=2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{{a}^{2}}+1=2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{{a}^{2}}=1\phantom{\rule{0ex}{0ex}}\Rightarrow x={a}^{2}$

Hence, the required values of *x* and *y* are ${a}^{2}$ and ${b}^{2}$ respectively.

(vii) Given pair of linear equations is

$\frac{2xy}{x+y}=\frac{3}{2},x+y\ne 0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x+y}{2xy}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{2xy}+\frac{y}{2xy}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2y}+\frac{1}{2x}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{4}{3}.....\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{xy}{2x-y}=\frac{-3}{10},x+y\ne 0,2x-y\ne 0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2x-y}{xy}=\frac{10}{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2x}{xy}-\frac{y}{xy}=-\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{y}-\frac{1}{x}=-\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{x}-\frac{2}{y}=\frac{10}{3}.....\left(2\right)$

Now putting $\frac{1}{x}=u$and $\frac{1}{y}=v$ in (1) and (2), the pair of equation becomes

$u+v=\frac{4}{3}.....\left(3\right)\phantom{\rule{0ex}{0ex}}u-2v=\frac{10}{3}.....\left(4\right)$

Substracting (4) from (3) we get

$3v=\frac{-6}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{-2}{3}$

Putting the 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.

#### Page No 27:

#### Question 10:

Find the solution of the pair of equations $\frac{x}{10}+\frac{y}{5}-1=0\mathrm{and}\frac{x}{8}+\frac{y}{6}=15$.

Hence, find λ, if *y* = λ*x* + 5.

#### Answer:

Given pair of equations is

$\frac{x}{10}+\frac{y}{5}-1=0.....\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{x}{8}+\frac{y}{6}=15.....\left(2\right)$

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

$x+2y=10.....\left(3\right)$

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

$3x+4y=360.....\left(4\right)$

On, multiplying (3) by 2 and then subtracting from (4), we get

$(3x-2x)+(4y-4y)=340\phantom{\rule{0ex}{0ex}}\Rightarrow x=340$

Put the value of *x* in (3), we get

$340+2y=10\phantom{\rule{0ex}{0ex}}\Rightarrow y=-165$

Given, *y* = *λx* + 5

$\Rightarrow -165=\lambda \times 340+5\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{-1}{2}$

Hence, the value of *λ *is $\frac{-1}{2}.$

#### Page No 27:

#### Question 11:

By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.

(i) 3*x* + *y* + 4 = 0

*x*– 2

*y*+ 4 = 0

(ii)

*x*– 2

*y*= 6

*x*– 6

*y*= 0

(iii)

*x*+

*y*= 3

*x*+ 3

*y*= 9

#### Answer:

(i) Given equations are

3*x* + *y* + 4 = 0 .....(1)

When *x* = $-$2, then *y* = 2; Point A

When *x* = 0, then *y* = $-$4; Point B

When *x* = $-$1, then *y* = $-$1; Point C

6*x* $-$ 2*y* + 4 = 0 .....(2)

When *x* = 1, then *y* = 5; Point P

When *x* = 0, then *y* = 2; Point Q

When *x* = $-$1, then *y* = $-$1; Point C

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* – 2*y* = 6 .....(1)

When *x* = 0, then *y* = $-$3

When *x* = 6, then *y* = 0

3*x* $-$ 6*y *= 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; Point A

When *x* = 3, then *y* = 0; Point B

3*x* + 3*y* = 9 .....(2)

When *x* = 0, then *y* = 3; Point A

When *x* = 3, then *y* = 0; Point B

$\therefore $ The given pair of linear equations is coincident and have infinitely many solutions.

#### Page No 27:

#### Question 12:

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

#### Answer:

The given pair of linear equations

$2x+y=4.....\left(1\right)\phantom{\rule{0ex}{0ex}}2x-y=4.....\left(2\right)$

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 $\u25b3$ABC = 2$\u25b3$Area of AOB

Area($\u25b3$ABC) = $2\times \left(\frac{1}{2}\times 4\times 2\right)$ = 8 square units.

Hence, the required area of $\u25b3$ABC is 8 square units.

#### Page No 27:

#### Question 13:

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

#### Answer:

Given pair of linear equations is

$x+y-2=0.....\left(1\right)\phantom{\rule{0ex}{0ex}}2x-y-1=0.....\left(2\right)$

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

${a}_{1}=1,{b}_{1}=1,{c}_{1}=-2\phantom{\rule{0ex}{0ex}}{a}_{2}=2,{b}_{2}=-1,{c}_{2}=-1$

$\Rightarrow \frac{{a}_{1}}{{a}_{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{b}_{1}}{{b}_{2}}=\frac{1}{-1}$

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, for *x* + *y* = 2 or *y* = 2 $-$ *x,*

if *x* = 2 then *y* = 0; Point A

if *x *= 0 then* y* = 2; Point B

and, if *x* = 1 then *y* = 1; Point E

And for 2*x* $-$ *y* $-$ 1 = 0 or *y* = 2*x* $-$ 1,

If *x* = 0 then *y *= $-$ 1; Point C

if *x* = $\frac{1}{2}$ then *y* = 0; Point D

and if *x *= 1 then *y* = 1; Point E

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* + 2*y* = 3, *x* + *y* = 2 and so on.

#### Page No 27:

#### Question 14:

If *x* + 1 is a factor of 2*x*^{3} + *ax*^{2} + 2*bx* + 1, then find the values of *a* and *b* given that 2*a* – 3*b *= 4.

#### Answer:

Given that (*x *+ 1) is a factor of 2*x*^{3} + *ax*^{2} + 2*bx* + 1.

Let *f*(*x*) = 2*x*^{3} + *ax*^{2} + 2*bx* + 1, then *f*($-$1) = 0. [We know that if $\left(x+\alpha \right)$ is a factor of *f*(*x*)= $a{x}^{2}+bx+c=0$ then *f*($-$α) = 0]

Then,*
f*($-$1) = 0

$\Rightarrow $$-$2 +

*a*$-$ 2

*b*+ 1 = 0

*$\Rightarrow $a*$-$ 2

*b*$-$ 1 = 0 …..(1)

Also, 2

*a*$-$ 3

*b*= 4 .....(2)

Now,

3

*b*= 2

*a*$-$ 4

$b=\frac{2a-4}{3}$

Putting the value of

*b*in (1), we get

$a-2\left(\frac{2a-4}{3}\right)-1=0$

$\Rightarrow $3

*a*$-$ 2(2

*a*$-$ 4) $-$ 3 = 0

$\Rightarrow $3

*a*$-$ 4

*a*+ 5 = 0

$\Rightarrow $

*a*= 5

Putting the value of

*a*in (1), we get

5 $-$ 2

*b*$-$ 1 = 0

$\Rightarrow $4 = 2

*b*

$\Rightarrow $

*b*= 2

Hence, the required values of

*a*and

*b*are 5 and 2 respectively.

#### Page No 27:

#### 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*.

#### Answer:

Given the angles of a triangle are *x*, *y *and 40°.

So, *x *+ *y* + 40° = 180° (Angle sum property)

⇒ *x* + *y* = 140° ..... (1)

Also, difference of angles = *x* $-$ *y *= 30° .....(2)

Adding (1) and (2), we get

2*x *= 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$\xb0$ and 55$\xb0$, respectively.

#### Page No 28:

#### 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?

#### Answer:

Let Salim and his daughter’s age be *x* and *y* years respectively.

According to the question,

*x $-$ *2 = 3(*y $-$* 2)

⇒* x $-$ *3*y* = $-$4 .....(1)

And,

(*x* + 6) = 2(*y *+ 6) + 4

⇒* x *+ 6 = 2*y* + 16

⇒ *x *– 2*y* = 10 .....(2)

Subtracting (1) from (2), we get

*y *= 14

Putting the value of *y* in (2), we get

*x *= 38

Hence, Salim and his daughter’s age are 38 years and 14 years respectively.

#### Page No 28:

#### 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.

#### Answer:

Let the present age of father and his two children be *x*, *y* and *z* respectively.

According to the question,

*x* = 2(*y* + *z*) .....(1)

And after 20 years,

(*x* + 20) = (*y* + 20) + (*z* + 20)

⇒* x* = *y* + *z* + 20

⇒ *y* + *z* = *x* $-$ 20 .....(2)

Substituting the value of (*y* +* z*) in (1), we get,

*x* = 2(*x *$-$ 20)

*x* = 40

Hence, the father’s age is 40 years.

#### Page No 28:

#### 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.

#### Answer:

Let the two numbers be *x* and *y*.

Then, by first condition ratio of these two numbers = 5 : 6

$\frac{x}{y}=\frac{5}{6}.....\left(1\right)$

And by second condition, then 8 is subtracted from each of the numbers, then ratio becomes 4 : 5

$\frac{x-8}{y-8}=\frac{4}{5}$

$\Rightarrow $5*x* – 4*y* = 8 .....(2)

Now, put the value of *y *in (2), we get

$5x-4\left(\frac{6}{5}\right)x=8$

$\Rightarrow \frac{x}{5}=8$

$\Rightarrow x=40$

Substituting the value of *x* in (1), we get

$y=\frac{6}{5}\times 40=48$

Hence, the required numbers are 40 and 48.

#### Page No 28:

#### 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.

#### Answer:

Let the number of students in halls A and B are *x *and *y*, respectively.

According to the question,

*x* – 10 = *y *+ 10

⇒* x *$-$ *y* = 20 .....(1)

And

(*x *+ 20) = 2(*y* $-$ 20)

⇒ *x *– 2*y* = $-$60 .....(2)

On subtracting (2) from (1), we get

(*x* $-$ *y*) $-$ (*x* $-$ 2*y*) = 80

*$\Rightarrow $y *= 80

On putting *y* = 80 in (1), we get

*x* – 80 = 20

⇒ *x* = 100

Hence, 100 students are in hall A and 80 students are in hall B.

#### Page No 28:

#### 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 ₹22 for a book kept for six days, while Anand paid ₹16 for the book kept for four days. Find the fixed charges and the charge for each extra day.

#### Answer:

Let the fixed charge be ₹*x* and additional charge for each day thereafter be ₹*y*.

According to the question,

*x* + 4*y* = 22 .....(1)

And,

*x* + 2*y* = 16 .....(2)

Now, subtracting (2) from (1), we get

2*y* = 6

*y* = 3

On substituting the value of *y* in (2), we get

*x *+ 2(3) = 16

*x* = 10

Hence, the fixed charge is ₹10 and the charge for each extra day ₹3.

#### Page No 28:

#### 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?

#### Answer:

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.

$\Rightarrow x\times 1-(120-x)\times \frac{1}{2}=90\phantom{\rule{0ex}{0ex}}\Rightarrow x-\frac{(120-x)}{2}=90\phantom{\rule{0ex}{0ex}}\Rightarrow x+\frac{x}{2}-60=90\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x}{2}=150\phantom{\rule{0ex}{0ex}}\Rightarrow x=100$

Hence, Jayanti answered 100 questions correctly .

#### Page No 28:

#### Question 22:

The angles of a cyclic quadrilateral ABCD are

∠A = (6*x* + 10)°, ∠B = (5*x*)°

∠C = (*x* + *y*)°, ∠D = (3*y* – 10)°

Find *x* and *y*, and hence the values of the four angles.

#### Answer:

We know that, by property of cyclic quadrilateral, sum of opposite angles = 180$\xb0$

∠A + ∠C = 180$\xb0$

$\Rightarrow $(6*x* + 10)° + (*x* + *y*)° = 180$\xb0$

So, 7*x* + *y* = 170 .....(1)

Similarly,

∠B + ∠D = 180$\xb0$

$\Rightarrow $(5*x*)° + (3*y* – 10)° = 180$\xb0$

So, 5*x *+ 3*y* = 190 .....(2)

On multiplying (1) by 3 and then subtracting (2) from it, we get

3(7*x* + *y*) – (5*x* + 3*y*) = 3(170) – 190

16*x* = 320

$\Rightarrow x=20\xb0$

Putting $x=20\xb0$ in (1), we get

7(20) + *y *= 170

$\Rightarrow y=30\xb0$

Substituting the values of *x *and *y, *we get

∠A = (6$\times 20$ + 10)° = 130° , ∠B = (5$\times 20$)° = 100°

∠C = ($20$ + 30)° = 50°, ∠D = (3$\times 30$ – 10)° = 80°

Hence, the values of the four angles i.e. ∠A, ∠B, ∠C, ∠D are 130°, 100°, 50° and 80° respectively.

#### Page No 33:

#### Question 1:

Graphically, solve the following pair of equations:

2*x* + *y* = 6

2*x* –* 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.

#### Answer:

Given equations are 2*x* + *y* = 6 and 2*x *$-$ *y* + 2 = 0

Finding points for equation 2*x* + *y* $-$ 6 = 0, we get for *x* = 0, *y* = 6 and for *y* = 0, *x* = 3.

Similarly, for equation 2*x* $-$ *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 $\u2206\mathrm{ACE}\mathrm{and}\u2206\mathrm{BDE}$ respectively where E(1, 4) is the intersection of the lines.

${A}_{1}$= Area of $\u2206\mathrm{ACE}$= $\frac{1}{2}\times \mathrm{AC}\times \mathrm{PE}$ =$\frac{1}{2}\times 4\times 4$ = 8

${A}_{2}$= Area of $\u2206\mathrm{BDE}$= $\frac{1}{2}\times \mathrm{BD}\times \mathrm{QE}$ =$\frac{1}{2}\times 4\times 1$= 2

${A}_{1}:{A}_{2}\phantom{\rule{0ex}{0ex}}=8:2\phantom{\rule{0ex}{0ex}}=4:1$

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.

#### Page No 33:

#### Question 2:

Determine, graphically, the vertices of the triangle formed by the lines

*y* = *x*, 3*y* = *x*, *x* + *y* = 8

#### Answer:

Given linear equations are

*y* = *x* .....(1)

3*y* = *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* = 3*y*,

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 $\u2206\mathrm{OQD}$ is formed by these lines. Hence, the vertices of the $\u2206\mathrm{OQD}$ formed by the given lines are O(0, 0), Q(4, 4) and D(6, 2).

#### Page No 33:

#### Question 3:

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

#### Answer:

Given equation of lines *x* = 3, *x* = 5 and 2*x* $-$ *y* $-$ 4 = 0.

Now, for line 2*x* $-$ *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 quadrilateral ABCD = $\frac{1}{2}$ × (Distance between parallel lines) × (Sum of parallel sides) [Since, quadrilateral ABCD is a trapezium]

$=\frac{1}{2}\times \mathrm{AB}\times (\mathrm{AD}+\mathrm{BC})$

= 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.

#### Page No 33:

#### Question 4:

The cost of 4 pens and 4 pencil boxes is ₹100. Three times the cost of a pen is ₹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.

#### Answer:

Let the cost of a pen be ₹*x* and the cost of a pencil box be ₹*y*.

According to the question,

4*x* + 4*y* = 100

⇒ *x* + *y* = 25 .....(1)

And

3*x* = *y* + 15

⇒ 3*x* $-$* y* = 15 .....(2)

Adding (1) and (2), we get

4*x* = 40

⇒ *x* = 10

By substituting the value of *x* in (1), we get

*y *= 25 $-$ 10 = 15

Hence, the cost of a pen and a pencil box are ₹10 and ₹15 respectively.

#### Page No 33:

#### Question 5:

Determine, algebraically, the vertices of the triangle formed by the lines

3*x* – *y* = 3

2*x* – 3*y* = 2

*x* + 2*y* = 8

#### Answer:

Given equation of lines are

3*x* $-$ *y* = 3 .....(1)

2*x* $-$ 3*y* = 2 .....(2)

*x* + 2*y* = 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*x $-$ *3*y*) $-$ (2*x $-$ *3*y*) = 9 $-$ 2

$\Rightarrow $7*x *= 7

*$\Rightarrow $x* = 1

Substituting the value of *x *in (1), we get

3 × 1$-$* y *= 3

*$\Rightarrow $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

(2*x* + 4*y*) $-$ (2*x* $-$ 3*y*) = 16 $-$ 2

*$\Rightarrow $*7*y* = 14

*$\Rightarrow $y* = 2

Substituting the value of *y* in (3), we get

*$\Rightarrow $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*x $-$ *2*y*) + (*x *+ 2*y*) = 6 + 8

$\Rightarrow $7*x *= 14

*$\Rightarrow $x* = 2

Substituting the value of *x* in (1), we get

3 × 2 $-$ *y* = 3

*$\Rightarrow $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).

#### Page No 33:

#### 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.

#### Answer:

Let the speed of the rickshaw and the bus be *x* and *y* km/h, respectively.

Now, she has taken time to travel 2 km by rickshaw, ${t}_{1}=\frac{2}{x}$ hr , $[\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}]$

Remaining distance travelled by bus = 12 km

Time taken by bus to cover remaining distance, ${t}_{2}=\frac{12}{y}$ hr

By first condition,

${t}_{1}+{t}_{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{x}+\frac{12}{y}=\frac{1}{2}.....\left(1\right)$

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, by second condition,

${t}_{3}+{t}_{4}=\frac{1}{2}+\frac{9}{60}=\frac{13}{20}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{4}{x}+\frac{10}{y}=\frac{13}{20}$ .....(2)

Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$ then (1) and (2) becomes

$2u+12v=\frac{1}{2}$ .....(3)

$4u+10v=\frac{13}{20}$ .....(4)

Multiplying (3) by 2 and then subtract (4) from (3), we get

$\left(4u+24v\right)-(4u+10v)=1-\frac{13}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{1}{40}$

Now, put the value of *v* in (3), we get

$2u+12\left(\frac{1}{40}\right)=\frac{1}{2}$

$2u=\frac{2}{10}$

$\Rightarrow u=\frac{1}{10}$

$\Rightarrow x=\frac{1}{u}=10$ km/h

$\Rightarrow y=\frac{1}{v}=40$km/h

Hence, the speed of rickshaw and the bus are 10 km/h and 40 km/h, respectively.

#### Page No 34:

#### 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.

#### Answer:

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, ${t}_{1}=\frac{40}{5+v}\mathrm{h}$

Time taken by person to cover 40 km upstream, ${t}_{2}=\frac{40}{5-v}\mathrm{h}$

By condition, ${t}_{2}=3\times {t}_{1}$

$\Rightarrow \frac{40}{5-v}=3\times \frac{40}{5+v}$

$\Rightarrow \frac{1}{5-v}=\frac{3}{5+v}$

$\Rightarrow 5+v=15-3v$

$\Rightarrow 4v=10$

$\Rightarrow v=2.5\mathrm{km}/\mathrm{h}$

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

#### Page No 34:

#### 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.

#### Answer:

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. ${t}_{1}+{t}_{2}=7$

$\Rightarrow \frac{30}{u-v}+\frac{28}{u+v}=7.....\left(1\right)$

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, ${t}_{4}+{t}_{3}=5$

$\Rightarrow \frac{21}{u+v}+\frac{21}{u-v}=5......\left(2\right)$

Let $x=\frac{1}{u+v}\mathrm{and}\mathrm{y}=\frac{1}{\mathrm{u}-\mathrm{v}}$

Now (1) and (2) becomes

$28x+30y=7.....\left(3\right)$

$\Rightarrow x+y=\frac{5}{21}.....\left(4\right)$

Multiplying (4) by 28 and substracting from (3), we get

$(28x-28x)+(30y-28y)=7-\frac{5}{21}\times 28$

$\Rightarrow 2y=7-\frac{20}{3}$

$\Rightarrow y=\frac{1}{6}$

Substituting value of *y* in (4), we get

$x+\frac{1}{6}=\frac{5}{21}$

$\Rightarrow x=\frac{5}{21}-\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{1}{14}$

As$,x=\frac{1}{u+v}\mathrm{and}y=\frac{1}{u-v}$

$\Rightarrow u+v=14\mathrm{and}u-v=6$

$\therefore u=10\mathrm{and}v=4$

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

#### Page No 34:

#### 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.

#### Answer:

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

Case I: Multiplying the sum of the digits by 8 and then subtracting 5, we get two digit number

$\Rightarrow 8\times (x+y)-5=10x+y\phantom{\rule{0ex}{0ex}}\Rightarrow 8x+8y-5=10x+y\phantom{\rule{0ex}{0ex}}$

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

Case II: Multiplying the difference of the digits by 16 and then adding 3, we get two digit number

$\Rightarrow 16\times (x-y)+3=10x+y\phantom{\rule{0ex}{0ex}}\Rightarrow 16x-16y+3=10x+y\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 6x-17y=-3$ .....(2)

Now, multiplying (1) by 3 and then subtracting from (2), we get

$\Rightarrow (6x-17y)-(6x-21y)=-3-(-15)\phantom{\rule{0ex}{0ex}}\Rightarrow 4y=12\phantom{\rule{0ex}{0ex}}\Rightarrow y=3$

Now, put the value of *y* in (1), we get

$2x-21=-5\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=16\phantom{\rule{0ex}{0ex}}\Rightarrow x=8$

The required two-digit number is

$10\times 8+3$

= $83$

Hence, the number is 83.

#### Page No 34:

#### 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.

#### Answer:

Let the cost of full first class fare be Rs *x *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

$\Rightarrow x+y=2530$ .....(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

$x+y+\frac{x}{2}+y=3810\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x}{2}+2y=3810\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 3x+4y=7620$ .....(2)

Now, multiplying (1) by 4 and then subtracting from (2), we get

$(3x+4y)-(4x+4y)=7620-10120\phantom{\rule{0ex}{0ex}}\Rightarrow -x=-2500\phantom{\rule{0ex}{0ex}}\Rightarrow x=2500$

On putting the value of *x* in (1), we get

$\Rightarrow 2500+y=2530\phantom{\rule{0ex}{0ex}}\Rightarrow y=30$

Hence, full first class fare from stations A to B is Rs 2500 and the reservation charges for a ticket is Rs 30.

#### Page No 34:

#### 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.

#### Answer:

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

$\left(100+8\right)\%\mathrm{of}x+(100-10)\%\mathrm{of}y=1008\phantom{\rule{0ex}{0ex}}\Rightarrow 108\%\mathrm{of}x+90\%\mathrm{of}y=1008\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 1.08x+0.9y=1008$ .....(1)

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

$\left(100+10\right)\%\mathrm{of}x+(100-8)\%\mathrm{of}y=1028\phantom{\rule{0ex}{0ex}}\Rightarrow 110\%\mathrm{of}x+92\%\mathrm{of}y=1028\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 1.10x+0.92y=1028$ .....(2)

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

$\Rightarrow 1.10x+0.92\times \frac{(1008-1.08x)}{0.9}=1028\phantom{\rule{0ex}{0ex}}$

$\Rightarrow $1.1 × 0.9*x *+ 0.92 × (1008 $-$ 1.08*x*) = 1028 × 0.9

$\Rightarrow $0.99x $-$ 0.9936*x* = 925.2 $-$ 927.36

$\Rightarrow $$-$0.0036x = $-$2.16

$\Rightarrow x=\frac{2.16}{0.0036}=600$

Substituting the value of *x* in (1), we get

$1.08\times 600+0.9y=1008\phantom{\rule{0ex}{0ex}}\Rightarrow 0.9y=1008-648\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{360}{0.9}=400$

Hence, the cost price of the saree and the list price (price before discount) of the sweater are Rs 600 and Rs 400 respectively.

#### Page No 34:

#### 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?

#### Answer:

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

$\Rightarrow \frac{(x\times 8\times 1)}{100}+\frac{(y\times 9\times 1)}{100}=1860$ $\left[\because \mathrm{Simple}\mathrm{Interest}=\frac{\mathrm{Principal}\times \mathrm{Rate}\times \mathrm{Time}}{100}\right]$

So, 8*x* + 9*y* = 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

$\Rightarrow \frac{(x\times 9\times 1)}{100}+\frac{(y\times 8\times 1)}{100}=20+1860$

So, 9*x* + 8*y* = 188000 .....(2)

Multiplying (1) by 9 and (2) by 8 and then subtracting them, we get

$\Rightarrow (72x+81y)-(72x+64y)=186000\times 9-188000\times 8$

$\Rightarrow $17*y* = 170000

$\Rightarrow y=10000$

Substituting the value of y in (1), we get

8*x* + 9(10000) = 186000

So *x* = 12000

Hence, she invested Rs 12000 and Rs 10000 in two schemes A and B respectively.

#### Page No 34:

#### 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.

#### Answer:

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

$\frac{2}{3}x+y=400\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+3y=1200.....\left(1\right)$

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

$x+\frac{4}{5}y=460\phantom{\rule{0ex}{0ex}}\Rightarrow 5x+4y=2300.....\left(2\right)$

On multiplying (1) by 4 and (2) by 3 and then subtracting them, we get

(8*x *+ 12*y*) – (15*x* + 12*y*) = 4800 – 6900

$\Rightarrow $$-$ 7*x* = $-$ 2100

$\Rightarrow $*x* = 300

Substituting the value of *x* in (1), we get

2(300) + 3*y* = 1200

3*y* = 600

$\Rightarrow $*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

Hence, he had 500 bananas.

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