SChand CLASS 9 Chapter 20 Coordinates and Graphs of Simultaneous Linear Equation TEST

 TEST

Question 1

Ans: $(1.21$ and $(3,8)$ are the end point of a diagonal of a square

So length of diagonal $=\sqrt{(3-1)^{2}+(8-2)^{2}}$

$=\sqrt{2^{2}+6^{2}}=\sqrt{4+36}=\sqrt{60}$

So Area of square $=\frac{(\text { diagonal })^{2}}{2}$

$\frac{\sqrt{40} 1^{2}}{2}=\frac{40}{2}=20$ sq units (b)

Question 2

Ans: Distance between $(0,-5)$ and $(x, 0)=13$

$\begin{aligned}&\Rightarrow \sqrt{(x-0)^{2}+(0+5)^{2}}=13 \\&\Rightarrow \sqrt{x^{2}+25}=13\end{aligned}$

squaring both sides.

$\begin{aligned}&x^{2}+25=169 \Rightarrow x^{2}=169-25-144 \\&\Rightarrow x^{2}=1 \pm 121^{2} \\&\text { So } x=\pm 12 \text { (d) }\end{aligned}$

Question 4

Ans: Distance between origin $(0.0)$ is

(9) $(2,-3)=\sqrt{2^{2}+3^{2}}=\sqrt{4+9} \sqrt{13}$

(b) $(6,0)=\sqrt{6^{2}+0^{2}}=\sqrt{36} 6$

(C) $(-2,-1)=\sqrt{(-2)^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}$

(d) $(3,5)=\sqrt{3^{2}+5^{2}}=\sqrt{9+25}=\sqrt{34}$

it is clear that point $(-2,-1)$ is nearest 10 $(0,0)$

option (C) is correct

Question 5

Ans: The coordinates of the each of a side ay square are $(4,-3)$ and $(-1,-5)$

so length of side $=\sqrt{(-1-4)^{2}+(-5+3)^{2}}$.

$\begin{aligned}&=\sqrt{(-5)^{2}+(-2)^{2}} \\&=\sqrt{25+4}=\sqrt{29} \\&\text { so Area of square }=\text { (side }^{2} \\&\left.=(\sqrt{29})^{2} \text { sq units }(d)\right)\end{aligned}$

Question 7

Ans: $2 x-y-1=0$ and $2 x+y=9$ 

$\Rightarrow x=\frac{y+1}{2}$

Giving some different values of $y$, we ret corresponding values of $x$ as given below.

$\begin{array}{|c|c|c|c|}\hline x & 1 & 2 & 0 \\\hline y & 1 & 3 & -1 \\\hline\end{array}$

Plot the points $(1,1)(2,3)$ and $(0,-1)$ on the graph and Join them to get a line.

Similarly $2 x+y=y$

$y=y-2 x$

$\begin{array}{|c|c|c|c|}\hline x & 4 & 5 & 6 \\\hline y & 1 & -1 & -3 \\\hline\end{array}$

Plot the points (4,1)$1(5,-1)$ and $(6,-3)$ on the graph and Join them to get another line.

These two lines intersect each other at point

$\begin{aligned}&\left(\frac{5}{2}, 4\right) \\&\text { so } x=\frac{5}{2}, y=4\end{aligned}$

(IMAGE TO BE ADDED)

Question 8

Ans: from the given graph.

Two points are given on it whose $x$ and $y$ are equal but in opposite signs it is through the origin

Equation will be $x=-y$

$\Rightarrow x+y=0$ (b)

Question 9

Ans: The distance between the point $(0,0)$ and point af intersection of $x=3$ and $y=4$

i.e. $\left(3,4)\right.$ will be $=\sqrt{(3-0)^{2}+(4-0)^{2}}$

$\begin{aligned}&=\sqrt{3^{2}+4^{2}}=\sqrt{9+16} \\&=\sqrt{25}=5 \text { units }\end{aligned}$

Question 10

Ans: The pointe $(-4,0),(4,0),(0,3)$ are given Now,

$\begin{aligned}A B &=\sqrt{(4+4)^{2}+(0+0)^{2}}=\sqrt{8^{2}+0^{2}} \\&=\sqrt{64}=8 \\B C &=\sqrt{(0-4)^{2}+(3-0)^{2}}=\sqrt{\left.(-4)^{2}+13\right)^{2}} \\&=\sqrt{16+9}=\sqrt{25}=5 \\& A C=\sqrt{10-4)^{2}+(3-0)^{2}}=\sqrt{(-4)^{2}+(3)^{2}} \\&=\sqrt{16+9}=\sqrt{25}=5 \\& \text { So } B C=A C\end{aligned}$

SO the given points are the vertices of an isosceles triangle (b)

SChand CLASS 9 Chapter 11 Rectilinear Figures TEST

  TEST

 Question 1

Sol : 

The given figure is trapezium if. A pair of two opposite side are parallel (b) option (b) is right 

 Question 2

Sol: If The given Figure is of a parallelogram. so Their opposite sides are equal and parallel

$\begin{aligned} \Rightarrow & x=3 \\ & 2 x+1=3 x-2 \Rightarrow 3 x-2 x=1+2 \\ \Rightarrow & x=3 \\ & \text { so } x=3 \end{aligned}$

option(e) is right 

 Question 3

Sol: In figure , ABCD is a rectangle whose diagonals bisect each other At O.

$\triangle A B D$ is a rignt triangle as $\angle A=90^{\circ}$. 

option (b) is rigut

 Question 4

(i) In figure ABCD is a parallelogram 

So Their opposite side are equal and parallel and opposite angle are equal 

$x+6=5 x-8 \Rightarrow 5 x-x=6+8$

$\Rightarrow 4 x=14 \Rightarrow x=\frac{14}{4}=3.5$

$14 y+6 y=180$

$20 y=180^{\circ} \Rightarrow y=\frac{180^{\circ}}{20^{\circ}}=9$

$6 y=a \Rightarrow 6 \times 9=a \Rightarrow 9=54^{\circ}$

$b=14 y=14 \times 9=126^{\circ}$

So $a=54, b=126^{\circ}$

(ii) P Q R S is a rectangle

(IMAGE TO BE ADDED)

opposite sides are equal and parallel 

Diagonals bisect Each other.

$c=330^{\circ}$

Similarly a=e 

e$=<1$

and $\angle L+33^{\circ}=90^{\circ}$

$\Rightarrow \angle L=90^{\circ}-33^{\circ}=57^{\circ}$

So $\begin{aligned} e=\angle 1 &=57^{\circ} \text { and } a=e=57^{\circ} \\ \text { if } O P &=O Q \\ \text { so } c &=\angle 2=33^{\circ} \\ \text { so } d &=180^{\circ}-c-\angle 2=180^{\circ}-30^{\circ}-35^{\circ} \\ &=180-66^{\circ}=114^{\circ} \\ & \text { and } b+d=180^{\circ} \\ \Rightarrow b &=180^{\circ}-d=180^{\circ}-114^{\circ}=66^{\circ} \end{aligned}$

Hence 

$a=57^{\circ}, b=66^{\circ}, c=33^{\circ}, d=194^{\circ}, e=57^{\circ}$

(iii) MNPQ is a rhombus in which all sides are equal and diagonals bisect each other at 90 and diagonals bisects opposite angles 

$b=53^{\circ}$

$a=d$

$e=53^{\circ}$

but $e+d=90^{\circ}$

$53^{\circ}+d=90^{\circ} \Rightarrow d=90^{\circ}-53^{\circ}=37^{\circ}$

$a=d=37^{\circ}$

$L c=90^{\circ}$

$a=37^{\circ}, b=53^{\circ}, \angle c=90^{\circ}, d=37^{\circ}, e=53^{\circ}$

 Question 5

Sol: In the figure ABCD is a kite in which BC = CD , AB = AD and diagonals AC and BC intersect at right angle at O and AC bisects the opposite angles 

$\angle O B C=58^{\circ}, \angle D A B=50^{\circ}$

(IMAGE TO BE ADDED)

(i) In $\triangle B C D \cdot B C=C O$

so $\angle O B C=\angle O D C=58^{\circ}$

So $\angle B C U=180^{\circ}-\angle O B C-\angle O D C$

$=180^{\circ}-58^{\circ}-58^{\circ}=180^{\circ}-116^{\circ}=64^{\circ}$

(ii) $\angle D A O=\frac{1}{2} \angle B A D=\frac{1}{2} \times 50^{\circ}=25^{\circ}$

(iii) $\angle O D A=90^{\circ}-\angle O A O=90^{\circ}-25^{\circ}=65^{\circ}$

(iv) $\angle A D C=\angle O O A+\angle O D C$

$=65^{\circ}+58^{\circ}=123^{\circ}$

 Question 6

Sol: In $A B C, P C$ and $Q R$ are mid segments $A B=B C$

So $P Q \| B C$ and $P Q=\frac{1}{2} \quad B C=B R$.............(i)

But AB =BC and PR Their mid points 

So PB = BR ...........(ii)

And QR||AB and QR =$\frac{1}{2} A B=P B$..........(ii)

and $Q R \| A B$ and $A R=\frac{1}{2} A B=P B$..........(iii)

From (i), (ii) and (iii)

$P B=B R=Q R=P Q$

SQ BPQR is a rhombus or a square 

But $\angle B \neq 90^{\circ}$

So $B P Q R$ is a rhombus.

SChand CLASS 9 Chapter 9 Mid Point and Intercept Theorems TEST

TEST

Question 1

Sol: 

In the given $\triangle A B C$

AD=DB and BE=EC

So D and E are mid point of AB and BC respectively 

So $D E \| A C$ and $D E=\frac{1}{2} A C$

(a) DE is mid segment of $\triangle A B C$

(b) DE is parallel to AC

(C) AD=BD

(d) DE is half of AC

(e) Twice of $E C=B C$

Question 2

Sol: In the given figure,

If $Y L=L X, Y M=M Z$ and $X N=N Z$

So L,M and N are the mid points of side XY, YZ

and XZ respectively

So $L M=\frac{1}{2} \times 2, M N=\frac{1}{2} x y$

XY =10cm LM =6cm, $\angle m N Z$=$25^{\circ}$

(a) $\mathrm{Nm}=\frac{1}{2} \times y=\frac{1}{2} \times 10 \mathrm{~cm}=5 \mathrm{~cm}$

(b)$x z=2 \times L m=2 \times 6=12 \mathrm{~cm}$

(c) $\quad N Z=\frac{1}{2} \times \times 2=\frac{1}{2} \times 12=6 \mathrm{~cm}$

(d) $\angle L M N=25^{\circ}$

(If LM||XZ an  MN is a transversal)

(e) $\angle Y \times z=\angle m N z=25^{\circ}$

(f)$\angle X L M=\angle X N M$ (opposite angle of a ||gm)

$=180^{\circ}-25^{\circ}=155^{\circ}$

Question 3

(IMAGE TO BE ADDED)

SOL: In $\triangle A B C$

$\mathrm{Lm} \| \mathrm{AC}$ AND $\angle m=\frac{1}{2} A C$

$\Rightarrow 32=\frac{1}{2} \times 4 n \Rightarrow 2 n=32$

$\Rightarrow n=\frac{32}{2}=16 \mathrm{~cm}$

So $n=16 \mathrm{~cm}$

Question 4

(IMAGE TO BE ADDED)

Sol: AE =EC and BD =DC 

So $D E \| A B$ and $D E=\frac{1}{2} A B$

$\Rightarrow 6 n=4 n+12$

$\Rightarrow 6 n-4 n=+12$

$\Rightarrow 2 n=12$

$n=\frac{12}{2}=6$

Question 5

Sol: In the given figure

CG,EH and FJ are mid segment of $\triangle A B D, \triangle G C D$ and $\triangle G H E$ Respectively

$A B=33 \mathrm{~cm}, \angle A B C=57^{\circ}$

(a)

$\begin{aligned}&C G=\frac{1}{2} A B \\&=\frac{1}{2} \times 33=16.5 \mathrm{~cm}\end{aligned}$

(b) 

$\begin{aligned} E H &=\frac{1}{2} D C \\ &=\frac{1}{2} \times \frac{1}{2} D B \\ &=\frac{1}{4} D B=\frac{1}{4} \times 44=11 \mathrm{~cm} \end{aligned}$

(C) 

$\begin{aligned} F J &=\frac{1}{2} G H=\frac{1}{2} \times \frac{1}{2} G C \\ &=\frac{1}{4} G C=\frac{1}{4} \times 16.5 \mathrm{~cm}=4.125 \mathrm{~cm} \end{aligned}$

(d) $\quad M \angle D C G=\angle D B A$ 

=$57^{\circ}$

(e) $\begin{aligned} M \angle G H E &=\angle G C D \\ &=37^{\circ} \end{aligned}$

(f) 

$\begin{aligned} M \angle F J H &=180^{\circ}-\angle G H E \\ &=180^{\circ}-57^{\circ}=123^{\circ} \end{aligned}$

SChand CLASS 9 Chapter 8 Triangles TEST

TEST

Question 1

(i)Among the given criterion 

SSA is not criterion other are criterion 

(ii)From the given figure, AC=FE

AB=FD

DC=DE

So $\triangle A B C \cong \triangle F D E$

Question 2

Sol:

(i) When x =18 , then 

$4 x-11=4 \times 18-11=72-11$

$=61 \Rightarrow R S=U T=61$

and $2 x=2 \times 18=36^{\circ}$

$\Rightarrow \angle S R T=\angle R T U=36^{\circ}$

$R T=R T \quad$ (common)

Hence $\triangle R S T \cong \triangle T U R$

(ii)In  the figure,

ABCD is a kite a which AB=DB

$\Rightarrow 3 x-9=3 \Rightarrow 3 x=3+9=12$

$\Rightarrow x=\frac{12}{3}=4$

$A C=D C$

$\Rightarrow 5=2 \times 4-3 \Rightarrow 5=8-3=5$

$B C=B C \quad$ (common)

So $\triangle A B C \cong \triangle D B C$

Question 3

(IMAGE TO BE ADDED)

Sol; In $\triangle A B C$

$D$ is point on $B C$ and $A D$ bisects $\angle B A C$

i.e. $\angle B A D=\angle C A D$

In $\triangle A D C$, then

Ext, $\angle A D B>\angle C A D$

So In $\triangle A B D$

$B A>B D$

Question 4

(IMAGE TO BE ADDED)

Sol: Two sides of $\triangle A B C$

Let BC =6cm and AB=2.6cm

Now BC-AB =6-2.6= 3.4cm

So third side must be greater than 3.4cm

If sum of any two sides is greater than third side 

So 3.2cm can be the third side (d)

Question 5

(IMAGE TO BE ADDED)

Sol: In $\triangle A B C, \angle B=30^{\circ}, \angle C=80^{\circ}$

Then $\angle A=180^{\circ}-\left(30^{\circ}+80^{\circ}\right)$

$=180^{\circ}-110^{\circ}=70^{\circ}$

Then $A B>B C>A C$   (c)

Question 6

Sol: In $\triangle A B C$ ,two sides are 4cm and 10cm

Let AB = 4CM and BC = 10cm

'a' is third side

If Sum of any two sides >third side

$10<4+a \Rightarrow 10-4<9 \Rightarrow 6>9$

and $a<4+10$

$\Rightarrow a<14$

So 6<a<14 (d)

SChand CLASS 9 Chapter 7 Logarithms TEST

 TEST

Question 1

The value of $\log_2 16$ is

(a) $\frac{1}{8}$

(b) 4

(c) 8

(d) 16

Sol :

(b) 4

$\log_2 16 = \log_2 2^4 = 4\log_2 2$ = 4×1 (∵ $log_2 a = 1$)

= 4

Question 2

If $a^x = b^y$ then

(a) $\log \frac{a}{b}=\frac{x}{y}$

(b) $\frac{\log a}{ \log b}=\frac{x}{y}$

(c) $\frac{\log a}{\log b}=\frac{y}{x}$

(d) None of these

Sol :

$a^x = b^y$

Taking log both sides,

$log_a x = b_by $

⇒ $x\log_a = y\log_b$

⇒ $\frac{\log a}{\log b}=\frac{y}{x}$

Question 3

If log 3 = 0.477 and $(1000)^x = 3$, then x equals

(a) 0.0159

(b) 0.0477

(c) 0.159

(d) 10

Sol :

log 3 = 0.477

$(1000)^x = 3$

Taking log of both sides

xlog 1000 = log3

⇒ x × 3 = log3 (∵ log 1000 = 3)

⇒ 3x = 0.477 

⇒ $x = \frac{0.477}{3}$

x = 0.159

Question 4

If $\log_{10}2 = 0.3010$, the value of log105 is

(a) 0.3241

(b) 0.6911

(c) 0.6990

(d) 0.7525

Sol :

$\log_{10} 2 = 0.3010$

$\log_{10} 5 = \log_{10} \left(\frac{10}{2}\right) = \log_{10} 10 – \log_{10} 2$

= 1 – 0.3010 = 0.6990 (c)

Question 5

If $\log_{10} 2 = 0.3010$, the value of $\log_{10} 80$ is

(a) 1.6020

(b) 1.9030

(c) 3.9030

(d) None of these

Sol :

(b) 1.9030

$\log_{10} 2 = 0.3010$

$\log_{10} 80 =\log_{10} 10 \times 2^3$

$= \log_{10} 10 + \log_{10} 2^3$

$= \log_{10} 10 + 3\log_{10} 2$

= 1 + 3×0.3010

= 1 + 0.9030 = 1.9030

Question 6

If $\log_{10} 7 = a$, then $\log_{10} \left(\frac{1}{70}\right)$ is equal to

(a) – (1 + a)

(b) $(1 + a)^{-1}$

(c) $\frac{a}{10}$

(d) $\frac{1}{10a}$

Sol :

(a) – (1 + a)

$\log_{10} 7 = a$

$\log_{10} \left(\frac{1}{70}\right) = \log_{10} 1 – \log_{10} 70$

$= 0 – \log_{10} (7 \times 10)$

$= 0 – \log_{10} 7 – \log_{10} 10$

= 0 – 0 – 1 = – (1 + a)

Question 7

If log 27 = 1.431, then the value of log 9 is

(a) 0.934

(b) 0.945

(c) 0.954

(d) 0.958

Sol :

(c) 0.954

log 27 = 1.431 

⇒ log 3³ = 1.431

⇒ 31og 3 = 1.431

⇒ log 3 $= \frac{1.431}{3}$ = 0.477

log 9 = log 3² = 2 log 2

= 2×0.477 = 0.954

Question 8

If $\log_{10}  5 + \log_{10}(5x + 1) = \log_{10}(x + 5) + 1$, then x is equal to

(a) 1

(b) 3

(c) 5

(d) 10

Sol :

(b) 3

$\log_{10} 5 + \log_{10}(5x + 1) = log_{10}(x + 5) + 1$

$\log_{10}5 (5x + 1) = \log_{10} (x + 5) + \log_{10} 10$

$\log_{10}(5x + 1) = \log_{10} 10 (x + 5)$

Comparing, we get

⇒ 5(5x + 1) = 10(x + 5)

⇒ 25x + 5 = 10x + 50

⇒ 25x – 10x = 50 – 5

⇒ 15x = 45 

⇒ $x = \frac{45}{15} = 3$

⇒ x = 3

Question 9

If $\log_x 4 = 0.4$, then the value of x is

(a) 1

(b) 4

(c) 16

(d) 32

Sol :

$\log_x 4 = 0.4$

⇒ $4 = x^{0.4}$

⇒ $x = 4^{1}{0.4}=(2^2)^{\frac{1}{0.4}}$

= $2^{2×\frac{1}{0.4}}=2^{\frac{1}{0.2}}$

= $2^{\frac{1}{5}}=2^{1×\frac{5}{1}}=2^5$

= 2×2×2×2×2 = 32

∴ x = 32

Question 10

The solution of $\log_π [\log_2 (\log_7 x)] = 0 $ is

(a) 2

(b) π²

(c) 72

(d) None of these

Sol :

$\log_π [\log_2 (\log_7 x)] = 0$

⇒ $\log_2 (log_7 x) = π° = 1$

⇒ $\log_7 x = 2^1 = 2$

⇒ $x = 7^2$

∴ $x = 7^2$

SChand CLASS 9 Chapter 6 Indices/Exponent TEST

 TEST

 Question 1

Determine whether each equation is true or false. Change the right side of the equation to make a true equation.

(i) $(2 a)^{-3}=\frac{2}{a^{3}}$

(ii) $\left(\left(a^{-1}\right)^{-1}\right)^{-1} =\frac{1}{a}$

(iii) $(2+3)^{-1}=2^{-1}+3^{-1}$

(iv) If $x \neq \frac{1}{3}$ , then $(3x – 1)^0 = (1 – 3x)^0$

Why is the condition $x \neq \frac{1}{3}$ given in part (iv) above?

Sol:

(i)$(2 a)^{-3}=\frac{2}{a^{3}}$

$(2 a)^{-3}=\frac{1}{(2 a)^{3}}=\frac{1}{8 a^{3}} \neq \frac{2}{a^{3}} \quad$ [it is not equal]

(ii) $\begin{aligned}\left(\left(a^{-1}\right)^{-1}\right)^{-1} &=\frac{1}{a} \\\left(\left(a^{-1}\right)^{-1}\right)^{-1} &=a\left(^{-1)} \times(-1) \times(-1)\right.\\ &=a^{-1}=\frac{1}{a}=\frac{1}{a} \end{aligned}$

[It is made true equation]

(iii) $(2+3)^{-1}=2^{-1}+3^{-1}$

$(2+3)^{-1}=5^{-1}=\frac{1}{5} .$

$=2^{-1}+3^{-1} .$

$=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$

$\begin{aligned} &=\frac{5}{6} \\ \because \frac{1}{5} & \neq \frac{5}{6} \end{aligned}$ it is not equal 

[It is not make proper equation]

(iv) $x \neq \frac{1}{3}$

$(3 x-1)^{0}=(1-3 x)^{\circ}$

$(3 x-1)^{\circ}=1$

$(1-3 x)^{0}=1 \quad\left(x^{\circ}=1\right)$

1=1 [It is make true equation]

According to Question if  $x=1 / 3$,

Then

$\left(3 x-1^{0}\right)=\left(3 \times \frac{1}{3}-1\right)^{0}$

$=(1-1)^{\circ}=0^{\circ}$

It is not make a proper equation which is not possible 

Simplify:

Question 2

$\frac{\left(5 x^{3} y^{-3} z\right)^{-2}}{y^{4} z^{-2}}$

Sol:

$\begin{aligned} & \frac{\left(5 x^{3} y^{-3} z\right)^{-2}}{y^{4} z^{-2}} \\=& \frac{\left(s^{-2}\right) x^{-6} y+1 z^{-2}}{y^{4} z^{-2}} \\=& \frac{1}{25} \frac{1}{x^{6}} \times y^{6-4} \cdot z^{-2+2} \\=& \frac{1}{25} \frac{y^{2}}{x^{6}} \times z^{0}=\frac{y^{2}}{25 x^{6}} \end{aligned}$ Ans 

 Question 3

$\left(\frac{8 a^{3} b^{-4}}{64 a^{-9} b^{2}}\right)^{2 / 3} $

Sol:

$\begin{aligned} &\left(\frac{8 a^{3} b^{-4}}{64 a^{-9} b^{2}}\right)^{2 / 3} \\=&\left[\frac{2 \times a^{3+9}}{8 \times b^{2+4}}\right]^{2 / 3} \\=&\left(\frac{a^{12}}{8 b^{6}}\right)^{2 / 3} \\=&\left(\frac{a^{12}}{2^{3} \times b^{6}}\right)^{2 / 3} \end{aligned}$

$\frac{a^{12 \times 2 / 3}}{2^{3 \times 2 / 3} \cdot b^{6 \times 2 / 3}}=\frac{a^{24 / 3}}{2^{6 / 3} \cdot b^{12 / 3}}$

$\frac{a^{8}}{2^{2} \cdot b^{4}} \Rightarrow \frac{a^{8}}{4 b^{4}}$

 Question 4

$-\sqrt[4]{16 a^{4} b^{8}}$

Sol:

$\begin{aligned}-\sqrt[4]{16 a^{4} b^{8}} &=-\left(2^{4} a^{4} b^{8}\right)^{1 / 4} \\ &=-\left(2^{a \times \frac{1}{4}} a^{4 \times \frac{1}{4}} b^{8 \times \frac{1}{4}}\right) \\ &=-\left(2^{1} a^{1} b^{2}\right) \\ &=-2 a b^{2} \text { Ans } \end{aligned}$

 Question 5

$\left[\frac{y^{2 / 3} \cdot y^{-5 / 6}}{y^{1 / 5}}\right]^{9}$

Sol:

=$\left[\frac{y^{2 / 3} \cdot y^{-5 / 6}}{y^{1 / 5}}\right]^{9}$

=$\left(y \frac{12-15-2}{18}\right)^{9}$

$=y^{-\frac{5}{2}}$

$=\frac{1}{y^{\frac{2}{5}}}$

 Question 6

$\left[\sqrt[3]{\sqrt{x^{6}}}\right.$

Sol:

$\begin{aligned} &\left[\sqrt[3]{\sqrt{x^{6}}}\right.\\=&\left[\left(x^{6}\right)^{1 / 2}\right]^{1 / 3} \\=& x^{6 \times 1 / 2} \times 1 / 3 \\=& x^{1}=x \end{aligned}$

 Question 7

Which of the following is (are) equivalent to $16\frac{–1}{2}$?

(a) – 8

(b) $\frac{1}{4}$

(c) – 4

(d) $4^{-1}$

Sol:

=$16^{-1\2}$

=$\left(4^{2}\right)^{\frac{-1}{2}}$

=$4^{2 \times(-1 / 2)}$

=$4^{-1}$

$=\frac{1}{4}$ option (b) and (d) both are correct

 Question 8

Which of the following is undefined?

(a) $– 25\frac{1}{2}$

(b) $25\frac{1}{2}$

(c) $– 25\frac{–1}{2}$

(d) $(-25)\frac{1}{2}$

Sol: 

option (d) $(-25)^{1 / 2}$ is correct

because- Square root of negative number is not defined

 Question 9

True or False?

(a) $\frac{a^{4n}}{a^n}=a^4$

(b) $\frac{1}{am−n}=d^{n−m}$

(c) $a^{-n}.a^n = 1$

(d) $\frac{a^n}{b^m}=\left(\frac{a}{b}\right)^{n−m}$

Sol :

(i)

Sol: $\begin{aligned} \frac{a^{4 n}}{a^{n}} &=a^{4 n-n} \\ &=a^{3 n} \neq a^{4} \end{aligned}$

It is false 

(ii)

$\begin{aligned} \frac{1}{a^{m-n}} &=a^{-(m-n)}=a^{-m+n} \\ &=a^{n-m} \\ &=a^{n-m} \text { it is true } \end{aligned}$

(iii) $a^{-n} \cdot a^{n}=1$

$a^{-n} \cdot a^{n}=a^{-n+n}$

$=a^{\circ}=1$ $=1$ is also true

(iv) $\frac{a^{n}}{b^{m}} \neq\left(\frac{a}{b}\right)^{a-n}$ 

It is not equal so it is false

 Question 10

(i) Solve : $(- 4.8)^k = 1$

(ii) $\sqrt[3]{\sqrt{0.000064}}$ is equal to

(a) 0.0002

(b) 0.002

(c) 0.02

(d) 0.2

Sol :

(i)

$\begin{aligned}&(-4 \cdot 8)^{k}=1 \\&(-4 \cdot 8)^{k}=(-4 \cdot 8)^{0}\end{aligned}$

comparing both sides of power

k=0

(ii) $\sqrt[3]{0.000064}$

$=(0.04 \times 0.04 \times 0.04)^{1 / 2 \times 1 / 3}$

$=(0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2)^{1 / 6}$

$=\left[(0.2)^{6}\right]^{1 / 6}=(0.2)^{6 \times \frac{1}{6}}$

$=(0.2)^{2}$

=0.2 Ans

option (d) is correct 

SChand CLASS 9 Chapter 5 Simultaneous Linear Equations TEST

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Question 1 

y + 2x = 5 , 3y – 5x = 4

Sol :

y + 2x = 5 ⇒ y = 5 – 2x

3y – 5x = 4

⇒ 3(5 – 2x) – 5x = 4

⇒ 15 – 6x – 5x = 4

⇒ – 11x = 4 – 15

⇒ – 11x = – 11

and y = 5 – 2x = 5 – 2 x 1 = 5 – 2 = 3

∴ x = 1, y = 3

---

Question 2 

$\frac{1}{2}x+2y=16$

$2x+\frac{1}{2}y=19$

Sol :

$\frac{1}{2}x+2y=16$...(i)

$2x+\frac{1}{2}y=19$...(ii)

Multiply (i) by 4 and (ii) by 1 , then subtracting

$\begin{aligned}2x+8y&=64\\ 2x+\frac{1}{2}y&=19\\ -\phantom{2x}-\phantom{8 y}&\phantom{=}-\phantom{19}\\ \hline \frac{15}{2}y &=45\end{aligned}$

$y=\frac{45 \times 2}{15}=6$

From (i) $\frac{1}{2}x+2\times 6=16$

⇒$\frac{1}{2}x=16-12$

⇒$\frac{1}{2}x=4$

⇒x=4×2=8

∴x=8 , y=6

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Question 3 

Which ordered pair is a solution of the system?

$\left\{ \begin{matrix} 2x-y=-2 \\ \frac{1}{3}y=x \end{matrix} \right.$

(a) (0, 2)

(b) (2, 6)

(c) (1, 3)

(d) (3, 8)

Sol :

(b) (2, 6)

2x – y = – 2, $\frac{1}{3}y=x$

⇒ y = 3x⇒ 2x – 3x = – 2 ⇒ – x = – 2 ⇒ x = 2

and y = 3x = 3 x 2 = 6

∴ x = 2, y = 6

∴ Order pair of solution is (2, 6)

---

Question 4 

Which of the following problems could be solved by finding the solution of the given system?

2x + 2y = 56, $y=\frac{1}{3}x$

(a) The area of a reactangle is 56 sq. unit. The width is one-third the length. Find the length of the rectangle.

(b) The area of a rectangle is 56 sq. unit. The length is one-third the perimeter. Find the length of the rectangle.

(c) The perimeter of a rectangle is 56 unit. The length is one-third more than the width. Find the length of the rectangle.

(d) The perimeter of a rectangle is 56 unit. The width is one-third the length. Find the length of the rectangle.

Sol :

(d) The perimeter of a rectangle is 56 unit. The width is one-third the length. Find the length of the rectangle.

2x + 2y = 56, $y=\frac{1}{3}x$

Here 2x + 2y = 56 ⇒ 2(x + y) = 56

i.e., perimeter of a rectangle whose length

and breadth x and y is 56

and breadth =$\frac{1}{3}$ length

∴ It is applies to (d)

Perimeter of rectangle = 56 unit

2(l + b) = 56

As breadth is one third of length

∴$2\left(x+\frac{x}{3}\right)=56$

$=\frac{4x}{3}=\frac{56}{2}$

⇒$x=\frac{56}{2} \times \frac{3}{4}$=21 unit

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Question 5 

What is the solution of the equation x – 7 = 0.9 and $11 (x + y)^{-1} = 2$?

(a) x = 3.2, y = 2.3

(b) x = 1, y = 0.1

(c) x = 2, y = 1.1

(d) x= 1.2, y = 0.3

Sol :

(a) x = 3.2, y = 2.3

x-y=0.9

⇒$11 (x + y)^{-1} = 2$

⇒$(x+y)^{-1}=\frac{2}{11}$

⇒$\frac{1}{x+y}=\frac{2}{11}$

⇒$x+y=\frac{11}{2}$

Now , $x+y=\frac{11}{2}$

$x-y=\frac{9}{10}$

Adding , we get

$2x=\frac{11}{2}+\frac{9}{10}=\frac{55+9}{10}=\frac{64}{10}$

$x=\frac{64}{10\times 2}=\frac{32}{10}=3.2$ and subtracting,

$2y=\frac{11}{2}-\frac{9}{10}=\frac{55-9}{10}=\frac{46}{10}$

$y=\frac{46}{10\times 2}=\frac{23}{10}=2.3$

∴x=3.2 , y=2.3

---

Question 6 

A number consists of two digits, whose sum is 10. If 18 is subtracted from the number, digits of the number are reversed. What is the product of the digits?

(a) 15

(b) 18

(c) 24

(d) 32

Sol :

(c) 24

Let unit digit of a two digit number = x

and ten digit = y

∴ x + 7 = 10 … (i)

and number = x + 10y

By reversing the digits, the number = y + 10x

∴ x + 10y – 18 = y + 10x

⇒ x + 10y – y – 10x = 18

⇒ – 9x + 97 = 18

⇒ – 9(x – 7)= 18 

⇒ $x-y=\frac{18}{-9}=-2$

x – y = – 2

Adding, we get

2x = 8 

⇒$x=\frac{8}{2}=4$

Subtracting,

27 = 12 

⇒ $y =\frac{12}{2}=6 $

∴ Product of digits = x × y = 4×6 = 24

---

Question 7 

If $\frac{2x-3y+1}{2}=\frac{x+4y+8}{3}=\frac{4x-7y+2}{5}$ , then what is (x+7) equal to?

(a) 3

(b) 2

(c) 0

(d) – 2

Sol :

(d) – 2

$\frac{2x-3y+1}{2}=\frac{x+4y+8}{3}=\frac{4x-7y+2}{5}$

$\frac{2x-3y+1}{2}=\frac{x+4y+8}{3}

⇒ 6x – 97 + 3 = 2x + 87 + 16

⇒ 6x – 97 – 2x – 87 = 16 – 3

⇒ 4x – 17y = 13

$=\frac{x+4y+8}{3}=\frac{4x-7y+2}{5}$

5x + 20y + 40 = 12x – 217 + 6

5x + 20y – 12x + 217 = 6 – 40

– 7x + 41y = – 34

7x – 41y = 3y … (i)

4x – 17y = 13 … (ii)

Multiply (i) by 4 and (ii) by

$\begin{matrix}28x-164y&=136\\28x-119y&=91\\ -\phantom{28x}+\phantom{119y}&\phantom{=}-\phantom{91} \\ \hline \phantom{28x}-45y&=45\end{matrix}$

$y=\frac{45}{-45}=-1$

From (ii),

⇒ 4x – 17×(- 1) = 13

⇒ 4x + 17 = 13 

⇒ 4x = 13 – 17 = – 4

∴ x + y = – 1 +(- 1) = – 1 – 1 = – 2

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Question 8 

A railway ticket for a child costs half the full fare but the reservation charge is the same on half tickets as much as on full ticket. One reserved first class full ticket for a journey between two stations is ₹ 362; one full and one half reserved first class tickets cost ₹ 554. What is the reservation charge?

(a) ₹18

(b) ₹ 22

(c) ₹ 38

(d) ₹ 46

Sol :

(b) ₹ 22

Let reservation charges = ₹ x per ticket

and let price of full ticket = ₹ y

and half ticket = ₹$\frac{1}{2}$

Now according to the question,

x + y = ₹ 362 … (i)

$2x+y+\frac{y}{2}= ₹ 554$

⇒ $2x + \frac{3}{2}y = ₹ 554$ … (ii)

Multiply (i) by $\frac{3}{2}$ and (ii) by 1

$\frac{3}{2}x+\frac{3}{2}y=362 \times \frac{3}{2}=543$

$2x+\frac{3}{2}y=554$

Subtracting (i) from (ii),

$\frac{1}{2}x=11$

⇒ x = 11×2 = 22

Cost of reservation = ₹ 22

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Question 9 

What is the sum of two numbers whose difference is 45 and the quotient of the greater number by the lesser number is 4?

(a) 100

(b) 90

(c) 80

(d) 75

Sol :

(d) 75

Let first number = x

and second number = y

∴ x – y = 45 … (i)

$\frac{x}{y} = 4 $

⇒ x = 4y … (ii)

From (i) 4y – y = 45

⇒ 3y = 45

$y = \frac{45}{3} = 15$

∴ x = 4 x 15 = 60

First number = 60

and second number = 15

Sum of two number = x + y

= 60 + 15 = 75

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Question 10 

Two numbers are in the ratio 2 : 3. If 19 is added to each number, they will be in the ratio 3 : 4. What is the product of the two numbers?

(a) 360

(b) 480

(c) 486

(d) 512

Sol :

(c) 486

Ratio in two number = 2 : 3

Let first number = x

and second number = y

Then, $\frac{1}{2} = \frac{1}{2}$ ⇒ 3x = 2y

⇒ $x = \frac{2}{3}y $… (i)

Adding 9 to each, we get

x + 9 and y + 9

∴ $\frac{x+9}{y+9}=\frac{3}{4}$ 

⇒ 4x + 36 = 3y + 27

= 4x – 3y = 27 – 36 = – 9 … (i)

⇒ $4 \times  \frac{2}{3}y – 3y = – 9$ [From (i)]

⇒ $ \frac{8}{3}y – 3y = – 9$

$\frac{8y−9y}{3} = – 9 $⇒ $\frac{−y}{3} = – 9$

⇒ y = – 9×(- 3) = 27

and $x = \frac{2}{3} \times 27 = 18$

∴ Numbers are 18, 27

Product of two numbers = 18×27

= 486

SChand CLASS 9 Chapter 4 Factorisation TEST

 TEST

Question 1

$ 8 x^{2} y^{3}-x^{5} $

Sol :

$\begin{aligned} =& 8 x^{2} y^{3}-x^{5} \\=& x^{2}\left(8 y^{3}-x^{3}\right) \\=& x^{2}\left[(2 y)^{3}-(x)^{3}\right] \\=& x^{2}(2 y-3)\left(y y^{2}+2 x y+x^{2}\right) \end{aligned}$

Question 2

$x^{2}+\frac{1}{x^{2}}+2-2 x-\frac{2}{x}$

Sol :

$\begin{aligned} =& x^{2}+\frac{1}{x^{2}}+2-2 x-\frac{2}{x} \\=&\left(x+\frac{1}{x}\right)^{2}-2\left(x+\frac{1}{x}\right) \\=&\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}-2\right) \end{aligned}$

Question 3

2x² – x – 6

Sol :

$=2 x^{2}-x-6$

$=2 x^{2}-4 x+3 x-6$

=2x(x-2)+3(x-2)

=(x-2)(2x+3)

Question 4

a³ – 0.216

Sol :

$\begin{aligned} =& a^{3}-0.226 \\=&(a)^{3}-(0.6)^{3} \\=&(a-0.6)\left(a^{2}+a \times 0.6+(0.6)^{2}\right] \\=&(a-0.6)\left(a^{2}+0.6 a+0.36\right) \end{aligned}$

Question 5

6x²y – xy – 2y

Sol :

$\begin{aligned} =& 6 x^{2} y-x y-2 y \\=& y\left[6 x^{2}-x-2\right) \\=& y\left[6 x^{2}-4 x+3 x-2\right] \\=& y[2 x(3 x-2)+2(3 x-2)\} \\=& y[2 x(3 x-2)+1(3 x-2)] \\=& y(3 x-2)(2 x+1) \end{aligned}$

Question 6

(x² – 3x)² – 8(x² – 3x) – 20

Sol :

$\left(x^{2}-3 x\right)^{2}-8\left(x^{2}-3 x\right)-20$

Let $x^{2}-3 x=y$, then

$\begin{aligned} =& y^{2}-8 y-20 \\=& y^{2}-10 y+2 y-20 \\=& y(y-10)+2(y-10) \\=&(y-10)(y+2) \end{aligned}$

Now,

$=\left(x^{2}-3 x-10\right)\left(x^{2}-3 x+2\right)$

$=\left\{x^{2}-5 x+2 x-10\right\}\left\{x^{2}-x-2 x+2\right\}$

$=\{x(x-5)+2(x-5)\}\{x(x-1)-2(x-1)\}$

$=(x-5)(x+2)(x-1)(x-2)$

Question 7

One of the factors of (x – 1) – (x² – 1) is

(a) x² – 1

(b) x + 1

(c) x – 1

(d) x + 4

Sol :

$=(x-1)-\left(x^{2}-1\right)$

=(x-1)-(x+1)(x-1)

=(x-1)[1-x+1]

So, x-1 is its factor 

option (C) is correct

Question 8

If $\frac{x}{y}+\frac{y}{x}=-1$ (x, y ≠ 0), then the value of x³ – y³ is

(a) 1

(b) – 1

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

(d) 0

Sol :

$=\frac{x}{y}+\frac{y}{x}=-1$

$=\frac{x^{2}+y^{2}}{x y}=-1$

$=x^{2}+y^{2}=-x y$

$=x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$....(formula)

=(x-y)×0=0

Option (d) is correct

Question 9

The product of $=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)$

(a) $x^4+\frac{1}{x^4}$

(b) $x^3+\frac{1}{x^3}-2$

(c) $x^{4}-\frac{1}{x^{4}}$

(d) $x^2+\frac{1}{x^2}+2$

Sol :

$=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)$

$=\left(x^{2}-\frac{1}{x^{2}}\right)\left(x^{2}+\frac{1}{x^{2}}\right)$

$=\left(x^{2}\right)^{2}-\left(\frac{1}{x^{2}}\right)^{2}=x^{4}-\frac{1}{x^{4}}$

Option (C) is correct

Question 10

If x – 2y= 11 and xy = 8, then the value of x³ – 8y³ is

(a) 1860

(b) 1600

(c) 1859

(d) 2000

Sol :

x-2y=11 , xy=8

x-2y=11

⇒$(x-2 y)^{3}=(11)^{3}$ {(i) using both sides}

⇒$x^{3}-8 y^{3}-3 x x \times(2 y)(x-2 y)=1331$

⇒$x^{3}-8 y^{3}-6 x y(x-2 y)=1331$

⇒$x^{3}-8 y^{3}-6 \times 8 \times 11=1331$

⇒$x^{3}-8 y^{3}-528=1331$

⇒$x^{3}-8 y^{3}=1331+528$

⇒$x^{3}-8 y^{3}=1859$

So, option (C) is correct

SChand CLASS 9 Chapter 3 Expansions TEST

 TEST

Question 1

The coefficient of x in the product (2 – 3x) (5 – 2x) is

(a) 19

(b) – 19

(c) 15

(d) 6

Sol :

=(2-3x)(5-2x)

term x has -4x-15x=-19x

Question 2

If $3x^4$ + kx² – 8 = (3x² – 2) (x² + 4) for all x, then the value of k is:

(a) – 2

(b) 12

(c) 10

(d) – 8

Sol :

$3 x^{4}+k x^{2}-8=\left(3 x^{2}-2\right)\left(x^{2}+4\right) $

$3 x^{4}+k x^{2}-8=3 x^{4}+12 x^{2}-2 x^{2}-8$

$3 x^{4}+k x^{2}-8=3 x^{4}+10 x^{2}-8 $

k=10

Question 3

The coefficient of x² in (3x + x³)$ \left(x + \frac{1}{x}\right)$ is

(a) 3

(b) 1

(c) 4

(d) 2

Sol :

$\left(3x+x^{3}\right)\left(x+\frac{1}{x}\right)$

$3 x^{2}+3+x^{4}+x^{2}=4 x^{2}+x^{4}+3$

$x^{2}=4$

Question 4

If a = 3 + b, prove that a³ – b³ – 9ab = 27.

Sol :

a=3+b

a-b=3

cubing both sides

$(a-b)^{3}=(3)^{3}$

$a^{3}-b^{3}-3 a b(a-b)=(9)^{3}$

$a^{3}-b^{3}-3 a b \times 3=27$

$a^{3}-b^{3}-9 a b=27$

Question 5

Simplify:

$\frac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3+(b-c)^3+(c-a)}$

Sol :

$\frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$

$\frac{3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)}{3(a-b)(b-c)(c-a)}$

$\frac{(a-b)(a+b)(b-c)(b+ c)(c- a)(c+a)}{(a-b)(b-c)(c-a)}$

(a+b)(b+c)(c+a)

Question 6

If $a+\frac{1}{(a+2)}=0$ , then the value of $(a+2)^3+\frac{1}{(a+2)^3}$ is

(a) 6

(b) 4

(c) 3

(d) 2

Sol :

$a+\frac{1}{(a+2)}=0$

$\frac{a \times(a+2)+1}{a+2)} \Rightarrow 0$

$a^{2}+2 a+1=0$

(a+1)=0

a=-1

Putting value of a in $(a+2)^{3}+\frac{1}{(a+2)^{2}}$

$(-1+2)^{3}+\frac{1}{(-1+2)^{3}}$

$(1)^{3}+\frac{1}{(1)^{3}}$

=1+1=2

Question 7

If $a+\frac{1}{a}+2=0$  , then the value of a $\left(a^{37}-\frac{1}{a^{100}}\right)$

a) 0

(b) – 2

(c) 1

(d) 2

Sol :

$a+\frac{1}{a}+2 \Rightarrow 0$

$a^{2}+1+2a \Rightarrow 0$

$(a-1)^{2}=0$

$(a-1) \Rightarrow 0$

a=1

Putting the value of a in $a^{37}-\frac{1}{a^{100}}$

$(-1)^{37}-\frac{1}{(-1)^{100}}$ 

${-1-\frac{1}{1}}$

-1-1=-2

Question 8

If (a – 1)² + (b + 2)² + (c + 1)² = 0 then the value of 2a – 3b + 7c is

(a) 12

(b) 3

(c) – 11

(d) 1

Sol :

$(a-1)^{2}+(b+2)^{2}+(c+1)^{2} \Rightarrow 0$

$(a-1)^{2} \Rightarrow 0$

a=1

$(b+2)^{2}=0$

b+2=0

b=-2

$(c+1)^{2} \Rightarrow 0$

c+1=0

c=-1

Now putting the value of a,b,c in 2a-3b+7c

=2×1-3×-2+7×-1

=2+6-7=1

Question 9

If ax + by = 3, bx – ay = 4 and x² + y² = 1, then the value of a² + b² is

(a) – 1

(b) – 25

(c) 1

(d) 25

Sol :

ax+by=3...(i)

bx-ay=4....(ii)

$x^{2}+y^{2} \Rightarrow 1$...(iii)

Squaring equation (i) and (ii) and adding

$(a x+b y)^{2}+(b x-a y)^{2} \Rightarrow 3^{2}+4^{2}$

$a^{2} x^{2}+b^{2} y^{2}+2 a b x y+b^{2} x^{2}+a^{2} y^{2}-2 a b x y \Rightarrow 9+16$

$\left(a^{2}+b^{2}\right) x^{2}+\left(a^{2}+b^{2}\right) y^{2} \Rightarrow 25$

$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \Rightarrow 25$

$\left(a^{2}+b^{2}\right) \times 1 \Rightarrow 25$ from equation (3)

$a^{2}+b^{2} \Rightarrow 25$

Question 10

If p + q = 10 and pq = 5, then the numerical value of $\frac{p}{q}+\frac{q}{p}$ will be:

(a) 22

(b) 18

(c) 16

(d) 20

Sol :

p+q⇒10 , p.q⇒5

p+q=10

Squaring both sides

$(p+q)^{2} \Rightarrow 10^{2}$

$p^{2}+q^{2}+2 p q \Rightarrow 100$

$p^{2}+q^{2}+2 \times 5 \Rightarrow 100$

$p^{2}+q^{2} \Rightarrow 100-10$

$p^{2}+q^{2} \Rightarrow 90$

Now $\frac{p}{q}+\frac{q}{p} \Rightarrow \frac{p^{2}+q^{2}}{p q}$

$\frac{90}{50}$

=1.8

SChand CLASS 9 Chapter 1 Rational number and Irrational number Test

 Test

---

Question 1

A number is an irrational number if and only if its decimal representation is

(a) non-terminating

(b) non-terminating and repeating

(c) non-terminating and non-repeating

(d) terminating

Sol :

(c) non-terminating and non-repeating

---

Question 2

Which of the following is an irrational number ?

(a) √29

(b) √441

(c) 0.5948

(d) $5.\sqrt{318}$

Sol :

(a) √29 is irrational number as 29 is not a perfect square.

---

Question 3

$(-2-\sqrt{3})(-2+\sqrt{3})$ when simplified is

(a) positive and irrational

(b) positive and rational

(c) negative and irrational

(d) negative and rational

Sol :

(b) positive and rational

$(-2-\sqrt{3})(-2+\sqrt{3})=(-2)^2-(\sqrt{3})^2$

=4-3=1

Which is positive and rational.

---

Question 4

If $\sqrt{6} \times \sqrt{15}=x\sqrt{10}$ , then the value of x is 

(a) 3

(b) ± 3

(c) √3

(d) √6

Sol :

⇒√6×√15=x√10

⇒$\sqrt{6 \times 15}$=x√10

⇒√90=x√10

⇒$\sqrt{9 \times 10}=x\sqrt{10}$

⇒3$\sqrt{10}=x\sqrt{10}$

Comparing, we get

∴ x = 3

---

Question 5

Two rational numbers between $\frac{2}{7}$ and $\frac{2}{14}$ are

(a) $\frac{1}{14}\text{ and }\frac{2}{14}$

(b) $\frac{1}{2}\text{ and }\frac{3}{2}$

(c) $\frac{3}{14}\text{ and }\frac{3}{7}$

(d) $\frac{5}{14}\text{ and }\frac{8}{14}$

Sol :

(d) $\frac{5}{14}\text{ and }\frac{8}{14}$

Two rational numbers between $\frac{2}{7} \text{ and } \frac{2}{14}$ are $\frac{5}{14}\text{ and }\frac{8}{14}$

∵$\frac{2}{7} \text{ and } \frac{5}{7}=\frac{4}{14} \text{ and } \frac{10}{14}$ and 5 and 8 lie between 4 and 10

---

Question 6

An irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$ is

(a) 0.75

(b) √6

(c) 0.7507500075000…

(d) 0.7512

Sol :

(c) 0.7507500075000…

An irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$ either √6 or 0.7507500075000....

∵ 0.75 and 0.7512 are rational number

Which does not line between $\frac{5}{7} \text{ and } \frac{7}{9}$

∴ 0.7507500075000… is irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$

---

Question 7

If √2=1.4142 , then the value of $\frac{7}{3+\sqrt{2}}$ correct to two decimal places is 

(a) 1.59

(b) 1.60

(c) 2.58

(d) 2.57

Sol :

(a) 1.59

√2=1.4142

$\frac{7}{3+\sqrt{2}}=\frac{7\times (3-\sqrt{2})}{(3-\sqrt{2})(3-\sqrt{2})}$

Rationalising denominator

$=\frac{7(3-\sqrt{2})}{(9-2)}=\frac{7(3-\sqrt{2})}{7}$

=3-√2

= 3 – 1.4142 = 1.5858 = 1.59

---

Question 8

Taking √3 as i.732 and √2=1.414 , the value of $\frac{1}{\sqrt{3}+\sqrt{2}}$ is 

(a) 0.064

(b) 0.308

(c) 0.318

(d) 2.146

Sol :

(c) 0.318

√3 = 1.732,  √2= 1.414

$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

(Rationalising denominator)

$\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}$

=1.732-1.414=0.318

---

Question 9

If $x=\sqrt{3}+\sqrt{2}$ , then the value of $\left(x+\frac{1}{x}\right)$ is

(a) 2

(b) 3

(c) $2\sqrt{2}$

 

(d) $2\sqrt{3}$

Sol :

(d) $2\sqrt{3}$

$x=\sqrt{3}+\sqrt{2}$

$\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

(Rationalising denominator)

$=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}$

$x+\frac{1}{x}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}$

---

Question 10

If $x=2+\sqrt{3}$ , then the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$ is 

(a) $3+\sqrt{3}$

(b) √6

(c) 2√6

(d) 6

Sol :

(b) √6

x=2+√3

$\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$

(Rationalising denominator)

$\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

∴$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$ and $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}+2$

=4+2=6

$\sqrt{x}+\frac{1}{\sqrt{x}}=\pm \sqrt{6}$

=√6

ML Aggarwal Solution Class 10 Chapter 22 Probability Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17

 Test

Question 1

A game consists of spinning an arrow which comes to rest at one of the regions 1, 2 or 3 (shown in the given figure). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.

Sol :

In a game,

No, the outcomes are not equally likely.

Outcome 3 is more likely to occur than the outcomes of 1 and 2.

Question 2

In a single throw of a die, find the probability of getting

(i) a number greater than 5

(ii) an odd prime number

(iii) a number which is multiple of 3 or 4.

Sol :

In a single throw of a die

Number of total outcomes = 6 (1, 2, 3, 4, 5, 6)

(i) Numbers greater than 5 = 6 i.e., one number

Probability $=\frac{1}{6}$

(ii) An odd prime number 2 i.e., one number

Probability $=\frac{1}{6}$

(iii) A number which is a multiple of 3 or 4 which are 3, 6, 4 = 3 numbers

Probability $=\frac{3}{6}=\frac{1}{2}$

Question 3

A lot consists of 144 ball pens of which 20 are defective and the others are good. Rohana will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that :

(i) She will buy it?

(ii) She will not buy it?

Sol :

In a lot, there are 144 ball pens in which defective ball pens are = 20

and good ball pens are = 144 – 20 = 124

Rohana buys a pen which is good only.

(i) Now the number of possible outcomes = 144

and the number of favourable outcomes = 124

$\therefore$ Probability of good pen will be,

$P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of possible outcome }}$

$=\frac{124}{144}=\frac{31}{36}$

(ii) Probability of not buying a defective pen will be $\mathrm{P}(\overline{\mathrm{E}})$

But $P(E)+P(\bar{E})=1$

$\therefore \frac{31}{36}+\mathrm{P}(\overline{\mathrm{E}})=1 $

$\Rightarrow \mathrm{P}(\overrightarrow{\mathrm{E}})=1-\frac{31}{36}=\frac{5}{36}$

Hence $P(\bar{E})=\frac{5}{36}$

Question 4

A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is

(i) acceptable to Varnika?

(ii) acceptable to the trader?

Sol :

Number of total mobiles = 48

Number of good mobiles = 42

Number having minor defect = 3

Number having major defect = 3

(i) Acceptable to Varnika = 42

Probability$=\frac{42}{48}=\frac{7}{8}$

(ii) Acceptable to trader = 42 + 3 = 45

Probability$=\frac{45}{48}=\frac{15}{16}$

Question 5

A bag contains 6 red, 5 black and 4 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is

(i) white

(ii) red

(iii) not black

(iv) red or white.

Sol :

Total number of balls = 6 + 5 + 4 = 15

Number of red balls = 6

Number of black balls = 5

Number of white balls = 4

(i) Probability of a white ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}=\frac{4}{15}$

(ii) Probability of red ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{6}{15}=\frac{2}{5}$

(iii) Probability of not black ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{15-5}{15}$

$=\frac{10}{15}$

$=\frac{2}{3}$

(iv) Probability of red or white ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{6+4}{15}$

$=\frac{10}{15}$

$=\frac{2}{3}$

Question 6

A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is:

(i) red or white

(ii) not black

(iii) neither white nor black

Sol :

Total number of balls in a bag = 5 + 8 + 7 = 20

(i) Number of red or white balls = 5 + 8 = 13

Probability of red or white ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{13}{20}$

(ii) Number of ball which are not black = 20 – 7 = 13

Probability of not black ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{13}{20}$

(iii) Number of ball which are neither white nor black

= Number of ball which are only red = 5

Probability of neither white nor black ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{5}{20}$

$=\frac{1}{4}$

Question 7

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :

(i) white or blue

(ii) red or black

(iii) not white

(iv) neither white nor black ?

Sol :

Number of total balls = 5 + 7 + 4 + 2 = 18

Number of white balls = 5

number of red balls = 7

number of black balls = 4

and number of blue balls = 2.

(i) Number of white and blue balls = 5 + 2 = 7

Probability of white or blue balls will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{7}{18}$

(ii) Number of red and black balls = 7 + 4 = 11

Probability of red or black balls will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{11}{18}$

(iii) Number of ball which are not white = 7 + 4 + 2 = 13

Probability of not white balls will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{13}{18}$

(iv) Number of balls which are neither white nor black = 18 – (5 + 4) = 18 – 9 = 9

Probability of ball which is neither white nor black will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{9}{18}=\frac{1}{2}$

Question 8

A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is

(i) an odd number

(ii) divisible by 2 or 3

(iii) prime number

(iv) not divisible by 10?

Sol :

In a box, there are 20 balls containing 1 to 20 number

Number of possible outcomes = 20

(i) Numbers which are odd will be,

1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10 balls.

Probability of odd ball will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{10}{20}=\frac{1}{2}$

(ii) Numbers which are divisible by 2 or 3 will be

2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13 balls

Probability of ball which is divisible by 2 or 3 will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{13}{20}$

(iii) Prime numbers will be 2, 3, 5, 7, 11, 13, 17, 19 = 8

Probability of prime number will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{8}{20}=\frac{2}{5}$

(iv) Numbers not divisible by 10 will be

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 = 18

Probability of prime number will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{18}{20}=\frac{9}{10}$

Question 9

Find the probability that a number selected at random from the numbers 1, 2, 3,……35 is a

(i) prime number

(ii) multiple of 7

(iii) multiple of 3 or 5.

Sol :

Numbers are 1, 2, 3, 4, 5,…..30, 31, 32, 33, 34, 35

Total = 35

(i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

which are 11

Probability of prime number will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{11}{35}$

(ii) Multiple of 7 are 7, 14, 21, 28, 35 which are 5

Probability of multiple of 7 will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{5}{35}=\frac{1}{7}$

(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12 ,15, 18, 20, 21, 24, 25, 27, 30, 33, 35.

Which are 16 in numbers

Probability of multiple of 3 or 5 will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{16}{35}$

Question 10

Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is

(i) divisible by 5

(ii) a number which is a perfect square.

Sol :

Number of cards which are marked with numbers

13, 14, 15, 16, 17,….to 59, 60 are = 48

(i) Numbers which are divisible by 5 will be

15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10

Probability of number divisible by 5 will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{10}{48}=\frac{5}{24}$

(ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers.

Probability of number which is a perfect square will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{4}{48}=\frac{1}{12}$

Question 11

The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has

(i) an odd number

(ii) a perfect square number.

Sol :

Cards in a box are from 14 to 99 = 86

No. of total cards = 86

One card is drawn at random

Cards bearing odd numbers are 15, 17, 19, 21, …, 97, 99

Which are 43

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{43}{86}$

$=\frac{1}{2}$

(ii) Cards bearing number which are a perfect square

= 16, 25, 36, 49, 64, 81

Which are 6

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{6}{86}$

$=\frac{3}{43}$

Question 12

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bags.

Sol :

Number of red balls = 5

and let number of blue balls = x

Total balls in the bag = 5 + x

and that of red balls $=\frac{5}{5+x}$

According to the condition,

$\frac{x}{5+x}=4 \times \frac{5}{5+x}$

$\frac{x}{5+x}=\frac{20}{5+x}$

x ≠ – 5

x = 20

Hence, number of blue balls = 20

and number of balls in the bag = 20 + 5 = 25

Question 13

A bag contains 18 balls out of which x balls are white.

(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?

(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be $\frac{9}{8}$ times that of probability of white ball coming in part (i). Find the value of x.

Sol :
Total numbers of balls in a bag = 18

No. of white balls = x

(i) One ball is drawn a random$=\frac{x}{18}$

(ii) If 2 more white balls an put, then number of white balls = x + 2

and probability is $\frac{9}{8}$ times

$=\frac{9}{8} \times \frac{x}{18}=\frac{x}{16}$

and number of balls $=18+2=20$

$\therefore \frac{x+2}{18+2}=\frac{x}{16} \Rightarrow \frac{x+2}{20}=\frac{x}{16}$

20x=16x+32 

$\Rightarrow 20 x-16 x=32$

$\Rightarrow 4 x=32 $

$\Rightarrow x=\frac{32}{4}=8$

$\therefore x=8$

Question 14

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :

(i) a red face card

(ii) neither a club nor a spade

(iii) neither an ace nor a king of red colour

(iv) neither a red card nor a queen

(v) neither a red card nor a black king.

Sol :

Number of cards in a pack of well-shuffled cards = 52

(i) Number of a red face card = 3 + 3 = 6

Probability of red face card will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{6}{52}=\frac{3}{26}$

(ii) Number of cards which is neither a club nor a spade = 52 – 26 = 26

Probability of card which’ is neither a club nor a spade will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{26}{52}=\frac{1}{2}$

(iii) Number of cards which is neither an ace nor a king of red colour

= 52 – (4 + 2) = 52 – 6 = 46

Probability of card which is neither ace nor a king of red colour will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{46}{52}=\frac{23}{26}$

(iv) Number of cards which are neither a red card nor a queen are

= 52 – (26 + 2) = 52 – 28 = 24

Probability of card which is neither red nor a queen will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{24}{52}=\frac{6}{13}$

(v) Number of cards which are neither red card nor a black king

= 52 – (26 + 2) = 52 – 28 = 24

Probability of cards which is neither red nor a black king will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{24}{52}=\frac{6}{13}$

Question 15

From pack of 52 playing cards, blackjacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting

(i) a red card

(ii) a face card

(iii) a diamond or a club

(iv) a queen or a spade.

Sol :

Total number of cards = 52

Black jacks, black kings and black aces are removed

Now number of cards = 52 – (2 + 2 + 2) = 52 – 6 = 46

One card is drawn

(i) No. of red cards = 13 + 13 = 26

∴Probability $=\frac{26}{46}=\frac{13}{23}$

(ii) Face cards = 4 queens, 2 red jacks, 2 kings = 8

∴Probability $=\frac{8}{46}=\frac{4}{23}$

(iii) a diamond on a club = 13 + 10 = 23

∴Probability $=\frac{23}{46}=\frac{1}{2}$

(iv) A queen or a spade = 4 + 10 = 14

∴Probability $=\frac{14}{46}=\frac{7}{23}$

Question 16

Two different dice are thrown simultaneously. Find the probability of getting:

(i) sum 7

(ii) sum ≤ 3

(iii) sum ≤ 10

Sol :

(i) Numbers whose sum is 7 will be (1, 6), (2, 5), (4, 3), (5, 2), (6, 1), (3, 4) = 6

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{6}{36}=\frac{1}{6}$

(ii) Sum ≤ 3

Then numbers can be (1, 2), (2, 1), (1, 1) which are 3 in numbers

∴Probability will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{3}{36}=\frac{1}{12}$

(iii) Sum ≤ 10

The numbers can be,

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, .6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),

(6, 1), (6, 2), (6, 3), (6, 4) = 33

Probability will be

$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome }}{\text { Number of possible outcome }}$

$=\frac{33}{36}=\frac{11}{12}$

Question 17

Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is

(i) 6

(ii) 12

(iii) 7

Sol :

Two dice are thrown together

Total number of events = 6 × 6 = 36

(i) Product 6 = (1, 6), (2, 3), (3, 2). (6, 1) = 4

Probability $=\frac{4}{36}=\frac{1}{9}$

(ii) Product 12 = (2, 6), (3, 4), (4, 3), (6, 2) = 4

Probability $=\frac{4}{36}=\frac{1}{9}$

(iii) Product 7 = 0 (no outcomes)

Probability $=\frac{0}{36}=0$

ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22

 Test

Question 1

Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x.

Sol :

Marks in English = 36

Marks in Civics = 44

Marks in Mathematics = 75

Marks in Science = x

Total marks in 4 subjects = 36 + 44 + 75 + x = 155 + x

average marks $=\frac{155+x}{4}$

But average marks = 50 (given)

$\frac{155+x}{4}=50$

⇒ 155 + x = 200

⇒ x = 200 – 155 = 45

Question 2

The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.

Sol :

Mean of 20 numbers =18

Total number = 18 × 20 = 360

By adding 3 to first 10 numbers,

The new sum will be = 360 + 3 × 10 = 360 + 30 = 390

New Mean $=\frac{390}{20}=19.5$

Question 3

The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.

Sol :

In first case,

Average height of 30 students = 150 cm

Total height = 150 × 30 = 4500 cm

Difference in copying the number = 165 – 135 = 30 cm

Correct sum = 4500 + 30 = 4530 cm

Correct mean $=\frac{4530}{30}$

=151 cm

Question 4

There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.

Sol :

Total students of a class = 50

No. of boys = 40

No. of girls = 50 – 40 = 10

Average weight of 50 students = 44 kg

Total weight = 44 × 50 = 2200 kg

Average weight of 10 girls = 40 kg

.’. Total weight of girls = 40 × 10 = 400 kg

Then the total weight of 40 boys = 2200 – 400 = 1800kg

Average weight of boys $=\frac{1800}{40}$

=45 kg

Question 5

The contents of 50 boxes of matches were counted giving the following results

No. of matches	41	42	43	44	45	46
No. of boxes	5	8	13	12	7	3

Calculate the mean number of matches per box.

Sol :

No. of matches (x)	No. of boxes (f)	f.x
41	5	205
42	8	336
43	13	559
44	12	528
45	7	315
46	5	230
Total	50	2173

Mean $=\frac{\sum f x}{\sum f}=\frac{2173}{50}$

=43.46

Question 6

The heights of 50 children were measured (correct to the nearest cm) giving the following results :

Height (in cm)	65	66	67	68	69	70	71	72	73
No. of children	1	4	5	7	11	10	6	4	2

Sol :

Calculate the mean height for this distribution correct to one place of decimal.

No. of matches (x)	No. of boxes (f)	f.x
65	1	65
66	4	264
67	5	335
68	7	476
69	11	759
70	10	700
71	6	426
72	4	288
72	2	146
Total	50	3459

Mean $=\frac{\sum f x}{\sum f}=\frac{3459}{50}$

=69.18=69.2

Question 7

Find the value of p for the following distribution whose mean is 20.6 :

Variate (xi)	10	15	20	25	35
Frequency(fi)	3	10	p	7	5

Sol :

Variate (xi)	Frequency(fi)	fixi
10	3	30
15	10	150
20	p	20p
25	7	175
35	5	175
Total	25+p	530+20p

Mean$=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{530+20 p}{25+p}$

But Mean=20.6 (given)

$\therefore \frac{530+20 p}{25+p}=20 \cdot 6 $

$\Rightarrow 530+20 p=515+20 \cdot 6 p$

$ \Rightarrow 20 \cdot 6 p-20 p=530-515$

$\Rightarrow 0-6 p=15 $

$ \Rightarrow  \frac{6}{10} p=15 $

$ \Rightarrow  p=\frac{15 \times 10}{6}=25$

Question 8

Find the value of p if the mean of the following distribution is 18.

Variate (x)	13	15	17	19	20+p	23
Frequency(f)	8	2	3	4	5p	6

Sol :

Variate (x)	Frequency (f)	f.x
13	8	104
15	2	30
17	3	51
19	4	76
20+p	5p	100p+5p2
23	6	138
Total	23+5p	399+100p+5p2

Mean$=\frac{\sum f x}{\sum f}=\frac{399+100 p+5 p^{2}}{23+5 p}$

But mean=18(given)

$\therefore \frac{399+100 p+5 p^{2}}{23+5 p}=\frac{18}{1}$

⇒399+100p+5p²=414+90p

$\Rightarrow \quad 5 p^{2}+100 p+399-90 p-414=0$

$ \Rightarrow 5 p^{2}+10 p-15=0$

$\Rightarrow \quad p^{2}+2 p-3=0 $

$ \Rightarrow \quad p^{2}+3 p-p-3=0$

$\Rightarrow \quad p(p+3)-1(p+3)=0 $

$ \Rightarrow \quad(p+3)(p-1)=0$

Either p+3=0, then p=-3 but it is not possible as it is negative

or p-1=0 , then p=1

Question 9

Find the mean age in years from the frequency distribution given below:

Age (in years)	25-29	30-34	35-39	40-44	45-49	50-54	55-59
No. of persons	4	14	22	16	6	5	3

Arranging the classes in proper form

Sol :

Class age (in years)	Mid value (xi)	No. of persons (fi)	fi xi
24.5-29.5	27	4	108
29.5-34.5	32	14	448
34.5-39.5	37	22	814
39.5-44.5	42	16	672
44.5-49.5	47	6	282
49.5-49.5	52	5	260
54.5-59.5	57	3	171
Total		70	2755

Mean$=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{2755}{70}$

=39.357

=39.36 years

Question 10

Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution :

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Students	2	4	5	16	20	10	6	8	4

Sol :

Calculate the mean height for this distribution correct to one place of decimal.

Marks	Students (fi)	Class Marks (xi)	fi.xi
10-20	2	15	30
20-30	4	25	100
30-40	5	35	175
40-50	16	45	720
50-60	20	55	1100
60-70	10	65	650
70-80	6	76	450
80-90	8	85	680
90-100	4	95	380
Total	75		4285

Mean$=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{4285}{75}$

=57.133

=57.1

Question 11

The mean of the following frequency distribution is 62.8. Find the value of p.

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	5	8	p	12	7	8

Sol :

Mean = 62.8

Class	Frequency (fi)	Class marks (x)	fi xi
0-20	5	10	50
20-40	8	30	240
40-60	p	50	50p
60-80	12	70	840
80-100	7	90	630
100-120	8	110	880
Total	40+p		2640+50p

Mean $=\frac{\sum f_{1} x_{i}}{\sum f_{i}}$

$\Rightarrow 62.8=\frac{2640+50 p}{40+p}$

$ \Rightarrow(40+p) \times 62.8=2640+5$

2512.0+62.8p=2640+50p

$\Rightarrow \quad 62.8 p-50 p=2640-2512$

$ \Rightarrow 12.8 p=128$

$p=\frac{128}{12.8}=\frac{128 \times 10}{128}=10$

Hence p=10

Question 12

The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.

Expenditure (in Rs)	140-160	160-180	180-200	200-220	220-240
No. of families	5	25	f1	f2	5

Sol 

Mean = 188,

No. of families = 100

Expenditure	Mid value (xi)	No. of persons (fi)	fi xi
140-160	150	5	750
160-180	170	25	4250
180-200	190	f1	190f1
200-220	210	f2	210f2
220-240	230	5	1150
Total		35+f1+f2=100	6150+190f1+210f2

35+f₁+f₂=100

$\Rightarrow f_{1}+f_{2}=100-35=65$..(i)

$\Rightarrow f_{1}=65-f_{2}$

and $\frac{6150+190 f_{1}+210 f_{2}}{100}=188$

$190 f_{1}+210 f_{2}=18800-6150=12650$

$190\left(65-f_{2}\right)+200 f_{2}=12650$

$12350-190 f_{2}+210 f_{2}=12650$

$20 f_{2}=12650-12350$

$20 f_{2}=300$

$f_{2}=\frac{300}{2}=15$

$\therefore f_{1}=65-15=50$

$f_{1}=50, f_{2}=15$

Question 13

The measures of the diameter of the heads of 150 screw is given in the following table. If the mean diameter of the heads of the screws is 51.2 mm, find the values of p and q

Diameter (in mm)	32-36	37-41	42-46	47-51	52-56	57-61	62-66
No. of screws	15	17	p	25	q	20	30

Sol :

Mean = 51.2

No. of screws = 150

Diameter (in mm)	No. of screws (fi)	Class Marks (xi)	fi xi
32-36	15	34	510
37-41	17	39	663
42-46	p	44	44p
47-51	25	49	1125
52-56	q	54	54q
57-61	20	59	1180
62-66	30	64	1920
Total	107+p+q		5498+44p+54q

107+p+q=150

p+q=150-107=43...(i)

Mean $=\frac{\sum f_{1} x_{i}}{\Sigma f_{i}} \Rightarrow \frac{5498+44 p+54 q}{150}$

=51.2

$\Rightarrow 5498+44 p+54 q=7680$

4p+54q=7680-5498

44p+54q=2182

22p+27q=1091...(ii)

Multiplying (i) by 27 and (ii) and by 1

27p+27q=1161...(iii)

22p+27q=1091...(iv)

Subtracting (iv) from (iii) we get

5p=70

$p=\frac{70}{5}=14$

But p+q=43

$\therefore q=43-p=43-14=29$

Hence p=14, q=29

Question 14

The median of the following numbers, arranged in ascending order is 25. Find x, 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46

Sol :

Here, n = 10, which is even

∴Median $=\frac{1}{2}\left[\frac{n}{2} \mathrm{th}+\left(\frac{n}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}\left[\frac{10}{2} \mathrm{th}+\left(\frac{10}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}$ (5th +6th term)

$=\frac{1}{2}(x+2+x+4)=\frac{2 x+6}{2}$

=x+3

But median is given=25

$\therefore x+3=25$

$ \Rightarrow x=25-3=22$

Question 15

If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.

Sol :

Arranging in ascending order, 3, 4, 5, x, 8, 9, 11,

Here n = 7 which is odd.

∴ Median $=\frac{n+1}{2}$ th term

$=\frac{7+1}{2}$= 4th term=x

∴ but median = 6

∴ x = 6

Question 16

Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 If 26 is replaced by 62, find the new median.

Sol :

Arranging the given data in ascending order

17, 26, 29, 32, 33, 34, 45, 56, 60

Here n = 9 which is odd

∴Median $=\frac{n+1}{2}$ th term

$=\frac{9+1}{2}=\frac{10}{2}$=5th term=33

(ii) If 26 is replaced by 62, their the order will be

17, 29, 32, 33, 34, 45, 56, 60, 62

Here 5th term is 34

∴ Median = 34

Question 17

The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12

Find

(i) the median

(ii) lower quartile

(iii) upper quartile

Sol :

Arranging the given data in ascending order:

1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23

Here n = 16 which is even.

(i) $\therefore$ Median $=\frac{1}{2}\left[\frac{16}{2}\right.$ th $\left.+\left(\frac{16}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}$(8th+9th) term

$=\frac{12+13}{2}=\frac{25}{2}$

=12.5

(ii) Lower quartile

$=\frac{1}{4} n=\frac{16}{4}$

=4th term

=6

(iii) Upper quartile

$=\frac{3}{4} n=\frac{3}{4} \times 16$

=12th term

=18

Question 18

Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9

Sol :

Here n = 9 which is odd.

∴Median $=\frac{n+1}{2}$th term

$=\frac{9+1}{2}=\frac{10}{2}$

=5th term=5

Here 5 occur maximum times

∴Mode = 5

Question 19

Calculate the mean, the median and the mode of the following distribution :

Age (in years)	12	13	14	15	16	17	18
No. of students	2	3	5	6	4	3	2

Sol :

Mean = 51.2

No. of screws = 150

Age (in years) (xi)	No. of screws (fi)	Class Marks (fi)	fi xi
12	2	2	24
13	3	5	39
14	5	10	70
15	6	16	90
16	4	20	64
17	3	23	51
18	2	25	36
Total	25		374

(i) Mean $=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{374}{25}$

=14.96

(ii) Here n=25 which is odd

∴$=\frac{n+1}{2}$ th term $=\frac{25+1}{2}=\frac{26}{2}$

=13 th term

=15

(iii) Here 15 occurs most i.e. in 6 times

∴Mode=15

Question 20

The daily wages of 30 employees in an establishment are distributed as follows :

Daily wages(in ₹)	0-10	10-20	20-30	30-40	40-50	50-60
No. of employees	1	8	10	5	4	2

Sol :

Estimate the modal daily wages for this distribution by a graphical method.

Daily wages (in ₹)	No. of employees
0-10	1
10-20	8
20-30	10
30-40	5
40-50	4
50-60	2

Taking daily wages on x-axis and No. of employees on the y-axis

and draw a histogram as shown. Join AB and CD intersecting each other at M.

From M draw ML perpendicular to x-axis, L is the mode

∴ Mode = Rs 23

Question 21

Using the data given below, construct the cumulative frequency table and draw the ogive. From the ogive, estimate ;

(i) the median

(ii) the inter quartile range.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	3	8	12	14	10	6	5	2

Also state the median class

Sol :

Marks	Frequency	c.f
0-10	3	3
10-20	8	11
20-30	12	23
30-40	14	37
40-50	10	47
50-60	6	53
60-70	5	58
70-80	2	60

Now plot the points (10,3),(20,11),(30,23)(40,37),(50,47),(60,53),(70,58),(80,60) on the graph and join them in free hand to form an ogive as shown.

Here n=60 which is an even number

(i) Median $=\frac{1}{2}\left[\frac{n}{2}\right.$ th $\left.+\left(\frac{n}{2}+1\right) \mathrm{th}\right]$

$=\frac{1}{2}\left[\frac{60}{2} t h+\left(\frac{60}{2}+1\right) t h\right]$ term

$=\frac{1}{2}(30+31)$

=30.5 th

Now take a point A(30-5) on y-axis. From A draw a line parallel to x-axis meeting the curve at P and from P, draw a perpendicular to x-axis meeting it in Q. Q is the median which is 35 and median class is 30-40

(ii) Lower quartile

$=\frac{n}{4}=\frac{60}{4}$

=15

Upper quartile

$=\frac{3}{4} n=\frac{3}{4} \times 60=45$

Now take points B(15) and C(45) on y-axis and from B and C draw lines parallel to x-axis meeting the curve at L and M respectively. From L and M, draw lines perpendicular to x-axis meeting it at E and F respectively. E and F are lower and upper quartile which are 22.3 and 47

∴Interquartile range$=Q_{3}-Q_{1}$

=47.0-22.3=24.7

Question 22

Draw a cumulative frequency curve for the following data :

Marks obtained	0-10	10-20	20-30	30-40	40-50
No. of students	8	10	22	40	20

Hence determine:

(i) the median

(ii) the pass marks if 85% of the students pass.

(iii) the marks which 45% of the students exceed.

Sol :

Marks obtained	No. of students	c.f
0-10	8	8
10-20	10	18
20-30	22	40
30-40	40	80
40-50	20	100

Now plot points (10,8),(20,18),(30,40),(40,80) and (50,100) on the graph and join them in free hand to form an ogive.

Here n=100 which is even

(i) $\therefore$ Median $=\frac{1}{2}\left[\frac{n}{2}\right.$ th $+\left(\frac{n}{2}+1\right)$ th $]$ term

$=\frac{1}{2}\left[\frac{100}{2} \mathrm{th}+\left(\frac{100}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}(50 t h+51 t h)=\frac{1}{2} \times 101=50 \cdot 5$

Now take a point A(50.5) on y-axis and from A, draw a line parallel to x-axis meeting the curve at P and from P , draw a perpendicular to x-axis meeting it at Q. Q is the median which is 32.5

(i) If 85% students pass, the pass marks will be 18

(ii) Marks which 45% of the students exceeds=34 marks