Zero is a rational or an irrational number?

Is (x+y)/2 is an irrational number or a rational number?

How to answer the following question?rnrnQ) There are two boxes. Which box requires the lesser amount of material to make? The measurements are as belowrna) height 30 cm, width 80 cm and length 10 cmrnb) height 20 cm,

In what ratio does the x axis divide the line segment joining the points (-4,-6) and (-1,7)? So, how to find the coordinate of the point of division?

How to answer the following question?

Ramesh is a cashier at Canara Bank. He has notes of denominations of Rs 100. 50 and 10 respectively. The ratio of the number of these notes are is  2: 3: 5 respectively.

What is 100 divided by 2?

How can we find the irrational numbers between two fractions

What is the multiplicative inverse of rational numbers

find two rationals between 0.5 and 0.55

Can u explain solving of rational numbers with appropriate property

## Is zero (0) a rational number ? justify your answer ?

Yes, zero is a rational number

zero can be expressed as    $\frac{0}{5},\frac{0}{26},\frac{0}{100}$  etc

which are in the form of, $\frac{p}{q}$  where p and q are integers and q $\neq$ 0.

## Write the simplest form of a rational number    $\frac{177}{413}$

$\frac{177}{413} = \frac{59\times 3}{59\times 7} = \frac{3}{7}$

## Is   $\frac{\sqrt{98}}{\sqrt{2}}$  a rational number or not ?

$\frac{\sqrt{98}}{\sqrt{2}} =\frac{ \sqrt{49}\times \sqrt{2}}{\sqrt{2}} =\sqrt{49}= 7$

so it is rational number.

## Identify an irrational number among the following numbers: 0.13 ,$0.13\overline{ 15}$ , $0.\overline{ 1315}$ , 0.3013001300013.....

0.13 is a terminating number. So, it is not an irrational number.

$0.13\overline{15}$ = 0.131515......, 15 is repeating continuously so it is not an irrational number.

$0.\overline{ 1315}$ = 0.13151315........is repeating continuously so it is not an irrational number.

0.3013001300013....., non terminating and non recurring decimal. Hence, it is an irrational number. So, 0.3013001300013 is an irrational number.

## Is the product of two irrational number is always an irrational number ?

No, it may be rational or irrational.

## Insert three rational numbers between $\frac{-1}{\: \: 3}$   and   $\frac{-2}{\: \: 3}$.

$\frac{-1}{\: \: 3} = \frac{-4}{12}$

$\frac{-2}{\: \: 3} = \frac{-8}{12}$

So three rational numbers are

$\frac{-5}{\: \: 3},\frac{-6}{12}\: \: and\: \: \frac{-7}{12}$

## Find two rational numbers between 4 and 5 .

$4 = \frac{4}{5} \times 5\: \: and \: \: 5 = \frac{5}{5}\times 5$

$4 = \frac{20}{5}\: \: and \: \: 5 = \frac{25}{5}$

The numbers are $\frac{21}{5}$ and $\frac{22}{5}$

## Express the rational numbers  $0 .\overline{9}$  in the form  $\frac{p}{q}$ ,  where p and q are integers and $q \neq 0$

Let  X  =  0.999............

10x  =  9.999.......

10x - x  =  (9.999....) – (0.999....)

9x  =  9

X = 1

## Calculate the irrational number 2 and 2.5

Since $\sqrt{5}= 2.236$

hence, the irrationl number between2 and 2.5 is $\sqrt{5}$

## Simplify the number   $\left ( \sqrt{2}+\sqrt{5} \right) ^{2}$

$\left ( \sqrt{2}+\sqrt{5} \right) ^{2}= \left (\sqrt{2} \right )^{2}+\left ( \sqrt{5} \right )^{2}$$+2\times \sqrt{2}\times \sqrt{5}$

$= 2+5+2\sqrt{10}$

$= 7+2\sqrt{10}\: \: \left ( irrational\: number \right )$

## Find the rational numbers between 0.121221222122221...and  0.141441444144441... in the p form, where p and q integers and $q \neq 0$

Two rational numbers between0.121221222122221...and 0.141441444144441...are 0.13 and 0.14

$\frac{13}{10} \: \: and\: \: \frac{14}{100}$

$\frac{13}{100}\: \: and\: \: \frac{7}{50}$

## Find four rational numbers between   $\frac{1}{5}$   and   $\frac{1}{6}$

Since LCM of 5 and 6 is 30

$\frac{1}{6} = \frac{1}{6} \times \frac{5}{5} = \frac{5}{30} \times \frac{5}{5} = \frac{25}{150}$

$\frac{1}{5} = \frac{1}{5} \times \frac{6}{6} = \frac{6}{30} \times \frac{5}{5} = \frac{30}{150}$

Hence , four rational numbers between   $\frac{1}{6}$   and   $\frac{1}{5}$

are $\frac{26}{150} , \frac{27}{150} , \frac{28}{150} , \frac{29}{150}$

## Find six rational numbers between 3 and 4.

let  a = 3 , and b = 4

Here, we find six rational numbers i,e n = 6

$d = \frac{b - a}{n + 1} = \frac{4 - 3}{6 + 1} = \frac{1}{7}$

1st rational number =   $a + d = 3 + \frac{1}{7} = \frac{22}{7}$

2nd rational number =  $a + 2d = 3 + \frac{2}{7} = \frac{23}{7}$

3 rd rational number =  $a + 3d = 3 + \frac{3}{7} = \frac{24}{7}$

4th rational number  =  $a + 4d = 3 + \frac{4}{7} = \frac{25}{7}$

5th rational number  =  $a + 5d = 3 + \frac{5}{7} = \frac{26}{7}$

6th rational number  =   $a + 6d = 3 + \frac{6}{7} = \frac{27}{7}$

So six rational numbers are $\frac{22}{7} , \frac{23}{7} , \frac{24}{7} , \frac{25}{7} , \frac{26}{7} , \frac{27}{7}$

## Insert three rational numbers between  and   $\frac{3}{5}$  and $\frac{5}{7}$

LCM of 5 and 7 is  35

$\frac{3}{5} = \frac{3}{5} \times \frac{7}{7} = \frac{21}{35}$

and     $\frac{5}{7} = \frac{5}{7} \times \frac{5}{5} = \frac{25}{35}$

so      $\frac{21}{35} < \frac{22}{35} < \frac{23}{35} < \frac{24}{35} < \frac{25}{25}$

The required three rational numbers are

$\frac{22}{35} , \frac{23}{35}\: \: and\: \: \frac{24}{35}$

## $\frac{\sqrt{147}}{\sqrt{75}}$  is not a rational number as $\sqrt{147}$  and $\sqrt{75}$ are not rational. State whether it is true or false. Justify your answer.

false

justification  :     $\frac{\sqrt{147}}{\sqrt{75}} = \sqrt{\frac{147}{75}} = \sqrt{\frac{49}{25}} = \frac{7}{5}$

Which is a rational number

## simplify   $\frac{6-4\sqrt{3}}{6+4\sqrt{3}}$   by rationalizing the denominator

$\frac{6-4\sqrt{3}}{6+4\sqrt{3}} \times \frac{6-4\sqrt{3}}{6-4\sqrt{3}}$

$= \frac{(6-4\sqrt{3})^{2})}{36-48}$

$= \frac{36+48-48\sqrt{3}}{-12}$

$= \frac{84-48\sqrt{3}}{-12} =-(7-4\sqrt{3})=4\sqrt{3}-7$

## Find three rational numbers between  $\frac{5}{7}$  and  $\frac{9}{11}$

Since LCM of 7 and 11 is 77

$\frac{5}{7} = \frac{5}{7}$$\times \frac{11}{11} = \frac{55}{77}$

$\frac{9}{11} = \frac{9}{11}\times \frac{7}{7} = \frac{63}{77}$

Hence three rational numbers between  $\frac{5}{7}$  and  $\frac{9}{11}$  are  $\frac{56}{77} , \frac{57}{77} , \frac{58}{77}$

## Rationalize the denominator  $\frac{1}{2\sqrt{7}+3\sqrt{3}}$

$\frac{1}{2\sqrt{7}+3\sqrt{3}} =\frac{1}{(2\sqrt{7}+3\sqrt{3})}\times \frac{(2\sqrt{7}-3\sqrt{3})}{2\sqrt{7}-3\sqrt{3}}$

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

$=\frac{2\sqrt{7}-3\sqrt{3}}{4\times 7-9\times 3}$$=\frac{2\sqrt{7}-3\sqrt{3}}{1}$$= \frac{2\sqrt{7}-3\sqrt{3}}{28-27} = 2\sqrt{7}-3\sqrt{3}$

## Rationalize the denominator of   $\frac{30}{5\sqrt{3}-3\sqrt{5}}$

$\frac{30}{5 \sqrt{3}-3\sqrt{5}} \times \frac{5\sqrt{3 }+3\sqrt{5}}{5\sqrt{3 }+3\sqrt{5}}$ $=\frac{30(5(\sqrt{3 }+3\sqrt{5})}{(5\sqrt{3 })^{2}-(3\sqrt{5})^{2}}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{30}$

$= 5\sqrt{3}+3\sqrt{5$

Alternative Method

$\frac{30}{5\sqrt{3}-3\sqrt{5}} = \frac{30}{(5\sqrt{3}-3\sqrt{5})}\times$ $\frac{(5\sqrt{3}+3\sqrt{5})}{(5\sqrt{3}+3\sqrt{5})}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{(5\sqrt{}3)^{2}-(3\sqrt{5})^{2}}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{75-45} =\frac{30(5\sqrt{3}+3\sqrt{5})}{30}$

$= 5\sqrt{3 }+3\sqrt{5}$

## Give two rational numbers whose  : (1)  Difference is a rational number (2)  Sum is a rational number (3)  Product is a rational number (4)  Division is a rational  number               Justify also.

Any example and verification of example

$Let\: \: \: m = \frac{4}{5},\: \: \: n = \frac{9}{2}$

$difference\: \: =\frac{9}{2}-\frac{4}{5}=\frac{37}{10}\: \: \: (Rational number )$

$sum\: \: =\frac{4}{5}+\frac{9}{2}=\frac{53}{10}\: \: \: (Rational number )$

product          =       4/5 x 9/2 = 36/10 (Rational number)

Division         =       9/2  $\div$ 4/5 = 45/8  ( Rational number)

## Rationalize the denominator of  $\frac{\sqrt{3}+\sqrt{2}}{5+\sqrt{2}}$

$\frac{\sqrt{3} +\sqrt{2}}{5+\sqrt{2} }= \frac{\sqrt{3} + \sqrt{2}}{5+ \sqrt{2}} \times \frac{5-\sqrt{2}}{5 -\sqrt{2}}$

$= \frac{5\sqrt{3} +5\sqrt{2} - \sqrt{6}-2}{25-2}$

$=\frac{5\sqrt{3}+5\sqrt{2} -\sqrt{6}-2}{23}$

## Given two rational numbers whose (1) difference is a rational number (2) sum is a rational number (3) product is a rational number (4) division is a rational number

Any example and verification of example

let m = 4/5, n= 9/2

$\: difference\: = \: \frac{9}{2}-\frac{4}{5}=\frac{37}{10}\: \: (\: rational \: \: number\: )$

$\: sum\: = \: \frac{4}{5}+\frac{9}{2}=\frac{53}{10}\: \: (\: rational \: \: number\: )$

$\: product\: = \: \frac{4}{5}\times \frac{9}{2}=\frac{36}{10}\: \: (\: rational \: \: number\: )$

$\: division\: = \: \frac{9}{2}\div \frac{4}{5}=\frac{45}{8}\: \: (\: rational \: \: number\: )$

## Find any two irrational numbers between 0.1 and 0.12.

Required to irrational number are :

i)  0.10100100010000......

ii) 0.1020020002000.......

## Find the values of a and b when   $\frac{5+\sqrt{6}}{5-\sqrt{6}} = a+b \sqrt{6}$

$\frac{5+\sqrt{6}}{5-\sqrt{6}} =\frac{5+\sqrt{6}}{5-\sqrt{6}} \times \frac{5+\sqrt{6}}{5+\sqrt{6}}$

$= \frac{(5+\sqrt{6})^{2}}{(5)^{2}-(\sqrt{6}^{2})}$

$= \frac{25+6+10\sqrt{6}}{25-6}$

$= \frac{31+10\sqrt{6}}{19}$

$\therefore a+b \sqrt{6}=\frac{31}{19} +\frac{10}{19}\sqrt{6}$

Compairing the rational and irrational parts  of both sides we get

$a =\frac{31}{19} \: \: b\: ,=\: \frac{10}{19}$

## Rationalize the denominator of $\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{4}}$

$\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{4}}\times \frac{(\sqrt{2}+\sqrt{3})+\sqrt{4}}{(\sqrt{2}+\sqrt{3})+\sqrt{4}}$  $=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{(\sqrt{2}+\sqrt{3})^{2}-{4}} = \frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{(2+3+2\sqrt{6})-4}$

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{1+2\sqrt{6}}\times \frac{1-2\sqrt{6}}{1-2\sqrt{6}}$

$=\frac{\sqrt{2}+ \sqrt{3}+\sqrt{4}-2\sqrt{12}-2\sqrt{18}-4\sqrt{6}}{1^{2}-(2\sqrt{6})^{2}}$

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-4\sqrt{3}-6\sqrt{2}-4\sqrt{6}}{1-24}$

$=\frac{-5\sqrt{2}-3\sqrt{3}+2-4\sqrt{6}}{-23}$

$=\frac{+5\sqrt{2}+3\sqrt{3}+4\sqrt{6}-2}{23}$

## if  $x =3-2\sqrt{2}$  then find the value of  $x^{4}-\frac{1}{x^{4}}$

Given $x\: = 3-2\sqrt{2}$

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

$\frac{1}{x}= \frac{(3+2\sqrt{2})}{9-8}= 3+2\sqrt{2}$

$\frac{1}{x^{2}}= (3+2\sqrt{2})^{2}=9+8+12\sqrt{2}=17+12\sqrt{2}$

$x^{2}= (3-2\sqrt{2})^{2}=9+8-12\sqrt{2}=17-12\sqrt{2}$

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

$=\left [ (17-12\sqrt{2})-(17+12\sqrt{2}) \right ] \left [ (17-12\sqrt{2}) +(17+12\sqrt{2})\right ]$

$= 17-12\sqrt{2}-17-12\sqrt{2}\left ( 17-12\sqrt{2}+17+12\sqrt{2} \right )$

$=(-24\sqrt{2})\times 34=-816\sqrt{2}$

## (i)Find the six rational  numbers bbetween 3 and 4  (ii) Which mathematical concept is used in this problem (iii) Which value is depticted in this question

(i) We known that between two rational numbers  x and y such that x< y there is a rational number $\frac{x+y}{2}$.

ie,       $3< \frac{7}{2}< 4$

Now a rational number between 3 and   $\frac{7}{2}< 4$   is :

$\frac{1}{2}\left ( 3+\frac{7}{2} \right )=\frac{1}{2}\times \left ( \frac{6+7}{2} \right )=\frac{13}{4}$

A rational nummber berween  $\frac{7}{2}\: \: and \: \: 4 \: \: is$

$\frac{1}{2}\left ( \frac{7}{2}+4 \right )=\frac{1}{2}\times \left ( \frac{7+8}{2} \right )=\frac{15}{4}$

$3< \frac{13}{4}< \frac{7}{2}< \frac{15}{4}< 4$

Further a rational number between 3 and  $\frac{13}{4}\: \: is :$

$\frac{1}{2}\left ( 3+\frac{13}{4} \right )=\frac{1}{2}\left ( \frac{12+13}{4} \right )=\frac{25}{8}$

A rational number between  $\frac{15}{4}\: \: and\: \: 4\: \: is :$

$\frac{1}{2}\left ( \frac{15}{4}+4 \right )=\frac{1}{2}\times \frac{15+16}{4}=\frac{31}{8}$

A rational number between  $\frac{31}{8}\: \: and\: \: 4\: \: is$

$\frac{1}{2}\left ( \frac{31}{8}+4 \right ) = \frac{1}{2}\times \left ( \frac{31+32}{8} \right )=\frac{63}{16}$

$\therefore 3< \frac{25}{8}< \frac{13}{4}< \frac{7}{2}< \frac{15}{4}< \frac{31}{8}< \frac{63}{16}< 4$

Hence , six rational numbers between 3 and 4 are

$\frac{25}{8}, \frac{13}{4},\frac{7}{2}, \frac{15}{4}, \frac{31}{8}\: \: and\: \: \frac{63}{16}$

(ii) Number syatem

(iii) Rationality is always welcomed

## Force applied on a body is directly proportional to the acceleration produced in the body. Write an equation  to express the situation and plot the graph of the equation taking the constant to be 5 units

Let F be the force applied and a be the acceleration produced.

$F\propto a$

$\Rightarrow F= ka,$where k is a constant

Replace  a by x and F by y.

y = k(x)

$\Rightarrow$ y =  xk

Here, k = 5

$\Rightarrow$ y = 5x

 X 0 1 -1 Y 0 5 -5

## The ratio of girls and boys in a class is  1 : 3. set up an equation between the student of a class  and boys  and then draw its graph. Also find the number of boys in a class of 40 students from the grpah.

Total students in the class  = y

Let the boys in the class = x, then equation between the students and the boys

$y =\frac{4}{3}x$

Now

 X 0 30 60 Y 0 40 80

Now, draw a graph between these points.

## Find the smallest natural number by which 1,200 should be multiplied so that the square root of the product is a rational number.

$1,200= 4\times 3\times (2\times 5)^{2}$

$=2^{4}\times 3\times 5^{2}$

The smallest natural number is 3.

## What is the condition  for the decimal expansion of a rational number to terminate? Explain with the help of an example.

The decimal expansion of a rational  number terminates, if the denominator of rational no $\frac{p}{q}$, When p and q  are co-primes and q can be expressed as 2m 5n . where m and n are non- negative integers.

eg-

$\frac{3}{10}=\frac{3}{2^{1}\times 5^{1}}=0.3$

## Find the smallest positive rational number by which  $\frac{1}{7}$ should be multiplied so that its decimal   expansion terminates after 2 places of decimal

Since  $\frac{1}{7}\times \frac{7}{100} = \frac{1}{100} = 0.01$

Thus smallest rational number is $\frac{7}{100}$

## What type of decimal expansion does a rational  number has? How can you distinguish it from decimal expansion of irrational number?

A rational number is either terminating or non- terminating repeating.

An irrational number is non-terminating and non- repeating.

## Show that $5\sqrt{6}$ is an irrational number.

Let  $5\sqrt{6}$  be a rational number, which can be put in the form $\frac{a}{b}$ , where   $b \neq 0$ , a and b   are co-prime

$5\sqrt{6}=\frac{a}{b}$

$\sqrt{6}=\frac{a}{5b}$

$\Rightarrow \sqrt{6}$= rational

But, we know that $\sqrt{6}$ is an irrational number.

Thus, our assumption is wrong.

Hence, 5$\sqrt{6}$ is an irrational number.

## Write the denominator of the rational number $\frac{257}{500}$  in the form 2m x 5n , where  m and n are non- negative integers. Hence write its decimal expansion without actual division.

Denominator  = 500

= 22 x 53

Decimal expansion,  $\frac{257}{500}=\frac{257\times 2}{2\times 2^{2}\times 5^{3}} = \frac{514}{10^{3}}$

$= 0.514$

## Express the number $0.3\overline{178}$ in the form of rational number $\frac{a}{b}$.

Let   $x =0.3 \overline{178}$

$\Rightarrow x = .3178178178....................$

$\Rightarrow 10,000x = 3178.178178....................$

$\Rightarrow 10x = 3.178178.........$

Substracting $9990x = 3175$

$\Rightarrow x = \frac{3175}{9990} = \frac{635}{1998}$

## Prove that $\sqrt{2}$ is an irrational number

Let $\sqrt{2}$ be a rational number

$\therefore \sqrt{2}=\frac{p}{q}$

Where p and q are co-prime integers and  $q \neq 0$

$\Rightarrow 2 =\frac{p^{2}}{q^{2}}$

$\Rightarrow p^{2} = 2q^{2}$

$\Rightarrow P^{2}$ is divisible by 2.

$\therefore$  P is divisible by 2.

Let p = 2r for some positive integer r

$\Rightarrow p^{2} = 4r^{2}$

$\therefore 2q^{2} = 4r^{2}$

$\Rightarrow q^{2} = 2r^{2}$

$\Rightarrow q^{2}$ is divisible by 2.

$\therefore$ q is divisible by 2.

From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes.

Hence, our assumption is false.

So, $\sqrt{2}$ is irrational

## Prove that $\sqrt{3}$  is an irrational number.

Let $\sqrt{3}$  be a rational number

$\sqrt{3}= \frac{a}{b}$

(a and b are integers and co-primes )

On squaring both the sides,

$3 = \frac{a^{2}}{b^{2}}$

$\Rightarrow 3b^{2}= a^{2}$

$\Rightarrow a^{2}$ is divisible by 3

$\therefore a$  is divisible by 3

We can write  a = 3c for some integer

$\Rightarrow a^{2}=9c^{2}$

$\Rightarrow 3b^{2}=9c^{2}$

$\Rightarrow b^{2}=3c^{2}$

$b^{2}$ is divisible by 3

b is divisible by 3.

From equations (i) ad (ii),  we get 3 as a factor of "a"  and  "b"

Which is contradicting the fact that a and b are co-primes. Hence our assumption that  $\sqrt{3}$ is a rational number, is false. So  $\sqrt{3}$ is an irrational number.

## if p is a prime number, then prove that $\sqrt{p}$ is irrational.

Let p be a prime number and if possible, let $\sqrt{p}$ be rational.

Let  $\sqrt{p}= \frac{m}{n}$  where m and n are  integers having no

common factor other than 1 and $n\neq 0$

Then,  $\sqrt{p}= \frac{m}{n}$

Squaring on both sides, we get

$\frac{(\sqrt{p})^{2}}{1}= \left ( \frac{m}{n} \right )^{2}$

$\Rightarrow \frac{p}{1} = \frac{m^{2}}{n^{2}}$

$\Rightarrow pn^{2} = m^{2}$

$\therefore$ P divides  m2 and p  divides m . [ $\because$ p divides pn2 ]

[$\because$ P is prime and p divides m2 $\Rightarrow$ p divides m]

Let              m = pq  for some integer q

on putting  m = pq  [in eq.(i) we get ]

pn2 = p2q2

n2 = pq2

p divides n2

and p divides n.

[ $\because$ p is prime and p divides n$\Rightarrow$ p divides n ]

Thus p is a common factor of m and n but this contraficts  the fact that m and n have no common factor other than 1.

The contradiction arises by assuming that $\sqrt{p}$ is a rational .

Hence, $\sqrt{p}$ is irrational.

## Prove that   $3+\sqrt{5}$ is an irrational number.

Let 3+$\sqrt{5}$ is  a rational number

$\therefore 3+\sqrt{5} = \frac{p}{q}, q\neq 0$

$3+\sqrt{5} = \frac{p}{q}$

$\Rightarrow \sqrt{5} = \frac{p}{q} -3$

$\Rightarrow \sqrt{5} = \frac{p-3q}{q}$

$\sqrt{5}$ is irrational  and   $\frac{p-3q}{q}$  is a rational

But the rational number cannot be equal to an irrational number.

$\therefore 3+\sqrt{5}$ is an irrational number.

## Prove that  $\sqrt{3}$  is an irrational number. Hence show that 7+2$\sqrt{3}$  is also an irrational number

If possible let $\sqrt{3}$  be a rational number.

(i)   $\therefore \frac{a}{b} = \sqrt{3}$, where a  and b are integers and co -primes

squaring both sides, we have

$\frac{ a^{2}}{b^{2}} = 3$

$\Rightarrow a^{2} = 3b^{2}$

$\Rightarrow a^{2}$ is divisible by 3

$\therefore$ a is divisible by 3    .............................(1)

We can write a = 3c for some c (integer)

(3c)2 = 3b2

$\Rightarrow 9c^{2} = 3b^{2}$

$\Rightarrow b^{2} = 3c^{2}$

$\Rightarrow b^{2}$ is divisible by  3.

$\Rightarrow b$ is divisible by 3   ...........................(2)

From eq (i) and (ii) we have,

3 is a factor a and b which is contradicting the fact that 'a' and 'b' are co-prime.

Thus our assumption that  $\sqrt{3}$  is rational numbers is wrong.

Hence, $\sqrt{3}$ is an irrational number.

(ii) Let us assume to the contrary that $7+2\sqrt{3}$  is a rational number.

$7+2\sqrt{3}=\frac{p}{q},\: q\: \neq 0\: and \: p\: ,\: q\epsilon \: Integer$

$7+2\sqrt{3}=\frac{p}{q}$

$2\sqrt{3}=\frac{p}{q}-7$

$2\sqrt{3}=\frac{p-7q}{q}$

$\sqrt{3}=\frac{p-7q}{2q}$

p- 7q and  2q both are integers hence $\sqrt{3}$  is a rational number.

But this contradicts the fact that is $\sqrt{3}$ is  irrational number. Hence is $7+2\sqrt{3}$ is an irrational number .

## If the ratio of the sum of first n terms of two AP is (7n+1) : (4n+27), find the ratio of their mth terms.

Let  a, A be the first terms and d,D be the common difference of two AP's

Then according to question

$\frac{S_{n}}{S^{'}_{n}}= \frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2A+(n-1)D]}=\frac{7n+1}{4n + 27}$

$\Rightarrow \frac{2a+(n-1)d}{2A+ (n-1)D} = \frac{7n+1}{4n+27}$

$\Rightarrow \frac{a+\left ( \frac{n-1}{2} \right )d}{A+\left ( \frac{n-1}{2} \right )D}= \frac{7n+1}{4n+27}$

Putting   $\frac{n-1}{2}= m-1$

$\Rightarrow n = 2m -1$

$\frac{a+(m-1)d}{A+(m-1)D} = \frac{7(2m-1)+1}{4(2m-1)+27}$

$\frac{a_{m}}{A_{m}} = \frac{14m-6}{8m+23}$

## Find the ratio in which the point P $\left ( \frac{3}{4}, \frac{5}{12} \right )$ divides the line segment joining the point  A  $\left ( \frac{1}{2}, \frac{3}{2} \right )$ and  B (2, -5)

Let  P divide AB in the ratio k :1

$\left ( \frac{1}{2},\frac{3}{2} \right )$

$\frac{ 2k +1/2}{k+1} = \frac{3}{4} \rightarrow 8k +2 = 3k +3$

$\rightarrow k = \frac{1}{5}$

Required ratio : 1: 5

## Find the ratio in which the point (-3, p) the line segment joining the points (-5, -4) and (-2, 3). Hence find the value of P.

Let x (-3, p) divides the join of A (-5,-4) and B (-2,3) in the ratio k :1

The coordinates of P  are  $\left [ \frac{-2k-5}{k+1}, \frac{3k-4}{k+1} \right ]$

But co-ordinates of P are (-3,p)

$\frac{-2k-5}{k+1} = -3, \Rightarrow k = 2$

$\frac{3k-4}{k+1}= p$

Substituting k = 2  gives

$p = \frac{2}{3}$

Ratio of division is 2 : 1 and

$p = \frac{2}{3}$

## Find the ratio in which the line 2x+3y- 5 = 0 divides the line segment joining the points (8,-9) and (2,1). Also find the co-ordinates of the point of division.

P (x,y) divides AB in the ratio  $m_{1}:m_{2}$

$A (8,-9): B (2,1)$

$x = \frac{2m_{1}+8m_{2}}{m_{1}+m_{2}}$

$y = \frac{ m_{1}-9m_{2}}{m_{1}+m_{2}}$

$2\left ( \frac{ 2m_{1}-8m_{2}}{m_{1}+m_{2}} \right )+3\left ( \frac{m_{1}-9m_{2}}{m_{1}+m_{2}} \right )-5 = 0$

$\therefore 2m_{1}-16m_{2}=0$

$\therefore m_{1}:m_{2}=8:1$

$x =\left ( \frac{2\times 8+8\times 1}{8+1} \right )= \frac{8}{3}$

$y = \left ( \frac{8\times 1-9\times 1}{8+1} \right )$

$= -\frac{1}{9}$

$\therefore P (x,y)= \left ( \frac{8}{3}, \frac{1}{9} \right )$

## In a trapezium ABCD, diagonals AC and BD  intersect at O. If AB  = 3CD, then find the ratio of areas of triangles COD and AOB

$\triangle AOB \sim \triangle COD$

$\frac{ar(\triangle COD)}{ar(\triangle AOB)} = \frac{CD^{2}}{AB^{2}}$

$= \frac{CD^{2}}{(3CD)^{2}} = \frac{CD^{2}}{9CD^{2}} = \frac{1}{9}$

ratio  = 1 : 9

## The curved surface area of a cylinder is  264 m2 and its volume 924m3. Find the ratio of its height to its  diameter.

Curved surface of cylinder  = $2\pi r$h

Volume of cylinder = $\pi r^{2}h$

$= \frac{\pi r^{2}h}{2\pi rh} =\frac{924}{264} \Rightarrow \frac{r}{2} = \frac{7}{2}$

$= r = 7 m$

$2 \pi rh = 264$

$\Rightarrow 2\times \frac{22}{7} \times 7\times h = 264$

$\Rightarrow h = 6 m$

$\therefore \frac{h}{2r} = \frac{6}{14}=\frac{3}{7}$

## The radii of two cylinders are in the ratio 2:3  and their height are in the ratio  5 : 3, find the ratio of their volume.

$\frac{Volume of I ^{st}cylinder}{Volume of II^{nd} cylinder } = \frac{\pi r_{1}^{2}h_{1}}{\pi r_{2}^{2}h_{2}}$

$= \left ( \frac{r_{1}}{r_{2}} \right )^{2}\times \frac{h_{1}}{h_{2}}$

$=\left ( \frac{2}{3} \right )^{2} \times \frac{5}{3}$

$= \frac{4}{9}\times \frac{5}{3} = \frac{20}{27}$

$= 20 :27$

## A metallic cylinder has radius 3 cm and height  5 cm. To reduce its weights, a conical hole is drilled in the cylinder. The conical hole  has a radius of  $\frac{3}{2}$  cm and its depth  $\frac{8}{9}$  cm calculate the ratio of the volume of metal taken out in a conical shape.

Volume of the cylinder = $\pi r^{2}h = \pi (3)^{2}5$

$= 45 \pi cm^{^{3}}$

Volume of conical hole  = $= \frac{1}{3} \pi r^{2}h = \frac{1}{3} \pi \left ( \frac{3}{2} \right )^{2} \times \frac{8}{9}$

$=\frac{2}{3} \pi cm^{3}$

Metal left in cylinder  =  $= 45 \pi -\frac{2}{3 } \pi = \frac{133 \pi }{3}cm^{3}$

Required ratio  =

$\frac{Volume of metal left }{Volume of taken out }$

$= \frac{\frac{133}{3} \pi }{\frac{2}{3} \pi } = 133: 2$

## Two supplementary angles are in the ratio 2:3 find the angles.

Let the two supplementary angle are 2x and 3x, then

$2x+ 3x = 180 ^{0}$

$\Rightarrow x= \frac{180^{0}}{9}=36^{0}$
Hence, the angles are 2x and 3x or 720 and 1080

## The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Let the meassure of the angles be 3x, 5x, 9x, 13x then

$3x\, +\, 5x\, +\, 9x\, +\, 13x =\, 360^{0}$

$30x \, = 360^{0}$                   [Angles of sum property of quadrilaterals]

$x = 12^{0}$

Angles are : 360, 600, 1080, 1560

## If angles of a quadrilateral are in ratio 1:2:3:4 Find the measure of all the angles of a quadrilateral.

Let the measure of the angles be x, 2x, 3x and 4x then,

x + 2x + 3x + 4x = 3600

x = 360

$\therefore$  Angles of a quadrilateral are 360, 720, 1080, 1440

## PQRS is a square. N and M are the midpoints of siders SR and QR respectively. O is a point on diagonal PR such that OP = OR. Show that ONRM is a  square. Also, find the ratio of ar (ORM ) and ar (PQRS).

Since OP = OR

$\therefore$  O is the midpoint of PR.

$In\,\, \Delta \, SRP$

O and N  are midpoints of siders PR and SR respectively

$\therefore$ by midpoint theorem,

$ON = \frac{1}{2}\, \, SP\, \, and\, \, ON\parallel PQ\, \, \, .................\, (1)$

$Similarly\, \, \, OM \left \| PQ\, \, \, \, ...........(2)$

Using (1) and (2)  we get

$ONRM\, \, is \, \, a\, \left \| ^{gm}$

$Now, \, \, ON = \frac{1}{2}\, \, SP$

$= \frac{1}{2} SR$

= NR

$In\, \, \left \| ^{gm} \, \,ONRM\, \,a\, \,pair\, \, of\, \, adjacent\, \, sides\, \, ON\, \, and\, \, NR\, \, are\, \, equal\, \, and\, \, \angle\, \, S = \angle N = 90^{0}$    (Corresponding angle as ON II PS)

$ONRM \, \, is \, \, a \, \, square$

Since OR is diagonal of a square

$ar(ORM) = \frac{1}{2}\, \, ar(ONRM ) ...............(3)$

$\because$  diagonal of an IIgm divides it into two congruent triangles)

$ar(ONRM ) NR \times RM$

$= \frac{1}{2}SR\times \frac{1}{2}RQ$

$= \frac{1}{4}(SR\times RQ)$

$= \frac{1}{4}(SR )^{2}$

$= \frac{1}{4}\, ar(PQRS ) .............. (5)$

Using (3) and (4) we get

$\frac{ ar(ORM)}{ar(PQRS)} = \frac{\frac{1}{2}ar(ONRM)}{4ar(PQRS)} = \frac{1}{8}$

$\therefore$ The ratio of  $ar (\Delta ORM) : ar (PQRS) = 1:8$

## Construct a triangle whose angles are in the ratio 1:3:5 and length of sides included by first and last angles is 6 cm.

Let angles x, 3x, and 5x

$\therefore x+3x+5x = 180^{0}$           (angles of sum property)

$9x = 180^{0}$

$x = 20^{0}$

$\therefore$ First angle  =  $20^{0}$

Second angle  =  600

Third angle    = 1000

Steps of construction:

(i)  Draw a line segment BC  = 6 cm

(ii)  At point A draw $\angle XAB = 20^{0}$  and at point B, draw $\angle YAB = 100^{0}$

(iii) XA and YB intersect at point C.

Thus, ABC is the required triangle.

## The sides of a triangle plot are in the ratio 4:5:6 and its perimeter is 150 cm, then find its sides.

Perimeter      =  150         .............................. (1)

Perimeter = 4x+5x+6X ............. (2)

$15x = 150$

$x = 10$

Sides are 40 cm, 50 cm and 60 cm.

## The volumes of two spheres are in the ratio of 64:27. Find their radii if the sum of their radii is 21 cm.

Let the radii of two spheres be $r_{1},r_{2}$

$\therefore r_{1}+r_{2}= 21$

$\Rightarrow r_{1}=21-r_{2}$

$Volume\, \, of\, \, 1\, \, sphere = \frac{4}{3}\, \pi\, r^{3}_{1}\, \, cm^{3}$

$Volume\, \, of\, \, II\, \, spehere = \frac{4}{3}\, \pi\, r^{3}_{2} \, \, cm^{3}$

$\frac{Volume \: of I \: spere }{Volume \: of \: II \: Sphere }\, =\, \frac{\frac{4}{3}\pi (21-r_{2} )^{3}}{\frac{4}{3}\pi r^{3}_{2}}$

$=\frac{64}{27}$

$\frac{(21-r_{2})^{3}}{r^{3}_{2}} = \frac{64}{27}$

$\frac{21-r_{2}}{r_{2}} = \frac{ 4}{3}$

$63-3 r_{2} = 4\, r_{2}$

$63=7\, r_{2}$

$\therefore r_{2} =9\, cm$

$\therefore r_{1} =12 cm$

## The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of surface area of the balloon in two cases.

$Original \, \, surface \, \, area\, \, of \, \, S_{1}\, = 4\, \pi \, r_{1}^{2}$

$= 4\pi \times 7\times 7 cm^{2}$

$New \, \, surface\, \, area\, \, S_{2}\, = 4\, \pi \, r_{1}^{2}$

$= 4\pi \times 14\times 14cm^{2}$

$S_{1}: S_{2} = \frac{4\pi \times 7\times 7}{4\pi\times 14\times 14}$

$= \frac{1}{4}$

$= 1:4.$

## Two cylinders have bases of same size. The diameter of each is 7 cm. If one of the cylinder is 10 cm high and the other is 20 cm high, then the ratio of their volumes is.....................

Let r denotes the radius of both cylinders l and h be their heights respectively.

$Ratio\, \, of \, \, their\, \, volumes \, =\, \frac{\pi r^{2}h}{\pi r^{2}h'} \, = \, \frac{h}{h'} \, =\, \frac{10}{20}$

= 1:2

## The radii of two right circular cylinder are in the ratio 2:3 and their heights are in the ratio 5:4, then the ratio of their volumes will be ..................

Let radii of the cylinders be 2x be 3x and heights be 5y and 3y respectively

$Ratio\, \, of\, \, volumes = \frac{\pi (2x)^{2}\times 5y}{\pi (3x)^{2}\times 3y}$

$= \frac{4x^{2}\times 5}{9x^{2}\times 3}$

= 20:27

## Mean of  15 observation is  23. If each observation is multiplied by 2, find the new mean.

If each observation under consideration is multiplied by 2, then new mean in obtained by multiplying same quantity (i.e., 2) in old mean

$\therefore$ New mean =  2 x old mean

=  2 x 23

=  46.

## Two supplementary angles are in ratio 2:7 Find the measure of angles.

$2x+7x= 180^{0} \Rightarrow x = 20^{0}$

So the angle are

$2x = 2\times 20^{0}$

= $40^{0}$

$7x = 7\times 20^{0}$

$= 140^{0}$

So two angles are $= 40^{0} \: \: and \: \: \: 140^{0}$

## What do we call a triangle if the angles are in the ratio 5:3:7 ?

Let the angles of the triangles are 5x , 3x and 7x, then

5x+3x+7x = 1800

$15x = 180^{0}$

Thus,         x = 120

$\therefore$ Angles are  $60^{0}, 36^{0}, 84^{0}$

$\therefore$ Each angles less than $90^{0}$

$\therefore$ The triangle is an acute angled triangle.

## The angles of a quadrilateral are in the ratio 2:3:6:7 the largest angle of the quadrilateral is ..................

Let the angles of the quadrilateral be $2x^{0}, 3x^{0}, 6x^{0},7x^{0}$

$\therefore 2x^{0}+3x^{0}+ 6x^{0}+7x^{0} = 360^{0}$   [Angles sum property of quadrilateral]

$\Rightarrow \: \: \: 180x = 360^{0}$

$\Rightarrow\: \: \: x = 20^{0}$

$\therefore$  Largest angle  = $7x^{0} = 140^{0}$

## The angles A,B,C, and D of a quadrilateral ABCD are in the ratio 2 : 4: 5 : 7. Find the measures of these angles. what type of quadrilateral is it.  GIve reasons?

Let the measures of the angles be 2x , 4x , 5x and 7x

$2x+4x+5x+7x= 360^{0}$  (Angles of sum of property )

$18x = 360^{0}$

$x = 20^{0}$

$\angle A = 40^{0}$

$\angle B = 80^{0}$

$\angle C = 100^{0}$

$\angle D = 140^{0}$

As $\angle A+\angle D = 180^{0} \: \: and \: \: \angle B+\angle C = 180^{0}$

$CD \left \| AB$

$\Rightarrow\: \: \: \: \: ABCD$  is a trapezium

## Two Consecutive angles of a parallelogram are in the ratio 1: 3, then what will be the smaller angles ?

Let the consecutive angles be x0 and  (3x)0

$\therefore \: \: x^{0}+3x^{0} = 180^{0}$

$4x^{0} = 180^{0}$

$x^{0} = 45^{0}$

Smaller angle $= x^{0}$

$= 45^{0}$

## If a triangle and a parallelogram are on same base and between same parallels, then the ratio of the area of the triangle to the area of a parallelogram is ................

Ratio of area of the triangle to the area of parallelogram is 1:2.

## If ratio of the corresponding sides of two similar triangles is  5: 6 , then find the ratio of their areas.

Let the  triangles be $\Delta ABC$   and  $\Delta DEF$

$\frac{ar (\Delta ABC)}{ar(\Delta DEF) }= \left (\frac{5}{6} \right )^{2} = \frac{25}{36}$

25 : 36

## In given figure in what ratio does P divides AB internally ?

P  divides AB internally in the ratio 4: 4  or  1 : 1

## To divide a line segment AB in the ratio 5 : 7  first AX is drawn, so that $\angle BAX$ is an acute angle and then at equal distance, points are marked on the ray AX, find the minimum number of these points

Minimum number of points marked on

AX= 5+7 = 12

## To divide a line segment  AB in the ratio  2 : 5 a ray AX is drawn such that

Minimum number of points marked will be  = 2+5= 7

## To find a point P on the line segment AB = 6 cm, such than  $\frac{AP}{ AB} = \frac{2}{5}$ , in which ratio the line segment AB is divided.

The line segment AB is divided in the ratio AP : PB = 2 : (5-2) = 2:3

## A segment AB is divided at point P such that $\frac{PB }{AB} = \frac{3}{7}$ then find the ratio  AP: PB

Here AB = 7, PB = 3

AP = AB - PB = 7-3 = 4

AP : PB = 4:3

## What is the ratio of division of the line segment B by the point P from A ?

The ratio of division of the line segment AB by the point P from A is AP :  AB =  3 : 5

## To divides the line segment AB in the ratio 2:3 a ray AX is drawn such that

Minimum number of  marks = 2+3 = 5

## Draw a line segment of length 7 cm. Find a point P on it which divides in the ratio 3 : 5.

Steps of construction :

1. Draw a line segment AB = 7 cm

2. Draw any ray AX making an acute angle with AB .

3. Draw the point $A_{1},A_{2}, A_{3}, A_{4}......................, A_{8}$ on AX such that

$AA_{1}= A_{1}A_{2} = A_{2}A_{3} = ............A_{7} A_{8}$

4.  Join BA8

5. Through the point A3, draw a line parallel to BA8

Then AP :  PB = 3:5

## Draw a line segment of length 5 cm and divide it in the ratio 3: 7

Steps of constructions:

1. Draw  a line segment A B = 5 cm

2. Draw any ray AX making an acute angle with AB

3. Draw the points $A_{1} , A_{2}, A_{3} ,................... A_{10}$ on AX such that

$AA_{1} =A_{1} A_{2} =A_{2} A_{3} ,................... A_{7}, A _{8}$

4. Join BA10

5. Through the point A3 draw a line parallel to BA10

Then AP : PB  = 3 : 7

## Divide a line segment of 7 cm length and divide externally in the ratio 3:5

Given : AB is a line segment of length 7 cm.

To construct: To divide a line segment of 7 cm length externally in the ratio 3: 5

Steps of construction :

1. Draw a line segment AB = 7 cm

2. Draw ray BX making an acute

3. Along BX, Mark off B1,B2,B3,B4 & B5. Join B2 to A.

4. Through B5 draw B5P II B2A intersecting BA at P.

5. The point P so obtained is the required point which divides AB externally in 3: 5.

## A cylinder, a cone and a hemisphere have same base and same height. Find the ratio of their columes.

Volume of cylinder : Volume of cone : Volume of hemisphere

$= \pi r^{2}h : \frac{1}{3} \pi r^{2}h : \frac{2}{3} \pi r^{3}$

$= \pi r^{2}h : \frac{1}{3} \pi r^{2}h : \frac{2}{3} \pi r^{2} \times h$

$= 1 : \frac{1}{}3 : \frac{2}{3}$

$\Rightarrow 3 : 1 : 2$

## What is the ratio of the total surface area of the solid hemisphere to the square of its radius

$\frac{Total \, \,surface \, \, area \, \, of \, \, hemisphere }{Square \, \, of \, \, its \, \, radius} = \frac{3\pi r^{2}}{r^{2}} = \frac{3\pi }{1}$

Total surface area of hemisphere : square of radius  =  $3\pi : 1$

## If the radius of the base of a right circular cylinder is havelled, keeping the height same, find the ratio of the volume of the reduced cylinder to that of orginal cylinder.

$\frac{Volume \, \, of \, reduced \, \, cyliner}{Volume\, \, of\, \, original \, \, cylinder }=\frac{\pi \times \left ( \frac{r}{2} \right )^{2}h}{\pi r^{2}h}$

$= \frac{1}{4} = 1 : 4$

## Volumes of two spheres are in the ratio 64: 27, find the ratio of their surface areas.

$\frac{Volume \, \, of \, \, I^{st} sphere}{Volume \, \, of \, \, II^{nd} sphere} = \frac{\frac{4}{3}\pi r_{1}^{3}}{\frac{4}{3}\pi r^{3}_{2}} = \frac{64}{27}$

$\frac{\pi r^{3}_{1}}{\pi r^{3}_{2}} = \frac{64}{27}$

$= \frac{r_{1}}{r_{2} }= \frac{4}{3}$

Ratio of their surface areas  =  Surface area of  Ist sphere

Surface area of IInd sphere

$\frac{4\pi r^{2}_{1}}{4\pi r^{2}_{2}} = \left ( \frac{r_{1}}{r_{2}} \right )^{2}$

$= \left ( \frac{4}{3} \right )^{2} = \frac{16}{9}$

$= 16 : 9$

## A cylinder and a cone have base radii 5 cm and 3 cm respectively and their respectively and their respective  heights are 4 cm and 8 cm. Find the ratio of their volumes.

Volume of cylinder  = $\pi (5)^{2} \times 4 cm ^{3}$

= 100$\pi cm ^{3}$

Volume of cone  = $\frac{1}{3}\pi \times 3^{2}\times 8$

$=24\pi$

$\therefore$ Required ratio = 100$\pi : 24 \pi$

= 25: 6

c.  4 : 5

## A mixture of milk and water is in the ratio of 3:2. Find the percentage of milk in the mixture.

Total = 3+2 = 5

Percentage of milk = 3/5 x 100

= 60%

3:6 ie., 1:2

3:1

## Lini scored 45 marks out of 50 and joy scored 70 out of 75. What is the ratio between those two girls?

$\frac{45}{50} \times 100 = 90$ % and $\frac{70}{75} \times 100 = 63.33$ %

## The angles of a quadrilateral are in a ratio of 1: 2: 3 : 4. The smallest angle is.................

The angles of a quadrilateral are in a ratio of 1:2:3:4. The smallest angles is  $36^{\circ}$

## Three numbers are in the ratio of 1:2:3 and the sum of their cubes are 4500. find the numbers.

Let the nos. are x, 2x,3x

$x^{3}+(2x)^{3} + (3x)^{3} = 4500$

$x^{3}+2x^{3} + 27x^{3} = 4500$

$36x^{3} = 4500$

$x^{3} =\frac{ 4500}{36} = \frac{500}{4} = 125$

$x = 5$

So, nos. are 5, 10 & 15

## Find the three rational numbers lying between 3 and 4.

LCM of 3 and 4 = 12

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

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

So rational nos. can be

$\frac{37}{12} , \, \, \frac{38}{12} ,\, \, \, \frac{39}{12}$

## The ratio between exterior angle and the interior angle of a regular polygon is 1:5. FInd the number of sides of the polygon.

For a regular polygon

$x+5x = 180^{0}$

$x = 30^{\circ}$

The sum of all exterior angles of the regular polygon = $360^{\circ}$

Hence  $n \times 30^{\circ} = 360^{\circ}$

$n = \frac{360^{\circ}}{30^{\circ}} = 12$

10m : 1 km

10 m : 1000 m

10 : 1000

1:100

## Find the 6 rational numbers between $\frac{-3}{2}$ and   ​$\frac{5}{3}$

$\frac{-3}{2} = \frac{-3 \times 3}{2\times 3 } = \frac{-9}{6}$

$\frac{5}{3} = \frac{5 \times 2}{3\times 2 } = \frac{10}{6}$

The rational numbers are $\frac{-8}{6}\: ,\: \frac{-7}{6}\: ,\: \frac{-6}{6}\: ,\: \frac{-5}{6}\: ,\: \frac{-4}{6}\: ,\: \frac{-3}{6}\: ,\:$$\frac{-2}{6}\: ,\frac{-1}{6}\: ,\: \frac{1}{6}\: ,\: \frac{2}{6}\: ,\: \frac{3}{6}\: ,\: \frac{4}{6}\: ,\: \frac{5}{6}\: ,\: \frac{6}{6}$ , $\frac{7}{5}\: ,\frac{8}{6}\: ,\frac{9}{6}$

## Find 3 rational numbers between 1/4 and 1/2

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

$\frac{1}{2}=\frac{1\times 8}{2\times 8}=\frac{8}{16}$

3 rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$ are $\frac{5}{16}\: ,\frac{6}{16}\: ,\frac{7}{16}$

## The ages of Hari and Harry are in the ratio 5:7  Four years from now the ratio of there will be 3:4. Find their present ages.

Let the ages of Hari & Harry are 5x, 7x

According to the problem,

$\frac{\left ( 5x+4 \right )}{\left ( 7x+4 \right )}=\frac{3}{4}$

$4\left ( 5x+4 \right )=3\left ( 7x+4 \right )$

$20x+16 = 21x+12$

$21x-20x=16-14$

$x=2$

Ages of Hari = $5x$ = $5\times 2$ = $10$

Ages of Harry = $7x$ = $7\times 2$ = $14$

20

1

13

19

24

35

## The coordinates of the vertices of a triangle are A(1,1) B(5,5) C(2,5) a. Write the coordinates of the midpoint D of AB. b. What is the length of CD? c. What are the coordinates of the point dividing the line CD in the ratio 2:1?

Given, the vertices of the,$\triangle ABC$

A (1,1) B (5,5) C(2,5)

a. Mid-point of D of AB

$\left ( \frac{x_{1} + x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$  (Formula)

Points A (1,1) B (5,5)

ie., D  =  $\left ( \frac{1 + 5 }{2}, \frac{1+5}{2} \right ) = (3,3)$

b. Length of CD points C(2,5) D (3,3) [Using distance formula]

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} = \sqrt{(3-2)^{2}+(3-5)^{2}}$

$= \sqrt{1+4} = \sqrt{5}$

c. Given CD = 2:1            (ie., m:n)

$\left ( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n} \right )$   (Formula)

$= \left ( \frac{2\times 3+1\times 2}{2+1}, \frac{2\times 3+1\times 5}{2+1} \right ) = \left ( \frac{6+2}{3}, \frac{6+5}{3} \right ) = \left ( \frac{8}{3}, \frac{11}{3} \right )$

## The length of a side of a square is 3 centimetres. What is the length of its diagonal?

Side of square = 3 cm

In an isosceles Right-triangle. Side are in the ratio, Side opposite to 450:  Side opposite to 450:  Side opposite to 900 = 1 : 1: $\sqrt{2}$

Here side opposite to 450 = 3 cm

Side opposite to 900 = 3$\sqrt{2}$ cm

Length of diagonal = 3$\sqrt{2}$ cm.

## See the pattern of the operations on the right a. Write the next line? b. Write the general form of all these, using algebra.

a. $\frac{1}{4}+\frac{1}{8} = \frac{3}{8}$

b. Generally this pattern can be written as

$n + \frac{1}{2n}=\frac{2n+1}{2n}$

22

1

## A ball is rolled up a slanted plank at 18m/second and the speed decreases every second by  a3m/s. When the speed becomes  0. It will start rolling downwards with speed increasing at the same rate. At what time does the speed become 0? If the speed is too be obtained as a non-negative number at all times, a) What is the algebraic equation connecting speed and time till 6 seconds? b) For the journey after 6 seconds, what is the algebraic equation connecting speed and time? c) If the speed downwards is taken as a negative number, what is the speed-time equation which holds for all times?

a) V =  u +at

1 sec         2sec       3sec        4sec       5 sec

18 m/s     15 m/s    12m/s       9m/s        6m/s

v = 18 + a x 6

v - 6a = 18

V is the find velocity a is the acceleration.

b) V = 3 + at

after 6 seconds

V = 3 + 6a

ie., V - 6a = 3

c) V = u+at

a is negative hence

v = u - at

## Ramu and Venu started a business. Ramu inested 50000 rupees and Venu invested 1, 50000 rupees. If they got 20% profit in a year, then  a. What is the total profit  b. What is the ratio of their investments? c.If the profit is dividing according to the ratio of their investments, how much amount each will get?

Ramu invested = RS. 50000/-;

Venu invested = Rs.1,50000/-.

Profit = 20%

a. Total investment  = 50000 + 1,50000 = 2,00000.

Total profit  = 200000 x 20/100 = 40000

b. Ratio of their investment = 50000 : 150000

= 5 : 15

= 1: 3

c. Ramu will get 40000 x 1/4 =  Rs. 10000/-

Venu will get  40000 x 3/4 =  Rs. 30000/-

## Divide a line of length 7 centimetres in the ratio 2:3.

AE : EB = 2 : 3

## See the pattern of the operations on the right  $1+\frac{1}{2}=\frac{3}{2}$ $\frac{1}{2}+\frac{1}{4} = \frac{3}{4}$ $\frac{1}{3}+\frac{1}{6} = \frac{3}{6}$ (i) Write the next line  (ii) Write the general form of all these using algebra.

(i) Next line : $\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$

(ii) $\frac{1}{n}+\frac{1}{2n}=\frac{2}{2n} + \frac{1}{2n} = \frac{3}{2n}$

## In the triangle ABC, AB = 9 cm, BC  = 12 cm. The line BD is the bisector of  $\angle B$ . What is the ratio of the area of $\triangle ABD$  and $\triangle BCD$ ?

AB = 9 cm, BC = 12 cm

Area of $\triangle ABD$ : area of $\triangle BCD = AB : BC = 9 : 12 = 3: 4$

## A rhombus with one side 3 metres and one angle 600 is given in the picture. Find the sum of the length of its diagonals correct to centimetres.

$\triangle ABD$ is an equilateral triangle.

$\angle A = 60 ^{\circ}$

$\therefore$ BD = 3 m

In $\triangle COD$ the angles are 300, 600 and 900

So its  sides are in the ratio $1 : \sqrt{3} : 2$

CD = 3 m

OD = 1.5 m, OC = 1.5$\sqrt{3}$ m

AC = $3\sqrt{3} m = 3\times 1.732$

=5.196 m = 5.20 m

Sum of the lengths of the diagonals  = 3 + 5.20 = 8.20 m

## In the figure, the centres A, B, C of the semicircles are on a line. The radii of the unshaded semicircles are in the ratio 1:3. The radius of the smallest semicircle is 2 centimetre. i) Find the area of the semicircle with centre at A. ii) Find the area of the semicircle with centre at B.  iii) Find the area of the shaded region.

Radius of the semicircle = 2 cm

Radius of the semicircle with centre C = 3 x 2 = 6 cm

Radius of the semicircle with centre B  = 2 x 6 = 12 cm

i. Area of the semicircle with centre A  $= \frac{1}{2}\, \, \pi \times 2\times2 = 2 \pi \, \, sq. cm$

ii. Area of the semicircle with centre B $= \frac{1}{2}\, \, \pi \times 12\times 12 = 6\times 12 \pi \, \, = 72\pi \, \, sq. cm$

iii. Area of the shaded portion $=72\pi - (2\pi +\frac{1}{2}\, \, \pi \times 6\times 6)$

$=72\pi - (2\pi +18\pi ) = 72\pi - 20\pi$

$=52\pi \, \, sq\, \, \, cm$

## In the given  figure$l\left \| m\left \| n$.From the figure , find the ratio of $(x+y):(y-x)$

$y=180^0 -(30^0 +20^0 )-130^0$

$l\left \| m\Rightarrow$$x+100^0 =180^0 \Rightarrow x=80^0$

$\therefore$$x+y=210^0 , y-x=50^0$

$\therefore$  $(x+y) :(y-x)=21:5$

## Prove that $\sqrt{5}$ is an irrational number and hence show that $2-\sqrt{5}$ is also an irrational number.

Let $\sqrt{5}$ be a rational number

$\therefore \sqrt{5} = \frac{a}{b}$

(a, b are co -prime integers and $b\neq 0$)

$\Rightarrow a = b \sqrt{5}$

$\Rightarrow a ^{2}= 5b^{2}$

$\Rightarrow$ 5 is a factor of $a^{2}$

$\Rightarrow$ 5 is a factor of a

let  a = 5c,

$\Rightarrow a^{2} = 25c^{2}$

$\Rightarrow 5b^{2} = 25c^{2}$

$\Rightarrow b^{2} = 5c^{2}$

$\Rightarrow$ 5 is a factor of $b^{2}$

$\Rightarrow$ 5 is a factor of b

5 is a common factor of a, b

But this contradicts the fact that a, b are co-primes

$\therefore \sqrt{5}$ is irrational

Let $2- \sqrt{5}$ be a rational

$\therefore 2- \sqrt{5} = a$

$\Rightarrow 2- a = \sqrt{5}$

2 - a is rational, so is $\sqrt{5}$

But $\sqrt{5}$ is not rational contradiction

$\therefore 2-\sqrt{5}$ is irrational.

## Prove that $\sqrt{2}$ is an irrational number. Hence show that $\frac{3}{\sqrt{2}}$ is alos an irrational number.

Let $\sqrt{2}$ be a rational number

$\sqrt{2} = \frac{a}{b}$

(a, b are co-prime integers and $b\neq 0$)

$a = \sqrt{2}b$

$a^{2} = 2b^{2}$

Squaring,

$\Rightarrow$ 2 divides a2

$\Rightarrow$ 2 divides a

So we can write a = 2c for some integer c,, substitute for $a, 2b^{2} = 4c^{2} , b^{2} = 2c^{2}$

This means 2 divides b2, so 2 divides b.

a and b have '2' as a common factor

But this contradicts that a, b have no commom factor other than 1.

Our assumption is wrong.

Hence, $\sqrt{2}$ is irrational.

Let $\frac{3}{ \sqrt{2}}$ be rational

$\frac{3}{ \sqrt{2}} = \frac{a}{b}$ where a and b are integers, $b \neq 0$

$\Rightarrow 3b = \sqrt{2a}$

$\Rightarrow \sqrt{2a} = \frac{3b}{a}$

$\frac{3b}{a}$ is rational but $\sqrt{2}$ is not rational

Our assumption is wrong

$\frac{3}{ \sqrt{2}}$ is rational.

## Show that there is no positive integer n, for which $\sqrt{n-1}+\sqrt{n+1}$ is rational.

Let us assume that there is a positive integer n for  which$\sqrt{ n-1} + \sqrt{ n + 1}$ is rational and equal to  $\frac{P}{Q}$,where  P and Q are positive integers ($Q\neq 0$)

$\sqrt{ n-1} + \sqrt{ n + 1} = \frac{P}{Q}$................(i)

$\Rightarrow \frac{P}{Q} = \frac{1}{\sqrt{n-1} + \sqrt{n+1}}$

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

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

$\Rightarrow \sqrt{n +1} - \sqrt{n-1} = \frac{2Q}{P}$................(ii)

Apply (i) + (ii) we get

$2 \sqrt{n+1} = \frac{P}{Q} + \frac{2Q}{P} = \frac{P^{2} + 2Q^{2}}{PQ}$

$\Rightarrow \sqrt{n+1} = \frac{P^{2}+2Q^{2}}{2PQ}$.............(iii)

Apply (i) and (ii) we get

$\sqrt{n-1} = \frac{P^{2} + 2Q^{2} }{2PQ}$.................(iv)

From (iii) and (iv), we can say $\sqrt{n +1}$ and $\sqrt{n -1}$ both are rational because P and Q both are  rational. But it is possible only when (n+1) and (n-1)  both  are perfect squares. But they differ by 2 and two perfect square never differ by 2. So both (n+1) and (n-1) cannnot be perfect squares, hence there is no positive integer n for which $\sqrt{n -1} + \sqrt{n+1}$ is rational.

## In the figure 0 is the centre and AB is a chord of the circle. if OA = 3 cm and $\angle AOB = 120^{\circ}$ then find the length of AB.

Draw OD from O, perpendicular to AB

$\angle ADO= 90^{\circ}$

$\angle A OD =6 0^{\circ}$$\therefore \angle OAD =30^{\circ}$

Angles of $\angle AOD =30^{\circ}, 60^{\circ}$ and $90^{\circ}$.

So sides opposite these angles are in the ratio $1 : \sqrt{3}:2$

Since AO = 3, OD = $\frac{3}{2} = 1.5,AD = \frac{3\sqrt{3}}{2}$

$AB = \frac{3\sqrt{3}}{2} \times 2 = 3\sqrt{3}$ cm

## In triangle ABC, $\angle A = 75^{\circ}, \angle C = 60^{\circ}$ a. What is the measure of $\angle B$ b. If AB = $5 \sqrt{2}$ what is the length of AC ? c. Find the ratio AB : BC :  AC.

a.  $\angle B = 180 - (75 + 60 ) = 45 ^{\circ}$

b. AD is drawn perpendicular to BC

In $\triangle ABD, \angle B = 45^{\circ}$

$\angle D = 90^{\circ},\therefore \angle BAD = 45^{\circ}$

Since angles of $\triangle ABD$ are $45^{\circ}, 45^{\circ}$ and $90^{\circ}$ its sides are in the ratio $1:1:\sqrt{2}$

$AB = 5\sqrt{2}, BD = 5, AD = 5 cm$

In $\triangle ADC, \angle C = 60^{\circ}, \angle D = 90^{\circ} \therefore \angle CAD = 90^{\circ}$

Angle of $\triangle ACD$ are $30^{\circ} ,60^{\circ}$ and $90^{\circ}$. So its sides are in the ratio $1 : \sqrt{3} : 2$

Since AD = 5, DC = $\frac{5}{\sqrt{3}}\, \, cm$

$AC = 2 \times \frac{5}{\sqrt{3}} = \frac{10}{\sqrt{3} } cm$

$BC = BD + DC = 5 +\frac{5}{\sqrt{3}} = 5 \left ( 1+ \frac{1}{\sqrt{3}} \right )$

AB : BC : AC = $5 \sqrt{2} : 5 \left ( 1+\frac{1}{\sqrt{3}} \right ) : \frac{10}{\sqrt{3}}$

$= \sqrt{2} : \left ( 1+\frac{1}{\sqrt{3}} \right ) : \frac{2}{\sqrt{3} } = \sqrt{6} : \left ( \sqrt{3} + 1 \right ) : 2$

## The coordinates of the vertices of a triangle are A(1,1), B(5,5), C(2,5). a. Write the coordinates of the midpoint D of AB. b. What is the length of CD ? c. What are the coordiaates of the point dividing the line CD in the ratio 2:1 ?

a. A(1,1), b(5,5), c(2,5)

Coordinates of D=$\left ( \frac{1+5}{2},\frac{1+5}{2} \right )=(3,3)$

b. Cd=$\sqrt{(2-3)^2 +(5-3)^2 }=\sqrt{1+4}=\sqrt{5}$

c. x coordinates  of the point dividing the line joining c(2,5) and D(3,3) in the ratio 2:1

$x_{1}+\frac{p}{p+q}(x_{2}-x_{1})=2+\frac{2}{3}(3-2)$

$=2+\frac{2}{3}\times 1=2+\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}$

y coordinates =$y_{1}+\frac{p}{p+q}(y_{2}-y_{1})$

$=5+\frac{2}{3}(3-5)=5+\frac{2}{3}\times -2$

$=5-\frac{4}{3}=\frac{15}{3}-\frac{4}{3}=\frac{11}{3}$

Coordinates of this point =$\left ( \frac{8}{3} ,\frac{11}{3}\right )$

## In the figure the radius of the circle centred at 0 is 6 units. Line Ab touches the circle at P and $\angle OAB=30^0$. a. Find the coordinates of the point A and P. b. Find teh equation ao AB.

a. In $\Delta OPA, \angle A=30^0 ,\angle P=90^0$ and $\angle O=60^0 .$

Since the angles are $30^0 ,60^ 0$ and $90^0 ,$ sides are in the  ratio $1:\sqrt{3}:2$

Since OP =6

PA =$6\sqrt{3}, OA=12$

Since OA =12, coordinates of A =(12,0)

Draw PQ perpendicular to OA.

Angles of $\Delta OPQ$ are $30^0 ,60^0$ and $90^0$.

Since OP=6,  OQ=3,  PQ=$3\sqrt{3}$

Coordinates of P=$(3,3\sqrt{3})$

b. Slope of AB=$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3\sqrt{3}-0}{3-12}$

$=\frac{3\sqrt{3}}{-9}=\frac{\sqrt{3}}{-3}=\frac{\sqrt{3}}{-\sqrt{3}\times \sqrt{3}}$

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

Let (x,y) be a point on the line AB.

Then slope of the line joining (x,y) and (12,0) will also be $\frac{-1}{\sqrt{3}}$

$\frac{0-y}{12-x}=\frac{-1}{\sqrt{3}},-1(12-x)=\sqrt{3}\times -y$

$-12+x=-\sqrt{3}y$

$x+\sqrt{3}y-12=0$

$x+\sqrt{3}y=12$.This is the equation of the line.

## In the figure, equation of the line joining the points A and B is x+2y=10. P and Q are points on this line. a. Find the coordinates of the points A and B. b. Find the coordinates of the points P that divides the line AB in the ratio 2:3. c. If AQ: BQ =2:3, find the coordinates of the point Q.

a. Equation of the line AB is x+2y=10

Since the y coordinates of the point A is zero, $x+2\times 0=10,x=10$

Coordinates of A = (10,0)

Since the x coordinates of the point B is sero, 0+2y=10,y=5

Coordinates of B =(0,5)

b. AP : PB =2:3

x coordinates of P =$x_{1}+\frac{p}{(p+q)}(x_{2}-x_{1})$

$=10+\frac{2}{5}(0-10)=10+\frac{2}{5}\times -10$

$=10+-6$

y coordinates of p =$y_{1}+\frac{p}{p+q}(y_{2}-y_{1})$

$=0+\frac{2}{5}(5-0)=0+2=2$

Coordinates of P=(6,2)

c. AQ: BQ=2:3

Then BA: AQ=1:2

if x is the x coordinates of Q.

$0+\frac{1}{3}(x-0)=10$

$\frac{1}{3}(x-0)=10$

$x-0=10\times 3=3 0,x=30$

If y is the y coordinates of Q.

$5+\frac{1}{3}(y-5)=0,5+\frac{(y-5)}{3}=0$

$3\times 5+\frac{3(y-5)}{3}=3\times 0$

15+y-5=0,y=5-15=-10

Coordinates of Q =(30,-10)