ML AGGARWAL CLASS 8 CHAPTER 5 Playing with Numbers Excercise 5.1

 Exercise 5.1

Question 1

(i) 89

Let given number be $a b$ i.e $10 a+b$

 ∴ 89=$10 \times 8+9$

(ii) $207=2 \times 100+0 \times 10+7$

(iii) $\quad 369=100 \times 3+10 \times 6+9$

Question 2

Sol:

 Given number = 34 

Number obtained by reverssing the digits = 43 

sum = 34 + 43 =77 = 7 x 11

i) when sum is divited by 11, quotient is 7 

ii) Whew sum is divided by 7 , quotient is 11

Question 3

Sol:

Given number =73

 

Number obtained by reversing the digits =37

Difference =73-37=36=$9 \times 4$

(i) Whew difference is divided by 9, quoptient is 4

(ii) When difference is divided by 4 (differen ot digits),

quotient is 9

Question 4

Sol:

given number =a b c

numbers obtained by reversing digits = b c a , c ab 

sum =a b c+b c a, c a b

i) When sum divided by 111 , quotient is (a+b+c)

ii when som divided by $(a+b+c),$ quctient is 111

iii) whew sum divided by 37, quotient is  3(a+b+c)

iv) when sum dividect by 3, quotient is 37(a+b+c)$

Question 5

Given number =843

number obtained by reversing digits =348

Difference $=843-348=495=99 \times 5$

i) When difrerence divided by 99, quotient =5

ii) when difference divided by 5, quotient =99

Question 6

Let the given number is ab

Sumof digit=11 = a+b=11 $\rightarrow(1)$

From given condition

$b a=a b-9$

$10 X b+a=10 X a+b-9$

$10(b-a)=(b-a)-9$

$10(b-a) - (b-a)=-9$

$9(b-a) = - 9$

b-a= -1------②

Solving (1) and (2)

$b-a=-1$

$b+a=11$

-------------

       2b = 10 

      b= 5

b + a =11

5 + a = 11

a = 6

Question 7

given 

let given number  is ab 

given condition 

ab - ba = 36

$10 a+b-(10 b+a)=36$

$10(a-b)+(b-a)=36$

$10(a-b)-(a-b)=36$

$9(a-b)=36$

$a-b=\frac{36}{9}$

a-b=4 

∴ Difference of two digits = 4 

Question 8

let given number be ab 

given condition 

sum of two digit number and the number 

obtained by reversing the digits is 55 

$a b+b a=3655$

$10 a+b+10 b+a=3655$

$10(a+b)+(b+a)=31655$

$11(a+b)=55$

$(a+b)=\frac{55}{11}$

$a+b=5$

∴ Sum of digits is 5 

Question 9

Let given3 digits number be abc 

Given   

     unit's digits ten's digit and hundred's digit 

are in the ratio  1 : 2 : 3

 ∴ c : b : a = 1 : 2 : 3---------- ①

and

 $\quad a b c-c b a=594 \rightarrow(2)$

$100 a+10 b+c-(100 c+10 b+a)=594$

$100(a-c)+10 b-10 b+(c-a)=594$

$100(a-c)+0-(a-c)=594$

$99(a-c)=594$

$a-c=\frac{594}{99}$

$a-c=6 \rightarrow(3)$

given $\quad c ; b: a=1: 2: 3$

let 

$c=1 \times x$

$b=2 \times x$

$a=3 \times x$

Substitube $a, c$ values in (3)

$\begin{aligned} 3 x-1 x &=6 \\ 2 x &=6 \\ x &=6 / 2 \\ x &=3 \end{aligned}$

$\begin{aligned} \therefore \quad a &=3 \times 3=9 \\ \ b &=2 \times 3=6 \\ c &=1 \times 3=3 \end{aligned}$

 ∴ given number is 963

Question 10

Let the given number be = abc 

given 

       c = a + 1 --------1

       b = a - 1 ---------2

given sum of original number and numbers 

 obtained by reversing cyclically is 2664 

 abc + bca + cab = 2664 --------3

but   abc + bca + cab = 111 X ( a +  b + c )--------4

 ∴ from eq 3 and eq 4 

$111 \times(a+b+c)=2664$

$a+b+C=\frac{2664}{111}$

$\quad a+b+c=24$

$a+a-1+a+1=24(\because$ From 1,2$)$♀

$3 a=24$

$a=\frac{24}{3}$

$a=8$

$c=a+1=8+1=9$

$b=a-1=8-1=7$

 ∴ given number is 879

Exercise 5.2

Question 1 

4  A 

3  5

--------

B  2

--------

A+5= we get a number whose unit's

digit in 2, A+5 should not exceed 14

$\begin{aligned} \therefore A+5 &=12 \\ A &=12-5 \\ A &=7 \end{aligned}$

4  7

3   5

-------

8   2

-------

B = 8

∴ A = 7 , B = 8 

Question 2

5     A 

7     9

---------

CB  3

---------

A+9 Should Not exceeD 18 . and for A+9= anumber

whose unit's place is 3 this:s possible when A=44

(daigram should be added)

∴ A = 4 , C = 1 , B = 3

Question 3

4  2  A 

2  A  5 

---------

A  0  2 

---------

A+5 sum should not exceed 14 and it is a

number whose Unit's digit is 2 This is possible

when A=7 

(daigram should be added)

A = 7 

satisfies given condition 

 A = 7 

Question 4

     A  A

+   A  A

-----------

   BA  8 

case(i)

 $\begin{aligned} A+A &=8 \\ 2 A &=8 \\ A &=4 \end{aligned}$

check 

        4  4 

        4  4 

-------------

         8  8 

but given the sum 

 as there digit number 

case (ii)

A + A 18 

2A = 18 

A = 9 

check :   9  9 

               9  9 

         -----1----------

               198

         ----------------

      A = 9 ,  B = 1 

      A = 9 ,   B = 1 

Question 5

  1  8  A 

+B  A  7

-----------

   C B   2 

-----------

A + 7 sum should not exceed 16 and it is a 

 

number whose unit, s digit is 2 

 ∴ A + 7 = 12 

A =  12 - 7 

 

A =  5

$1+8+A \Rightarrow 1+8+5=13$

1+B=C

C=1+4

C=5

 ∴ A = 5 , B = 4 , C = 5 

Question 6

 A  2  1  B 

+1  C  A  B 

--------------

   B 4   9   6 

 

B +  B  sum should not exceed 18 and its units digits is 6 

case (i) 

B + B = 6 

2B = 6 

B = 3 

 Then 1 + A = 9 

A = 8 

A + 1 = B 

8 + 1 = B 

B = 9 but we got B = 3 earlier 

so B ≠ 3

case (ii) 

B+B=16

2 B=16

$B=\frac{16}{2}$

B=8

7   2  1   8 

1   2  7   8 

--------------

 8  4   9   6 

A + 1 = B 

A   + 1 = 8 

A  = 8 - 1 

 A = 7      A = 7 

 ஃ A = 7  B = 8  C = 2 

Question 7

B  3  4  5

C  9  B  A 

--------------

8   B  A  2 
--------------

 Sum 5 + A = 12 

A = 12 - 5 

A  = 7

1 + 4  + B = A 

1 +  4  + B = 7 

B = 7 - 5

B = 2 

$\begin{aligned} B+C &=8 \\ 2+C &=8 \\ C &=8-2 \end{aligned}$

c=6 

$A=7, \quad B=2, C=6$

Question 8

$\begin{array}{r}A B \\(-) B 6 \\\hline 4 7 \\\hline\end{array}$

$\begin{aligned} B-6 &=7 \\ B &=7+6 \\ B &=13 \end{aligned}$

means  B = 3 , because B should be single digit 

$\begin{aligned} A-B-1 &=4 \\ A-3-1 &=4 \\ A-4 &=4 \\ A &=8 \\ \therefore \quad A=8, \quad B &=3 \end{aligned}$

Question 9

Sol :

$\begin{array}{r}2 A \\\times 3 A \\\hline B 7 A\end{array}$

A x A = A means A should be 1 or 5 or 0 

case (i) A = 0 

$\begin{array}{r}20 \\30 \\\hline 600 \\\hline\end{array}$ 

 ≠ B7A 

∴ This is not given number so A  ≠ 0

case (ii) A = 1 

$\begin{array}{l}21 \\31 \\\hline 21 \\\hline 23 \\\hline 251 \\\hline\end{array}$

 ≠ B71

∴ B ≠ 1

 

case (iii)

$\begin{array}{l}25 \\35 \\\hline 125 \\75 \\\hline 875\end{array}$

∴ B = 8 

$\therefore \quad A=5, \quad B=8$

Question 10

$\begin{array}{r}A B \\A B \\\hline 6 A B\end{array}$

B should either 0 , 1 or  5 

case (i) B = 0 

$\begin{array}{r}A 0 \\B 0 \\\hline 00 \\A B O \\\hline A B B \quad 0 \end{array}$

≠ 6AB

∴ A ≠ 0

case (ii) B = 1

(DIAGRAM SHOULD BE ADDED )

 ≠ 6AB 

case (iii)

B = 5 

A 5 

A 5 

------

X25   =  6AB   (∵ may be any number)

A = 2 (∵from rules by squar number)

Question 11

$\begin{array}{l}A A \\4 A \\\hline 9 A 4\end{array}$

A = 2

A = 2 is correct 

Question 12

(DIAGRAM SHOULD BE ADDED )

Question 13

(DIAGRAM SHOULD BE ADDED )

Question 14

(DIAGRAM SHOULD BE ADDED )

Exercise 5.3

Question 1

(i) 87035

it is divisible by 5 

(ii) 75060

It is divisible by both 5 and 10 

(iii) 9685

It is divisible by 5 

iv. 10730

It is divisible by both 5 and 10 

Question 2

i. 67894

It is divisible by '2'

ii. 5673244

It is divisible by both 2 and 4

iii. 9685048

It is divisible by 2 , 4 and 8

iv. 75379

It neither divisible by 4 nor by 8 

Question 3

i. 45639

sumot digits  = 4 + 5 + 6 + 3 + 9 =27

27 is divisible by both 3 and 9

∴ 45639 is divisible by both 3 or 9 

ii. 301248

Sum of digits = 3 + 0 + 1 + 2 + 4 + 8 = 18

18 is divisible by 9

∴ . 301248 is divisible by both 3 or 9 

iii). 567081

Sum of digits =5+6+1+0+8+1=27

27 is divisible by 9

∴ 567081 is divisible by either 3 or by 9 

iv.  345903

sum of digits =3+4+5+9+0+3=24

24 is divisibe by only'3'

∴ 345903  is divisibly by '3'only.

v. 345046

sum of digits =3+4+5+0+4+6=22

22 is neither divisible by 3 nor by 9

∴ 345046 is neither divisible by 3 nor by 9 

Question 4

i. 10835

Sum of odd daplace digits =5+8+1=14

Sum of even place digits =3+0=3

Disfference =14-3=11

∴ 10835 is divisible by 11

ii. 380237

Sum of odd place digits =1+2+8=17

Sum of even place digits =3+0+3=6

Difference =17-6=11

∴ 380237 is divisible by 11

iii. 504670

Sum of odol place digits =0+6+0=6

Sum of even place digits =7+4+5=16

Difference = 16 - 6 = 10 

∴ z504670 is not divisible by 11 

iv. 28248

Sum of odd place digits =8+2+2=12

Sum of even place digits =4+8=12

difference =12-12=0

 ∴ 28248 is divisible by 11 

Question 5

i. 15414

15414 is divisible by '2' because units place contain '4'

1+5+4+1+4=15 divisible by 3

  ∴ 15414 in divisible by 3

  ∴ 15414 is divisible by 6

ii. 213888

213888 is divisible by 2

Sum of digits =2+1+3+8+8+8=30 divisible by'3

  ∴ 213888 is  also divisible by 3

  ∴ 213888 is divisible by ' 6 '

iii. 469876

469876 is divisible by 2

Sumot digits =4+6+9+8+7+6=40 not divisible by '3'

  ∴ 469876 in not divisibie by 3

  ∴ 469876 in also not divisible by 6 '.

Question 6

i. 4618894875

Sum of digits of alternative blocks

875+618=1493  and 894+4=898

Dibterence $=595$ whicl is divisible by 7

   ∴ 4618894875 is divisible by 7 

ii. 3794856

Sum of digits of alternative blocks

856+3=859 and 794

difference = 65 which is not divisible by 7 

iii. 39823

sum of digits of alternative blocks

823 and 39

Difference 823-39=784 whichin divisible by 7

 ∴ 3794856 is not divisible by 7 

iii. 39823

sum of digits of alternative blocks

823 and 39

Difference 823-39=784 whichin divisible by 7

 ∴  39823 is divisible by ' 7 '.

Question 7

i. Given number = 34x 

it 34x is multiple of 3 then (3+ 4 + x ) should be multiple of '3'

3+ 4+ x = 9+x

so to make sum of digits multiple of 3 

x should be 0 , 3 , 6 , 9 

ii) 74 x 5284 is multiple of '3 'so

7+4+x+5+2+844=30+x

So to make sum of digits multiple of 3

x should be 0,3,6,9 .

Question 8

Given number 43Z3 

If 42 z 3 in multiple of  then (4+2+z+3)

Should be multiple of 9

4+2+z+3=9+z

To make sumob digits multiple of a

should be 0,9 .

   

Question 9

(i)  49*2207

If a number is divisble by 9 then its sum of digits 

should be divisble by 9 

4 + 9 + * + 2 + 2+ 0 + 7 = 24+*

The near by multiple of 9 is 27 so in place of the digit must be 27- 24 = 3

ii) 5938*623

"If a number is divisible by 9 then it's sum of digits

Should be divisible by 9

5 + 9 + 5 + 8 + * + 6 + 2 + 3= 36 + *

36 is multiple of 9 so * should be either 0 or 9 

Question 10

i) 97*542

As unit's place is 2 , number in divisible by 2

irrespective of digit in * 'Place.

if a number divisible by 3 then its sum of digits should be divisble by 3

9+7-1 x+5+4+2=27+x

∴ 27 in a multiple of 3 so in place of '*' Should be

either 0,3,6, or 9

In place of * 0 , 3 , 6 , 9 number makes given 

number divisible by 6 (As it is divisible by both 2 and 3)

(ii)  709*94

As units digit in 4 given number divisible by

2  irrespective of number  in place of *

To make given number divisible by ,3 the sum of digits should be divisible by 3 

7+0+9+*+9+4=29+*

To make divisible $29, *$ should be either 1,4,7 

$\therefore$ In place ot *, 1,4 or 7 digits make the given number divisible by 6. (As it is divisible by 2 \  and 3

Question 11

(i) given number $64 * 2456$

Sum of digits in odd places =6+4+*+6=16+*

Sum of digits in even places =5+2+4=11

Dilterence $=16+*-11=5+*$

* Should bí Becouse to make given number

divisible by 11 , the diffrerence in sum of digits in even

and odd place is either 0 or mulliple of 11 

∴ should be 6

(ii) 86*6194

sum of digits in even places = 9 + 6+ 6= 21

sum of digits in odd places = 4+ 1 + * + 8 = 13+ *

so the difference should be either 0 or multiple 

of 11 to be divisible by 11

8 - * =0

* = 8

∴ to make 86*6194 divisible by 11 , in place of * digits must be place 

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ML AGGARWAL CLASS 8 CHAPTER 4 Cubes and Cube Roots Excercise 4.1

 

Exercise 4.1

Question 1

(i) 648

Sol: Expersing it in to prime factors 

$\begin{array}{r|l}2&648\\ \hline 2& 324 \\ \hline 2&162\\ \hline 3& 81\\ \hline 3& 9 \\ \hline 3&27\\ \hline 3& 9 \\ \hline 3&3\\ \hline &1\end{array}$

648 = 2 x 2 x 2 x 3 x 3 x 3 x 3 

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

Since 3 is left after grouping in triplets 

∴ 648 is not perfect cube 

(ii) 8640 

Expressing it in to prime factors 

  $\begin{array}{r|l}2&8640\\ \hline 2& 4320 \\ \hline 2&2160\\ \hline 2& 1080\\ \hline 2& 540 \\ \hline 2&270\\ \hline2&135 \\ \hline 3&27\\ \hline 3&9\\ \hline 3&3\\ \hline&1\end{array}$

∴ Since 5 is left after grouping in triplets 

8640 is not a perfect cube 

(iii) $729=9 \times 9 \times 9=9^{3}$ is a perfect cube

(iv) $\quad 8000=20 \times 20 \times 20=20^{3}$ is a perfect cuse

Question 2 

(i) 1728

Expressing it into prime factors 

$\begin{array}{r|l}2&1728\\ \hline 2& 864 \\ \hline 2&432\\ \hline 2& 216\\ \hline 2& 108 \\ \hline 2&54\\ \hline 3& 27 \\ \hline 3&9\\ \hline 3&3\end{array}$

$1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

$=2^{3} \times 2^{3} \times 3^{3}$=$=(2 \times 2 \times 3)^{3}$

$=12^{3}$

$\therefore 12^{3}= 1728$ is a perfect cube

And 1728 is the cube of number 12

(ii) 5832

Expressing it into prime factors

$\begin{array}{l|l}2 & 5832 \\\hline 2 & 2916 \\\hline 2 & 1458 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline\end{array}$

$5832=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$\begin{aligned}=2^{3} \times 3^{3} \times 3^{3} &=(2 \times 3 \times 3)^{3} \\ &=18^{3} \end{aligned}$

$\therefore 18^{3}=5832$ is a perfect cube

And 5832 is the cube of number 18.

(iii) 13824  

Expressing it into prime factors

$\begin{array}{c|c}2 & 13824 \\\hline 2 & {6912} \\\hline 2 & 3452 \\\hline 2 & 1728 \\\hline 2 & 864 \\\hline 3 & 432 \\\hline 3 & 144 \\\hline 2 & {78} \\\hline 3 & {16} \\\hline 2 & {8} \\\hline 2 &{4} \\\hline 2 & 2 \\ 2 &  \hline\end{array}$

$13824=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

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

$=(2 \times 2 \times 2 \times 3)^{3}$

$13824=24^{3}$ is a perfect cube

(iv) 35937

$\begin{array}{l|l}3 & 35932 \\\hline 3 & 11979 \\\hline 3 & 3993 \\\hline 3 & 1331 \\\hline 3 & 447 \\\hline & 149\end{array}$

$=3 \times 3 \times 3 \times 3 \times 3 \times 149$

∴ It is not a perfect cube 

Question 3

(i) 243 

Expressing it into prime factors

$\begin{array}{l|l}3 & 243 \\\hline 3 & 81 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline 3 & 3 \\\hline\end{array}$

$243=3 \times 3 \times 3 \times 3 \times 3$

If we multiply abore number with 3 

then it becomes = $3 \times 3 \times 3 \times 3 \times 3 \times 3$

=$3^{3} \times 3^{3}=9^{3}=729$, perfect cube

∴ Therefore the smallest number 3 is to be multiplied to 

make the number a perfect cube.

(ii) 3072

Exprssing it into prime factors 

$\begin{array}{l|l}2 & 3072 \\\hline 2 & 1536 \\\hline 2 & 768 \\\hline 2 & 384 \\\hline 2 & 192 \\\hline 2 & 96 \\\hline 2 & 48 \\\hline 2 & 24 \\2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline\end{array}$

$3072=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

If we multiply the above number with $2 \times 2 \times 3 \times 3$ i.e 36 it will become

$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 1$

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

$(2 \times 2 \times 2 \times 2 \times 3)^{3}$

$48^{3}=110592$ i.e $3672 \times 36$

∴ Therefore the smallest number 36 is to be multiplied with 3072 to make the number a perfect cube.

(iii) 11979

Expressing it in to prime factors

$\begin{array}{l|l}3 & 11979 \\\hline 3 & 3993 \\\hline 11 & 1331 \\\hline 11 & 121 \\\hline 11 & 11 \\\hline &1\end{array}$

$11979=3 \times 3 \times 11 \times 11 \times 11$

In the above, Prime factors 3 occure twice 11 occures thrice . Therefore the smallest number by which the given number must be multiplied so that the product is a perfect cube i.e 3

Then product = $3 \times 3 \times 3 \times11 \times 11$ 

$=3^{3} \times 11^{3}=33^{2}, 35937=11979 \times 3$

(iv) 19652

Expressing it into prime factors

$\begin{array}{l|l}2 & 19652 \\\hline 2 & 9826 \\\hline 17 & 4913 \\\hline 17 & 289 \\\hline & 17\end{array}$

$19672=2 \times 2 \times 17 \times 17 \times 17$

2 occurs twice , 17 occure thrice therefore the smallest number by which given number must be multiplied So that product is a perfect cube is 2

Then product =$2 \times 2 \times 2 \times 17 \times 17 \times 17$

$2^{3} \times 17^{3}=34^{3}=39,304 = 19652 \times 2$

Question 4

(i) 1536

Expressing it in to prime factors 

$\begin{array}{l|l}2 & 1536 \\\hline 2 & 768 \\\hline 2 & 384 \\\hline 2 & 192 \\\hline 2 & 96 \\\hline 2 & 48 \\2 & 24 \\\hline 2 & {12} \\\hline 2 & 6\end{array}$

$1536=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

1536 is not a perfect cube

To make it perfect cube, we should divide the given number by 3, then the prime factorisation of the quotient will not contain 3.

In that case 

        $1536 \div 3=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=512$

Which is a perfect cube 

So the smallest number by which 1536 must be divided So that quotient is a perfect cube is 3

(ii) 10985

Expressing it into prime factors, we have

$\begin{array}{l|l}5 & 10985 \\\hline 13 & 2197 \\\hline 13 & 169 \\\hline 13 & 13 \\\hline &1\end{array}$

10985= $5 \times 13 \times 13 \times 13$

10985 is not a perfect cube.

To make it perfect cube, we should divide the given number by 5 ,then the prime factorisation of the quotient will not contain 5.

In that case 

 $10985 \div 5=13 \times 13 \times 13=2197$ , Which is a perfect cube 

So, the smallest numbr by which 10985 must be divided 

So that quotient is a perfect cube is 5

(iii) 28672

Expressing it into prime factors

$\begin{array}{l|l}2 & 28672 \\\hline 2 & 14336 \\\hline 2 & 7168 \\\hline 2 & 3584 \\\hline 2 & 1792 \\\hline 2 & 896 \\\hline 2 & 448 \\\hline 2 & 224 \\\hline 2 & 112 \\\hline 2 & 56 \\\hline 2 & 28 \\\hline 2 & 14 \\\hline & 7\end{array}$

$28172=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7$

28672 is not a perfect cube 

To make it a perfect cube, we should divide the given number by 7

Then the prime factorisation will not '7'

In that case $28672 \div 7=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times \times 2$

$=4096$ is a perfect cube

So the smallest number by which given number must be divided 

So that product will become perfect cube is ' 7'

(iv) 13718

Expressing it in the prime factors

$\begin{array}{l|l}2 & 13718 \\\hline 19 & 689 \\\hline 19 & 361 \\\hline 19 & 19 \\\hline & 1 \\\end{array}$

$13718=19 \times 19 \times 19 \times 2 .$

It is not a perfect cube 

To make it a perfecet cube, we should divide the given number by 2 , then prime factorisation will no contain '2'

In that case $13718 \div 2=19 \times 19 \times 19=6855$ is a perfect cube 

So the smallest number 2 must be diivided from given number to make it perfect cube.

Question 5

The volume occupied by one Cuboid is $3 \times 3 \times 5=45$

45 is not a perfect cube 

In order to make it a cube, the number which is to multiplied is $45 \times 3 \times 5 \times 5$ i.e $3 \times 5 \times 5=75$ is to be multiplied in order to make a cube.

So total number of cuboids are needed to form a Cube are 75.

Question 6

Given Surface area of a cubical box is $486 \mathrm{~cm}^{2}$ 

We have , volume of Cubical box is (side) $^{3}$ and

Surface area of Cubical box is $6 \times(\text { side })^{2}$

i.e Let side of a box is 'a' cm

$6 a^{2}=486 \Rightarrow a^{2}=\frac{486}{6}$

$a^{2}=81 = 9 \times 9$

$a=9 \mathrm{~cm}$

Volume of a Cubical box is $a^{3}=9^{3}=729 \mathrm{~cm}^{3}$

Question 7

(i) $125=5 \times 5 \times 5=5^{3}$ , Cube of odd natural number 

$\begin{array}{l|l}5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline & 1 \\\end{array}$

(ii) 

 $\begin{array}{l|l}2 & 512\\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 & 64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\2 & 2 \\\hline\end{array}$   

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$=2^{3} \times 2^{3} \times 2^{3}=8^{3}$

Cube of even natural number

(iii) $1000=10 \times 10 \times 10=10^{3}$ , Cube of even natural number 

(iv)$2197=13 \times 13 \times 13=13^{3}$, Cube of odd natural number 

(v) 

$\begin{aligned} 4096=4 \times 4 \times 4 \times 4 \times 4 \times 4 &=4^{3} \times 4^{3} \\ &=16^{3}\end{aligned}$ Cube of even natural number 

(vi) $6859=19 \times 19 \times 19=$$19^{3}$ Cube of odd nuatural number.

Question 8

(i) 231 , unit's digit of cube of number is 1

(ii) 358, One's digits of cube of number is 2

(iii) 419 One's digits of cube of number is 9

(iv)725 One 's digits of cube of number is 5

(v)854 One's digits of cube of number is 4

(vi)987 One's digits of cube is 3

(vii)752 One's digits of cube is 8

(viii)893 One's digits of cube is 7.

Question 9

i) $(-13)^{3}=-13\times-13 x-13=(-13)^{3}=-2197$

(ii) $\left(3 \frac{1}{5}\right)^{3}=\left(\frac{16}{5}\right)^{3} \cdot \frac{16 \times 16 \times 16}{5 \times 5 \times 5}=\frac{4096}{125}$

(iii) $\left(-5 \frac{1}{7}\right)^{3}=\left(-\frac{36}{7}\right)^{3} = \frac{-36 x-36 x-36}{7 \times 7 \times 7}= \frac{-46656}{343}$

Exercise 4.2

Question 1

(i) 12167 

Expressing it in to prime factors

$\begin{array}{l|l}23 & 12167 \\\hline 23 & 529 \\\hline &23\end{array}$

$12167=23 \times 23 \times 23$

Hence, Cube root of 12167 is 23

(ii)35937

Expressing it in to prime factors

$\begin{array}{l|l}33 & 35937 \\\hline 33 & 1089\\\hline &33\end{array}$

$35937=33 \times 33 \times 33$

Hence, cube root of 35937 is 33

(iii) 42875

Expressing it in to prime factors

$\begin{array}{l|l}35 & 42875 \\\hline 35 & 1225\\\hline &35\end{array}$

$42875=35 \times 35 \times 35$

Hence, Cube root of 43875is 35

(iv) 21952

Expressing it in to prime factors

$\begin{array}{l|l}2 & 21952 \\\hline 2 & 10976 \\\hline 2 & 5488 \\\hline 2 & 2744 \\\hline 2 & 1372 \\\hline 2 & 686 \\\hline 7 &343 \\\hline 7 & 49 \\\hline&7\end{array}$

$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7$

$=(2 \times 2 \times 7)^{3}=(28)^{3}$

Hence, cube root of 21952 is 28

(v) 373248

Expressing it into prime factors

$\begin{array}{l|l}2 & 373248 \\\hline 2 & 186624 \\\hline 2 & 93312 \\\hline 2 & 46656 \\\hline 2 & 23328 \\\hline 2 & 11664 \\\hline 2 & 5832 \\\hline 2 & 2916 \\\hline 2 & 1458 \\\hline 3 &729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3& 27 \\\hline 3&9\\\hline &3\end{array}$

$333248=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$=(2 \times 2 \times 2 \times 3 \times 3)^{3}$

$=72^{3}$

Hence, Cube root of 373248 is 72

(vi)32768

Expressing it in to prime factors 

$\begin{array}{l|l}2 & 32768 \\\hline 2 & 16384 \\\hline 2 & 8192 \\\hline 2 & 4096 \\\hline 2 & 2048 \\\hline 2 & 1024 \\\hline 2 & 512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 &64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2& 4 \\\hline 2&2\\\hline &2\end{array}$

$32768=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$=(2 \times 2 \times 2 \times 2 \times 2)^{3}$

$=32^{3}$

Hence , Cube root of 32768 is 32.

(vii)262144

Expressing it in to prime factors 

$\begin{array}{l|l}2 & 262144 \\\hline 2 & 131072 \\\hline 2 & 65536 \\\hline 2 & 32768 \\\hline 2 & 16384 \\\hline 2 & 8192 \\\hline 2 & 4096 \\\hline 2 & 2048 \\\hline 2 & 1024 \\\hline 2 &512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2& 64 \\\hline 2&32\\\hline 2&16 \\\hline 2& 8\\\hline 2&4\\\hline 2&2\\\hline &1\end{array}$

$262144=\underbrace{2 \times 2 \times 2} \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$=(2 \times 2 \times 2 \times 2 \times 2 \times 2)^{3}$

$=64^{3}$

Hence, Cube root of 262144 is 64

(viii) 157464

Expressing it into prime factors

$\begin{array}{l|l}2 & 157464 \\\hline 2 & 78732 \\\hline 2 & 39366 \\\hline 3 & 19683 \\\hline 3 & 6561 \\\hline 3 & 2187 \\\hline  & 48 \\\hline 2 & 24 \\2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline\end{array}$

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

$=(2 \times 3 \times 3 \times 3)^{3}$

$=54^{3}$

Hence , Cube root of 157464 is 54

Question 2 

(i) 19683

$\frac{19}{\text { second group }}$ 

$\frac{683}{\text { First group }}$

First gruop decides the unit digits of required cube root 

So, The number 683 ends with 3 . we know that 3 comes 

at units place of a number only when it's when it's cube root 

end in 7 

Now take second group 19 , then it will decide the ten's digit of required cube root

Now that $2^{3}=8$ and $3^{3} = 27$ ,Also $8<19<27 .$

We take the one's place of smaller number 8 as ten's digit of required cube root (i.e 2)

Therefore $\sqrt[3]{19683}=27$ 

(ii) 59319 

$\frac{59}{\text { second group }}$

$\frac{319}{\text { first group }}$

First group decides the one's digits of required Cube root 

The number 319 ends with 9. we know that 9 comes at 

Unit's place of a number only when its cube root ends in 9 

Now second group decides the ten's digits of required cube root 

59 lies in between $3^{3}=27$ and $4^{3}=64$. We take one's 

place of smaller number 27 as the ten's digits of required cube 

root

So $\sqrt[3]{59319}=27$

(iii) 85184

$\frac{85}{\text {second group }}$

$\frac{184}{\text { First group }}$

184 ends with 4 . we know that 4 comes at unit's 

place of number only when its cube root ends in 4

Second group decides ten's digits 

i.e 85 lies in between $4^{3}=64$ and $5^{3}=125$

we know that one's place of smaller number 64 as 

ten's digits of required Cube root 

So $\sqrt[3]{85184}=44$

(iv) 148877

$\frac{148}{\text { second group }}$

$\frac{877}{\text { First group }}$

Step:1 First form group of three digits starting from rightmost 

digit (i.e unit's digits ) of number 

Step:2 First group decides unit's digits of required root 

The number 877 ends with 7. we know that 7 comes at 

unit's place of a number only when it's cube root ends in

3

So the unit digit of required Cube root is 3

Step:3 If no group is left then number obtained is the cube root of given number 

But if second group exits (in this case 148) then it will decide the ten's digit of required 

cube root 

Now take second group i.e 148

We know that $5^{3}=125$ and $6^{3}=216$ . Also $125<148<216$ 

We take one's place of the smaller number 725 as the ten's digit of required 

cube root (i.e 5)

Step: 4 If no group is left then the digit obtained in step 2 and 

step 3 decides the cube root of given number 

i.e $\sqrt[3]{148877}=53$.

Question 3

(i) 250047

Expressing it into to prime factors 

-250047= $7\times -7\times -7 \times -3 \times -3 \times-3\times -3\times -3\times -3$ 

= $(-7\times -3\times -3)^{3}$

$=(-63)^{3}$

Hence , Cube root of -250047 is -63

(ii) $\frac{-64}{1331}$

Expressing 64 and 1331 in to prime factors 

$\begin{aligned} 64 &=4 \times 4 \times 4=4^{3} \\ 1331 &=11 \times 11 \times 11=11^{3} \end{aligned}$

$\frac{-64}{1331}=\frac{(-4)^{3}}{(11)^{3}}=\left(\frac{-4}{11}\right)^{3} \Rightarrow \sqrt[3]{\frac{-6 y}{1331}}=\frac{-4}{11}$

(iii) $4 \frac{17}{27}=\frac{125}{27}$

Expresssing 125 and 27 it into prime factors 

$125= 5 \times 5 \times 5=5^{3}$

$27=3 \times 3 \times 3=3^{3}$

Hence,$\sqrt[3]{\frac{125}{27}}=\sqrt[3]{\left(\frac{5}{3}\right)^{3}}=\frac{5}{3} .$

(iv)$5 \frac{1182}{2197}=\frac{12167}{2197}$

$12167=23 \times 23 \times 23=23^{3}$

$2193=13 \times 13 \times 13=13^{3}$

$\frac{12167}{2197}=\frac{23^{3}}{13^{3}} =\left(\frac{23}{13}\right)^{3}$

Hence, Cube root of $5 \frac{1182}{2197}$ is $\frac{23}{13}$

Question 4

(i) $\sqrt[3]{512 \times 729}$

Expressing it into prime factors 

 $512=8 \times 8 \times 8=8^{3}$

$729=9 \times 9 \times 9=9^{3}$

$512 \times 729=8^{3} \times 9^{3}$= $(8 \times 9)^{3}=72^{3}$

Hence, $\sqrt[3]{512 \times 729}$ = $\sqrt[3]{72^{3}}$= 72

(ii) $\sqrt[3]{(-1331) \times(3335)}$

Expressing it into prime factors 

$\begin{array}{l|l}11 & 1131 \\\hline 11 & 121\\\hline11&11\\\hline &1\end{array}$

$\begin{array}{l|l}5 & 3375\\\hline 5 & 675 \\\hline 5 & 135 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline 3 & 3\\\hline&1\end{array}$

$-1331=(-11)^{3}$

$\begin{aligned} 3375 \times 5 \times 5 \times 3 \times 3 \times 3=&(5 \times 3)^{3} \\ &=15^{3} \end{aligned}$ 

$-1331 \times 3375=(-11 \times 15)^{3}$

Hence $\sqrt[3]{(-1331 \times 3] 75)}=\sqrt[3]{(-11 \times 15)^{3}}$ = $-11 \times 15$

$=-165$

Question 5

(i) 0.003375

$\sqrt[3]{0.003335}=\sqrt[3]{\frac{3375}{1000000}}$

$=\sqrt[3]{\frac{15 \times 15 \times 15}{100 \times 100 \times 100}}$

$=\frac{15}{100}=0.15$

(ii) 19.683

$\sqrt[3]{19.683}=\sqrt[3]{\frac{19683}{1000}}$

$=\sqrt[3]{\frac{27 \times 27 \times 27}{6 \times 10 \times 10}}$\

$=\frac{27}{10}=2.7$

$\begin{array}{l|l}3 & 19683 \\\hline 3& 6561 \\\hline 3 & 2187 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3 &27 \\\hline 3 & 9 \\\hline&3\end{array}$

=$3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

=$(3 \times 3 \times 3)=27^{3}$

Question 6

$\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.664}$

$27=3 \times 3 \times 3 \Rightarrow \sqrt[3]{27}: \sqrt[3]{3 \times 3 \times 3}=3$

$\sqrt[3]{0 \cdot 008}=\sqrt[3]{\frac{8}{1000}}=\sqrt[3]{\frac{2 \times 2 \times 2}{10 \times 10 \times 10}}=\frac{2}{10}=0.2$

$\sqrt[3]{0.064}=\sqrt[3]{\frac{64}{1000}}=\sqrt[3]{\frac{4 \times 4 \times 4}{10 \times 10 \times 10}}=\frac{4}{10}=0.4$

$\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}, 3+0.2+0.4=3.6$

Question 7

Sol: 6561

Expressing it into prime factors

$\begin{array}{l|l}3 & 6561 \\\hline 3& 2187 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 &81 \\\hline 3 & 27 \\\hline 3 &9 \\\hline 3 & 3 \\\hline&1\end{array}$

$6561=3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$=27 \times 27 \times 9$

9 is left in the above expansion, if we multiply 

the above number with 3 i.e $9\times 3$ = 27

i.e it becomes 

$6561 \times 3=27 \times 27 \times 9 \times 3=27 \times 27 \times 27=19683-27^{3}$

So the smallest number 3 must be multiplied to become 

the number a perfect cube 

Cube root of 19683 = 27

Question 8

Sol: 8748

Expressing it in to prime factors

$8748=3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 2 \times 2$

$=27 \times 27 \times 3 \times 2 \times 2$

If we divide above numbr by $\frac{4}{9}$

then it becomes $=\frac{27 \times 27 \times 3 \times 4}{(4 / 9)}$

$27 \times 27 \times 27$ =$27^{3}$ 

$=19683$

$\frac{8748 \times 9}{4}=19683=27^{3}$

Question 9

Givne Volume of cubical box = $21952 \mathrm{~m}^{3}$

We know that Volume of cube = (Side) $^{3}$

Let side length of cube = a

$a^{3}=21952$ 

$\begin{array}{l|l}4 & 21952 \\\hline 4& 5488 \\\hline 4 &1372 \\\hline 7 &343 \\\hline 7 &49 \\\hline \\\hline&7\end{array}$

$=4 \times 4 \times 4 \times 7 \times 7 \times 7$

$a^{3}=(4 \times 7)^{3}$

$a=28 m$

∴ Length of side of box = 28 m

Question 10

Let the three number be $3 x, 4 x, 5 x$, then

$(3 x) \times(4 x)(5 x)=480$

$60 x^{3}=480$

$x^{3}=\frac{480}{60}=8$

$x^{3}=8=2^{3}$

∴ $x=2 \Rightarrow 3 x=6, \quad 4 x=8,5 x=10$

Question 11

Let the two numbers are $4 x, 5 x$, Then

$(5 x)^{3}-(4 x)^{3}=61$

$125 x^{3}-64 x^{3}=61$

$61 x^{3}=61$

$x^{3}=\frac{61}{61}=1$

$x^{3}=1 \Rightarrow x=1$

∴ The number are 4, 5

Question 12

Let the cube root of smaller number be x

Given 

$8^{3}-x^{3} =387$

$512-x^{3}=387$

$x^{3}=512-387=125$

$x^{3}=125=5 \times 5 \times 5=5^{3}$

$x^{3}=5^{3}$

$x=5$

∴ Therefore the smaller number is 5 

Cube of this number is 125

Continue Reading →

ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.1

 Exercise 3.1

Question 1

(i) 729

Sol :

Prime fractorisation

$\begin{array}{r|l}3&729\\ \hline 3& 243 \\ \hline 3&81\\ \hline 3& 27\\ \hline 3& 9 \\ \hline 3&3\\ \hline&1\end{array}$

⇒729=3×3×3×3×3×3

∴$729=27^{2}$

because 729 can be expressed as product of pairs of equal prime factors. 

(ii) 5488

Sol:

$\begin{array}{r|l}2& 5488\\ \hline 2& 2744 \\ \hline 2&1372\\ \hline 2& 686\\ \hline 7& 343 \\ \hline 7&49\\ \hline 7&7\\ \hline&1\end{array}$

⇒5488= 2×2×2×2×7×7×7

Since 7 left unpaired 5488 in not a perfect square

 

(iii)1024 

Sol:

 $\begin{array}{r|l}2&1024\\ \hline 2& 512 \\ \hline 2&256\\ \hline 2& 128\\ \hline 2& 64 \\ \hline 2&32\\ \hline2&16 \\ \hline 2&8\\ \hline 2&4\\ \hline 2&2\\ \hline&1\end{array}$

 ⇒1024= 2x2x2x2x2x2x2x2x2x2x2 

since 1024 is expressed as the product of pairs of equal prime number so it is a perfect square 

(iv) 243

Sol:

$\begin{array}{r|l}3&243\\ \hline 3& 81 \\ \hline 3&27\\ \hline 3& 9\\ \hline 3& 3 \\ \hline&1\end{array}$

$243=3 \times 3 \times 3 \times 3 \times 3$

As ' 3 ' let unpaired, So 243 is not a square (perfect)

Question 2 

(i) 1296

 $\begin{array}{r|l}2&1246\\ \hline 2& 648 \\ \hline 2&324\\ \hline 2& 162\\ \hline 3& 81 \\ \hline 3&27\\ \hline3&9 \\ \hline 3&3\\ \hline&1\end{array}$

⇒1296=2x2x2x2x3x3x3x3

since 1296 is expressed as the product of pairs of equal prime numbers so it is a perfect square 

1296= $2^{2} \times 2^{2} \times 3^{2} \times 3^{2}$

1296= $(2 \times 2 \times 3 \times 3)^{2}=36^{2}$

∴ 1296 is square of 36 

 

(ii) 1784

Sol:

$\begin{array}{r|l}2&1784\\ \hline 2& 892 \\ \hline 2&446\\ \hline 223& 223\\ \hline&1\end{array}$

1784= $=2 \times 2 \times 2 \times 223$

As 1784  can not be expressed as product of pairs of

 equal prime factors, so is not a perfect square 

(iii) 3025 

Sol:

$\begin{array}{r|l}5&3025\\ \hline 5& 605 \\ \hline 11&121\\ \hline&1\end{array}$

$3025=5 \times 5 \times 11 \times 11$

Since 3025 can be expressed as the product of pairs of

equal prime factors.

$3025=(5 \times 11)^{2}=55^{2}$

Hence, 55 is a number whose square is 3025

(iv) 3969

Sol: 

$\begin{array}{r|l}3&3969\\ \hline 3& 1323 \\ \hline 3&441\\ \hline 3& 147\\ \hline 3& 49 \\ \hline 7&7\\ \hline&1\end{array}$

$3969=3 \times 3 \times 3 \times 3 \times 7 \times 7$

3969 Can be expressed as product of pairs of

equal prime numbers.

$3969=3^{4} \times 3^{2} \times 7^{2}$

$3969=(3 \times 3 \times 7)^{2}$

$3969=63^{2}$

Hence, 63 is the number whose square is 3969

Question 3 

1008

Sol:    $\begin{array}{r|l}2& 1008\\ \hline 2& 504 \\ \hline 2&252\\ \hline 2& 126\\ \hline 7& 63 \\ \hline 3&9\\  \hline 3&3\\ \hline&1\end{array}$

$1008=2 \times 2 \times 2 \times 2 \times 7 \times 3 \times 3$

Since' 7 ' is left unpaired, so to make 1008 a

Perfect square it should be muttiplied by 7

Question 4 

5808

Sol:

⇒ $\begin{array}{r|l}2& 5808\\ \hline 2& 2904 \\ \hline 2&1452\\ \hline 2& 726\\ \hline 3& 363 \\ \hline 11&121\\  \hline 11&11\\ \hline&1\end{array}$

$5808=2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11$

Since ' 3' let unpaired. To make 5808 a perfect square

it should be divided by '3'.

so divide 5808 by '3'

$\frac{5808}{3}=\frac{2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11}{3}$

$1936=(2 \times 2 \times 11)^{2}=(44)^{2}$

so 44 is a number whose square is 1936

Exercise 3.2

Question 1

(i) 2

(ii) 13

(iii) 27

(iv) 88

(v) 243

Question 2 

(i) 1

(ii) 4

(iii) 1

(iv) 9 

(v) 6

(vi) 5

(vii) 9

(viii) 4

(ix) 0

(x) 6

Question 3 

(i) 567

567 has '7' in its unit's place. a perfect square 

Should have 1,4,5,6,9,0 in it's unit's place.

so 567 is not a persect square.

(ii) 2453

2453 has '3' in it's unit's place. But a pertect square

should have 0,1,4,5,6,9 in it's unit's place.

So 2453 is not a persect square.

(iii) 5298

5298 has 8 in it's unit's place. But a perfect square

should have 0,1,4,5,6,9 in it's unit's place.

so 5298 is not a persect square.

(iv)4692

46292 has 2 in it's unit's place. But a perfect square

Should have 0,1,4,5,6,9 in it's unit's place

so 46292 is not a pertect square.

(v) 74000

74000 has 0 in it's unit's place but it has

odd no.of zero's and 740  is not a perfect square

so 74000 is not a perfeet square.

Question 4 

(i) 573

square of 573 is a odd number because ,If a number 

has 3 in the units place , then its square and in '9'

(ii) 4096

Square of 4096 is a even number because, If a rumber

has ' 6 ' in the units place, Then its square ends in ' 6 '

iii) 8267

Squar of 8267 is a odd number becouse, If a number

has 7 in the Units place, Then its square ends in ' 9

iv) 37916

square of 37916 is a even number becaise if a number 

has ' 6 ' in the Units place, then it square ends in ' 6 '

Question 5

i. 12 and 13

There are  2n non-square numbers betweew the squares of

two conseclutive numbers n and n+1

∴ natural numbers between 12 and $(12+1)=2 \times 12=24$

hence , there are 24 natural number between $12^{2}$ and $13^{2}$ 

(ii) 90 and 91

There are 2n non-square rumbers betweow the Squares

of two consective numbers n and n+1

∴ Natural numbers between 90 and 91=2 \times 90= 180

Hence, There are 180 natural numbers betwecn $90^{2}$ and $91^{2}$

Question 6

(i) $1+3+5+7+9+11+13=7^{2}=49$

(ii) $1+3+5+7+9+11+13+15+17+\cdots+29=15^{2}=225$

sum of first 'n' odd numbers = $n^{2}$

Question 7

(i) 64

$64-1=63 ; 63-3=60 ; 60-5=55$

$55-7=48: \quad 48-9=39 ; \quad 39-11=28$

$28-13=15 ; \quad 15-15=0$

∴ $64=1+3+5+7+9+11+13+15=8^{2}$

(ii) 121 

$121-1=120 ; 120-3=117 ; \quad 117-5=112 ; 112-7=105 ;$

$105-9=96 ; 96-11=85 ; 85-13=72 ; 72-15=57 ;$

$57-17=40 ; 40-19=21 ; \quad 21-21=0$

∴ $121=1+3+5+7+9+11+13+15+17+19+21=11^{2}$

Question 8

(i) $19^{2}=361$

"we can exrpress the square of any odd number greater

than 1 as the sum of two consective natural numbers."

First number $=\frac{19^{2}-1}{2}=180$

Second number $=\frac{19^{2}+1}{2}=181$

$19^{2}=361=180+181$

(ii) $33^{2}=1089$

First number $=\frac{33^{2}-1}{2}=544$

Second number $=\frac{33^{2}+1}{2}=545$

$33^{2}=1089=544+545$

(iii) $47^{2}=2209$

First number $=\frac{47^{2}-1}{2}=1104$

Second number $=\frac{47^{2}+1}{2}=1105$

$47^{2}=2209=1104+1105$

Question  9

(i) $31^{2}=(30+1)^{2}=(30+1)(30+1)$

$=30(30+1)+1(30+1)$

$=900+30+30+1$

$31^{2}=961$

(ii) $42^{2}=(40+2)^{2}=(40+2)(40+2)$

$=40(40+2)+2(40+2)$

$=1600+80+80+4$

$42^{2}=1764$

(iii) $86^{2}=(80+6)^{2}=(80+6)(80+6)$

$=80(80+6)+6(80+6)$

$=6400+480+480+36$

$86^{2}=7396$

(iv) $94^{2}=(90+4)^{2}=(90+4)(90+4)$

$=90(90+4)+4(90+4)$

$=8100+360+360+16$

$94^{2}=8836$

Question 10

(i) 45

Comparing with a5 where a = 4 

$4^{-1}(a 5)^{2}=a(a+1)$ hundreds +25

$45^{r}=4(4+1)$ hundreds +25

$=20$ hundreds +25

 

$45^{2}=2025$

(ii) 305

Comparing with a5 where a =30

$\left(a_{5}\right)^{2}=a(a+1)$ hundreds +25

$(305)^{2}=30(30+1)$ hundreds +25

 ⇒930 hundred +25

$(305)^{2}$= 93025

(iii) 525 

Comparing with a5 where a = 52

$(a 5)^{2}=a(a+1)$ hundreds +25

$(525)^{2}=52(52+1)$ humdreds +25

⇒ 2756 hundreds +25 

⇒$(525)^{2}=275625$

Question 11 

(i) 8 

Given number = 8 

⇒let us assume $m^{2}-1=8$

⇒$m^{2}=9$

⇒m = 3

Remaining two numbers of pythagorean triplet are

$m^{2}+1,2 m$

$3^{2}+1, 2 \times 3$

10 , 6

The required triplet (6,8,10) with one number 

(ii) 15

Givew number =15

Let us assume $m^{2}-1=15$

$m^{2}=16$

m = 4

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$16+1 \quad ,2 \times 4$

$17 \quad ,  8$

∴ The required triplet (8,15,17) with one number as 15 

(iii) 63 

Givew number 63

Let us assume $m^{2}-1=63$

$m^{2}=64$

$m=8$ 

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$8^{2}+1 \quad 2 \times 8$

$65-16$

∴ We required triplet (16,63,65) with one rumber '63'

(iv) 80 

given number 80 

let us aasume $m^{2}-1=80 \Rightarrow m^{2}=81$

$m=9$

Remainig two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$q^{2}+1,2 \times 9$

82,18

∴ The required triplet (18,80,82) wits one number '80'

Question 12 

$21^{2}=$ 441

$201^{2}=$ 40401

$2001^{2}=$ 4004001

$20001^{2}=$ 40004001

$200001^{2}=$ 4000400001

Question 14

$7^{2}=$ 49

$67^{2}=$ 4489

$667^{2}=$ 444889

$6667^{2}=$ 44448889

$66667^{2}=$ 4444488889

$666667^{2}=$ 444444888889

Exercise 3.3

Question 1

(i) 121 

given number = 121

$121-1=120 ; 120-3=117 ; 117-5=112 ; 112-7=105$

$105-9=96 ; 96-11=85 ; \quad 85-13=72 ; 72-15=57$

$57-17=40 ; \quad 40-19=21 ; \quad 21-21=0$

∴ $121$ is a perfect squars 

we have done '11 Substractions

Hence, square root of 121 is $11 \Rightarrow \sqrt{121}$ = 11

(ii) 55 

given number = 55

$55-1=54 ; 54-3=51 ; 51-5=46 ; 46-7=39$

$39-9=30 ; 30-11=29 ; 29-13=16 ; 16-15=1$

$1-17=-16$

∴ 55 is not a perfect square   

(iii) 36 

Given number  =36

$36-1=35 ; 35-3=32 ; 32-5=27 ; 27-7=20$

$20-9=11 ; \quad 11-11=0$

$\ \quad 36$ is a perbect square

we have done 6 subtractions

Hence, square root of 36 i'e $\sqrt{36}$= 6

iv) 90

Givew number = 90

90-1=89 ; 89-3=86 ; 86-5=81 ; 81-7=74: 74-9=65

65-11=54 ; 54-13=41 ; 41-15=25 ; 25-15=10 ; 10-17=-7

∴ 90 is not a pertect square

 Question 2

(i) 784

Given number = 784 

⇒ $\begin{array}{r|l}2&784\\ \hline 2& 392 \\ \hline 2&196\\ \hline 2& 98\\ \hline 7& 49 \\ \hline 7&7\\ \hline&1\end{array}$

784 = 2 x 2 x 2 x 2 x 7 x7

$\sqrt{784}=\sqrt{2^{2} \times 2 \times 7^{2}}$

$\sqrt{784}=2 \times 2 \times 7=28$

(ii) 441

Given number = 441 

 $\begin{array}{r|l}3&441\\ \hline 3& 147\\ \hline 7&49\\ \hline 7& 7\\  \hline&1\end{array}$

441= 3 x 3x 7 x 7 

$\sqrt{441}=\sqrt{3^{2} \times 7^{2}}$

$\sqrt{441}=3 \times 7=21$

(iii) 1849

Given number = 1849

 $\begin{array}{r|l}43&1849\\ \hline 43& 43\\ \hline&1\end{array}$

1849 = 43 x 43

$\sqrt{1849}=\sqrt{43 \times 43}$

$\sqrt{1849}=43$

(iv) 4356

Givew number = 4356 

 $\begin{array}{r|l}2&4356\\ \hline 2&2178 \\ \hline 3&1089\\ \hline 3& 368\\ \hline 11& 121 \\ \hline 11&11\\ \hline&1\end{array}$

4356 = 2 x 2 x 3 x 3 x 11 x 11

$\sqrt{4356}=\sqrt{2^{2} \times 3^{2} \times 11^{2}}$

$\sqrt{4356}=2 \times 3 \times 11=66$

(v) 6241 

Given number = 6241

$\begin{array}{r|l}79&6241\\ \hline 79& 79\\ \hline&1\end{array}$

 6241 = 79 x 79

$\sqrt{6241}=\sqrt{79^{2}}=79$

(vi) 8836 

Givew number = 8836

 $\begin{array}{r|l}2&8836\\ \hline 2& 4418\\ \hline 47&2209\\ \hline 47&47\\  \hline&1\end{array}$

8836 = 2 x 2 x 47 x 47

$\sqrt{8836}=\sqrt{2^{2} \times 47^{2}}$

$\sqrt{8836}=2 \times 47=94$

(vi)  8281 

Given number = 8281 

 $\begin{array}{r|l}7&8281\\ \hline 7& 1183\\ \hline 13&169\\ \hline 13&13\\  \hline&1\end{array}$

8281 =7 x 7 x 13 x 13 

$\sqrt{8281}=\sqrt{7^{2} \times 13^{2}}$

$\sqrt{8281}=7 \times 13=91$ 

(viii) 9025

 $\begin{array}{r|l}5&9025\\ \hline 5& 1805\\ \hline 19&361\\ \hline 19&19\\  \hline&1\end{array}$

9025 = 5 x 5 x 19 x19

$\sqrt{9025}=\sqrt{5^{2} \times 19^{2}}$

$\sqrt{9025}=95$

 Question 3

(i)  $9 \frac{67}{121}=\frac{1156}{121}$

Sol: 

 ⇒ $\begin{array}{r|l}2&1156\\ \hline 2& 578\\ \hline 17&289\\ \hline 17&17\\  \hline&1\end{array}$

⇒$1156=2 \times 2 \times 17 \times 17$

⇒$9 \frac{67}{121}=\frac{2 \times 2 \times 17 \times 17}{11 \times 11}$

⇒$\sqrt{9 \frac{67}{121}}=$ $\sqrt{\frac{2 \times 2 \times 17 \times 17}{11 \times 11}}$

$=\frac{2 \times 17}{11}$

⇒ $\sqrt{9 \frac{67}{121}}=\frac{34}{11}$

(ii) $17 \frac{13}{36}=\frac{625}{36}$

Sol:

$\begin{array}{r|l}5&625\\ \hline 5& 125\\ \hline 5&25\\ \hline 5&5\\  \hline&1\end{array}$

$625=5 \times 5 \times 5 \times 15$

$17 \frac{13}{36}=\frac{5 \times 5 \times 5 \times 5}{6 \times 6}$

$\sqrt{17 \frac{13}{36}}=\sqrt{\frac{5 \times 5 \times 5 \times 5}{6 \times 6}}$

$=\frac{5 \times 5}{6}$

$\sqrt{17 \frac{13}{36}}$ = $\frac{25}{6}$

(iii) $1.96=\frac{196}{100}$

Sol:

$1.96=\frac{2 \times 2 \times 7 \times 7}{10 \times 10}$

$\begin{array}{r|l}2&196\\ \hline 2& 98\\ \hline 7&49\\ \hline 7&7\\  \hline&1\end{array}$

$196=2 \times 2 \times 7 \times 7$

$1.96=\frac{2 \times 2 \times 7 \times 7}{10 \times 10}$

$\sqrt{1.96}=\frac{2 \times 7}{10}=1.4$

(iv)  0.0064

Sol: 

$0.0064=\frac{64}{10000}$

$\begin{array}{r|l}2&64\\ \hline 2&32 \\ \hline 2&16\\ \hline 2& 8\\ \hline 2& 4 \\ \hline 2&1\\ \hline&1\end{array}$

$0.0064=\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{10 \times 10 \times 10 \times 10}$

$\sqrt{0.0064}=\sqrt{\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{10 \times 10 \times 10 \times 10}}$

$=\frac{2 \times 2 \times 2}{10 \times 10}=0.08$

$\sqrt{0.0064}=0.08$

Question 4

(i) Given number =588

Expressing in prime factors

$\begin{array}{r|l}2&588\\ \hline 2&294 \\ \hline 7&147\\ \hline 3& 21\\ \hline 7& 7 \\ \hline&1\end{array}$

$588=2 \times 2 \times 7 \times 7 \times 3$

Since ' 3 ' left unpaired, so to

make   588 it should multiplied 

by ' 3 '

 

$588 \times 3=2 \times 2 \times 7 \times 7 \times 3 \times 3$

$1764=2^{2} \times 7 \times 3^{2}$

$\sqrt{1764}=\sqrt{2^{2} \times 7^{2}+3^{2}}$

$\sqrt{1764}=2 \times 7 \times 3=42$

(ii) Given number =720

Expressing in prime factors

 $\begin{array}{r|l}2& 720\\ \hline 2& 360 \\ \hline 2&180\\ \hline 2& 90\\ \hline 3& 45 \\ \hline 3&15\\  \hline 5&5\\ \hline&1\end{array}$

$720=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$

Since 5 left unpaired, to make 

720 perfect square it should be

multiplied by 5

$3600=2^{2} \times 3^{2} \times 5^{2} \times 2^{2}$

$\sqrt{3600}= \sqrt{2^{2} \times 2^{2} \times 3^{2} \times 5^{2}}$

$\sqrt{3600}=2 \times 2 \times 3 \times 5=60$

(iii)  Sol:

Given number 2178

Expressing in prime factors

$\begin{array}{r|l}2&2178\\ \hline 3&1089 \\ \hline 3&363\\ \hline 11& 121\\ \hline 11& 11 \\ \hline&1\end{array}$

\

$2178=2 \times 3 \times 3 \times 11 \times 11$

since 2' left unpaired to make 2178  a product square it should be multiplied by 2

$\begin{aligned} 2178 \times 2 &=2 \times 2 \times 3 \times 3 \times 11 \times 11 \\ 4356 &=2^{2} \times 3^{2} \times 11^{2} \end{aligned}$

$\sqrt{4356}=\sqrt{2^{7} \times 3^{2} \times 11^{2}}$

= 2 x 3 x 11

$\sqrt{4356}=66$

(iv) Givew number =3042

Expressing in prime factors 

$\begin{array}{r|l}2&2178\\ \hline 3&1089 \\ \hline 3&363\\ \hline 11& 121\\ \hline 11& 11 \\ \hline&1\end{array}$

$3042=2 \times 3 \times 3 \times 13 \times 13$

 

since '2' left unpaired so to make 3042 a perfect square it should be multiplied by ' 2 '

$3042 \times 2=2 \times 2 \times 3 \times 3 \times 13 \times 13$

$$6084=2^{2} \times 3^{2} \times 13^{2}$

$\sqrt{6084}=2 \times 3 \times 13=78$

(v) 6300

Given number =6300

Expressiny in prime factors

$\begin{array}{r|l}2 & 6300 \\ \hline 2 & 3150 \\ \hline 5 & 1575 \\ \hline 5 & 315 \\ \hline 7 & 63 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline&1\end{array}$

$6300=2 \times 2 \times 5 \times 5 \times 7 \times 3 \times 3$ 

 since '7' left unparied , so to make 6300 a perfect square , it should be multiplied by '7' 

$6300 \times 7=2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 3 \times 3$

$\sqrt{44100}=\sqrt{2^{2} \times 5^{2} \times 7^{2} \times 3^{2}}$

$\sqrt{44100}=2 \times 5 \times 7 \times 3$

$\sqrt{44100}=210$

Question 5

(i)   Given number 1872 expressing in prime factors 

$\begin{array}{r|l}2 &1872 \\ \hline 2 & 936 \\ \hline 2 & 468 \\ \hline 2&234 \\ \hline 3 & 117 \\\hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1\end{array}$ 

 Since 13 left unpaired, so to make 1872 a perfect square it should be divided by 13

$\frac{1872}{13}=\frac{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 13}{13}$

$144=2 \times 2 \times 2 \times 2 \times 3 \times 3$

$\sqrt{144}=\sqrt{2^{2} \times 2^{2} \times 3^{2}}$

$\sqrt{144}=2 \times 2 \times 3=12$

(ii)  Given number 2592  expressing in prime number

$\begin{array}{r|l}2 &2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2&324 \\ \hline 2 & 162 \\ \hline 2 & 81 \\ \hline 3 & 27 \\ \hline3 & 9 \\ \hline3 & 3 \\ \hline & 1\end{array}$ 

2592 =  $2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

Since ' 2 ' lets unpaired, so to

make 2592 a prefect square, it

Should be divided ' 2 '

$\frac{2592}{2}=\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3}{2}$

sqrt{2^{2}

$1296=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$1

$\sqrt{1296}=\sqrt{2^{2} \times 2^{2} \times 3^{2} \times 3^{2}}$

$\sqrt{1296}=36$

(iii)  Given number 3380 expressing interms of prime factors 

$\begin{array}{r|l}2&3380\\ \hline 2&1690 \\ \hline 5&845\\ \hline 13& 169\\ \hline 13& 13 \\ \hline&1\end{array}$

since '5' is left unpaired so to make a perfect square it should be divided by 5 

$3380 \div 5=\frac{2 \times 2 \times 5 \times 13 \times 13}{5}$

$676=2 \times 2 \times 13 \times 13$

$\sqrt{676}=\sqrt{2 \times 2 \times 13 \times 13}$

$\sqrt{676}=2 \times 13 \in 26$

(iv) 16244

Expressing in terms of prime factors

16244

$\begin{array}{r|l}2 &16244 \\ \hline 2 & 8122 \\ \hline11 & 4061 \\ \hline 3&371 \\ \hline 3 & 117 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1\end{array}$ 

Question 8 

Sol: let breadth of rectangle = x .m 

given 

        length of rectangle is equal to 4 times the breadth 

∴ L = 4 . x 

area of rectangle = 1936 

l x b = 1936 

$\begin{aligned} 4 x \times x &=1936 \\ x^{2} &=\frac{1936}{4} \\ x^{r} &=484 \\ x &=22 \mathrm{~m} \end{aligned}$

hence breadth of rectangle = x = 22m 

length of rectangle $=4 x=4 \times 22=88 \mathrm{~m}$

Question 9

Sol :

let the no.of columns  = x

givew no.of rows equal to no of column 

ஃ no of rows = x

Total students equal to 2000 but 64 studers

could not be accomndated in These rows columns

$\begin{aligned} \therefore \quad x \times x &=2000-64 \\ x^{2} &=1936 \\ x &=\sqrt{1936} \\ x &=44 \end{aligned}$

hence , no of rows = 44

$\begin{array}{r|l}2 &1936 \\ \hline 2 & 868 \\ \hline2 &434 \\ \hline 2&242 \\ \hline 11 & 121 \\ \hline 11 & 11\\ \hline & 1\end{array}$

$1936=2 \times 2 \times 2 \times 2 \times 11 \times 11$

$\sqrt{1936}=2 \times 2 \times 11=44$

(v) Given number 61347 

expressing interms of prime factors 

$\begin{array}{r|l}3&61347\\ \hline 11&20449 \\ \hline 11&1859\\ \hline 13& 169\\ \hline 13& 13 \\ \hline&1\end{array}$

61347 = $=3 \times 11 \times 11 \times 13 \times 13$

since 3 left unpaired so to make 61347 a perfect square it should be divided by 3

$61347 \div 3=\frac{3 \times 11 \times 11 \times 13 \times 13}{3}$

$20449=11 \times 11 \times 13 \times 13$

$\sqrt{20449}=\sqrt{11^{2} \times 13^{2}}$

$\sqrt{20449}=11 \times 13={143}$

(6) let no of rows of plants in garden = x 

given each row contains as many plants as the no of rows 

no of plants in each row = x 

total no of plants = $x \times x=x^{2}$

given Total no.ot plants in garden =4225

$\begin{aligned} \therefore \quad x^{2} &=4225 \\ x &=\sqrt{4225} \\ & x=65 \end{aligned}$

hence, no.of rows in garden = 65

no.ot plants in a row $=65$

Question 10 

 Let no of students = x 

given contribution of such student = no of students 

∴ contribution of each students = $\bar{₹} x$

Total collected for Pcnic =22304

∴ $x \times x=2304$

$x^{2}=2304$

$x=\sqrt{2304}$

$x=48$

Question 11

Let number is 15 times the other 

second number = 15.x

product of two number = 7260 

$\begin{aligned} 15 x \cdot x &=7260 \\ x^{2} &=\frac{7260}{15} \\ x^{2} &=484 \\ x &=\sqrt{484} \\ x &=22 \end{aligned}$

$\begin{array}{r|l}2&484\\ \hline 2&242 \\ \hline 11&121\\ \hline 11& 11\\ \hline&1\end{array}$

$484=2 \times 2 \times 11 \times 11$

$\sqrt{484}=2 \times 11=22$

Question 12 

Sol : Given number are in ratio of 2 : 3 : 5 

let number be 2x , 3x , 5x 

given sum of squares of numbers $=950$

$(2 x)^{2}+(3 x)^{2}+(5 x)^{2}=950$

$4 x^{2}+9 x^{2}+25 x^{2}=950$

$x^{2}=\frac{950}{38}$

$x^{2}=25$

$x=5$

$\begin{aligned} \text { Hence, } & \text { Numbers are } 2 x, 3 x, 5 x \\ & 10,15,25 \end{aligned}$

 

Question 13

Sol:

Perimeters of two Squares $=60 \mathrm{~m}, 144 \mathrm{~m}$

$P_{1}=60 \mathrm{~m} ; P_{2}=144 \mathrm{~m}$

Perimeter let length of sides of square are = $x_{1}, x_{2}$

$\begin{aligned} \therefore P_{1} &=4 x_{1} \\ 60 &=4 x_{1} \\ x_{1} &=15 \mathrm{~m} \end{aligned}$

$P_{2}=4 x_{2}$

$144=4 x_{2}$

$x_{2}=36 m$

$\begin{aligned} \text { Areas } A_{1} &=x_{1}^{2} \\ A_{1} &=15^{2} \\ A_{1} &=225 \mathrm{~m}^{2} \end{aligned}$

Area $\begin{aligned} A_{2} &=x_{2}^{2} \\ A_{2} &=36^{2} \\ A_{2} &=1296 \mathrm{~m}^{2} \end{aligned}$

let a square of side x with area ' $A_{1}+A_{2}$'

$A=A_{1}+A_{2}$

$x^{2}=225+1296$

$x^{2}=1521$

$\begin{array}{r|l}3&1521\\ \hline 3&507 \\ \hline 13&169\\ \hline 13& 13\\ \hline&1\end{array}$

$\begin{aligned} 1521 &=3 \times 3 \times 13 \times 13 \\ \sqrt{1521} &=3 \times 13=39 \end{aligned}$

$x^{2}=1521$

$x_{2} \sqrt{1521}$

$x=39 \mathrm{~m}$

Perimeter oF square 

$\begin{aligned} P &=4 x \\ &=4 \times 39 \\ P &=156 \mathrm{~m} \end{aligned}$

∴ Hence, perimeter =156 m

EXERCISE 3.4

Question 1

(i) Given number 2401

(diagram should be added  )

$\therefore \sqrt{2401}=49$

(ii) 4489

(diagram should be added  )

$\therefore \sqrt{4489}=67$

(iii)  106929

(diagram should be added  )

$\therefore \sqrt{106929}=327$

(iv) given number 167281

(diagram should be added)

$\therefore \sqrt{167281}=409$

v) givew number 53824

$\therefore \sqrt{213444}=462$

$\therefore \sqrt{53824}=232$

vi) given number 213444

$\therefore \sqrt{213444}=462$

$\therefore \sqrt{213444}=462$

Question 2

i) Given number 81=2 (even)

         ⇒   The number of digits in its square root = $\frac{2}{2}=1$

ii) Given number $169=3($ odd $)$

⇒The number of digits in its square root $=\frac{3+1}{2}=2$

iii) Given number $4761=4$ (even)

⇒the number of digits in its square root $=\frac{4}{2}=2$

iv) Given number $27889=5$ (odd)

⇒ The number of digits in its square root $=\frac{5+1}{2}=3$

V) Givew number $525625=6($ even $)$

∴ The  number of digits in its square root $\frac{6}{2}=3$

Question 3

Sol :

(i) Given number 51.84

(diagram should be added  )

$\therefore \sqrt{51.84}=7.2$

(ii) 42.25

(diagram should be added  )

$\sqrt{42.25}=6.5

(iii)  Given number 18.4041

(diagram should be added  )

$\sqrt{18.4041}=4.29$

(iv) Given number 5.774409 

(diagram should be added  )

$\therefore \sqrt{5.774409}=2.403$

Question 4

(i) 645.8 

(diagram should be added  )

$\therefore \sqrt{645.8}=25.412 \approx 25.41$ (correct to 2 decimals )

(ii) 107.45

(diagram should be added  )

$\sqrt{107 \cdot 45}=10.365 \approx 10.36$

(iii) Given number 5.462

(diagram should be added  )

$\therefore \sqrt{5.462}=2.337 \approx 2.34$ (corrected to '2 'decimals)

(iv) Given number 2 

(diagram should be added  )

$\sqrt{2}=1.414 \approx 1.41$

(v)  Given number 3 

(diagram should be added  )

$\sqrt{3}=1.732 \approx 1.73$ (Corrected to 2 decimals $)$

Question 5

(i) $\frac{841}{1521}$

(diagram should be added  )

$=\frac{\sqrt{841}}{\sqrt{1521}}=\frac{29}{39}$

(ii)  $8 \frac{257}{529}=\frac{4489}{529}$

(diagram should be added  )

$\sqrt{8 \frac{257}{529}}=\sqrt{\frac{4489}{529}}$

$\sqrt{8 \frac{257}{529}}=\frac{67}{23}$

(iii) $16 \frac{169}{441}=\frac{7225}{441}$

(diagram should be added  )

$\sqrt{16 \frac{169}{441}}=\frac{\sqrt{7225}}{\sqrt{441}}=\frac{85}{21}$

Question 6

(i) Given nunmber 2000

⇒(diagram should be added  )

⇒ Hence , the least number that must be subtracted from 

2000 so as to make it a perfect square is 64 

∴ Requred perfect square numbers =2000 - 64

= $1936=44^{2}$

(ii) Given number 984

⇒(diagram should be added  )

⇒ Hence , the least number that must be subtracted 

  

from 984 so as to make it a perfect square is 23 

∴ Requred perfect square numbers = 984 - 23 = 961 = $=31^{2}$

iii) Givew number 8934

⇒(diagram should be added  )

⇒ Hence, the least number that must be subtracted

from 8934 so as to make it a perfect square in 98

∴ The required Square number 8934-98 = 8836=$94^{2}$

iv) Givew number 11021

⇒(diagram should be added  )

⇒ Hence , the least number that must be subtracted 

from 11021 so as to make it a perfect square is 205 

 The required square number 11021 - 205 = 10816 = $104^{2}$

Question 7 

(i) Given number 1750 

⇒(diagram should be added  )

⇒ 1750\rangle$(41)^{2} \Rightarrow$ Remainder $=69$

⇒$(42)^{2}=1764$

⇒ $\therefore$ Required number =1764-1750=14

⇒ Hence, the least number that must be added to 1750

So as to make it a perfect square is 14

(ii) Givew number 6412

⇒(diagram should be added  )

⇒$6412>(80)^{2}$

=$81^{2}=6561$

⇒ $\therefore$ Required number $=6561-6412=149$

⇒ Hence, the least number That must be added to 6412

So as to make it a perfect square is 149

(iii)  givew number 6598

⇒(diagram should be added  )

⇒ $6598>(81)^{2}$

=$(82)^{2}=6724$

$\therefore$ Required number $=6(82)^{2}-6598=126$

⇒ hence , the minimum number that must be added to 6598 so as to make it a perfect square is 126

(iv) Givew number 8000

⇒(diagram should be added  )

 

⇒ $8000>89^{2}$

⇒$90^{2}=8100$

⇒ $\therefore$ Required number $=90^{2}-8000=100$

⇒ hence , the minimum number that must be added to 

 

8000 so as to make it a perfect square is 100

Question 8 

Smallest four digit number =  1000 

⇒(diagram should be added  )

⇒ $1000>31^{2}$

⇒ $32^{2}$ will be next perfect square

⇒ $32^{2}=1024$

⇒ Hence , 1024 is smallest four digit number which perfect square

Question 9 

Greatest six digit number = 999999 

⇒ (diagram should be added  )

⇒ To make 999999 a perfect square , we have to subtract 1998 from 999999

⇒ The required number = 998001 

⇒ hence , 998001 is greatest six digit number which is a perfect square 

Question 10 

(i) AB = 14 cm 

Bc = 48 cm

 

according to pythagorus theorem 

⇒ $A C^{2}=A B^{2}+B C^{2}$

⇒$14^{2}+48^{2}$

⇒ $A C^{2}=2500$

⇒ $A C=\sqrt{2500}$

⇒ $A C=50 \mathrm{~cm}$

(ii) $A C=37 \mathrm{~cm}, B C=35 \mathrm{Cm}, A B=?$

 

⇒ According to pythagorus theorem

⇒ $A C^{2}=A B^{2}+B C^{2}$

⇒ $37^{2}=A B^{2}+35^{2}$

⇒ $1369=A B^{2}+1225$

⇒ $A B^{2}=144$

⇒ $A B=12 \mathrm{~cm}$

Question 11 

Total plants = 1400 

let no . of rows = x 

no. of columns = x 

⇒ (diagram should be added  )

$x^{2}=1400$

$1400>(37)^{2}$

$38^{2}=1444$

So To make 1400 a perfect square, we have add

minimum of 44

$\therefore 44$ plants needed more.

Question 12

⇒ Total no of students = 1000 

⇒ let no of row = no of columns = x

⇒ Total students rows x columns = 1000

⇒ (diagram should be added  )

⇒ $x \times x=1000$

$x^{2}=1000$

$x=\sqrt{1000}$

So Remainder $=39$

⇒ hence 39 children will be left out 

Question 13

⇒  (diagram should be added  )

⇒ Distance that amit walk while retuning 

⇒ AC

 In $\triangle A B C$

⇒ According to pythagorus theorem

⇒ $A C=A B^{2}+B C^{2}$

⇒$A C^{2}=16^{2}+63^{2}$

⇒$A C^{2}=4225$

⇒$A C=65 m$

ஃ Hence amit walks 65c while returing to his house

Question 14

Sol: (diagram should be added  )

⇒ Length of  ladder = 6m 

height of wall = 4.8m

In $\triangle A B C$

According Pythagorus tbeorem

⇒ $A C^{2}=A B^{n}+B C^{2}$

$B^{2}=4 \cdot 8^{2}+B C^{2}$

$B C^{N}=12.96$

$B C=\sqrt{12.96}$

$B C=3.6 \mathrm{~m}$

⇒ Hence, Distance between wall ond foot of ladder

is 3.6 m

⇒

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