(i) 36 - 1 = 35
(ii) 35 - 3 = 32
(iii) 32 - 5 = 27
(iv) 27 - 7 = 20
(v) 20-9 = 11
(vi) 11 - 11 = 0
Since from 36 we subtracted successive odd numbers starting from 1 and obtained 0 at the 6th step, therefore \(\sqrt { 36 } \) = 6

Number of above average students=\({1\over6}\) of the class student
Number of average students= \({1\over4}\)of the class student
\(\therefore\) Number of below average students  \(=1-[{1\over6}+{1\over4}]\) of the class student
\(=-[{2+3\over12}]=1-{5\over12}={7\over12}\)of the class student
Since, number of students in the class = 48
\(\therefore\)Number of below average students
\(={7\over12}\times48=28\)
So, Number of below average students= 28

Converting \(\frac { 1 }{ 4 } \) and \(\frac { 1 }{ 2 } \) rational numbers with the same denominations , we have
\(\frac { 1 }{ 4 } \times \frac { 8 }{ 8 } =\frac { 8 }{ 32 } \) and \(\frac { 1 }{ 2 } \times \frac { 16 }{ 16 } =\frac { 16 }{ 32 } \)
Five rational numbers between \({8\over32}=({1\over4})and{16\over32}=({1\over2})are{9\over32},{10\over32},{11\over32},{12\over32},{13\over32}\)

Let the number of currency notes of denominations
Rs 100, Rs. 50 and Rs. 10 be 2x, 3x and 5x, respectively.
The amount she has from Rs.100 notes
= 2x x 100 = 200x
The amount she has from Rs. 50 notes = 3x x 50 = 150x
The amount she has from Rs. 10 notes = 5x x 10 = 50x
According to the question,
Total amount = 400000
\(\Rightarrow\) 200x + 150x + 50x = 400000
\(\Rightarrow\) 400x = 400000
\(\Rightarrow\) \(x=\frac { 400000 }{ 400 } =1000\)
\(\therefore\) Number of Rs.100 notes = 2x = 2 x 1000 = 2000
N umber of Rs. 50 notes = 3x = 3 x 1000 = 3000
and number of Rs. 10 notes = 5x = 5 x 1000 = 5000
Hence, she has 2000, 3000 and 5000 notes of denominations Rs. 100, Rs.  50 and Rs. 10, respectively

Let the three consecutive integers be x, (x + 1)and (x + 2).
According to the question,
Sum of consecutive integers = 51
\(\Rightarrow\) x + (x + 1)+ (x + 2) = 51
\(\Rightarrow\) x + x + l + x + 2 = 51
\(\Rightarrow\) 3x + 3 = 51 \(\Rightarrow\) 3x = 51 - 3  [transposing 3 to RHS]
\(\Rightarrow\)  3x = 48 \(\Rightarrow\) x = \(\frac { 48 }{ 3 } =16\) [dividing both sides by 3]
\(\therefore\) First integer = x = 16
second integer = x + 1= 16 + 1= 17
and third integer = x + 2 = 16 + 2 = 18
Hence, the required integers are 16, 17 and 18.

Total number of students = 6250
Number of students forming a square = 6250 - 9
= 6241
Thus, 6241 students form a big square which has number of rows equal to the number of students in each row.
Let the number of students in each row be x, then the number of rows will be x.
Therefore, x x x = 6241 \(\Rightarrow\) x = \(\sqrt { 6241 } \) = 79
Hence, there are 79 students in each row of the square formed.
The value depicted here is that exercise is essential for good health. It makes our body fit and strong.

Volume of the cubical tank
=91125 m³
Let the length of each side of cubical tank be x m.
We know that, Volume of a cube
= (Side)³
SO, (x)³ =91125
x = \(\sqrt [ 3 ]{ 91125 } \)
x = \(\sqrt [ 3 ]{ \begin{matrix} \underline { 3\times 3\times 3 } \times \underline { 3\times 3\times 3 } \times \underline { 5\times 5\times 5 } \\ x=3\times 3\times 5 \end{matrix} } \)
x = 3 x 3 x 5
x = 45 m
Hence, the height of the tank will be 45 m.

(b) The value depicted here is that Aditya is conscious environment. He saves water by reserving rain water.

\(41\over12\)

Let the number of Rs. 5 coins be x.
Then, the number of Rs. 2 coins = 3x
The total number of coins is 160 .
The number of coins of Rs. 1= 160 - (x + 3x)
= (160 - 4x)
The amount that I have from Rs.  5 coins = 5 x x = 5x
The amount that I have from Rs.  2 coins = 2 x 3x = 6x
The amount that I have from Rs.  1 coins
= 1 x (160 - 4x) = 160 - 4x
According to the question,
Total amount = 300
\(\Rightarrow\) 5x + 6x + (160 - 4x) = 300
\(\Rightarrow\) 5x +6x +160 - 4x = 300
\(\Rightarrow\) 7x + 160 = 300   [transposing 160 to RHS]
\(\Rightarrow\) 7x = 300 -160  [transposing 160 to RHS]
\(\Rightarrow\) 7x = 140
\(\Rightarrow\) x = \(\frac { 140 }{ 7 } \) = 20   [dividing both sides by 7]
Number of Rs. 5 coins = x = 20
Number of Rs. 2 coins = 3x = 3 x 20 = 60
and number of Rs.1 coins = 160 - 4x = 160 - 4 x 20
= 160 - 80 = SO
Hence, I have SO,60 and 20 coins of denomination Rs.1, Rs. 2 and Rs 5, respectively.

We find the mean of the given rational numbers.
As given in the above example, the mean is \(\frac{3}{8} \text { and } \frac{1}{4}<\frac{3}{8}<\frac{1}{2} \text {. }\)
We now find another rational number between \(\frac{1}{4} \text { and } \frac{3}{8}\) For this, we again find the mean of \(\frac{1}{4} \text { and } \frac{3}{8}\). That is, \(\left(\frac{1}{4}+\frac{3}{8}\right) \div 2=\frac{5}{8} \times \frac{1}{2}=\frac{5}{16}\)
Now find the mean of \(\frac{3}{8} \text { and } \frac{1}{2}\). We have, \(\left(\frac{3}{8}+\frac{1}{2}\right) \div 2=\frac{7}{8} \times \frac{1}{2}=\frac{7}{16}\) 
Thus we get \(\frac{1}{4}<\frac{5}{16}<\frac{3}{8}<\frac{7}{16}<\frac{1}{2}\)
Thus, \(\frac{5}{16}, \frac{3}{8}, \frac{7}{16}\) are the three rational numbers between \(\frac{1}{4} \text { and } \frac{1}{2}\)

Assume the length of the rectangle to be x cm.
The perimeter of the rectangle = 2 x (length + width)
\( =2 \times\left(x+2 \frac{3}{4}\right) \)
\(=2\left(x+\frac{11}{4}\right) \)
The perimeter is given to be 13 cm. Therefore,
\(2\left(x+\frac{11}{4}\right)=13\)
\( x+\frac{11}{4} =\frac{13}{2} \)
\(x =\frac{13}{2}-\frac{11}{4} \)
\(=\frac{26}{4}-\frac{11}{4}=\frac{15}{4}=3 \frac{3}{4}\)

We have, 128
Resolving 128 into prime factors, we get
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
The prime factor 3 does not appear in a group of three. So, 128 is not a perfect cube.
Thus, we will divide the number by 3.
∴ 128 ÷ 2 =64 = 2 x 2 x 2 x 2 x 2 x 2
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

We have, 2700
Resolving 2700 into prime factors,
we get
2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5
Clearly, the prime factors of 2 and 5 do not appear in groups of three (triples).
So, 2700 is not a perfect cube.

Given, two consecutive numbers are 12 and 13.
Here, n = 12 and n + 1= 13
We know that, total numbers lie between n² and (n+1)² = 2n
\(\therefore\) The total numbers lie between squares of 12 and 13
= 2 x 12 = 24
Hence, the total numbers lie between square of 12 and 13 is 24.