Let the radius of the base and height of the right circular cylinder be r units and h units respectively.
Then,
Volume of the cylinder = πr²h cubic units
Curved surface area of the cylinder = 2πrh square units
If volume = curved surface area, then πr²h = 2πrh ⇒ r = 2
Hence, it is possible only when radius of the base is 2 units.

Let the side of the square be a units.
Then, area of the square = a² square units
According to the question,
a² < 6a
⇒ a² - 6a < 0
⇒ a (a-6)<0
⇒ a - 6<0  |∴ a>0
⇒ a < 6
⇒ a = 1,2,3,4,5
Hence, the sides of the square may be 1, 2, 3, 4, or 5 units.

Let the number of units of the side of the square be x.
Then,
area of the square = x² square units
perimeter of the square = 4x units
According to the question,
x² = 4x
⇒ x²  -4x = 0
⇒ x(x - 4) = 0
⇒ x = 0, 4
x = 0 is inadmissible
∴ x =  4
Hence, the number of units of the side of the square is 4.

Let the number of bees in the swarm be x. Then,
Number of bees settled on a blossom of Kadamba = \(\frac{x}{5}\).
Number of bees settled on a flower of Silindhiri = \(\frac{x}{3}\).
Number of bees flew to the bloom of Kutaja = 3(\(\frac{x}{3}\)-\(\frac{x}{5}\))
∴Number of bees left in the swarm = x-{\(\frac{x}{5}\)+\(\frac{x}{3}\)+3(\(\frac{x}{3}\)-\(\frac{x}{5}\))}
=x-(\(\frac{x}{5}\)+\(\frac{x}{3}+x-​​\frac{3x}{5}\))
=x-\(\frac{3x+5x+15x-9x}{15}\)
=\(x-\frac{14x}{15}=\frac{15x-14x}{15}=\frac{x}{15}\)
According to the question,
\(\frac{x}{15}=10\)
⇒ x = 10 x 15  I Multiplying both sides by 15
⇒ x = 150
Hence, the number ofbees in the swarm was 150.
Check
\(\frac{150}{5}=30\)
\(\frac{150}{3}=50\)
3(50 - 30) = 3(20) = 60
30 + 50 + 60 = 140
150 - 140 = 10
Hence, the result is verified.

(i) Let the awards given in 1958 be x% of the awards given in 1993. Then,
x% of 81=28
⇒ \(\frac{x}{100} \times 81=28\)
⇒ \(x=\frac{28 \times 100}{81}\)
= 34.5 (approx.)
Hence, the awards given is 1958 is 34.5 per cent of the awards given in 1993.
(h) Let the awards given in 1993 be x% of the awards given in 1958. Then,
x% of 28 = 81
⇒ \(\frac{x}{100} \times 28=81\)
⇒\(x=\frac{81 \times 100}{28}=\frac{81 \times 25}{7}\)
⇒ \(\frac{2025}{7}\)=289 (approx.)
Hence, the award given is 1993 is 289 per cent of the awards given in 1958.

Let the volume of water required to make 221 cm³ of ice be V cm³. Then,
\(V+\frac{4}{100}V=221\)
\(\Rightarrow V+\frac{V}{25}=221\)
\(\Rightarrow \frac{26V}{25}=221\)
\(\Rightarrow V=\frac{221 \times 25}{26}\)
\(\Rightarrow V=\frac{17 \times 25}{2}\)
\(\Rightarrow V=\frac{425}{2}\)
\(\Rightarrow V=221\frac{1}{2} cm^{3}\)
Hence, the required volume of water is \(221\frac{1}{2} cm^{3}\).

(i) a² + b² = 3² + 4² = 9 + 16 = 25 = 5² = c²
(ii) a² + b² = 5² + 12² = 25 + 144 = 169 = 13² = c²
More such triplets
(i) For, m = 4, n = 3, we get
a = m² - n² = 4² - 3² = 16 - 9 = 7
b = 2mn = 2(4) (3) = 24
c = m² + n² = 4² + 3² = 16 + 9 = 25
We find, a²+ b² = 7² + 24²
= 49 + 576 = 625
= 252 = c²
(ii) For, m = 5, n = 4, we get
a = m² - n² = 5² - 4² = 25 - 16 = 9
b = 2mn. =2(5)(4)=40
c = m² + n² = 5²+ 4²= 25 + 16 = 41
We find, a² + b² = 9² + 40²
= 81 + 1600 = 1681
= 412 = c².

Let (10x + y) be a 2-digit number, it is given that when we subtract the number formed by reversing its digits, the result so obtained is a perfect cube.
The number formed by reversing its digits is (10y + x).
Difference = (10x + y) - (10y + x)
= 10x + y -10 y - x = 10x - x + y -10 y
= 9x - 9y = 9(x - y)
It is given that, 9(x - y) is a perfect cube.
Since, 9(x - y) = 3 x 3(x - y)
If x - y = 3
Then,9(x-y) = 3 x 3 x 3 =(3)³
Hence, all such pairs whose difference of digits is 3, satisfy this condition.
Here, we take x > y
If x = 9 and y = 6, then x - y = 9 - 6 = 3
The number obtained = 9 x 10 + 6 = 90 + 6 = 96
If x = 8 and y = 5, then x - y = 8 - 5 = 3
The number obtained = 8 x 10 + 5 = 80 + 5 = 85
If x = 7 and y = 4, then x - y = 7 - 4 = 3
The number obtained = 7 x 10 + 4 = 70 + 4 = 74
If x = 6 and y = 3, then x - y = 6 - 3 = 3
The number obtained = 6 x 10 + 3 = 60 + 3 = 63
If x = 5 and y = 2, then x - y = 5 - 2 = 3
The number obtained = 5 x 10 + 2 = 50 + 2 = 52
If x = 4 and y = 1,then x - y = 4 -1 = 3
The number obtained = 4 x 10 +1= 41
If x = 3 and y = 0, then x - y = 3 - 0 = 3
The number obtained = 3 x 10 + 0 = 30
The numbers which satisfy the given condition are 96,85,74,63,52,41,30

\(\begin{matrix} \quad \quad A\ B \\ +\quad 3\ 7 \\ \_ \_ \_ \_ \_ \_ \_\_ \\ \quad 6\ A \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found.
Studying the addition in one's column, we have B + 7 and we get A from this, i.e. a number whose one's digit is A.
Also, from the addition of ten's columns, we have A +3 and we get 6 from this. Therefore, A must be 0, 1, 2 and 3
If A = 0, then puzzle becomes
\(\begin{matrix} \quad \quad 0\quad B \\ +\quad 3\quad 7 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 6\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
It is not possible because first number AB becomes B 
i.e. one digit number.
If A = 1, then puzzle becomes
\(\begin{matrix} \quad \quad 1\quad B \\ +\quad 3\quad 7 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 6\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Then, B + 7 gives 1, so B must be 4 and sum in ten's column is
1+ 1 + 3 = 5 \(\neq\) 6 , so it is not possible.
If A = 2, then puzzle becomes
\(\begin{matrix} \quad \quad 2\quad B \\ +\quad 3\quad 7 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 6\quad 2 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Then, B + 7 gives 2, so B must be 5 and sum in ten's column is 1+ 2 + 3 = 6.
So, it is correct.
Therefore, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 2\quad 5 \\ +\quad 3\quad 7 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 6\quad 2 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 2 and B = 5

\(\begin{matrix} \quad 1\ A \\ \times \ \ \ A \\ \_ \_ \_ \_ \_ \_ \_ \\ \quad 9\ A \\ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have one letter A whose value is to be found. 
Since, the one's digit of A x A = A, so it must be 1, 5 or 6.
When A = 1, then
\(\begin{matrix} \quad 1\quad 1 \\ \times \ \ \ \ \ \ \ 1 \\ \_ \_ \_ \_ \_ \_ \_ \\ \quad 1 \quad1 \\\_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
But the product is 9A, so A = 1 is not possible.
When A = 5, then
\(\begin{matrix}\quad 1\quad 5 \\ \times \quad \quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 7\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
But the product is 9A, so A = 5 is not possible.
When A = 6, then
\(\begin{matrix} \quad 1\quad 6 \\ \times \quad \quad 6 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 9\quad 6 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A =6