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 \(221\frac{1}{2} cm^{3}\).

0, 2, 4, 6, and 8

x = 0 or 3 or 6 or 9

(i) 10 x 34 + 5
(ii) 10 x 27 + 5

(a) Total amount received as pocket money is given by 2x + 3x + 5x + 7x
\(\Rightarrow\)  (2+ 3+ 5+ 7)x = 680
\(\Rightarrow\) 17x = 680 \(\Rightarrow\) x = 40
Hence, each of them receives amount Rs 80, Rs120, Rs 200, and Rs 280 as pocket money, respectively.
(b) The boys are very sensible and know the importance of saving money. They have interest in reading inspirational books like autobiographies of great personalities.

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

A = 8 and B = 9

A = 8 B = 1 and C = 3

\(\begin{matrix} \quad \quad A\ B \\ +\quad 3\ 7 \\ \_ \_ \_ \_ \_ \_ \_\_ \\ \quad 6\ A \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
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

We have, if 
    2 B
+  A B  
    8 A   
In column II, we see that sum of 2 and A is equal to 8, A cannot be a 2-digit number. Clearly, A =6.
\(\because\)  2 + 6 = 8
Now, we study the column I.
Since, B + B = 6, \(\Rightarrow\) 2B = 6 \(\Rightarrow\) B =3
Hence, A = 6 and B =3.
\(\therefore\)    23
+  63   
    86