(i) Required number of students = 8 + 32 = 40
(ii) Here, We notice that classes are continuous but class-size is not the same for all the classes. We notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of rectangle (adjusted frequency) for each class.
Length of rectangle (adjusted frequency) =\(\frac { Frequency\ of\ Class }{ Width\ of\ class } \times Minimum\ class-size\)

Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies as heights.
Histogram representing marks obtained by students in unit test of Mathematics.

(iii) Hardwork and Dilligence.

Required histogram is as follows:

(i)

(ii) Reproduce health conditions is the major cause of women's ill health and death worldwide.
(ii) Lack of proper diet, lack of advised exercises.

Number of observations (n) = 10, which is even,
\(\therefore\) Median
\(=\frac { { \left( \frac { n }{ 2 } \right) }^{ th }observation+{ \left( \frac { n }{ 2 } +1 \right) }^{ th }observation }{ 2 } \)
\(\Rightarrow 65=\frac { 5^{ th }observation+6^{ th }observation }{ 2 } \)
\(\Rightarrow\)   \(65 = \frac {x+\left(x+2\right)}{2}\)
\(\Rightarrow\)  2x + 2 = 130
\(\Rightarrow\)  2x = 130 - 2 = 128 
\(\Rightarrow\)  \(x = \frac {128}{2}=64\)

According to the question,
\(\frac {21+16+21+x+29+15}{6}=23\)
\(\Rightarrow\)  105 + x = 138
\(\Rightarrow\)    x = 33

According to the question,
\(\frac {10+12+18+13+x+17}{6}=15\) 
\(\Rightarrow\)  x + 70 = 90
\(\Rightarrow\)   x = 20

We shall first make the class intervals continuous. Then, the modified table is as follows: