Frequency distribution table:
| Class Interval | Frequency | Cumulative Frequency (less than type) |
| 0–10 | 10 | 10 |
| 10–20 | \(x\) | 10 + \(x\) |
| 20–30 | 25 | 35 + \(x\) |
| 30–40 | 30 | 65 + \(x\) |
| 40–50 | \(y\) | 65 + \(x + y\) |
| 50–60 | 10 | 75 + \(x + y = n\) |
Given: \(n = 100\)
From the condition:
\(75 + x + y = 100\)
⇒ \(x + y = 25\) ----(i)
Again, since the **median = 32**,
the median class is (30–40)
To find median:
\[
\text{Median} = l + \left[\frac{n}{2} - cf}{f} \right] \times h
\]
Where,
\(l = 30\),
\(n = 100\),
\(cf = 35 + x\),
\(f = 30\),
\(h = 10\)
So,
\[
\text{Median} = 30 + \left[\frac{50 - (35 + x)}{30} \right] \times 10
= 30 + \frac{15 - x}{3}
\]
Given median = 32
So,
\[
30 + \frac{15 - x}{3} = 32
\Rightarrow \frac{15 - x}{3} = 2
\Rightarrow 15 - x = 6
\Rightarrow x = 9
\]
Substitute into (i):
\(9 + y = 25\)
⇒ \(y = 16\)
∴ Required values are \(x = 9\), \(y = 16\)
(Answer)