If the median of the following frequency table is 28.5, find the values of $x$ and $y$ where the sum of frequencies is 80. 
Step 1: Given that total frequency = 80.
\[ 5 + x + 20 + 15 + y + 5 = 80 \Rightarrow x + y = 35 \quad \text{...(1)} \] Step 2: Median formula.
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - c.f.}{f}\right) \times h \] Step 3: Find median class.
Median = 28.5, $\Rightarrow$ lies in 20–30 class. So, \[ l = 20, \; h = 10, \; f = 20, \; N = 80, \; \frac{N}{2} = 40 \] Cumulative frequency before median class: $5 + x = (5 + x)$.
Step 4: Substitute values.
\[ 28.5 = 20 + \left(\frac{40 - (5 + x)}{20}\right) \times 10 \]
Step 5: Simplify.
\[ 28.5 - 20 = \frac{(35 - x)}{2} \] \[ 8.5 = \frac{35 - x}{2} \Rightarrow 17 = 35 - x \Rightarrow x = 18 \]
Step 6: Find $y$.
From (1): $x + y = 35 \Rightarrow 18 + y = 35 \Rightarrow y = 17$.
Step 7: Conclusion.
\[ \boxed{x = 18, \; y = 17} \]
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be:
The mean of the following table will be:
| Class-interval | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 |
|---|---|---|---|---|---|
| Frequency (f) | 3 | 1 | 5 | 4 | 7 |
From the following data, the modal class of the table will be:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency (f)} & 11 & 21 & 23 & 14 & 5 \\ \hline \end{array} \]
Find the Mean from the Following Table
Given data:
\[ \begin{array}{|c|c|} \hline \text{Class-interval} & \text{Frequency (f)} \\ \hline 0-10 & 3 \\ 10-20 & 10 \\ 20-30 & 11 \\ 30-40 & 9 \\ 40-50 & 7 \\ \hline \end{array} \]