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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 1 · Q32
(a) The median of the following data is 137. Find the values of x and y, given that total of frequencies is 68.
The median of the following data is 137. Find the values of x and y , given that total of frequencies is 68.
| Class | Frequency |
| $\displaystyle 65$-$\displaystyle 85$ | $\displaystyle 4$ |
| $\displaystyle 85$-$\displaystyle 105$ | $\displaystyle 5$ |
| $\displaystyle 105$-$\displaystyle 125$ | x |
| $\displaystyle 125$-$\displaystyle 145$ | $\displaystyle 20$ |
| $\displaystyle 145$-$\displaystyle 165$ | $\displaystyle 14$ |
| $\displaystyle 165$-$\displaystyle 185$ | y |
| $\displaystyle 185$-$\displaystyle 205$ | $\displaystyle 4$ |
Find mean and mode of the following distribution :
| Class | Frequency |
| $\displaystyle 0$-$\displaystyle 10$ | $\displaystyle 3$ |
| $\displaystyle 10$-$\displaystyle 20$ | $\displaystyle 6$ |
| $\displaystyle 20$-$\displaystyle 30$ | $\displaystyle 11$ |
| $\displaystyle 30$-$\displaystyle 40$ | $\displaystyle 10$ |
| $\displaystyle 40$-$\displaystyle 50$ | $\displaystyle 13$ |
| $\displaystyle 50$-$\displaystyle 60$ | $\displaystyle 3$ |
| $\displaystyle 60$-$\displaystyle 70$ | $\displaystyle 4$ |
Marking-scheme solution
| Class | \(\displaystyle f\) | cf |
| $\displaystyle 65$-$\displaystyle 85$ | $\displaystyle 4$ | $\displaystyle 4$ |
| $\displaystyle 85$-$\displaystyle 105$ | $\displaystyle 5$ | $\displaystyle 9$ |
| $\displaystyle 105$-$\displaystyle 125$ | x | \(\displaystyle 9+\mathrm{x}\) |
| $\displaystyle 125$-$\displaystyle 145$ | $\displaystyle 20$ | \(\displaystyle 29+\mathrm{x}\) |
| $\displaystyle 145$-$\displaystyle 165$ | $\displaystyle 14$ | \(\displaystyle 43+\mathrm{x}\) |
| $\displaystyle 165$-$\displaystyle 185$ | y | \(\displaystyle 43+\mathrm{x}+\mathrm{y}\) |
| $\displaystyle 185$-$\displaystyle 205$ | $\displaystyle 4$ | \(\displaystyle 47+\mathrm{x}+\mathrm{y}\) |
Given, Median = $\displaystyle 137$
∴ Median Class is $\displaystyle 125$-145.
\[125+\left[\frac{\dfrac{68}{2}-(9+x)}{20}\right] \times 20=137
\]
\(\displaystyle \Rightarrow \mathrm{x}=13\)
Also, \(\displaystyle 47+\mathrm{x}+\mathrm{y}=68\)
\[\therefore 47+13+y=68 \Rightarrow y=8
\]
| Class | \(\displaystyle \boldsymbol{x}_{\boldsymbol{i}}\) | \(\displaystyle f_{i}\) | \(\displaystyle \boldsymbol{u}_{\boldsymbol{i}}=\frac{\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{3 5}}{\mathbf{1 0}}\) | \(\displaystyle f_{i} u_{i}\) |
| $\displaystyle 0$-$\displaystyle 10$ | $\displaystyle 5$ | $\displaystyle 3$ | -$\displaystyle 3$ | -$\displaystyle 9$ |
| $\displaystyle 10$-$\displaystyle 20$ | $\displaystyle 15$ | $\displaystyle 6$ | -$\displaystyle 2$ | -$\displaystyle 12$ |
| $\displaystyle 20$-$\displaystyle 30$ | $\displaystyle 25$ | $\displaystyle 11$ | -$\displaystyle 1$ | - $\displaystyle 11$ |
| $\displaystyle 30$-$\displaystyle 40$ | $\displaystyle 35$ | $\displaystyle 10$ | $\displaystyle 0$ | $\displaystyle 0$ |
| $\displaystyle 40$-$\displaystyle 50$ | $\displaystyle 45$ | $\displaystyle 13$ | $\displaystyle 1$ | $\displaystyle 13$ |
| $\displaystyle 50$-$\displaystyle 60$ | $\displaystyle 55$ | $\displaystyle 3$ | $\displaystyle 2$ | $\displaystyle 6$ |
| $\displaystyle 60$-$\displaystyle 70$ | $\displaystyle 65$ | $\displaystyle 4$ | $\displaystyle 3$ | $\displaystyle 12$ |
| Total | $\displaystyle 50$ | -$\displaystyle 1$ |
Mean \(\displaystyle =35+\frac{(-1)}{50} \times 10=34.8\)
Modal class is $\displaystyle 40$-50.
\[\begin{aligned}
\text { Mode } & =40+\left(\frac{13-10}{2 \times 13-10-3}\right) \times 10 \\
& =42.3 \text { (approx.) }
\end{aligned}
\]
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