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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 3 · Q34
(a) Find mean and mode of the following frequency distribution:
Find mean and mode of the following frequency distribution :
| Class : | $\displaystyle 10$-$\displaystyle 30$ | $\displaystyle 30$-$\displaystyle 50$ | $\displaystyle 50$-$\displaystyle 70$ | $\displaystyle 70$-$\displaystyle 90$ | $\displaystyle 90$-$\displaystyle 110$ | $\displaystyle 110$-$\displaystyle 130$ | $\displaystyle 130$-$\displaystyle 150$ |
| Frequency : | $\displaystyle 6$ | $\displaystyle 8$ | $\displaystyle 12$ | $\displaystyle 10$ | $\displaystyle 14$ | $\displaystyle 11$ | $\displaystyle 9$ |
If the median of the distribution given below is $\displaystyle 28.5$, find the values of $\displaystyle x$ and y .
| Class : | $\displaystyle 0$-$\displaystyle 10$ | $\displaystyle 10$-$\displaystyle 20$ | $\displaystyle 20$-$\displaystyle 30$ | $\displaystyle 30$-$\displaystyle 40$ | $\displaystyle 40$-$\displaystyle 50$ | $\displaystyle 50$-$\displaystyle 60$ | Total |
| Frequency : | $\displaystyle 5$ | $\displaystyle x$ | $\displaystyle 20$ | $\displaystyle 15$ | y | $\displaystyle 5$ | $\displaystyle 60$ |
Marking-scheme solution
| Class | Frequency | \(\displaystyle \boldsymbol{x}_{\boldsymbol{i}}\) | \(\displaystyle \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}}\) |
| $\displaystyle 10$-$\displaystyle 30$ | $\displaystyle 6$ | $\displaystyle 20$ | $\displaystyle 120$ |
| $\displaystyle 30$-$\displaystyle 50$ | $\displaystyle 8$ | $\displaystyle 40$ | $\displaystyle 320$ |
| $\displaystyle 50$-$\displaystyle 70$ | $\displaystyle 12$ | $\displaystyle 60$ | $\displaystyle 720$ |
| $\displaystyle 70$-$\displaystyle 90$ | $\displaystyle 10$ | $\displaystyle 80$ | $\displaystyle 800$ |
| $\displaystyle 90$-$\displaystyle 110$ | $\displaystyle 14$ | $\displaystyle 100$ | $\displaystyle 1400$ |
| $\displaystyle 110$-$\displaystyle 130$ | $\displaystyle 11$ | $\displaystyle 120$ | $\displaystyle 1320$ |
| $\displaystyle 130$-$\displaystyle 150$ | $\displaystyle 9$ | $\displaystyle 140$ | $\displaystyle 1260$ |
| Total | $\displaystyle 70$ | $\displaystyle 5940$ |
\[\begin{aligned}
\therefore \text { Mean } & =\bar{x}=\frac{5940}{70} \\
& =84.8 \text { (approx.) }
\end{aligned}
\]
\[\begin{aligned}
& \text { Modal Class }=90-110 \\
& \therefore \text { Mode }=90+\frac{14-10}{28-10-11} \times 20
\end{aligned}
\]
\[\begin{aligned}
& =90+\frac{4 \times 20}{7} \\
& =101.4 \text { (approx.) }
\end{aligned}
\]
| Class | Frequency | cf |
| $\displaystyle 0$-$\displaystyle 10$ | $\displaystyle 5$ | $\displaystyle 5$ |
| $\displaystyle 10$-$\displaystyle 20$ | \(\displaystyle x\) | \(\displaystyle x+5\) |
| $\displaystyle 20$-$\displaystyle 30$ | $\displaystyle 20$ | \(\displaystyle x+25\) |
| $\displaystyle 30$-$\displaystyle 40$ | $\displaystyle 15$ | \(\displaystyle x+40\) |
| $\displaystyle 40$-$\displaystyle 50$ | \(\displaystyle y\) | \(\displaystyle x+y+40\) |
| $\displaystyle 50$-$\displaystyle 60$ | $\displaystyle 5$ | \(\displaystyle x+y+45\) |
| Total | $\displaystyle 60$ |
Median Class \(\displaystyle =20-30\)
\[\therefore 28.5=20+\frac{10}{20}\left(\frac{60}{2}-x-5\right)
\]
\(\displaystyle \Rightarrow x=8\)
As, \(\displaystyle x+y+45=60\)
\[\Rightarrow 8+y=15
\]
\(\displaystyle \therefore y=7\)
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.