CBSE 2025 · Region 4 · Set 2 · Q32 · 5 marks
The following table shows the number of traffic challans issued in the month of April by the traffic police : Number of Challans Number of Days $\displaystyle 0$-$\displaystyle 10$ $\displaystyle 3$ $\displaystyle 10$-$\displaystyle 20$ $\displaystyle 5$ $\displaystyle 20$-$\displaystyle 30$ $\displaystyle 10$ $\displaystyle 30$-$\displaystyle 40$ $\displaystyle 9$ $\displaystyle 40$-$\displaystyle 50$ $\displaystyle 2$ $\displaystyle 50$-$\displaystyle 60$ $\displaystyle 1$ Total $\displaystyle 30$
Find the 'mean' and 'mode' of the above data.
| Number of Challans | Number of Days |
| $\displaystyle 0$-$\displaystyle 10$ | $\displaystyle 3$ |
| $\displaystyle 10$-$\displaystyle 20$ | $\displaystyle 5$ |
| $\displaystyle 20$-$\displaystyle 30$ | $\displaystyle 10$ |
| $\displaystyle 30$-$\displaystyle 40$ | $\displaystyle 9$ |
| $\displaystyle 40$-$\displaystyle 50$ | $\displaystyle 2$ |
| $\displaystyle 50$-$\displaystyle 60$ | $\displaystyle 1$ |
| Total | $\displaystyle 30$ |
Marking-scheme solution
| Number of Challans | Number of days \(\displaystyle \left(f_{i}\right)\) | Class Mark \(\displaystyle \left(x_{i}\right)\) | \(\displaystyle f_{i} x_{i}\) |
| $\displaystyle 0$-$\displaystyle 10$ | $\displaystyle 3$ | $\displaystyle 5$ | $\displaystyle 15$ |
| $\displaystyle 10$-$\displaystyle 20$ | $\displaystyle 5$ | $\displaystyle 15$ | $\displaystyle 75$ |
| $\displaystyle 20$-$\displaystyle 30$ | $\displaystyle 10$ | $\displaystyle 25$ | $\displaystyle 250$ |
| $\displaystyle 30$-$\displaystyle 40$ | $\displaystyle 9$ | $\displaystyle 35$ | $\displaystyle 315$ |
| $\displaystyle 40$-$\displaystyle 50$ | $\displaystyle 2$ | $\displaystyle 45$ | $\displaystyle 90$ |
| $\displaystyle 50$-$\displaystyle 60$ | $\displaystyle 1$ | $\displaystyle 55$ | $\displaystyle 55$ |
| Total | \(\displaystyle \sum f_{i}=30\) | \(\displaystyle \sum f_{i} x_{i}=800\) |
\[\begin{aligned}
\text { Mean } & =\frac{800}{30} \\
& =\frac{80}{3} \text { or } 26.67 \text { or } 27 \text { (approx.) }
\end{aligned}
\]
Modal class is $\displaystyle 20$-$\displaystyle 30$
\[\begin{aligned}
\text { Mode } & =20+\frac{10-5}{2 \times 10-5-9} \times 10 \\
& =\frac{85}{3} \text { or } 28.3 \text { or } 28 \text { (approx.) }
\end{aligned}
\]
Let the length of shortest side be \(\displaystyle x\) m
∴ length of longest side \(\displaystyle =(x+4) \mathrm{m}\) and length of third side \(\displaystyle =(x+2) \mathrm{m}\)
Now, \(\displaystyle (x+4)^{2}=x^{2}+(x+2)^{2}\)
\(\displaystyle \Rightarrow x^{2}-4 x-12=0\)
\(\displaystyle \Rightarrow(x-6)(x+2)=0\)
\(\displaystyle \Rightarrow x=6\)
∴ sides are $\displaystyle 6$ m, $\displaystyle 8$ m and $\displaystyle 10$ m
Area \(\displaystyle =\frac{1}{2} \times 6 \times 8=24 \mathrm{~m}^{2}\)
Perimeter \(\displaystyle =6+8+10=24 \mathrm{~m}\)
Difference \(\displaystyle =0\)
\(\displaystyle \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}\)
\(\displaystyle \Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}\)
Simplifying, we get \(\displaystyle 2 x^{2}-19 x+42=0\)
\(\displaystyle \Rightarrow(x-6)(2 x-7)=0\)
\(\displaystyle \Rightarrow x=6\) or \(\displaystyle x=\frac{7}{2}\)
As, \(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5}\)
\(\displaystyle \Rightarrow \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}\)
⇒ \(\displaystyle \angle \mathrm{C}=\angle \mathrm{R}\)
(i)
In △ ADC and \(\displaystyle \triangle \mathrm{PSR}\),
\(\displaystyle \angle \mathrm{ADC}=\angle \mathrm{PSR}=90^{\circ}\)
and \(\displaystyle \angle \mathrm{C}=\angle \mathrm{R}\)
\(\displaystyle \therefore \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR}\)
(ii)
\(\displaystyle \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5}\)
\(\displaystyle \Rightarrow \frac{4}{\mathrm{PS}}=\frac{3}{5}\)
\(\displaystyle \Rightarrow \mathrm{PS}=\frac{20}{3} \mathrm{~cm}\)
(iii)
\(\displaystyle \frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{PQR})}=\frac{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{AD}}{\dfrac{1}{2} \times \mathrm{QR} \times \mathrm{PS}}\)
\[=\frac{3}{5} \times \frac{3}{5}=\frac{9}{25}
\]
\(\displaystyle \therefore \operatorname{ar}(\triangle \mathrm{ABC}): \operatorname{ar}(\triangle \mathrm{PQR})=9: 25\)
Correct statement
Join AF intersecting line \(\displaystyle m\) at G
In \(\displaystyle \triangle \mathrm{ACF}, \mathrm{BG} \| \mathrm{CF}\)
\[\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AG}}{\mathrm{GF}}
\]
In \(\displaystyle \Delta \mathrm{FDA}, \mathrm{GE} \| \mathrm{AD}\)
\[\Rightarrow \frac{\mathrm{EF}}{\mathrm{DE}}=\frac{\mathrm{GF}}{\mathrm{AG}} \text { or } \frac{\mathrm{DE}}{\mathrm{EF}}=\frac{\mathrm{AG}}{\mathrm{GF}}
\]
From, (i) and (ii), we get \(\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}\)
StatisticsMean of Grouped DataApplylong_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.