Mathematics - Functions Question with Solution | TestHub

$\begin{array}{ccc}\text{ Marks }& \text{ No. of s tudents }& \text{ c.f. }\\ 10-20& 12& 12\\ 20-30& 30& 42\\ 30-40& ?& 42+\mathrm{x}\\ 40-50& 65& 107+\mathrm{x}\\ 50-60& ?& 107+\mathrm{x}+\mathrm{y}\\ 60-70& 25& 132+\mathrm{x}+\mathrm{y}\\ 70-80& 18& 150+\mathrm{x}+\mathrm{y}\\ \text{ Total }& 229& \end{array}$ $\begin{array}{rl}& \frac{\mathrm{n}}{2}=\frac{229}{2}=114.5,\text{ Median }=46\\ & \therefore \text{ Median class }=40-50\\ & \therefore \ell =40,\text{ c.f. }=42+\mathrm{x},\mathrm{f}=65,\mathrm{h}=10\\ & \text{ Median }=\ell +\left(\frac{\frac{\mathrm{n}}{2}-\text{ c.f. }}{\mathrm{f}}\right)\times \mathrm{h}\\ & 46=40+\frac{114.5-(42+\mathrm{x})}{65}\times 10\\ & \text{ or }46-40=\frac{(114.5-42-\mathrm{x})}{13}\times 2\\ & 6=\frac{(72.5-\mathrm{x})}{13}\times 2\text{ or }78=145-2\mathrm{x}\\ & 2\mathrm{x}=145-78=67\text{ or }\mathrm{x}=\frac{67}{2}=33.5\end{array}$ $\therefore \mathrm{x}=34(\because$ Number of students cannot be in fraction) Now $\Sigma f_i=29\therefore \mathrm{x}+\mathrm{y}+150=229$ $x+y=229-150=79$ ...(i) Putting the value of $x$ in (i), we get $34x+y=79\Rightarrow y=79-34=45$ $\therefore \mathrm{x}=34,\mathrm{y}=45$