Question

Candidates of four schools appear in a mathematics test. The data were as follows.

Schools	No. of candidates	Average Score.
I	$60$	$75$
II	$48$	$80$
III	Not available	$55$
IV	$40$	$50$

If average score of the candidates of all the four schools is $66$, find the number of candidates that appeared from school II.

Answer 1

Solution:-

Given,

Average score of candidates of I school $\left({\overline{x}}_1\right)=75$

Average score of candidates of II school $\left({\overline{x}}_2\right)=80$

Average score of candidates of III school $\left({\overline{x}}_3\right)=55$

Average score of candidates of IV school $\left({\overline{x}}_4\right)=50$

Average score of the candidates of all the four schools $\left({\overline{x}}_{mean}\right)$ = 66

As we know that,

$\overline{x}=\frac{\text{Sum of score of all candidates}}{\text{Total number of students}}$

$\text{School I}\Rightarrow$

$\therefore {\overline{x}}_1=\frac{\text{Sum of scores of all candidates in school I}\left(S_1\right)}{\text{Total number of candidates in school I}\left(n_1\right)}$

$\Rightarrow 75=\frac{S_1}{60}$

$\Rightarrow S_1=4500$

$\text{School II}\Rightarrow$

$\therefore {\overline{x}}_2=\frac{\text{Sum of scores of all candidates in school II}\left(S_2\right)}{\text{Total number of candidates in school II}\left(n_2\right)}$

$\Rightarrow 80=\frac{S_2}{48}$

$\Rightarrow S_2=3840$

$\text{School III}\Rightarrow$

$\therefore {\overline{x}}_3=\frac{\text{Sum of scores of all candidates in school III}\left(S_3\right)}{\text{Total number of candidates in school III}\left(n_3\right)}$

Let the number of candidates in school III be n.

$\Rightarrow 55=\frac{S_3}{n}$

$\Rightarrow S_3=55n$

$\text{School IV}\Rightarrow$

$\therefore {\overline{x}}_{IV}=\frac{\text{Sum of scores of all candidates in school IV}\left(S_4\right)}{\text{Total number of candidates in school IV}\left(n_4\right)}$

$\Rightarrow 40=\frac{S_4}{50}$

$\Rightarrow S_4=2000$

$\because {\overline{x}}_{mean}=\frac{S_1+S_2+S_3+S_4}{n_1+n_2+n_3+n_4}$

$66=\frac{4500+3840+55n+2000}{60+48+n+40}$

$66=\frac{10340+55n}{148+n}$

$\Rightarrow 66(148+n)=10340+55n$

$\Rightarrow 9768+66n=10340+55n$

$\Rightarrow 66n-55n=10340-9768$

$\Rightarrow 11n=572$

$\Rightarrow n=\frac{572}{11}$

$\Rightarrow n=52$

Hence, the number of candidates that appeared from $SchoolIIIare52.$