An exercise in formatting
Variables
Let's call the category that contains all people the category 0 and we forget category 4. this smplifies the wording of the following definitions:
- $f_i$: number of persons playing football only and $i$ is the highest category they are member of
- $h_i$: number of persons playing hockey only and $i$ is the highest category they are member of
- $b_i$: number of persons playing football and hockey and $i$ is the highest category they are member of
- $n_i$: number of persons playing neither football nor hockey and $i$ is the highest category they are member of
Equations
- In a group of 200 people, number of people having at least primary education (assuming - Category I): number of people having at
least middle school education (Category II): number of people having at least high school education (Category III) are in the ratio 7 : 3 : 1
$$
\begin{eqnarray}
(f_0+h_0+b_0+n_0) + (f_1+h_1+b_1+n_1) \\
+ (f_2+h_2+b_2+n_2) + (f_3+h_3+b_3+n_3) &=&200 \\
((f_1+h_1+b_1+n_1)+(f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) \\
: \\
((f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) \\
: \\
(f_3+b_3+n_3) &=& 7 : 3: 1
\end{eqnarray}
$$
the latter equation means
$$
\begin{eqnarray}
((f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) \\
: \\
(f_3+h_3+b_3+n_3) &=& 3: 1 \\
((f_1+h_1+b_1+n_1)+(f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) \\
: \\
(f_3+h_3+b_3+n_3) &=& 7 : 1
\end{eqnarray}
$$
and the meaning of these ratios is
$$
\begin{eqnarray}
((f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) &=& 3 (f_3+h_3+b_3+n_3) \\
((f_1+h_1+b_1+n_1)+(f_2+h_2+b_2+n_2)+ \\
(f_3+h_3+b_3+n_3)) &=& 7 (f_3+h_3+b_3+n_3) \\
\end{eqnarray}
$$
so the first statement can be expressed as
$$
\begin{eqnarray}
(f_0+h_0+b_0+n_0) + (f_1+h_1+b_1+n_1) \\
+ (f_2+h_2+b_2+n_2) + (f_3+h_3+b_3+n_3) &=&200 \tag{1}\\
((f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) &=& 3(f_3+h_3+b_3+n_3) \tag{2}\\
((f_1+h_1+b_1+n_1)+(f_2+h_2+b_2+n_2)\\
+(f_3+h_3+b_3+n_3)) &=& 7 (f_3+h_3+b_3+n_3) \tag{3}\\
\end{eqnarray}
$$
- Out of these, 90 play football and 60 play hockey.
$$
\begin{eqnarray}
(f_0+f_1+f_2+f_3)+(b_0+b_1+b_2+b_3)&=&90 \tag{4} \\
(h_0+h_1+h_2+h_3)+(b_0+b_1+b_2+b_3)&=&60 \tag{5} \\
\end{eqnarray}
$$
- Also, 5 in category III and one-fourth each in categories I and II do not play any game.
$$
\begin{eqnarray}
n_3&=&5 \tag{6} \\
4 n_1&=& (f_1+h_1+b_1+n_1) \tag{7} \\
4 n_2&=& (f_2+h_2+b_2+n_2) \tag{8} \\
\end{eqnarray}
$$
- In each of the above categories, the number of people who play only hockey equals the number of people who play only football.
$$
\begin{eqnarray}
f_1&=&h_1 \tag{9} \\
f_2&=&h_2 \tag{10} \\
f_3&=&h_3 \tag{11} \\
\end{eqnarray}
$$
- Two persons each in categories I and II and one person in category III play both the games. Two persons who play both the games are uneducated (category 0).
$$
\begin{eqnarray}
b_0&=&2 \tag{12} \\
b_1&=&2 \tag{13} \\
b_2&=&2 \tag{14} \\
b_3&=&1 \tag{15} \\
\end{eqnarray}
$$
- Five persons in category III play only hockey.
$$
\begin{eqnarray}
h_3=5 \tag{16} \\
\end{eqnarray}
$$
Questions
- How many people have middle school education?
$$(f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)$$
- How many high school educated people do not play football?
$$h_3+n_3$$
- How many people having middle school, but not high school, education
play only football?
$$f_2$$
- How many people who completed primary school could not finish middle
school?
$$f_1+h_1+b_1+n_1$$
- How many uneducated people play neither hockey nor football?
$$n_0$$
Calculations
We have 16 variables and 16 linear equations. If these equations are linearly independant the euationa system has exactly one solution tuple. We have additional
requirements to our solution: the numbers have to be nonnegative and integers. I used maxima to solve the equations and get the answers to the queries.
(%i1) solve([
(f_0+h_0+b_0+n_0) + (f_1+h_1+b_1+n_1)
+ (f_2+h_2+b_2+n_2) + (f_3+h_3+b_3+n_3) =200,
((f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) = 3 * (f_3+h_3+b_3+n_3),
((f_1+h_1+b_1+n_1)+(f_2+h_2+b_2+n_2)
+(f_3+h_3+b_3+n_3)) = 7* (f_3+h_3+b_3+n_3) ,
(f_0+f_1+f_2+f_3)+(b_0+b_1+b_2+b_3)=90,
(h_0+h_1+h_2+h_3)+(b_0+b_1+b_2+b_3)=60,
n_3=5,
4*n_1= (f_1+h_1+b_1+n_1),
4*n_2= (f_2+h_2+b_2+n_2),
f_1=h_1,
f_2=h_2,
f_3=h_3,
b_0=2,
b_1=2,
b_2=2,
b_3=1,
h_3=5],
[f_0,f_1,f_2,f_3,h_0,h_1,h_2,h_3,n_0,n_1,n_2,n_3,b_0,b_1,b_2,b_3]
);
(%o1) [[f_0 = 44,f_1 = 23,f_2 = 11,f_3 = 5,h_0 = 14,h_1 = 23,h_2 = 11,h_3 = 5,
n_0 = 28,n_1 = 16,n_2 = 8,n_3 = 5,b_0 = 2,b_1 = 2,b_2 = 2,b_3 = 1]]
(%i2) [(f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3), h_3+n_3,
f_2, f_1+h_1+b_1+n_1, n_0],%[1];
(%o2) [48,10,11,64,28]
Answers
- How many people have middle school education?
$$48$$
- How many high school educated people do not play football?
$$10$$
- How many people having middle school, but not high school, education
play only football?
$$11$$
- How many people who completed primary school could not finish middle
school?
$$64$$
- How many uneducated people play neither hockey nor football?
$$28$$