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?
$28$