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  1. 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
  2. Out of these, 90 play football and 60 play hockey.
  3. Also, 5 in category III and one-fourth each in categories I and II do not play any game.
  4. In each of the above categories, the number of people who play only hockey equals the number of people who play only football.
  5. 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 IV).
  6. Five persons in category III play only hockey.

Assumption: middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under the four categories described above.

Now the questions are,

  • How many people have middle school education?

  • How many high school educated people do not play football?

  • How many people having middle school, but not high school, education play only football?

  • How many people who completed primary school could not finish middle school?

  • How many uneducated people play neither hockey nor football?

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@down voters, can you tell me why you down-vote my question? –  harish.raj Dec 24 '12 at 20:02
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may be because this is not really helpful for the research. I don't know though, I am not in the group of down voters anyway. –  Deepak Dec 25 '12 at 0:36
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I have no idea what you're saying. –  Alexander Gruber Dec 25 '12 at 10:29
    
I apologize. Please look at the question now. may be it is clear now. –  harish.raj Dec 25 '12 at 11:43
    
Your statement is contradictory now. In the first statement you say that category $I$ is people with at least a middle school education, and in statement 3 you say that category $I$ is people with only a middle school education. Which is it? –  Alexander Gruber Dec 29 '12 at 3:39
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4 Answers

up vote 1 down vote accepted
+50

First of all the idea here is that you give some indication that you've tried the problem before posting it, i.e. point out places where you get stuck or areas of confusion. That makes it easier to write a helpful answer. Not doing that may also be why you've been downvoted.

Having said that, here's how I approach the problem. If I've interpreted your statement correctly, Category III is a subset of Category II is a subset of Category I (i.e. anyone who has "at least high school education" also has "at least middle school education").

Statement 5 jumps out as a good starting point just because it has actual numbers in it. And it tells us that one person in category III plays both games and two people in categories I and II play both. So this means that one guy with at least a high school education plays both, and one guy with at least a middle school education (but not a high school education) plays both. Add in the last bit of 5 and we see that two uneducated people play both. So in total we have 4 (different!) people playing both sports.

If I'm interpreting it correctly, statement 6 tells us that exactly 5 people in category III play only hockey, so by statement 4 exactly 5 people in category III play only football. And by statement 3, 5 people in category III play no game. So we've solved category III: 1 person plays both sports, 5 people play only hockey, 5 play only football, and 5 people play nothing. And we have 1 + 5 + 5 + 5 = 16 people in category III.

Using Statement 1, for every person who finished high school there are two people who finished middle school but not high school. So there are 32 people who finished middle school but not high school. And by similar logic there are 64 people who finished primary school but not middle school.

At this point the questions you've listed should be at least approachable by similar means. Hope that helped!

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This seems a bit more complicated than just drawing a Venn diagram. I would try to diagram each item out algebraically instead. For example,

  1. Group of 200 people, 4 total categories, 3 of them in a ratio of 7:3:1. We can write $$ w + x + y + z = 200 $$ where $w = $ number of people in Category IV, $x = $ number of people in Category I, $y = $ number of people in Category II, $z = $ number of people in Category III. You can also write $$ x = 3y = 7z $$ based off of the ratio given. (Do you see why?) So that's two equations.
  2. 90 play football and 60 play hockey. People can either only play football, only play hockey, play both, or play none. If we denote only football with $f$, only hockey with $h$, both with $b$, and none with $n$, then we get that $$ f + h + 2b = 150. $$ (Do you see why?)

The next 4 items lead us to relationships between our variables $w, x, y, z$ for the number of people in each category, and our variables $f, h, b, n$ for the number of people who play various combinations of the two sports. From there you can get the answer by solving for the equations. Hope that helps.

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I'm not sure the first equality holds given that, if I'm interpreting the definitions of categories correctly, any person who is in category III is in categories I and II as well. –  lamb_da_calculus Dec 31 '12 at 6:02
    
Ah, I interpreted it as a person can only be in one of the four categories. In that case the calculations would be different. –  august Jan 3 '13 at 17:16
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For problems like this, I would draw a Venn-diagram for the sports and split each of the four pieces of the Venn diagram into four smaller pieces representing education. Some pieces can be filled in immediately (four instance, we mark down five people in the hockey-only diagram in the category-three area, as well as in e football-only area by part 4). Fill in as many exact numbers as you can. Then, give a name (i.e, $a,b,c$, etc.) to each area that is not filled in, and use rules 1-5 to write equations governing them (I.e. if $a$ is the number of uneducated people playing hockey only and $b$ is the number of uneducated people playing football only, $a=b$). Then, solve these equations using substitution. This idea will work for other, more complicated logical puzzles as well. Good luck! Happy new year!

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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

  1. 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 $$ \begin{eqnarray} (f_0+h_0+b_0+n_0) + (f_1+h_1+b_1+n_1) \\ + (f_2+h_2+b_2+n_2) + (f_3+h_3+b_3+n_3) &=&200 \tag{1}\\ ((f_2+h_2+b_2+n_2)+(f_3+h_3+b_3+n_3)) &=& 3(f_3+h_3+b_3+n_3) \tag{2}\\ ((f_1+h_1+b_1+n_1)+(f_2+h_2+b_2+n_2)\\ +(f_3+h_3+b_3+n_3)) &=& 7 (f_3+h_3+b_3+n_3) \tag{3}\\ \end{eqnarray} $$
  2. Out of these, 90 play football and 60 play hockey. $$ \begin{eqnarray} (f_0+f_1+f_2+f_3)+(b_0+b_1+b_2+b_3)&=&90 \tag{4} \\ (h_0+h_1+h_2+h_3)+(b_0+b_1+b_2+b_3)&=&60 \tag{5} \\ \end{eqnarray} $$
  3. Also, 5 in category III and one-fourth each in categories I and II do not play any game. $$ \begin{eqnarray} n_3&=&5 \tag{6} \\ 4 n_1&=& (f_1+h_1+b_1+n_1) \tag{7} \\ 4 n_2&=& (f_2+h_2+b_2+n_2) \tag{8} \\ \end{eqnarray} $$
  4. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. $$ \begin{eqnarray} f_1&=&h_1 \tag{9} \\ f_2&=&h_2 \tag{10} \\ f_3&=&h_3 \tag{11} \\ \end{eqnarray} $$
  5. 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). $$ \begin{eqnarray} b_0&=&2 \tag{12} \\ b_1&=&2 \tag{13} \\ b_2&=&2 \tag{14} \\ b_3&=&1 \tag{15} \\ \end{eqnarray} $$
  6. 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?

$$28$$

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