How do you find the minimum percentage who have studied all four subjects?

Question

At college, 70% students studied Maths, 75% students studied English, 85% studied french and 80% studied german. What percentage at least must have studied all 4?

Answer 1

Consider two subjects first. Suppose a certain fraction take subject A and another fraction take B then we can imagine shading in a bar with those fractions:

Then we are free to 'shift' these two colours so as to minimise the overlap.

Once you can solve this for two subjects then JiK has explained how to solve it for an arbitrary number.

Answer 2

The question is:

What percentage at least must belong to all of these four groups:

• Studied maths
• Studied English
• Studied French
• Studied German

You can reword this question as:

What percentage at least must belong to all of these three groups:

• Studied both maths and English
• Studied French
• Studied German

and further:

What percentage at least must belong to all of these two groups:

• Studied all of maths, English, and French
• Studied German

Answer 3

Imagine all students arriving at this college initially intended to study all four subjects: math, English, French, and German. But after careful consideration, $30\%$ of them decided to drop math; $25\%$ of them decided to drop English; $15\%$ of them decided to drop French; and $20\%$ of them decided to drop German.

Then the maximum number of students still taking all four courses would be achieved if everyone who dropped English, French, or German happened to be one of the same $30\%$ of students who dropped math--that is, when the number of students who decided not to take a course is as few as possible. The minimum number occurs when the number of students dropping a course is as many as they possibly can be, which occurs when each "drop" decision is performed by a different student (if possible). That is, everyone who drops a course stays in the other three.

In this case, it could be that the set of students can be partitioned into the $30\%$ of them who dropped math, the $25\%$ who dropped English, the $15\%$ who dropped French, the $20\%$ who dropped German, and the remaining $x\%$ who decided to study all four subjects; and so the answer is $x\%$.

This partition is only possible, of course, when the sizes of the four subsets, added together, total less than $100\%.$ This fortunately is the case with this particular problem. If this were not the case, then there are necessarily some students who dropped at least two of the four subjects, and the number of students who have taken all four subjects can be as low as zero.

Answer 4

Most of us would tend to add up the number of students taking each course and try figure out the least common inclusion. I don't know if it can be done this way. I think this problem needs an indirect approach for solution. Think about exclusions instead of inclusions. When we say 70% students take Maths, it also means 30% do not take Maths. By the same rule, 25% do not take English, 15% do not take French and 20% do not take German. If we add up these exclusions (assuming these non-takers are mutually exclusive to maximize their count) we get a figure of 90%. So there is a possibility that 90% students do not take at least one of the subjects. It may be highly unlikely on a probability perspective, but still is mathematically possible. The more important point here is that it is mathematically impossible to increase the number of non-takers with the given data. So the answer will be AT LEAST 10% of the students must have taken all four subjects.

Answer 5

Using the Generalized Principle of Inclusion-Exclusion, we compute:

$N(0)=100$
$N(1)=70+75+80+85=310$
$N(2)=x_1+x_2+x_3+x_4+x_5+x_6=X$
$N(3)=y_1+y_2+y_3+y_4= Y$
$N(4)=Z$

$100\binom{0}{0}-310\binom{1}{0}+X\binom{2}{0}-Y\binom{3}{0}+Z\binom{4}{0}= 0 $ Study Nothing
$\hphantom{100\binom{0}{1}-}310\binom{1}{1}-X\binom{2}{1}+Y\binom{3}{1}-Z\binom{4}{1}=D$ Study One subject
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+}X\binom{2}{2}-Y\binom{3}{2}+Z\binom{4}{2}=C$ Study two subjects $\hphantom{100\binom{0}{1}-310\binom{1}{1}+X\binom{2}{2}-}Y\binom{3}{3}-Z\binom{4}{3}=B$ Study three subjects
$\hphantom{100\binom{0}{1}-310\binom{1}{1}+X\binom{2}{2}-Y\binom{3}{3}+}Z\binom{4}{4}=A$ Study four subjects.

$x_i$ = study any two subjects

$y_i$ = study any three subjects

$100 - 310 + X - Y + Z = 0 => 100 - 310 + C + 3B + 6A - B - 4A + A = 0 =>C + 2B + 3A = 210$ ... (1)

$310 - 2X + 3Y - 4Y = D => D = 310 - 2(C+3B+6A) + 3(B+4A) - 4A = 310 - 2C - 3B - 4A$

$X - 3Y + 6Z = C => X = C +3Y - 6Z = C + 3(B+4A) - 6A = C + 3B +6A$

$Y - 4Z = B => Y = B + 4A$

$Z = A$

After working backwards, we have gotten to a point we have reduced it to three variables,namely, percentage of students who study two subjects (C), percentage of students who study three subjects (B) and finally percentage of students who study four subjects (A) with following equation $ C + 2B + 4A = 210$. If we have C and B we could get A. But I still wonder how you could get the minimum A with just four numbers in the question.

Thanks

Satish