# word problem with percentages

Could you please explain to me how to solve this problem?

In a city, $45\%$ of the population read magazine $A$, $55\%$ read magazine $B$ and $40\%$ read magazine $C$. $30\%$ read $A$ and $B$, $15\%$ read $A$ and $C$, $15\%$ read $B$ and $C$ and $10\%$ read all three magazines. How many percent read no magazine at all?

[The answer is apparently $20\%$.]

• Solve it for a city that has $100$ inhabitants. – drhab Sep 3 '15 at 8:36
• I tried, but I just don't get how the percentages are related to each other. I don't get how "30 people of all 100 people (Read A and B)+ 10 of all (Read A and B and C) + 15 of all (read A and C)" can add up to 45% in total for reading A. I guess I get the relations wrong, but I don't understand how. – Saiai Sep 3 '15 at 8:55
• As you can find in the answer of @calculus the answer lies in applying inclusion/exclusion. So have a look at that link. – drhab Sep 3 '15 at 8:59
• I've noticed that this problem has been incorrectly transcribed on various webpages (eg. spiegel.de/quiztool/quiztool-64253.html). The correct version of the problem has 25% reading B and C, not the 15% you've written down. If you use 25%, you'll get the correct answer. FYI, rather than doing all the tricky notation, I just scribbled out a three-set venn diagram to get the answer. Edit: This website explains a very similar problem, if you'd like to use this venn diagram technique yourself: civilservicereview.com/2014/12/… – Nathan Sep 3 '15 at 10:00
• Thanks, Nathan, that solves my problem completely! – Saiai Sep 4 '15 at 8:28

Hint: You can apply the inclusion-exclusion-principle.

$P(A \cup B \cup C) =P(A)+P(B)+P(C)-P(A\cap B)-P(A \cap C)-P(C\cap B)+P(A\cap B\cap C )$

$1-P(A\cup B\cup C )$. But I have a result, which is different from 20%.

• Also my result differs. – drhab Sep 3 '15 at 8:45
• @drhab In that case it is 100 percent sure, that 20% is the wrong solution. – callculus Sep 3 '15 at 8:53
• @calculus: thanks! I understand the maths here. The task is from a newspaper, so it could be that they gave an wrong solution with 20%. – Saiai Sep 3 '15 at 9:03
• @Saiai If there are no other options given, then I would say yes. – callculus Sep 3 '15 at 9:08

This answer is in the same line as the answer of @Calculus and works with numbers in stead of probabilities.

Let it be that the city has $100$ inhabitants and denote $n\left(A\right)=45$, $n\left(B\right)=55$ et cetera.

As you will understand $n(A)$ stands for the number of persons that read magazine $A$

With inclusion/exclusion we find:

$$n\left(A\cup B\cup C\right)=$$$$n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(A\cap B\right)-n\left(A\cap C\right)-n\left(B\cap C\right)+n\left(A\cap B\cap C\right)$$

Next to that we have:

$$n\left(A^{c}\cap B^{c}\cap C^{c}\right)=100-n\left(A\cup B\cup C\right)$$