In a town, 45% population read magazine A, 55% read magazine B, 40% read magazine C, 30% read magazines A and B, 15% read magazines B and C, 25% read magazines A and C; and 10% read all the three magazines. What percentage do not read any magazine?

The correct answer is 20% (Option 3).

Step-by-Step Explanation:

Let the total population of the town be represented by the universal set, where total percentage = 100%.

Let the percentage of people reading each magazine be denoted as:
P(A) = percentage of people who read magazine A = 45%
P(B) = percentage of people who read magazine B = 55%
P(C) = percentage of people who read magazine C = 40%

The percentages of people reading combinations of two magazines are:
P(A ∩ B) = percentage of people who read both A and B = 30%
P(B ∩ C) = percentage of people who read both B and C = 15%
P(A ∩ C) = percentage of people who read both A and C = 25%

The percentage of people reading all three magazines is:
P(A ∩ B ∩ C) = 10%

To find the percentage of people who read at least one magazine, we use the Principle of Inclusion-Exclusion for three sets:

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

Now, substitute the given values into the formula:

$P(A\cup B\cup C)=45+55+40-30-15-25+10$

Simplifying step-by-step:
Sum of individual percentages: 45 + 55 + 40 = 140%
Subtract double intersections: 140 - 30 - 15 - 25 = 70%
Add triple intersection: 70 + 10 = 80%

So, 80% of the population reads at least one magazine.

Now, to find the percentage of people who do not read any magazine, subtract the percentage of people reading at least one magazine from the total percentage (100%):

$\text{Percentage not reading any magazine}=100\%-P(A\cup B\cup C)$

$\text{Percentage not reading any magazine}=100\%-80\%=20\%$

Therefore, 20% of the town's population does not read any magazine.