Question

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy news-paper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspaper, find the number of families, which buy
(1)A only
(2) B only
(3) none of A, B, and C.

Answer 1

Take first total families as 100% then make a Venn diagram to get a clear picture of the data given and then solve to get the desired results.

Complete step-by-step solution:
So let’s represent number of families which take A as n(A), B as n(B), C as n(C), both A and B as $n\left( A\cap B \right)$, both A and C as $n\left( A\cap C \right)$, both B and C as $n\left( B\cap C \right)$, all A, B and C as $n\left( A\cap B\cap C \right)$, only A as n(A only) , only B as n(B only), only C as n(C only).
As per the given data we can write,
$\begin{aligned} & n\left( A \right)=40\% \\\ & n\left( B \right)=20\% \\\ & n\left( C \right)=10\% \\\ & n\left( A\cap B \right)=5\% \\\ & n\left( B\cap C \right)=3\% \\\ & n\left( A\cap C \right)=4\% \\\ & n\left( A\cap B\cap C \right)=2\% \\\ \end{aligned}$
Now we can write,
$\begin{aligned} & only\left( A \right)=n(A)-n\left( A\cap B \right)-n\left( A\cap C \right)+n\left( A\cap B\cap C \right) \\\ & \Rightarrow only\left( A \right)=40\%-5\%-4\%+2\% \\\ & \Rightarrow only(A)=33\% \\\ \end{aligned}$
Similarly, we can write the other two as,

& only\left( B \right)=n(B)-n\left( A\cap B \right)-n\left( B\cap C \right)+n\left( A\cap B\cap C \right) \\\ & \Rightarrow only\left( B \right)=20\%-5\%-3\%+2\% \\\ & \Rightarrow only(B)=14\% \\\ \end{aligned}$$ $\begin{aligned} & only\left( C \right)=n(C)-n\left( B\cap C \right)-n\left( A\cap C \right)+n\left( A\cap B\cap C \right) \\\ & \Rightarrow only\left( C \right)=10\%-3\%-4\%+2\% \\\ & \Rightarrow only(C)=5\% \\\ \end{aligned}$ Let’s represent the above data as a Venn diagram,  The values are written in the Venn Diagram in the form of %. For finding the percentage of families, which does not take any the newspaper, we will subtract that percentage of families who takes at least one newspaper from $100\%$ so we get, So the percentage of families who does not read any of A, B, C is =$100\% -33\% -14\%-5\%-3\%-2\%-2\%-1\%$ =$40\%$ Hence 33% of families read-only A, 14% of families read-only B, 40% of families do not read any of them. But in the question we are asked to find in number of families which are, n(only A)=$\dfrac{33}{100}\times 10000=3300$ n(only B)= $\dfrac{14}{100}\times 10000=1400$ n(not any of A, B, C)= $\dfrac{40}{100}\times 10000=4000$ So 3300 families read only A, 1400 families read only B and 4000 families read none of them. Hence, the answer is (1) 3300 (2) 1400 and (3)4000. **Note:** Students generally get confused between A and only A, B and only B, C and only C. Actually A means people can read only A or A and B or A and C or A and B and C. The same goes for B and C also.