Question

In a town 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper 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 newspapers, find the number of families which buy
A. A only
B. B only
C. None of A, B and C.

Answer 1

Hint: First of all, find the set of families who buy newspaper A, B, C, A and B, B and C, C and A and all the three newspapers by converting the given percentages to the number of people. So, that we can find the solution of the given problem by using this data.

Complete step-by-step answer:
iven total number of families in a town \[n\left( U \right) = 10000 = 100\% \]
The set of families buy newspaper A is given by \[n\left( A \right) = 40\% {\text{ of 10000}} = \dfrac{{40}}{{100}} \times 10000 = 4000\]
The set of families buy newspaper B is given by \[n\left( B \right) = 20\% {\text{ of 10000}} = \dfrac{{20}}{{100}} \times 10000 = 2000\]
The set of families buy newspaper C is given by \[n\left( C \right) = 10\% {\text{ of 10000}} = \dfrac{{10}}{{100}} \times 10000 = 1000\]
The set of families buy newspaper A and B is given by \[n\left( {A \cap B} \right) = 5\% {\text{ of 10000}} = \dfrac{5}{{100}} \times 10000 = 500\]
The set of families buy newspaper B and C is given by \[n\left( {B \cap C} \right) = 3\% {\text{ of 10000}} = \dfrac{3}{{100}} \times 10000 = 300\]
The set of families buy newspaper C and A is given by \[n\left( {C \cap A} \right) = 4\% {\text{ of 10000}} = \dfrac{4}{{100}} \times 10000 = 400\]
The set of families buy all the three newspapers are given by \[n\left( {A \cap B \cap C} \right) = 2\% {\text{ of 10000}} = \dfrac{2}{{100}} \times 10000 = 200\]
A. A only
The set of families buy newspaper A only is given by
\[
   \Rightarrow n\left( {A{\text{ only}}} \right) = n\left( A \right) - \left[ {n\left( {A \cap B} \right) + n\left( {A \cap C} \right)} \right] + n\left( {A \cap B \cap C} \right) \\
   \Rightarrow n\left( {A{\text{ only}}} \right) = 4000 - \left[ {500 + 400} \right] + 200 \\
   \Rightarrow n\left( {A{\text{ only}}} \right) = 4000 - 900 + 200 \\
   \Rightarrow n\left( {A{\text{ only}}} \right) = 4000 - 700 \\
  \therefore n\left( {A{\text{ only}}} \right) = 3300 \\
\]
Thus, the number of families who buy newspaper A only is 3300.
B. B only
The set of families buy newspaper B only is given by
\[
   \Rightarrow n\left( {B{\text{ only}}} \right) = n\left( B \right) - \left[ {n\left( {A \cap B} \right) + n\left( {B \cap C} \right)} \right] + n\left( {A \cap B \cap C} \right) \\
   \Rightarrow n\left( {B{\text{ only}}} \right) = 2000 - \left[ {500 + 300} \right] + 200 \\
   \Rightarrow n\left( {B{\text{ only}}} \right) = 2000 - 800 + 200 \\
   \Rightarrow n\left( {B{\text{ only}}} \right) = 2000 - 600 \\
  \therefore n\left( {B{\text{ only}}} \right) = 1400 \\
\]
Thus, number of families buy newspaper B only is 1400
C. None of A, B and C
The set of families buy none of the newspapers are given by
\[ \Rightarrow n\left( {{\text{None of A, B and C}}} \right) = n\left( U \right) - n\left( {A \cup B \cup C} \right)\]
We know that \[n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right)\]
By using this formula, we have
\[
   \Rightarrow n\left( {{\text{None of A, B and C}}} \right) = n\left( U \right) - \left[ {n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right)} \right] \\
   \Rightarrow n\left( {{\text{None of A, B and C}}} \right) = 10000 - \left[ {4000 + 2000 + 1000 - 500 - 300 - 400 + 200} \right] \\
   \Rightarrow n\left( {{\text{None of A, B and C}}} \right) = 10000 - \left[ {6000} \right] \\
  \therefore n\left( {{\text{None of A, B and C}}} \right) = 4000 \\
\]
Thus, the number of families who buy none of A, B and C are 4000.

Note:The intersection of two sets \[A\] and \[B\], denoted by \[A \cap B\], is the set containing all elements of \[A\] that also belong to \[B\] (or equivalently, all elements of \[B\] that also belong to \[A\]). The union of two sets \[A\] and \[B\], denoted by \[A \cup B\] is the set of all elements that are found in \[A\] OR \[B\](or both)