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 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, then number of families which buy newspaper A only is:
A) 3100
B) 3300
C) 2900
D) 1400

Answer 1

Hint:
We will convert the given percentages into number of families by using the formula: $x\% {\text{ of y}} = \dfrac{x}{{100}} \times y$ and then we will use the method of Venn diagram to calculate the number of families which buy newspaper A only.

Complete step by step solution:
We are told that in a town, there are 10,000 families.
There are families who buy different newspapers A, B, C and they are given in percentages. We will convert them into the number of families by using the formula: $x\% {\text{ of y}} = \dfrac{x}{{100}} \times y$ . Here, x is the percentage of families and y is the total number of families.
40% families buy newspaper A. Converting it into number of families, we get
$ \Rightarrow n(A) = 40\% = \dfrac{{40}}{{100}} \times 10000 = 4000$
20% families buy newspaper B. Therefore, the number of families that buy B are:
$ \Rightarrow n(B) = 20\% = \dfrac{{20}}{{100}} \times 10000 = 2000$
10% families buy newspaper C. Therefore, the number of families that buy newspaper C are:
$ \Rightarrow n(C) = 10\% = \dfrac{{10}}{{100}} \times 10000 = 1000$
5% families buy both A and B newspapers. Therefore, the number of families that buy both A and B are:
$ \Rightarrow n(A \cap B) = 5\% = \dfrac{5}{{100}} \times 10000 = 500$
4% families buy both A and C. Therefore, the number of families that buy both A and C are:
$ \Rightarrow n(A \cap C) = 4\% = \dfrac{4}{{100}} \times 10000 = 400$
3% families buy both B and C. Therefore, the number of families that buy B and C are:
$ \Rightarrow n(B \cap C) = 3\% = \dfrac{3}{{100}} \times 10000 = 300$
2% families buy all the three newspapers A, B and C. Therefore, the number of families that buy A, B and C are:
$ \Rightarrow n(A \cap B \cap C) = 2\% = \dfrac{2}{{100}} \times 10000 = 200$
Now, with this data, we can draw the Venn diagram as:
Number of families that buy only A and B not C =$n(A \cap B) - n(A \cap B \cap C)$ = 500 – 200 = 300
Number of families that buy only A and C not B = $n(A \cap C) - n(A \cap B \cap C)$ = 400 – 200 = 200
Number of families that buy both B and C but not A = $n(B \cap C) - n(A \cap B \cap C)$ = 300 – 200 = 100

From the Venn diagram, we can see that the 3300 families buy only newspaper A.

Hence, option (B) is correct.

Note:
Note: In this question, you may get confused in conversion of percentages into number of families and then in sketching the Venn diagram. You can also solve this question by using the formula: the number of families that buy only newspaper A = $n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$ = $4000 - 500 - 400 + 200 = 3300$.