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 newspaper, find the number of families which buy (1) A only (2) B only (3) none of A, B, C
(a) (1) 3000, (2) 1800, (3) 4600
(b) (1) 3300, (2) 1400, (3) 4000
(c) (1) 3500, (2) 1600, (3) 3800
(d) none

Answer 1

Hint: In this question, we first need to draw the venn diagram using the given conditions so that we get a clear picture of what we have to find. Then subtracting the number of families reading A & C , A & B and then adding the families reading all three gives the value of A only. In the same way subtract families reading B & C, B & A and then add those reading three to get the value of B only. Now, subtracting the whole things inside the venn diagram from 10000 gives the families reading none.

Complete step-by-step answer:
Now, the total number of families in the town are 10,000
Now, given that 40% of them buy newspaper A
\[\Rightarrow \dfrac{40}{100}\times 10000\]
\[\Rightarrow 4000\]
Thus, 4000 families read A
Now, given that 20% of them buy newspaper B
\[\Rightarrow \dfrac{20}{100}\times 10000\]
\[\Rightarrow 2000\]
Thus, 2000 families buy B
Now, given that 10% of them buy newspaper C
\[\Rightarrow \dfrac{10}{100}\times 10000\]
\[\Rightarrow 1000\]
Thus, 1000 families buy C
Now, given that 5% of them buy newspaper A & B
\[\begin{align}
  & \Rightarrow \dfrac{5}{100}\times 10000 \\
 & \Rightarrow 500 \\
\end{align}\]
Thus, 500 families buy both A & B.
Now, given that 3% of them buy newspaper B & C
\[\begin{align}
  & \Rightarrow \dfrac{3}{100}\times 10000 \\
 & \Rightarrow 300 \\
\end{align}\]
Thus, 300 families buy both B & C.
Now, given that 4% of them buy newspaper A & C
\[\begin{align}
  & \Rightarrow \dfrac{4}{100}\times 10000 \\
 & \Rightarrow 400 \\
\end{align}\]
Thus, 400 families buy both A & C.
Now, given that 2% of them buy newspaper A & B & C
\[\begin{align}
  & \Rightarrow \dfrac{2}{100}\times 10000 \\
 & \Rightarrow 200 \\
\end{align}\]
Thus, 200 families buy all A & B & C
Now, let us draw the venn diagram

(1) Now, let us find the number of families that buy only A
\[\Rightarrow A-\left( A\cap B \right)-\left( A\cap C \right)+\left( A\cap B\cap C \right)\]
Now, on substituting the respective values we get,
\[\Rightarrow 4000-500-400+200\]
Now, on further simplification we get,
\[\Rightarrow 3300\]
Thus, 3300 families buy only A
(2) Now, let us find the number of families that buy only B
\[\Rightarrow B-\left( B\cap C \right)-\left( B\cap A \right)+\left( A\cap B\cap C \right)\]
Now, on substituting the respective values we get,
\[\Rightarrow 2000-300-500+200\]
Now, on further simplification we get,
\[\Rightarrow 1400\]
Thus, 1400 families buy only B
(3) Let us now find the families that buy none of A, B, C
\[\Rightarrow 10000-\left( A+B+C-\left( A\cap B \right)-\left( B\cap C \right)-\left( C\cap A \right)+\left( A\cap B\cap C \right) \right)\]
Now, on substituting the respective values we get,
\[\Rightarrow 10000-\left( 4000+2000+1000-500-300-400+200 \right)\]
Now, on further simplification we get,
\[\Rightarrow 10000-6000\]
Now, this can be further written as
\[\Rightarrow 4000\]
Thus, 4000 families buy none of the newspaper
Hence, the correct option is (b).

Note: It is important to note that we need to draw the venn diagram first and then add or subtract the things accordingly by observing the venn diagram. Here, when we find only A after subtracting the things common to A & B, A & C we need to add \[A\cap B\cap C\] because we subtracted the term twice but we need to subtract only once.
It is also important to note that while doing the calculations we should not neglect any of the terms or consider it multiple times because it changes the result completely.