In a car agency one day 120 cars were decorated with three different accessories viz , power window, AC and - Maths - Sets

Question

In a car agency one day 120 cars were decorated with three
different accessories viz , power window, AC and music system 80
cars were decorated with power windows 84 cars were decorated with
AC and 80 cars were decorated with music systems What is the
minimum and maximum number of cars which were decorated with of
three accessories?
(A) 10, 61 (B) 10, 45 (C) 25, 35 (D) None of these

Mayur Pisode · Accepted Answer

Dear Student,

Assuming the question is what is the minimum and maximum number of cars which were decorated with all of the three accessories.

Let A denote the set of cars with power windows
Let B denote the set of cars with an AC
Let C denote the set of cars with a music system

Here , A = 80 , B = 84 and C = 80

We know,

|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|

120 = 80 + 84 + 80

- |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|

|A \cap B| + |B \cap C| + |C \cap A| - |A \cap B \cap C|

= 244 - 120

|A \cap B| + |B \cap C| + |C \cap A| - |A \cap B \cap C|

= 124

Let

X \cap Y

denote the elements in both sets $X$ and $Y$ but not in set $Z$ for three sets $X$ , $Y$ , $Z$ . Then we have,

|A \cap B| + |B \cap C| + |C \cap A| - |A \cap B \cap C|

=

|A \cap B| + |B \cap C| + |C \cap A| + 2 |A \cap B \cap C|

Hence,

|A \cap B| + |B \cap C| + |C \cap A| + 2 |A \cap B \cap C|

= 124
The maximum value is

\frac{124}{2}

= 62 when

|A \cap B| = ||B \cap C|| = 0 a n d |C \cap A| = 1

Regards