CAT 2025 Lesson : Set Theory - 4 Set Venn Diagrams

A class of $150$ students were asked if they liked or disliked each of the $4$ colours - red, blue, green and black. It was noted that $80, 70, 90$ and $60$ students liked the colours red, blue, green and black respectively. $30$ students did not like any of the $4$ colours, $40$ students liked exactly $2$ of the colours and $10$ students liked exactly $3$ of the colours. How many students liked all $4$ colours?

Solution

Let Red, Blue, Green and Black be the sets comprising of students who liked these respective colours. Let P, Q, R and S be the number of students who liked exactly $1, 2, 3$ and $4$ colours respectively. Let $X$ be the number of students who did not like any of the colours.

Total students $=$ P $+$ Q $+$ R $+$ S $+$ X $= 150$
⇒ P $+ 40 + 10 + S + 30 = 150$ ⇒ P $+$ S $= 70$ -----(1)
$n$(Red) $+$ $n$(Blue) $+ n$(Green) $+ n$(Black) $=$ P $+$ $2$Q $+$ $3$R $+$ $4$S
⇒ $80 + 70 + 90 + 60 =$ P $+ 80 + 30 + 4$S ⇒ P $+ 4$S $= 190$ -----(2)
Subtracting (1) from (2) ⇒ $3$S $=$ $120$ ⇒ S $=$ $40$

Answer: $40$

In a club of $42$ members, each of them played at least one of chess, carrom, rummy and poker. $18, 22, 24$ and $28$ members played chess, carrom, rummy and poker respectively. A total of $20$ members played exactly one game and number of people who played only chess, only carrom, only rummy and only poker are in increasing order of an arithmetic progression where all the terms are even. No member plays exactly $2$ games. $12$ members play chess & carrom, $12$ play chess & rummy, and $14$ play chess & poker.

(I) How many members play all $4$ games?
(II) How many members do not play chess but play carrom or rummy?
(III) How many members play chess, carrom and rummy?

Solution

Case I: $20$ members play exactly $1$ game and $0$ members play exactly $2$ games. Let P, Q, R and S be the number of people who play exactly $1$, exactly $2$, exactly $3$ and exactly $4$ games respectively.

P $+$ Q $+$ R $+$ S $= 42$ ⇒ $20 + 0 +$ R $+$ S $= 42$
⇒ R $+$ S $= 22$ -----(1)

P $+ 2$Q $+ 3$R $+ 4$S $= 18 + 22 + 24 + 28$
⇒ $3$R $+ 4$S $= 72$ -----(2)

Solving (1) and (2), we get R $= 16$ and S $= 6$

Case II: $6$ players play all $4$ games. So, the centre square that is common to all $4$ sets is filled with $6$. There are $6$ regions that are common to exactly $2$ of the $4$ sets. These are to be filled with $0$, as there is no member who plays exactly $2$ games. Number of players who play exactly $1$ game is $\bm{20}$. As the $4$ regions are in AP with common difference $= 2$, the 1st and 4th terms are $a$ and $a + 6$. Average of $4$ terms $= \dfrac{a + a + 6}{2} = \dfrac{20}{4}$ ⇒ $a = 2$

∴ Members playing only chess, only carrom, only rummy and only poker are $2, 4, 6$ and $8$ respectively. Let $p, q, r$ & $s$ be the number of players who play exactly $3$ games pertaining to the regions shown above.

From the question, we can form the following equations. P $+$ Q $+$ R $+$ S $=$ $16$      [$16$ members play exactly $3$ games] P $+$ R $+$ 6 $= 12$ ⇒ P $+$ R $= 6$      [$12$ play chess & carrom] Q $+$ R $+$ 6 $= 12$ ⇒ Q $+$ R $= 6$      [$12$ play chess & rummy] P $+$ Q $+$ 6 $= 14$ ⇒ P $+$ Q $= 8$      [$14$ play chess & poker] Solving the $4$ equations above, we get P = 4, Q = 4, R = 2, S = 6

∴ Members who do not play chess but play carrom or rummy $= 6 + 4 + 6 =$ $\bm{16}$

Case III: Number of members playing chess, carrom and rummy $= 6 + 2 =$ $\bm{8}$

Answer: (I) $6$; (II) $16$; (III) $8$