4. Venn Diagrams with 3 Sets

The following is a Venn Diagram with $3$ sets - A, B and C. We typically write the name of the set just outside that circle in capital letters. In small letters from $a$ to $h$ are the numbers of elements in each of the $8$ different regions formed. For instance, $d$ is the number of elements that are in sets A and B but not in set C.

$n$(A) $= a + d + f + g$
$n$(B) $= b + d + e + g$
$n$(C) $= c + e + f + g$

Let P, Q and R be the number of elements in exactly 1 set, exactly 2 sets and all 3 sets respectively.

P $= a + b + c$
Q $= d + e + f$
R $= g$

$n$(A $\cup$ B $\cup$ C) $= a + b + c + d + e + f + g$
⇒ $\bm{n}$(A $ \ \bm{\cup} \ $ B $\ \bm{\cup} \ $ C) $=$ P + Q + R $\space\space\space$ (Property 1)

$n$(A) $+ n$(B) $+ n$(C) $= a + b + c + 2(d + e + f) + 3g$
⇒ $\bm{n}$(A) $+$ $\bm{n}$(B) $+$ $\bm{n}$(C) $=$ P + 2Q + 3R $\space\space\space$ (Property 2)

Note that $n$(u) = $n$(A $\cup$ B $\cup$ C) $+ h$
⇒ $\bm{n}$(U) $=$ P + Q + R + h $\space\space\space$ (Property 3)

Example 7

A survey was conducted in a neighbourhood of $700$ residents to assess the readership of $3$ newspapers - The Hindu (TH), The Indian Express (IE) and The Times of India (TOI). Every resident reads at least one of these papers. At the end of the survey it was found that $240$ read TH, $400$ read IE, $500$ read TOI and $100$ read all $3$ newspapers.

(I) How many read exactly $1$ paper?
(II) How many read at least $2$ of these newspapers?

Solution

Let P, Q and R be the number of people who read exactly $1$, exactly $2$ and all $3$ newspapers respectively.
As $100$ residents read all $3$ newspapers, R = 100.

As every resident reads at least one of the newspapers, $n$(U) $=$ P $+$ Q $+$ R $=$ $700$
⇒ P $+$ Q $= 600$ -----(1)
$n$(TH) $+$ $n$(IE) $+$ $n$(TOI) $= 240 + 400 + 500$
⇒ P $+$ $2$Q $+$ $3$R $= 1140$
⇒ P $+$ $2$Q $= 840$ -----(2)

Solving (1) and (2), we get P = 360 & Q = 240

Case I: Number of residents who read exactly $1$ paper $=$ P $= 360$
Case II: Number of people reading at least $2$ newspapers $=$ Q $+$ R $= 340$

Answer: (I) $360$; (II) $340$

Example 8

Every student plays at least one of $3$ sports - cricket, football and hockey. $40 \%$ play cricket, $44 \%$ play football and $60 \%$ play hockey. $14 \%$ play cricket and football, $20 \%$ play football and hockey, while $16 \%$ hockey and cricket.
(I) What percentage of the students play all sports?
(II) If x is the number of students who play hockey but not football, then what percent of x play cricket?
(III) What percent of students play hockey or cricket but not both?

Solution

Let C, F and H represent the sets of students who play Cricket, Football and Hockey respectively.
Let P, Q and R be the number of people who play exactly $1$, exactly $2$ and all $3$ sports respectively.
Let $100$ be the total number of students. We can now write the percentages as absolute values.

As every student plays at least one of the sports,
$n$(U) $= n$(C $\cup$ F $\cup$ H) $= 100$
⇒ P $+$ Q $+$ R $= 100$ -----(1)

P $+$ $2$Q $+$ $3$R
$= n$(C) $+$ $n$(F) $+$ $n$(H)
$= 40 + 44 + 60$
⇒ P $+$ $2$Q $+$ $3$R$ = 144$ -----(2)

As $14$ play cricket & football ⇒ $d + g = 14$ -----(4)
As $20$ play football & hockey ⇒ $e + g = 20$ -----(5)
As $16$ play hockey & cricket ⇒ $f + g = 16$ -----(6)

Adding these, d $+$ e $+$ f $+$ $3$g $= 50$
⇒ Q $+$ $3$R $= 50$ -----(3)
Solving equations (1), (2) & (3), we get P = 62, Q = 32 & R = 6.
R$ = \bm{g = 6}$
Substituting these in (4), (5) and (6), we get $\bm{d = 8, e = 14, f = 10}$

Substituting these values,
$\bm{a} = 40 - d - f - g = \bm{16}$
$\bm{b} = 44 - d - e - g = \bm{16}$
$\bm{c} = 60 - e - f - g = \bm{30}$

Case I: $6 \%$ of the students play all games.
Case II: Number of students who play hockey but not football $= x = c + f = 30 + 10 = 40$
Number of students in these who play cricket $= f = 10$
Percent of Cricket players in $x = \dfrac{10}{40} \times 100 = 25 \%$
Case III: Students playing hockey or cricket but not both $= a + d + c + e = 68 \%$

Answer: (I) $6 \%$; (II) $25 \%$; (III) $68 \%$

Example 9

There are $200$ students studying in Josh College of Management, which has $3$ clubs - A, B and C. Students are allowed to be a member in more than $1$ club. $20 \%$ of the students are a member in all clubs. $30$ students are not a member of Club A alone. Clubs B and C have $100$ and $120$ students as members respectively.

(I) How many students are not members in clubs B or C?

(1) $30$            (2) $40$            (3) $50$            (4) Cannot be determined

(II) How many students are not in club A?

(1) $30$            (2) $40$            (3) $50$            (4) Cannot be determined

(III) What percentage of the students who passed in C did not pass in B?

(1) $41.7 \%$            (2) $58.3 \%$            (3) $66.7 \%$            (4) Cannot be determined

(IV) At most how many students are in exactly one of A or C?

(1) $100$            (2) $130$            (3) $160$            (4) Cannot be determined

Solution

This case, despite requiring a
$3$-set venn diagram, has very little information.
Therefore, we shall start by drawing a venn diagram
with the $8$ different regions marked with variables from $a$ to $h$ as follows.

Members in all $3$ clubs
$=$ $\bm{g}$ $= \dfrac{20}{100} \times 200 =$ $\bm{40}$

If $30$ members are not in Club A alone,
then they are definitely members in both Clubs B and C. Therefore, $\bm{e} =$ $\bold{30}$

B $=$ $b + d + e + g = 100$ ⇒ $ b + d + 30 + 40 = 100$
⇒ $\bm{b + d}$ $= \boldsymbol{30}$

C $= c + f + e + g = 120$ ⇒ $c + f + 30 + 40 = 120$
⇒ $\bm{c + f}$ $= \boldsymbol{50}$

U $= a + b + c + d + e + f + g + h = 200$
⇒ B $+ a + f + c + h = 200$
⇒ $100 + a + 50 + h = 200$
⇒ $\bm{a + h}$ $= \boldsymbol{50}$

(I) To select students who are not members in clubs B or C, we need to identify elements not coming under the circles B or C, i.e. $\bm{a + h}$ $= \boldsymbol{50}$

(II) Members who are not in club A $= b + e + c + h = b + c + h + 30$
As we cannot solve for $b, c$ or $h$, this value cannot be determined.

(III) $120$ students passed in C. Of this, number of students who did not pass in B $= c + f = 50$
∴ Of the students who passed in C, the $ \%$ who did not pass in B $= \dfrac{50}{120} \times 100 \%$ $= \boldsymbol{41.7 \%}$

(IV) Students in exactly $1$ of A or C $=$ Students in A but not in C $+$ Students in C but not in A $= a + d + c + e = a + d + c + 30$

From the $3$ equations we formed earlier ⇒ $b + d = 30, c + f = 50$ and $a + h = 50$, maximum possible values of $a, d$ and $c$ are $50, 30$ and $50$ respectively.

∴ Maximum number of students in exactly $1$ of A or C $= a + d + c + 30 = 50 + 30 + 50 + 30 =$ $\boldsymbol{160}$

Answer: (I) (3) $50$; (II) (4) Cannot be determined; (III) (1) $41.7 \%$; (IV) (3) $160$