Sets

Let the total number of students be $100\%$. Let $E$ and $M$ be the sets of students who passed in English and Mathematics.

$n(E)=75\%$, $n(M)=55\%$, $n(\overline{E\cup M}) = 5\%$, so $n(E\cup M) = 95\%$.

$n(E\cap M) = 75\% + 55\% - 95\% = 35\%$, and this equals 21 students.

(i) Only English $= 40\%$, both $= 35\%$, only Mathematics $= 20\%$, neither $= 5\%$.

 U
+---------------------------------------+
|     E               M                 |
|   +-----------+-----------+           |
|   |    40%    |    35%    |   20%     |
|   |           |  (21)     |           |
|   +-----------+-----------+           |
|                                       |
|   outside both = 5%                   |
+---------------------------------------+

(ii) $35\%$ of the total $= 21$, so total $= 21 \times \dfrac{100}{35} =$ 60 students.

(iii) Passed only in Mathematics $= 55\% - 35\% = 20\%$ of $60 =$ 12 students.

Check: $24 + 21 + 12 + 3 = 60$.

Let $A$ be the set of students who passed in Social Studies and $B$ the set who passed in Mathematics. Let the total number of students be $n(U)$.

(ii) Failed in both $= 5\%$, so passed in at least one subject $= 100 - 5 = 95\%$.

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ $$95\% = 80\% + 75\% - n(A \cap B) \Rightarrow n(A \cap B) = 60\%$$

Given $n(A \cap B) = 300$, so $60\%$ of the total is 300:

$$n(U) = \frac{300 \times 100}{60} = 500$$

Number of students appeared $= 500$.

(i) Numbers in each region (out of 500):

• passed in both: $60\%$ of 500 $= 300$
• passed in Social Studies only: $(80 - 60)\% = 20\%$ of 500 $= 100$
• passed in Mathematics only: $(75 - 60)\% = 15\%$ of 500 $= 75$
• failed in both: $5\%$ of 500 $= 25$

Check: $100 + 300 + 75 + 25 = 500$.

 U, n(U) = 500
+-------------------------------------+
|     A           B                   |
|   +-----------+---+-----------+     |
|   |    100    |300|    75     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 25                 |
+-------------------------------------+

(A = Social Studies, B = Mathematics.)

(iii) Number who passed only in Mathematics $=$ 75.

Let $A$ be the set of students who passed in Nepali and $B$ the set who passed in Science. Let the total number of students be $n(U)$.

(ii) Failed in both $= 5\%$, so passed in at least one subject $= 100 - 5 = 95\%$.

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ $$95\% = 80\% + 75\% - n(A \cap B) \Rightarrow n(A \cap B) = 60\%$$

Given $n(A \cap B) = 300$, so $60\%$ of the total is 300:

$$n(U) = \frac{300 \times 100}{60} = 500$$

Number of students appeared $= 500$.

(i) Numbers in each region (out of 500):

• passed in both: $60\%$ of 500 $= 300$
• passed in Nepali only: $(80 - 60)\% = 20\%$ of 500 $= 100$
• passed in Science only: $(75 - 60)\% = 15\%$ of 500 $= 75$
• failed in both: $5\%$ of 500 $= 25$

Check: $100 + 300 + 75 + 25 = 500$.

 U, n(U) = 500
+-------------------------------------+
|     A           B                   |
|   +-----------+---+-----------+     |
|   |    100    |300|    75     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 25                 |
+-------------------------------------+

(A = Nepali, B = Science.)

(iii) Number who passed only in Science $=$ 75.

(a) $n(\overline{B \cup C}) = 20$.

(b) $n(B \cup C) = 120 - 20 = 100$. Both $= 60 + 55 - 100 = 15$.

Cricket only $= 45$; basketball only $= 40$.

 U, n(U) = 120
+-------------------------------------+
|     C           B                        |
|   +-----------+---+-----------+     |
|   |     45    | 15  |    40     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 20                  |
+-------------------------------------+

(Only C = 45; both = 15; only B = 40; neither = 20.)

(c) Cricket only $= 45$ students.

(d) Cricket only : basketball only $= 45 : 40 = 9 : 8$. Cricket-only students are 5 more than basketball-only students.

(a) $n(S \cap M) = 350$.

(b) S only $= 550 - 350 = 200$; M only $= 600 - 350 = 250$.

 U
+-------------------------------------+
|     S           M                        |
|   +-----------+---+-----------+     |
|   |     200   | 350 |    250    |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 200                 |
+-------------------------------------+

(Only S = 200; both = 350; only M = 250; neither = 200.)

(c) Total $= 200 + 350 + 250 + 200 = 1000$ people.

(d) Only sports : only music $= 200 : 250 = 4 : 5$.

(a) $n(S \cap M) = 35$.

(b) S only $= 55 - 35 = 20$; M only $= 60 - 35 = 25$.

 U
+-------------------------------------+
|     S           M                        |
|   +-----------+---+-----------+     |
|   |     20    | 35  |    25     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 20                  |
+-------------------------------------+

(Only S = 20; both = 35; only M = 25; neither = 20.)

(c) Total $= 20 + 35 + 25 + 20 = 100$ students.

(d) Only Science : only Mathematics $= 20 : 25 = 4 : 5$.

Let $U$ be all the people surveyed, $F$ those who like folk songs and $M$ those who like modern songs.

$n(F) = 70\%$, $n(M) = 60\%$, neither $= 10\%$ of $n(U)$.

(ii) People liking at least one kind $= 100\% - 10\% = 90\%$.

$n(F \cap M) = 70\% + 60\% - 90\% = 40\% \text{ of } n(U)$

Given $n(F \cap M) = 4000$, so $40\%$ of $n(U) = 4000$:

$n(U) = \frac{4000 \times 100}{40} = 10000$

The total number of people in the survey is $10000$.

(i) Venn diagram (with $n(U) = 10000$):

• Folk only $= 70\% - 40\% = 30\%$ of $10000 = 3000$
• Both $= 4000$
• Modern only $= 60\% - 40\% = 20\%$ of $10000 = 2000$
• Neither $= 10\%$ of $10000 = 1000$

 U, n(U) = 10000
+-------------------------------------+
|     F           M                   |
|   +-----------+---+-----------+     |
|   |    3000   |4000|   2000   |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 1000               |
+-------------------------------------+

Check: $3000 + 4000 + 2000 + 1000 = 10000$.

(iii) The number who like folk songs only is $3000$.

Let $F$ = folk songs and $M$ = modern songs, with total $= 100\%$.

Both $= 30\%$, folk only $= 65 - 30 = 35\%$, modern only $= 50 - 30 = 20\%$.

Neither $= 100 - (35 + 30 + 20) = 15\%$.

(i) Venn diagram (percentages):

 U (100%)
+-----------------------------------+
|     F               M             |
|   +--------+------+--------+      |
|   |  35%   | 30%  |  20%   |      |
|   +--------+------+--------+      |
|                                   |
|   outside both = 15%              |
+-----------------------------------+

(ii) $15\%$ of the total $= 450$:

$\text{Total} = \frac{450\times 100}{15} = 3000$

3000 people took part in the survey.

(iii) Folk songs only $= 35\%$ of $3000 = 1050$ people.

Let $C$ be the set of people who like curd and $M$ the set who like milk, with $n(C)=70\%$, $n(M)=60\%$ and neither $=20\%$.

(ii) People liking at least one: $n(C\cup M)=100\%-20\%=80\%$.

$n(C\cap M)=n(C)+n(M)-n(C\cup M)=70\%+60\%-80\%=50\%.$

Given $50\%$ of the total $=550$:

$\text{Total}=550\times\frac{100}{50}=\mathbf{1100}\text{ people}.$

(i) Curd only $=70\%-50\%=20\%$, milk only $=60\%-50\%=10\%$. In numbers:

• Curd only $=20\%$ of $1100=220$
• Both $=550$
• Milk only $=10\%$ of $1100=110$
• Neither $=20\%$ of $1100=220$

U = 1100
+--------------------------------------+
|   C                M                  |
|  +-----------+-----------+            |
|  |    220    |    550    |   110      |
|  |           |           |            |
|  +-----------+-----------+            |
|                          neither: 220 |
+--------------------------------------+

Check: $220+550+110+220=1100$.

(iii) The number of people who like curd only is 220.

Let the total number of people be $N$, and let $A$ and $O$ be the sets who like apple and orange.

$n(A) = 60\%$, $n(O) = 70\%$, $n(A \cap O) = 400$, $n(\overline{A \cup O}) = 10\%$.

(ii) People liking at least one $= 100\% - 10\% = 90\%$ of $N$.

$n(A \cup O) = n(A) + n(O) - n(A \cap O)$ $90\% N = 60\% N + 70\% N - 400 \Rightarrow 400 = 40\% N \Rightarrow N = 1000$

The total number of people is $1000$.

(iii) $n(A) = 600$, so apple only $= 600 - 400 = 200$.

(i) Venn diagram (values from the above):

 U, n(U) = 1000
+-------------------------------------+
|     A           O                   |
|   +-----------+---+-----------+     |
|   |    200    |400|    300    |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 100                |
+-------------------------------------+

(Apple only $= 200$; both $= 400$; orange only $= 700 - 400 = 300$; none $= 100$. Check: $200 + 400 + 300 + 100 = 1000$.)

The number who like apple only is $200$.

(a) $n(U) = 450$.

(b) Let $both = x$. Total: $144 + x + 216 + 63 = 450 \Rightarrow x = 450 - 423 = 27$.

 U, n(U) = 450
+-------------------------------------+
|     M           S                        |
|   +-----------+---+-----------+     |
|   |     144   | 27  |    216    |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 63                  |
+-------------------------------------+

(Only M = 144; both = 27; only S = 216; neither = 63.)

(c) Number who liked mathematics $= n(M) = 144 + 27 = 171$.

(d) Both $= 27$, none $= 63$.

Percentage less $= \frac{63 - 27}{63} \times 100\% = \frac{36}{63} \times 100\% = 57.14\%$.

So the number who liked both subjects is about $57.14\%$ less than the number who liked neither.

(a) $n(I)= 320$.

(b) Let $n(I \cap A) = x$. Then I only $= 320 - x$, A only $= 225 - x$, and nobody is outside.

 U, n(U) = 500
+-------------------------------------+
|     I           A                        |
|   +-----------+---+-----------+     |
|   |    320-x  |  x  |   225-x   |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 0                   |
+-------------------------------------+

(Only I = 320-x; both = x; only A = 225-x; neither = 0.)

(c) Since $n(I \cup A) = n(U) = 500$:

$(320 - x) + x + (225 - x) = 500 \Rightarrow 545 - x = 500 \Rightarrow x = 45$.

I only $= 320 - 45 = 275$; A only $= 225 - 45 = 180$.

Exactly one type $= 275 + 180 = 455$ youths.

(d) Android only $= 180$, iPhone only $= 275$.

Difference $= 95$; $\frac{95}{275} \times 100\% = 34.55\%$.

The number of youths who liked Android only is about $34.55\%$ less than those who liked iPhone only.

(a) $n(F \cap V) = 20$.

(b) Let football only $= 2k$, volleyball only $= k$. Then

$2k + 20 + k + 10 = 120 \Rightarrow 3k = 90 \Rightarrow k = 30$.

So football only $= 60$, volleyball only $= 30$.

 U, n(U) = 120
+-------------------------------------+
|     F           V                        |
|   +-----------+---+-----------+     |
|   |     60    | 20  |    30     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 10                  |
+-------------------------------------+

(Only F = 60; both = 20; only V = 30; neither = 10.)

(c) $n(F) = 60 + 20 = 80$ students like football.

(d) $n(V) = 30 + 20 = 50$, $n(F) = 80$.

Percentage less $= \frac{80 - 50}{80} \times 100\% = 37.5\%$.

Volleyball lovers are $37.5\%$ fewer than football lovers.

(a) $n(\overline{M \cup E}) = 30$.

(b) Let $n(M \cap E) = x$. Students liking at least one subject $= 200 - 30 = 170$.

$n(M \cup E) = n(M) + n(E) - n(M \cap E) \Rightarrow 170 = 120 + 110 - x \Rightarrow x = 60$.

M only $= 120 - 60 = 60$; E only $= 110 - 60 = 50$.

 U, n(U) = 200
+-------------------------------------+
|     M           E                        |
|   +-----------+---+-----------+     |
|   |     60    | 60  |    50     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 30                  |
+-------------------------------------+

(Only M = 60; both = 60; only E = 50; neither = 30.)

(c) Exactly one subject $= 60 + 50 = 110$ students.

(d) Both subjects $= 60$ and double of those liking none $= 2 \times 30 = 60$. They are equal, so I agree with Ram's statement.

(a) $n(\overline{M \cup S}) = 12$.

(b) Let Maths only $= 4k$, Science only $= 3k$, both $= 5$, none $= 12$.

$4k + 5 + 3k + 12 = 80 \Rightarrow 7k = 63 \Rightarrow k = 9$.

Maths only $= 36$, Science only $= 27$.

 U, n(U) = 80
+-------------------------------------+
|     M           S                        |
|   +-----------+---+-----------+     |
|   |     36    |  5  |    27     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 12                  |
+-------------------------------------+

(Only M = 36; both = 5; only S = 27; neither = 12.)

(c) $n(M) = 36 + 5 = 41$ students.

(d) At least one subject $= 36 + 5 + 27 = 68$. At most one subject $= 80 - 5 = 75$ (all except those taking both).

Ratio $= 68 : 75$.

(a) $n(\overline{N \cup E}) = 15$.

(b) Students in at least one competition $= 205 - 15 = 190$.

$n(N \cap E) = 140 + 100 - 190 = 50$.

N only $= 140 - 50 = 90$; E only $= 100 - 50 = 50$.

 U, n(U) = 205
+-------------------------------------+
|     N           E                        |
|   +-----------+---+-----------+     |
|   |     90    | 50  |    50     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 15                  |
+-------------------------------------+

(Only N = 90; both = 50; only E = 50; neither = 15.)

(c) Exactly one language $= 90 + 50 = 140$ students.

(d) At least one $= 190$. At most one $= 205 - 50 = 155$.

Ratio $= 190 : 155 = 38 : 31$. So the number taking part in at least one is greater than the number taking part in at most one.

(a) $n(I \cap A) = 10$.

(b) Youths liking iPhone or Android $= 80 - 10 = 70$.

Let $n(I) = 3k$ and $n(A) = 2k$. Then $n(I \cup A) = 3k + 2k - 10 = 70 \Rightarrow 5k = 80 \Rightarrow k = 16$.

$n(I) = 48$, $n(A) = 32$. I only $= 48 - 10 = 38$; A only $= 32 - 10 = 22$.

 U, n(U) = 80
+-------------------------------------+
|     I           A                        |
|   +-----------+---+-----------+     |
|   |     38    | 10  |    22     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 10                  |
+-------------------------------------+

(Only I = 38; both = 10; only A = 22; neither = 10.)

(c) iPhone only $= 38$ youths.

(d) Difference $= 38 - 22 = 16$.

$\frac{16}{22} \times 100\% = 72.73\%$.

The number liking iPhone only is about $72.73\%$ more than the number liking Android only.

(a) $n(\overline{B \cup C}) = 100$.

(b) Let both $= x$: $100 + x + 60 + 100 = 360 \Rightarrow x = 100$.

 U, n(U) = 360
+-------------------------------------+
|     B           C                        |
|   +-----------+---+-----------+     |
|   |     100   | 100 |    60     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 100                 |
+-------------------------------------+

(Only B = 100; both = 100; only C = 60; neither = 100.)

(c) Students who like both games $= 100$.

(d) In the second survey the 100 students who liked no game now like cricket, so every student likes at least one game.

$n(B \cup C) = 100 + 100 + (60 + 100) = 360$.

Number liking at least one game $= 360$.

(a) $n(\overline{A \cup O}) = 23$.

(b) Let both $= x$: $75 + x + 45 + 23 = 160 \Rightarrow x = 17$.

 U, n(U) = 160
+-------------------------------------+
|     A           O                        |
|   +-----------+---+-----------+     |
|   |     75    | 17  |    45     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 23                  |
+-------------------------------------+

(Only A = 75; both = 17; only O = 45; neither = 23.)

(c) $n(A) = 75 + 17 = 92$ people like apple.

(d) $n(O) = 45 + 17 = 62$.

$n(A) = 92 > n(O) = 62$; apple lovers exceed orange lovers by 30. Ratio $92 : 62 = 46 : 31$.

(a) $n(\overline{H \cup A}) = 30$.

(b) $n(H \cup A) = 120 - 30 = 90$.

$n(H \cap A) = 70 + 50 - 90 = 30$. H only $= 40$; A only $= 20$.

 U, n(U) = 120
+-------------------------------------+
|     H           A                        |
|   +-----------+---+-----------+     |
|   |     40    | 30  |    20     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 30                  |
+-------------------------------------+

(Only H = 40; both = 30; only A = 20; neither = 30.)

(c) Only one business $= 40 + 20 = 60$ households.

(d) If 10 of the "other work" households start homestay, $n(H) = 70 + 10 = 80$ and $n(A) = 50$.

Ratio $= 80 : 50 = 8 : 5$.

(a) $n(T \cup C) = 450 - 50 = 400$.

(b)
First find $n(T \cap C)$: $n(T \cap C) = n(T) + n(C) - n(T \cup C) = 200 + 250 - 400 = 50.$

Tea only $= n(T) - n(T \cap C) = 200 - 50 = 150$.
Coffee only $= n(C) - n(T \cap C) = 250 - 50 = 200$.
Neither $= 50$.

Venn diagram:

U (n(U) = 450)
+-----------------------------------+
|   +-----------+   +-----------+   |
|  /    T       \ /      C      \  |
| |    150      | 50  |    200   | |
|  \            \     /          /  |
|   +------------+---+-----------+  |
|                                   |
|         50                        |
+-----------------------------------+

(Only T = 150; $T \cap C$ = 50; only C = 200; neither = 50.)

(c) Tea only $= 150$ people.

(d) Both $= 50$, coffee only $= 200$.
Ratio $= 50 : 200 = 1 : 4$.
So the number of people who like both tea and coffee is one-fourth of the number who like coffee only.

(a) $n(J \cap B) = 20$.

(b) J only $= 60 - 20 = 40$; B only $= 40 - 20 = 20$; both $= 20$; neither $= 10$.

 U
+-------------------------------------+
|     J           B                        |
|   +-----------+---+-----------+     |
|   |     40    | 20  |    20     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 10                  |
+-------------------------------------+

(Only J = 40; both = 20; only B = 20; neither = 10.)

(c) Total students $= n(U) = 40 + 20 + 20 + 10 = 90$.

(d) The 10 students now consider Janakpur suitable, so J only $= 40 + 10 = 50$; B only $= 20$.

Ratio $= 50 : 20 = 5 : 2$.

(a) $n(\overline{E \cup M}) = 20$.

(b) Let both $= x$. E only $= 60 - x$, M only $= 50 - x$.

 U, n(U) = 100
+-------------------------------------+
|     E           M                        |
|   +-----------+---+-----------+     |
|   |    60-x   |  x  |   50-x    |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 20                  |
+-------------------------------------+

(Only E = 60-x; both = x; only M = 50-x; neither = 20.)

(c) $(60 - x) + x + (50 - x) + 20 = 100 \Rightarrow 130 - x = 100 \Rightarrow x = 30$.

E only $= 30$, M only $= 20$. Exactly one subject $= 30 + 20 = 50$ students.

(d) If 30 students like neither, $n(E \cup M) = 70$. With $n(E) = 60$, $n(M) = 50$ unchanged:

$n(E \cap M) = 60 + 50 - 70 = 40$.

The number liking both subjects increases by 10 (from 30 to 40).

(a) $n(G \cap T) = 9$.

(b) G only $= 34 - 9 = 25$; T only $= 24 - 9 = 15$; both $= 9$.

 U, n(U) = 50
+-------------------------------------+
|     G           T                        |
|   +-----------+---+-----------+     |
|   |     25    |  9  |    15     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = x                   |
+-------------------------------------+

(Only G = 25; both = 9; only T = 15; neither = x.)

(c) From the diagram: $25 + 9 + 15 + x = 50 \Rightarrow 49 + x = 50 \Rightarrow x = 1$.

1 student did not like any stream.

(d) Yes, the diagram changes. If 24 students like both streams, then all of $T$ ($n(T) = 24$) lies inside $G$, so $T \subset G$. T only becomes $0$, and the two circles become one inside the other (G only $= 10$, none $= 16$).

(a) $n(\overline{S \cup E}) = 10$.

(b) Let both $= x$: $30 + x + 40 + 10 = 100 \Rightarrow x = 20$.

 U, n(U) = 100
+-------------------------------------+
|     S           E                        |
|   +-----------+---+-----------+     |
|   |     30    | 20  |    40     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 10                  |
+-------------------------------------+

(Only S = 30; both = 20; only E = 40; neither = 10.)

(c) $n(E) = 40 + 20 = 60$ students like economics.

(d) $30\%$ of the total $= \frac{30}{100} \times 100 = 30$. But $n(E) = 60 \ne 30$, so no, $n(E)$ is $60\%$ of the total students.

(a) $n(\overline{P \cup L}) = 10$.

(b) Let both $= x$: $20 + x + 15 + 10 = 50 \Rightarrow x = 5$.

 U, n(U) = 50
+-------------------------------------+
|     P           L                        |
|   +-----------+---+-----------+     |
|   |     20    |  5  |    15     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 10                  |
+-------------------------------------+

(Only P = 20; both = 5; only L = 15; neither = 10.)

(c) $n(P) = 20 + 5 = 25$ tourists liked Pokhara.

(d) Only one place $= 20 + 15 = 35$; neither $= 10$. Since $35 > 10$, yes, more tourists like only one place than neither.

(a) $n(F \cup V) = 20$ (every student plays at least one game).

(b) Let both $= x$: $(14 - x) + x + (12 - x) = 20 \Rightarrow 26 - x = 20 \Rightarrow x = 6$.

F only $= 8$, V only $= 6$.

 U, n(U) = 20
+-------------------------------------+
|     F           V                        |
|   +-----------+---+-----------+     |
|   |      8    |  6  |     6     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 0                   |
+-------------------------------------+

(Only F = 8; both = 6; only V = 6; neither = 0.)

(c) Exactly one game $= 8 + 6 = 14$ students.

(d) The two are from the football-only group, so F only becomes $6$. Hence $n(F)$ changes (from 14 to 12), and with it $n(F \cup V)$ (from 20 to 18) and $n(F \text{ only})$. $n(V)$ and $n(F \cap V)$ stay the same, because the two students were not in volleyball.

(a) $n(O \cap M) = 75$.

(b) O only $= 80 - 75 = 5$; M only $= 85 - 75 = 10$; both $= 75$; neither $= 50$.

 U
+-------------------------------------+
|     O           M                        |
|   +-----------+---+-----------+     |
|   |      5    | 75  |    10     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 50                  |
+-------------------------------------+

(Only O = 5; both = 75; only M = 10; neither = 50.)

(c) Total $= 5 + 75 + 10 + 50 = 140$ people.

(d) Only one fruit $= 5 + 10 = 15$; both $= 75$. So the number liking both fruits is 5 times the number liking only one fruit (ratio $15 : 75 = 1 : 5$).

(a) $n(\overline{M \cup E}) = 6$.

(b) Let both $= x$: $11 + x + 13 + 6 = 45 \Rightarrow x = 15$.

 U, n(U) = 45
+-------------------------------------+
|     M           E                        |
|   +-----------+---+-----------+     |
|   |     11    | 15  |    13     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 6                   |
+-------------------------------------+

(Only M = 11; both = 15; only E = 13; neither = 6.)

(c) $n(M) = 11 + 15 = 26$ students like Mathematics.

(d) If every student who likes English also likes Mathematics, then $E \subset M$, so $M \cap E = E$. $n(E) = 13 + 15 = 28$, so $n(M \cap E)$ becomes $28$ (increases by 13 from 15). (Here the question's $M \cap N$ means $M \cap E$.)

(a) $n(\overline{N \cup E}) = 25$.

(b) Let both $= x$: $85 + x + 65 + 25 = 250 \Rightarrow x = 75$.

 U, n(U) = 250
+-------------------------------------+
|     N           E                        |
|   +-----------+---+-----------+     |
|   |     85    | 75  |    65     |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 25                  |
+-------------------------------------+

(Only N = 85; both = 75; only E = 65; neither = 25.)

(c) $n(N) = 85 + 75 = 160$ students like Nepali.

(d) $n(E) = 65 + 75 = 140$. Nepali lovers (160) are 20 more than English lovers (140); ratio $160 : 140 = 8 : 7$.

(a) People who like to sing only: $n(S \cap D^c)$, which is the same as $n(S) - n(S \cap D)$ (or $n(S-D)$), and its value is $60$.

(b) Let the number who like both be $x$. Then the number who like neither is $3x$ (ratio $3:1$). Total: $60 + x + 100 + 3x = 320 \Rightarrow 4x = 160 \Rightarrow x = 40$. So both $= 40$ and neither $= 3 \times 40 = 120$.

 U, n(U) = 320
+-------------------------------------+
|     S               D               |
|   +-----------+---+-----------+     |
|   |     60    | 40|    100    |     |
|   +-----------+---+-----------+     |
|                                     |
|   outside both = 120                |
+-------------------------------------+

(c) Number who like to sing $= n(S) = 60 + 40 = 100$.

(d) Percentage who like neither $= \dfrac{120}{320} \times 100\% = 37.5\%$.

Let $M$ and $S$ be the sets of people who like Mathematics and Science. Here $n(U)=150$, $n(M\cap S)=45$ and neither $=30$.

$n(M\cup S)=150-30=120$

Let $n(M)=2k$ and $n(S)=3k$. Using $n(M\cup S)=n(M)+n(S)-n(M\cap S)$:

$120 = 2k+3k-45 \Rightarrow 5k=165 \Rightarrow k=33$

So $n(M)=66$ and $n(S)=99$.

Venn diagram:

 U, n(U) = 150
+--------------------------------------+
|    M            S                    |
|  +----------+----+----------+        |
|  |    21    | 45 |    54    |        |
|  +----------+----+----------+        |
|                                      |
|   outside both = 30                  |
+--------------------------------------+

Mathematics only $= n(M)-n(M\cap S) = 66-45 = 21$.

Check: $21+45+54+30=150$.

Hence 21 people liked Mathematics only.

(a) $n(I\cup A)=n(U)-\text{none}=300-25=275$.

(b) Let $n(I\cap A)=x$. Using $n(I\cup A)=n(I)+n(A)-n(I\cap A)$:

$275=150+200-x \Rightarrow x=75$

So I only $=150-75=75$, A only $=200-75=125$.

 U, n(U) = 300
+--------------------------------------+
|    I            A                    |
|  +----------+----+----------+        |
|  |    75    | 75 |   125    |        |
|  +----------+----+----------+        |
|                                      |
|   outside both = 25                  |
+--------------------------------------+

Check: $75+75+125+25=300$.

(c) The number who liked I-phone only $=150-75=$ 75.

(d) Both phones $=75$; neither phone $=25$. The number who like both is $3$ times the number who like neither, that is, $50$ more (or $200\%$ more).

(a) $n(\overline{F\cup V})=30$.

(b) Let $n(F\cap V)=x$. People who like at least one game $=200-30=170$, so

$n(F\cup V)=n(F)+n(V)-n(F\cap V)\Rightarrow 170=120+85-x\Rightarrow x=35.$

Football only $=120-35=85$; volleyball only $=85-35=50$.

 U, n(U) = 200
+--------------------------------------+
|    F                V                |
|  +----------+--------+----------+    |
|  |    85    |   35   |    50    |    |
|  +----------+--------+----------+    |
|                       none = 30      |
+--------------------------------------+

(c) Number who like volleyball only $=n(V)-n(F\cap V)=85-35=50$.

(d) Both games: 35 people; neither game: 30 people. So the number liking both games is greater, by $35-30=5$.

Tea only $=28$, both $=12$, so those who like tea $=28+12=40$.

Tea drinkers are twice the milk drinkers, so milk drinkers $=\dfrac{40}{2}=20$.

Milk only $=20-12=8$.

U (52):   [ Tea only: 28 ( Both: 12 ) Milk only: 8 ]    Neither: x

$28+12+8+x=52\Rightarrow x=4$

People who like milk only $=8$; people who like neither drink $=4$.

Let $U$ = voters (35000), $X$ = voters for candidate $X$, $Y$ = voters for candidate $Y$. $n(X)=18000$, $n(Y)=14000$, $n(X\cap Y)=2000$.

(i) Only $X$ $= 18000-2000 = 16000$, only $Y$ $= 14000-2000 = 12000$.

$n(X\cup Y) = 16000+2000+12000 = 30000$, so outside both circles $= 35000-30000 = 5000$.

U = 35000
  X                 Y
 ( 16000 [ 2000 ] 12000 )        5000 (outside)

In the Venn diagram: $X$ only $=16000$, $X\cap Y = 2000$, $Y$ only $=12000$, outside both circles $=5000$.

(ii) People who did not cast a vote $= n(U)-n(X\cup Y) = 35000-30000 =$ 5000

(iii) Each voter had to vote for a single candidate, so the 2000 who voted for both are invalid.

Valid votes $= 16000+12000 =$ 28,000

(i) Let $C$ be the set of students who like cricket and $B$ the set who like basketball.

$n(U)=180$, $n(C\text{ only})=50$, $n(B\text{ only})=30$, and neither $=50$.

Let $x$ students like both. Then $50+x+30+50=180$, so $x=50$.

U (180)
+-------------------------------------+
|     C                B              |
|   +---------+---------+             |
|   |   50    |   50    |   30        |
|   |  (only) | (both)  | (only)      |
|   +---------+---------+             |
|                                     |
|                       50 (neither)  |
+-------------------------------------+

The sets $C$ and $B$ overlap; the regions hold 50 (only cricket), 50 (both), 30 (only basketball) and 50 (neither).

(ii) Students who like cricket: $n(C)=50+50=100$. Students who like basketball: $n(B)=30+50=80$.

$n(C):n(B)=100:80=5:4$

(a) $n(U)=150$, $n(A)=130$, $n(B)=125$ and $n(A\cup B)^c = 15$.

(b) Students who passed at least one subject $= n(A\cup B)=150-15=135$.

$$n(A\cap B)=n(A)+n(B)-n(A\cup B)=130+125-135=120$$

So 120 students passed in both subjects.

(c) Students who passed in at least one subject $=n(A\cup B)=\mathbf{135}$.

(d) If everyone who passed optional mathematics also passed compulsory mathematics, then $B\subset A$, so the set $B$ is drawn completely inside the set $A$, and both lie inside the universal set $U$.

 U
+------------------------------+
|   A                          |
|  +----------------+          |
|  |    B           |          |
|  |  +--------+    |          |
|  |  |        |    |          |
|  |  +--------+    |          |
|  +----------------+          |
+------------------------------+

(a) $n(E-D)=n(E\text{ only})=76$.

(b) Let $n(E\cap D)=x$. Then the Venn diagram has: only $E=76$, both $=x$, only $D=12$.

 U, n(U) = 100
+--------------------------------+
|    E               D           |
|  +---------+---+---------+     |
|  |   76    | x |   12    |     |
|  +---------+---+---------+     |
|   outside both = ?             |
+--------------------------------+

(c) $n(E)=76+x$ and $n(D)=12+x$. Since $n(E)=5\,n(D)$:

$$76+x=5(12+x)\Rightarrow76+x=60+5x\Rightarrow4x=16\Rightarrow x=4$$

People liking at least one vehicle $=76+4+12=92$.

People who like neither $=100-92=\mathbf{8}$.

(d) At least one vehicle: 92; neither: 8. The number who like at least one vehicle is greater by $92-8=84$ (it is $11.5$ times the number who like neither).