Let the percentage of people who are in favour of exactly one party be ‘a’, percentage of people who are in favour of exactly two parties be ‘b’.
Percentage of people who are in favour of all three parties = 5%
a + b + 5 = 68 => a + b = 63
a + 2b + 15 = 40 + 30 + 20 = 90 => a + 2b = 75
=> b = 75 - 63 = 12
=> a = 63 - 12 = 51
So, percentage of people who are in favour of at least two parties = b + c = 12% + 5% = 17%
So, number of people who are in favour of at least two parties = 17% of 500 = (17/100) * 500 = 85

For finding the maximum possible percentage of employees who drink both the beverages, there should be maximum possible overlap between tea and coffee drinkers

Maximum possible overlap = min (78, 64) = 64

Let ‘a’ be the percentage of residents who read only one newspaper. Let ‘b’ be the percentage of residents who read exactly two newspapers. Let ‘c’ be the percentage of residents who read all three newspapers.

a + b + c = 100
a + 2b + 3c = 60 + 80 + 55 = 195
=> b + 2c = 95
To find the maximum percentage of residents who read exactly two newspapers, ‘b’ has to be maximum and ‘c’ has to be minimum. If c = 0, then b = 95, in which case a = 5.

To find the minimum percentage of residents who read exactly two newspapers, ‘b’ has to be minimum. Since b + 2c = 95, if b = 0, c = 47.5, in which case a = 52.5

So, the minimum percentage is 0% and the maximum percentage is 95%

Let the number of students who pass in exactly one subject be a, exactly 2 be b, exactly 3 be c, exactly 4 be d and exactly 5 be e.

a+b+c+d+e = 50
a+2b+3c+4d+5e = 25+27+29+33+35 = 149

We have to get the maximum number of students who passed in at least three subjects, i.e. we have to maximize c+d+e.

From the 2 equations, we have, b+2c+3d+4e = 99.
So, b+2(c+d+e)+d+2e = 99. b =1 and c = 49 and d = e = a = 0 => c + d + e = 49 is the maximum value.

From the information given in the question, the following Venn Diagram can be constructed:

So, in order to maximize the number of families that own only a bike, we can put the remaining 44 families in ‘only bike’ region. Similarly, in order to minimize the number of families that own only a bike, we can put the remaining 44 families in ‘only scooter’ region.

So, the maximum number of families that own only a bike is 44 and the minimum number of families that own only a bike is 0.

So, sum = 44 + 0 = 44

Let the number of students in the college be 30x.
So, $$30\times\ \frac{2}{3}x\ =\ 20x$$ students study English, $$\frac{30x}{5}=6x$$ students study Telugu and $$\frac{30x}{10}=3x$$ students study both the subjects. So, the number of students studying only English will be $$20x-3x=17x$$, and the number of students studying only Telugu will be $$6x-3x=3x$$.

So, the total number of students studying at least one language = $$17x+3x+3x=23x$$

So, the required probability = $$\frac{23}{30}$$

Thus, the correct option is D.

Assuming A is the number of students who play exactly 1 game, B is the number of students who play exactly 2 games and C is the number of students who play exactly 3 games. 

A+B+C = 100;

A+2B+3C = 60+84 +72 = 216.

3*first equation - second equation => 2A+B = 84.

IF A=0, B=84.

Therefore, min. of C = 16%.

Alternate explanation:

Assuming A is the number of students who play exactly 1 game, B is the number of students who play exactly 2 games and C is the number of students who play exactly 3 games.

A+B+C = 100

A+2B+3C = 60+84+72 = 216

Second Equation-2*First Equation = C-A = 16

=> C= A+16, the minimum value of A can be 0. Hence the minimum value of C will be 16.

Let the number of students who took exactly one exam be ‘x’.
The number of students who took exactly two exams be ‘y’ and the number of students who took exactly three exams be ‘z’.
So, x + y + z = 200
x + 2y + 3z = 112 + 160 + 128 = 400
From these two equations, we get, y + 2z = 200
To minimize ‘z’, we have to maximize ‘y’. If z = 0, y = 200. In this case, x = 0
So, the minimum number of students who took all the three tests is 0.

From the Venn Diagram, the number of people who listened to at least one genre of music = 9+3+3+5+9+2+10 = 41
So, number of people who did not listen to any of the three genres of music = 46 - 41 = 5

Let the number of people in the group who drink exactly one beverage be ‘a’, the number of people who drink exactly two beverages be ‘b’, the number of people who drink exactly three beverages be ‘c’, the number of people who drink exactly four beverages be ‘d’. Let the number of people who do not drink any beverage be ‘e’.
So, a + b + c + d + e = 1000 → (1)
a + 2b + 3c + 4d = 400 + 500 + 300 + 200 = 1400 → (2)
It is given that d = 40, c = 90 and b = 120
So, from equation (2), a = 1400 - 2*120 - 3*90 - 4*40 = 730
From equation (1), a + e = 1000 - 120 - 90 - 40 = 750
=> e = 20
So, the number of people who do not drink any beverage = 20

Let the number of people who eat pizza and hotdog but not burger be x.
So, adding all the values, we have 93 - x + 12 + 30 + x + 65 + 18 + 67 - x = 270
=> x = 15
So, the number of people who eat pizza and hotdog but not burger is 15.
If these 15 people stop eating pizza and instead start eating burger, the condition in the question is satisfied. However, in this rearrangement, the number of people who eat only pizza is not affected. So, 93 - x = 93 - 15 = 78 is the maximum number of people who eat only pizza.

The number of people playing exactly one sport = a, exactly 2 sports = b and all three sports = c.
a+2b+3c = 30+40+52 = 122. a+b+c = 100.
Subtracting, we get,b+2c = 22.
Subtracting this from first equation, we get a-c = 78.
If a = 78, c = 0 and b = 22. If b = 0, a = 89 and c = 11.
Therefore, max. value of a = 89

Let x, y, z denote families with 1, 2 and 3 pets.

Hence, x+y+z=160 and x+2y+3z=230.

Hence, y+2z=70. The max value of z is when y=0.

Thus max z, 2z = 70 => z=35. Option b) is the correct answer.

Let the number of students who like both = x.
Number of students who like only Sunrisers Hyderabad is 240-x and number of students who like only Mumbai Indians is 280-x.
Sum = 240-x+x+280-x = 520 - x = 400 - number of students who do not follow the sport.
=> x = 120+ number of students who do not follow the sport. x is maximized when we use the maximum no. of students who do not follow the sport in the above equation.
Hence max value for x = 120 + 40 = 160.

If the largest and smallest numbers are 14 and 1, the other elements can be between 2-13. Each of these 12 numbers can either be in the subset or not. Hence, there are two possibilities with each of the 12 numbers. Hence, the number of ways of having the other elements is $$2^{12}$$.
Similarly, for 13 and 2, the number of ways is $$2^{10}$$
Total is $$2^{12}$$ + $$2^{10}$$ + $$2^{8}$$ + $$2^{6}$$ + $$2^{4}$$ + $$2^{2}$$ + $$2^{0}$$ = 5461

Let the number of people who read exactly one newspaper be a, those who read exactly two newspapers be b and those who read exactly 3 newspapers be c. From the given information, a+2b+3c = 47 and c = 2 => a+2b = 41. Since 2b is even, a has to be odd. From the options, a can be both 19 and 21.

Let a, b and c be the number of students who chose exactly 2, exactly 3 and all 4 sports respectively.
=> 2a + 3b + 4c = 2000
=> a + b + c = 800
=> b + 2c = 400
Here, c can be a maximum of 200.

From the Venn Diagram, the number of people who participated in none of the activities = 46 - 41 = 5

For minimum overlap between the two sets the union of the two sets should be as large as possible. Hence assume that the union of the two sets is 100%.

So the minimum overlap between the two sets is 78+64-100 = 42%

Let x, y, z denote families with 1, 2 and 3 pets.

Hence, x+y+z=160 and x+2y+3z=230.

Hence, y+2z=70 and x-z=90 => x=90+z.

The minimum possible value of x is when z is minimum.

Hence, when z=0, y=70 and x=90. Hence, minimum value of x is 90.

Assuming A is the number of students who play exactly 1 game, B is the number of students who play exactly 2 games and C is the number of students who play exactly 3 games. 

A+B+C = 100;

A+2B+3C = 60+84 +72 = 216.

3*first equation - second equation => 2A+B = 84.

IF A=0, B=84.

Therefore, min. of C = 16%.

Alternate explanation:

Assuming A is the number of students who play exactly 1 game, B is the number of students who play exactly 2 games and C is the number of students who play exactly 3 games.

A+B+C = 100

A+2B+3C = 60+84+72 = 216

Second Equation-2*First Equation = C-A = 16

=> C= A+16, the minimum value of A can be 0. Hence the minimum value of C will be 16.