CRE 4 - Venn Diagram based Optimization | LR - Venn Diagram
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Answer the next 2 questions based on the information given below:
In a class of 40 students, 20 like cricket and 25 like football.
What is the maximum number of students who like both the sports.
Answer: 20
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Explanation :
Out of total students 20 like cricket and 25 like football.
∴ Maximum students who like both the sports can be 20 (lesser of cricket or football).
Hence, 20.
Note: Students who do not like any of the sport = 20 + 25 – 20 = 25 students.
Workspace:
What is the least number of students who like both the sports.
Answer: 5
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Explanation :
Out of total students 20 like cricket and 25 like football.
Let x be the number of students who like bot the sports.
∴ Number of students who like only cricket = 20 - x
∴ Number of students who like only football = 25 - x
⇒ Number of students who like at least one of the sport = (20 - x) + (25 - x) + x ≤ 40
⇒ 5 ≤ x
∴ Minimum value of x = 5.
Alternately,
Let the number of students who,
like exactly 1 sport = a
like exactly 2 sports = b
like none of the 2 sports = n
⇒ a + b + n = 40 …(1)
and, a + 2b = 20 + 25 = 45 …(2)
(2) – (1)
⇒ b – n = 5
⇒ b = 5 + n
Minimum value of b = 5 when n = 0.
Hence, 5.
Workspace:
Answer the next 3 questions based on the information given below.
In a group of 130 students, 60 like Cricket, 50 like Football, 45 like TT and 90 like exactly one of the three games.
What is the maximum possible number of students who like at least two of the three games?
- (a)
26
- (b)
32
- (c)
34
- (d)
48
Answer: Option B
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Explanation :
Let the number of people liking
exactly 1 game = a (all orange areas together) = 90
exactly 2 games = b (all green areas together)
all three games = c (blur area)
none of the three games = n
⇒ n + a + b + c = 130
⇒ n + b + c = 40 ...(1)
Also, When we add those who like Cricket (60), Football (45) and TT (45), we add 'a' once, 'b' twice and 'c' thrice.
∴ a + 2b + 3c = 60 + 50 + 45
⇒ 2b + 3c = 65 ...(2)
Now, to maximise 'b + c', we should maximise the variable with lesser coefficient and to minimise 'b + c', we should maximise the variable with higher coefficient.
Here, we need to maximise 'b + c', hence we will maximise 'b' and minimise 'c'.
For, 'b' and 'c' to be integers, least value 'c' can take is 1.
Hence, 2b + 3 = 65
⇒ b = 31
∴ Maximum possible value of 'b + c' = 31 + 1 = 32.
Hence, option (b).
Workspace:
What is the maximum possible number of students who like none of the games?
- (a)
40
- (b)
32
- (c)
18
- (d)
8
Answer: Option C
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Explanation :
Let the number of people liking
exactly 1 game = a (all orange areas together) = 90
exactly 2 games = b (all green areas together)
all three games = c (blur area)
none of the three games = n
⇒ n + a + b + c = 130
⇒ n + b + c = 40 ...(1)
To maximise 'n', we need to minimise 'b + c'.
Also, when we add those who like Cricket (60), Football (45) and TT (45), we add 'a' once, 'b' twice and 'c' thrice.
∴ a + 2b + 3c = 60 + 50 + 45
⇒ 2b + 3c = 65 ...(2)
Now, to maximise 'b + c', we should maximise the variable with lesser coefficient and to minimise 'b + c', we should maximise the variable with higher coefficient.
Here, we need to minimise 'b + c', hence we will minimise 'b' and maximise 'c'.
For, 'b' and 'c' to be integers, least value 'b' can take is 1.
Hence, 2 + 3c = 65
⇒ b = 21
∴ Minimium possible value of 'b + c' = 1 + 21 = 22.
⇒ n + b + c = 40
⇒ n + 22 = 40
⇒ n = 18
Hence, option (c).
Workspace:
What is the maximum possible number of students who like Football but like neither Cricket nor TT?
- (a)
29
- (b)
39
- (c)
49
- (d)
50
Answer: Option C
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Explanation :
Consider the solution to first quetion of this set.
Since, the least number of people liking all three sports = c = 1.
⇒ Maximum number of people liking only Football = 50 - 1 = 49.
Hence, option (c).
Workspace:
Of the students who like TT, the number of students who like football is 5 more than those who like Cricket. What is the difference between the number of students who like only Cricket and those who like only Football?
- (a)
5
- (b)
15
- (c)
25
- (d)
39
Answer: Option B
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Explanation :
Consider the diagram above.
⇒ p + r + c + x = 60 ...(1)
⇒ q + r + c + x + 5 = 50 ...(2)
(1) - (2)
⇒ p - q - 5 = 10
∴ p - q = 15
Hence, option (b).
Workspace:
Answer the next 2 questions based on the information given below.
According to a survey conducted among 1000 people in a multiplex on a binge watching weekend, 400 participants bought tickets for Avengers, 500 participants bought tickets for Interstellar and 600 participants bought tickets for Eat Pray Love. Each person had purchased tickets for at least one movie.
What is the minimum number of people who bought tickets for all the three movies?
- (a)
400
- (b)
500
- (c)
600
- (d)
None of these
Answer: Option D
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Explanation :
All of them buy at least 1 ticket.
Let number of people buying tickets for exactly
1 movie = a
2 movies = b
3 movies = c
⇒ a + b + c = 1000 ...(1)
Also, when we add number of tickets bought for Avengers (400), Interstellar (500) and Eat Pray & Love (600) we would add 'a' once, 'b' twice and 'c' thrice.
⇒ a + 2b + 3c = 400 + 500 + 600
⇒ a + 2b + 3c = 1500 ...(2)
(2) - (1)
⇒ b + 2c = 500
The least possible value of 'c' here is 0.
In this case b = 500 [from (2)] and a = 500 [from (1)]
A possible case with these values is as follows:
Hence, option (d).
Workspace:
What is the maximum number of people who bought tickets for all the three movies?
- (a)
250
- (b)
600
- (c)
400
- (d)
None of these
Answer: Option A
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Explanation :
All of them buy at least 1 ticket.
Let number of people buying tickets for exactly
1 movie = a
2 movies = b
3 movies = c
⇒ a + b + c = 1000 ...(1)
Also, when we add number of tickets bought for Avengers (400), Interstellar (500) and Eat Pray & Love (600) we would add 'a' once, 'b' twice and 'c' thrice.
⇒ a + 2b + 3c = 400 + 500 + 600
⇒ a + 2b + 3c = 1500 ...(2)
(2) - (1)
⇒ b + 2c = 500
The highest possible value of 'c' here is 250.
In this case b = 0 [from (2)] and a = 750 [from (1)]
This case is as follows:
Hence, option (d).
Workspace:
Answer the next 2 questions based on the information given below.
According to a survey, at least 70% participants like apples, at least 75% participants like bananas and at least 80% participants like cherries.
What is the minimum percentage of people who like all the three fruits?
- (a)
25
- (b)
70
- (c)
41
- (d)
Cannot be determined
Answer: Option A
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Explanation :
[Note: We will be dealing in % only]
Let the number of people liking
exactly 1 fruit = a
exactly 2 fruits = b
exactly 3 fruits = c
exactly no fruit = n
⇒ a + b + c ≤ 100 ...(1)
To minimise 'c' we will assume least number of people like each of the three fruits.
∴ 70% people like apples,
75% people like bananas, &
80% people like cherries.
Also, a + 2b + 3c = 70 + 75 + 80
a + 2b + 3c = 225 ...(2)
Here, we need to minimise 'c'.
Let's check if 'c' can be 0 or not. If 'c' = 0, then a + 2b must be equal to 225.
From (1), if we take maximum value of 'b' = 100, then a = 0 ⇒ a + 2b = 200.
∴ c = 0, is not possible.
When b takes the highest possible value of 100,
a + 2b + 3c = 200.
We need to increase this sum, hence we will decrease the value of 'b' and increase the value of the variable with higher coefficient i.e., 'c'.
∴ If b = 99 and c = 1 and a remains 0.
⇒ a + 2b + 3c = 201
∴ By shifting 1 unit from 'b' to 'c' we could increase the value of a + 2b + 3c by 1 unit.
We need to increase the value of a + 2b + 3c from 200 to 225.
⇒ We will have to shift 25 units from 'b' to 'c'
∴ b = 75 and c = 25 and a remains 0.
Now, a + 2b + 3c = 225
∴ Least possible value of c = 25.
Hence, option (a).
Workspace:
What is the maximum percentage of people who like all the three fruits?
- (a)
80
- (b)
70
- (c)
41
- (d)
100
Answer: Option D
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Explanation :
The questions states that at least 70% people like apples. Hence, it is possible that all 100% people like apples.
Same is the case for bananas and cherries as well.
Hence, assuming that all 100% people like apples, 100% people like bananas and 100% people like cherries.
⇒ all 100% of people like all three fruits.
Hence, option (d).
Workspace:
Answer the next 2 questions based on the information given below.
A survey was conducted among a group of 50000 people about their preferred newspapers. 90% liked Times of India, 85% liked Economic Times, 80% liked Hindustan Times, 82% liked Mumbai Mirror, 75% liked Daily News and Analysis and 4000 did not like any of these newspapers.
What is the minimum number of people who liked all the five newspapers?
- (a)
44
- (b)
37500
- (c)
22000
- (d)
75
Answer: Option C
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Explanation :
Number of people liking
TOI = 90% of 50,000 = 45,000
ET = 85% of 50,000 = 42,500
HT = 80% of 50,000 = 40,000
MM = 82% of 50,000 = 41,000
DNA = 75% of 50,000 = 37,500
None of the newspapers = 4,000
Let the number of people liking
exactly 1 of these newspapers = a
exactly 2 of these newspapers = b
exactly 3 of these newspapers = c
exactly 4 of these newspapers = d
exactly 5 of these newspapers = e
none of these newspapers = n = 4,000
⇒ a + b + c + d + e + n = 50,000
⇒ a + b + c + d + e = 46,000 ...(1)
Also, a + 2b + 3c + 4d + 5e = 45000 + 42500 + 40000 + 41000 + 37500
⇒ a + 2b + 3c + 4d + 5e = 2,06,000 ...(2)
We need to minimise e. Let us check if e can be 0 or not.
From (1), maximum possible value of d = 46000. Assuming d = 46000, then a = b = c = e = 0
⇒ a + 2b + 3c + 4d + 5e = 1,84,000
Since, we need to increase this sum from 1,84,000 to 2,06,000,
we will have to decrease the value of d and increase the value of variable with highest coefficient i.e., e.
If d = 45,999, e = 1, then the sum will become 1,84,001
i.e., for transfer of 1 unit from 'd' to 'e', sum increases by 1 unit.
We need to increase the sum by 22,000 units.
∴ We need to transfer 22,000 units from 'd' to 'e'.
⇒ Least possible value of e = 22,000
Hence, option (c).
Workspace:
What is the maximum percentage of people who liked three out of the five newspapers?
- (a)
24
- (b)
12000
- (c)
34000
- (d)
68
Answer: Option A
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Explanation :
Number of people liking
TOI = 90% of 50,000 = 45,000
ET = 85% of 50,000 = 42,500
HT = 80% of 50,000 = 40,000
MM = 82% of 50,000 = 41,000
DNA = 75% of 50,000 = 37,500
None of the newspapers = 4,000
Let the number of people liking
exactly 1 of these newspapers = a
exactly 2 of these newspapers = b
exactly 3 of these newspapers = c
exactly 4 of these newspapers = d
exactly 5 of these newspapers = e
none of these newspapers = n = 4,000
⇒ a + b + c + d + e + n = 50,000
⇒ a + b + c + d + e = 46,000 ...(1)
Also, a + 2b + 3c + 4d + 5e = 45000 + 42500 + 40000 + 41000 + 37500
⇒ a + 2b + 3c + 4d + 5e = 2,06,000 ...(2)
We need to maximise 'c'.
From (1), maximum possible value of c = 46000. Assuming c = 46000, then a = b = d = e = 0
⇒ a + 2b + 3c + 4d + 5e = 1,38,000
Since, we need to increase this sum from 1,38,000 to 2,06,000,
we will have to decrease the value of 'c' and increase the value of variable with highest coefficient i.e., 'e'.
If c = 45,999, e = 1, then the sum will become 1,38,002
i.e., for transfer of 1 unit from 'c' to 'e', sum increases by 2 units.
We need to increase the sum by 68,000 units.
∴ We need to transfer 34,000 units from 'c' to 'e'.
⇒ Highest possible value of 'c' = 46,000 - 34,000 = 12,000
∴ Maximum % of people liking exactly 3 newspapers = 12,000/50,000 × 100% = 24%
Hence, option (a).
Workspace:
Answer the next 5 questions based on the information given below:
In a group of 100 people 65 can speak Hindi, 55 can speak Tamil and 60 can speak English.
What is the minimum number of people who can speak exactly one language?
Answer: 20
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Explanation :
Let number of people who can speak
Exactly one of the three languages = a
Exactly one of the three languages = b
Exactly one of the three languages = c
None of the three languages = n
⇒ a + b + c + n = 100 …(1)
Also, a + 2b + 3c = 65 + 55 + 60
⇒ a + 2b + 3c = 180 …(2)
Now, we need to minimize a.
Let’s see if a can be 0.
From (1), if a = 0 and even if b = 100 and c = 0
a + 2b + 3c = 200
But we want this sum to be 180 [from (2)].
So we will have to reduce the value of b and increase the value of the variable with smaller coefficient i.e., a.
Now, if even decrease b by 1 unit and increase a by 1 unit.
a + 2b + 3c = 199 i.e., the sum decreases by 1 unit.
∴ To decrease the sum from 200 to 180, we need to decrease it by 20 units, hence we need to transfer 20 units from b to a.
∴ Least possible value of a = 20.
Following is a possible Venn diagram for this case.
Hence, 20.
Workspace:
What is the minimum number of people who can speak at least one language?
Answer: 65
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Explanation :
Let the number of people speaking exactly one language = a
Let the number of people speaking exactly two languages = b
Let the number of people speaking exactly three languages = c
Let the number of people speaking none of the three languages = n
⇒ a + b + c + n = 100 …(1)
Also, a + 2b + 3c = 65 + 55 + 60
⇒ a + 2b + 3c = 180 …(2)
(2) – (1)
⇒ b + 2c – n = 80 …(3)
Number of people speaking at aleast one language = a + b + c
To mimise the number of people speaking at least one language we need to minimize a + b + c.
Since 65 people speak Hindi, a + b + c cannot be less than 65.
Hence, 65.
Workspace:
What is the maximum number of people who can speak exactly one language?
Answer: 60
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Explanation :
Let number of people who can speak
Exactly one of the three languages = a
Exactly one of the three languages = b
Exactly one of the three languages = c
None of the three languages = n
⇒ a + b + c + n = 100 …(1)
Also, a + 2b + 3c = 65 + 55 + 60
⇒ a + 2b + 3c = 180 …(2)
Now, we need to maximise a.
Let’s see if a can be 100.
From (1), if a = 100 then b = 0 and c = 0
⇒ a + 2b + 3c = 100
But we want this sum to be 180 [from (2)].
So we will have to reduce the value of a and increase the value of the variable with highest coefficient i.e., c.
Now, if we decrease a by 1 unit and increase c by 1 unit.
a + 2b + 3c = 102 i.e., the sum increases by 2 units.
∴ To increase the sum from 100 to 180, we need to increase it by 80 units, hence we need to transfer 40 units from a to c.
∴ Maximum possible value of a = 100 - 40 = 60.
Following is the venn diagram for this case.
Hence, 60.
Workspace:
What is the minimum number of people who can speak exactly two languages?
Answer: 0
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Explanation :
Let number of people who can speak
Exactly one of the three languages = a
Exactly one of the three languages = b
Exactly one of the three languages = c
None of the three languages = n
⇒ a + b + c + n = 100 …(1)
Also, a + 2b + 3c = 65 + 55 + 60
⇒ a + 2b + 3c = 180 …(2)
Now, we need to minimize b.
Let’s see if b can be 0.
From (1), if b = 0 and highest possible value of c = 55, then highest value of a = 45.
⇒ a + 2b + 3c = 210
But we want this sum to be 180 [from (2)].
So we will have to reduce the value of c and increase the value of the variable with smaller coefficient i.e., a or n.
Now, if we decrease c by 1 unit and increase n by 1 unit.
a + 2b + 3c = 207 i.e., the sum decreases by 3 units.
∴ To decrease the sum from 210 to 180, we need to decrease it by 30 units, hence we need to transfer 10 units from c to n.
∴ Value of a = 45, c = 45, b = 0 and n = 10.
Following is a possible Venn diagram for this case.
Hence, 0.
Workspace:
What is the maximum number of people who can speak exactly two languages?
Answer: 90
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Explanation :
Let number of people who can speak
Exactly one of the three languages = a
Exactly one of the three languages = b
Exactly one of the three languages = c
None of the three languages = n
⇒ a + b + c + n = 100 …(1)
Also, a + 2b + 3c = 65 + 55 + 60
⇒ a + 2b + 3c = 180 …(2)
Now, we need to maximise b.
Let’s see if b can be 100.
From (1), if b = 100 and then a = 0 and c = 0.
⇒ a + 2b + 3c = 200
But we want this sum to be 180 [from (2)].
So we will have to reduce the value of b and increase the value of the variable with smaller coefficient i.e., a or n.
Now, if even decrease b by 1 unit and increase n by 1 unit.
a + 2b + 3c = 198 i.e., the sum decreases by 2 units.
∴ To decrease the sum from 200 to 180, we need to decrease it by 20 units, hence we need to transfer 10 units from b to n.
∴ Value of a = 0, c = 0, b = 90 and n = 10.
Following is the Venn diagram for this case.
Hence, 0.
Workspace:
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