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bogdanovich [222]
3 years ago
6

Tickets for a concert cost $2 for children, $3 for students, and $4 for adults. Ticket sales totaled $522 and 177 people attende

d the concert. Twice as many students as adults attended. Find how many of each type of ticket were sold. Assume that everyone who bought a ticket attended the concert.
Mathematics
1 answer:
snow_lady [41]3 years ago
3 0

Answer:

Children = 51

Adults = 42

Students= 84

Step-by-step explanation:

Children = $2

Students = $3

Adults = $4

Total sales = $522

Total people who attended = 177

Adults =a

Students = s= 2a

Children = c

c+s+a = 177 (1)

2c+3s+4a=522 (2)

Substitute s=2a into the equations

c + 2a + a = 177

c + 3a = 177 (3)

2c + 3(2a) + 4a = 522

2c + 6a + 4a = 522

2c + 10a = 522 (4)

c + 3a = 177 (3)

2c + 10a = 522 (4)

Multiply (3) by 2

2c + 6a = 354 (3b)

2c + 10a = 522

Subtract (3b) from (4)

10a - 6a = 522 - 354

4a = 168

Divide both sides by 4

a= 168/4

= 42

a= 42

s= 2a

= 2(42)

= 84

s= 84

Substitute the value of s and a into (1)

c+s+a = 177 (1)

c + 84 + 42 = 177

c + 126 = 177

c = 177 - 126

= 51

c=51

Children = 51

Adults = 42

Students= 84

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Answer:

The number of students that like only two of the activities are 34

Step-by-step explanation:

Number of students that enjoy video games, A = 38

Number of students that enjoy going to the movies, B = 12

Number of students that enjoy solving mathematical problems, C = 24

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Here we have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) -n(A∩C) + n(A∩B∩C)

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Also the number of student that like only one activity is found from the following equation;

n(A) - n(A∩B) - n(A∩C) + n(A∩B∩C) + n(B) - n(A∩B) - n(B∩C) + n(A∩B∩C) + n(C) - n(C∩B) - n(A∩C) + n(A∩B∩C) = 30

n(A) + n(B) + n(C) - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 3·n(A∩B∩C) = 30

38 + 12 + 24 - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 24 = 30

- 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) = -68

n(A∩B) + n(B∩C) + n(A∩C) = 34

Therefore, the number of students that like only two of the activities = 34.

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