SOLUTION: an art museum sold 63 tickets in 30 minutes. The ratio of adult to child tickets was 5:1. The ratio of adult to senior tickets was 5:3. How many of each kind of ticket were sold?

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: an art museum sold 63 tickets in 30 minutes. The ratio of adult to child tickets was 5:1. The ratio of adult to senior tickets was 5:3. How many of each kind of ticket were sold?      Log On


   

Question 1151939: an art museum sold 63 tickets in 30 minutes. The ratio of adult to child tickets was 5:1. The ratio of adult to senior tickets was 5:3. How many of each kind of ticket were sold?
Found 4 solutions by MathLover1, ikleyn, MathTherapy, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
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let adult tickets be x, child tickets y, and senior tickets z=>
x%2By%2Bz=63
if the ratio of adult to child tickets was 5%3A1, we have
x%3Ay=5%3A1
x=5y........eq.1
if the ratio of adult to senior tickets was 5%3A3, we have
x%3Az=5%3A3
3x=5z
x=5z%2F3........eq.2
from eq.1 and eq.2 we have
5y=5z%2F3
y=5z%2F%283%2A5%29
y=z%2F3..........eq.2a


if art museum sold 63 tickets, we have
x%2By%2Bz=63......substitute x from eq.2 and y from eq.2a
5z%2F3%2Bz%2F3%2Bz=63..........both sides multiply by 3 and solve for z
5z%2Bz%2B3z=63%2A3
9z=189
z=189%2F9
z=21
go to x=5z%2F3........eq.2,substitute z
x=%285%2A21%29%2F3
x=5%2A7
x=35

y=z%2F3
y=21%2F3
y=7
answer:
35 adult tickets were sold,
7 child tickets were sold
21 senior tickets were sold

Answer by ikleyn(52832) About Me  (Show Source):
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.

            It can be solved mentally,  without using equations. 


The total tickets is 63.


The ratio of child to adults is 1:5.

The ratio of senior to adults is 3:5.


It means that for every 5 adults there are 1 child and 3 seniors.


Hense, you can group tickets in sets, in a way that every set contain 5 adult tickets, 1 child and 3 senior ticket; 
in all, 5 + 1 + 3 = 9 tickets in each set.


Then the number of sets is 63/9 = 7.


ANSWER.  5*7 = 35 adult tickets;  7 child tickets; and 3*7 = 21 senior tickets.



Answer by MathTherapy(10555) About Me  (Show Source):
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an art museum sold 63 tickets in 30 minutes. The ratio of adult to child tickets was 5:1. The ratio of adult to senior tickets was 5:3. How many of each kind of ticket were sold?
Let number of children, adults, and seniors be C, A, and S, respectively

Since children to adults, or C:A = 1:5, it follows that: 

Since adults to seniors, or A:S = 5:3, then matrix%281%2C3%2C+A%2FS%2C+%22=%22%2C+5%2F3%29

                                            matrix%281%2C3%2C+5C%2FS%2C+%22=%22%2C+5%2F3%29 ------ Replacing A with 5C

                                            5S = 3(5C) ---- Cross-multiplying

                                            matrix%281%2C5%2C+S%2C+%22=%22%2C+3%285C%29%2F5%2C+%22=%22%2C+3C%29

With total number of attendees being 63, we then get: C + A + S = 63, or C + 5C + 3C = 63

                                                                         9C = 63

                                                                         C, or

Answer by greenestamps(13203) About Me  (Show Source):
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The formal algebraic solutions by tutors @MathTherapy and @MathLover1 using ratios are valid, but a lot more work than necessary.

The mental solution by tutor @ikleyn is by far the easiest way to solve the problem.

If you want to formalize her solution, you get an algebraic solution with much less work than the other algebraic solutions you have seen.

adult:senior= 5:3 and adult:child = 5:1 means adult:senior:child = 5:3:1

So let the numbers of tickets be
5x = adults
3x = seniors
x = children

The total number of tickets was 63:

5x%2B3x%2Bx+=+63
9x+=+63
x+=+7

The numbers of tickets were
adults = 5x = 35
seniors = 3x = 21
children = x = 7