Word Problem

Question

At a college function, a total of 180 tickets were sold. The tickets were priced at $5, $10, and $15, and the money collected that day was $1,900. The sum of the numbers of $5 and $15 tickets sold was twice the number of $10 tickets sold. Find the number of each type of ticket sold for the college function.

Answer 1

A = number of $5 tickets sold

B = number of $10 tickets sold

C = number of $15 tickets sold

 

The total number of tickets sold gives us equation 1.

A + B + C = 180

 

The total amount of money collected gives us equation 2.

5A +10B + 15C = 1900

 

The relationship between the number of each type of ticket sold gives us equation 3.

A + C = 2B

 

Substitute equation 3 into equation 1 as follows:

rewrite A + B + C = 180 as A + C + B = 180.  Since we know what A + C is from equation 3, substitute and get:

2B + B = 180

3B = 180

B = 60

 

Now, substitute the value of B  in equations 2 and 3

5A + 10(60) + 15C = 1900 equation 2

A + C = 2(60)

 

Rewrite

5A + 15C = 1300  (new equation 2)

A + C = 120 (new equation 3)

 

Using elimination, solve:

5A + 15C = 1300

-5A -5C = -600

Add these 2 equations together to get

10C = 700

Therefore C = 70

Substitute C back into one of your equations (say equation 3) to get

A + 70 = 120

A = 50

 

A=50, B=60, C=70

 

Answer 2

First lets determine the number of $10 tickets sold.

let that me n

so number of $5 and $15 tickets sold combined is 2n

so, n + 2n = 180

or, n=60

therefore, number of number of $5 and $15 tickets sold combined is 120

total money collected by selling $10 tickets is 10*60=600

therefore, total money collected by selling $5 and $15 tickets is 1900-600= $ 1300

let the number of $5 tickets sold be x,

therefore, number of $15 tickets sold is 120-x

therefore,

5x + 15(120-x) = 1300

5x + 1800 - 15x= 1300

-10x = -500

x=50

so number of $5 tickets sold = 50 (total collection = $ 250)

number of $15 tickets sold = 120-50 = 70 (total collection = $ 1050)

number of $ 10 tickets sold = 60 (total collection = $ 600)

all the collection adds up to 1900.

Answer 3

In this type of problem you need to turn each of the relations they give you into an equations.

(i'll use "a" as the number of $5 tickets sold, b as $10, and c as $15)

 

At a college function, a total of 180 tickets were sold. (total tickets sold is the sum of the individual tickets sold)

a+b+c=180

 

the money collected that day was $1,900 (total money made is the amount of each ticket sold times the value of that ticket)

a*5+b*10+c*15=1900

 

The sum of the numbers of $5 and $15 tickets sold was twice the number of $10 tickets sold (this is directly turning their sentence into an equation, where the number of # of $5+# of $15=a+c, and twice the # of $10=2b)

a+c=2b

 

Now you have three equations and three variables

a+b+c=180

a*5+b*10+c*15=1900

a+c=2b

I'll leave the algebra to you. You can post again if you need more help but you'll use subsitution to get it down to two equations with two variables (make sure its the same two variables in each), then use substitution again to solve for one of the variables. At that point its basically solved, you just have to plug back into an equation with 2 variables to get the second, and then plug back into one an equation with three variables to get the third.

Answer 4

Let each ticket amount be a different variable, for example:

a = # of $5 tickets sold

b = # of $10 tickets sold

c = # of $15 tickets sold

 

Then we can set up some equations using the given information:

total of the tickets sold is 180, so                         a + b + c = 180

total amount of money collected was $1900, so    5a + 10b + 15c = 1900

5 and 10 together was twice 15, so                      a + b = 2c

 

You can then use either substitution or elimination on 2 pairs of the equations to get down to 2 equations with 2 variables.  And then one more elimination or substitution to solve for a, b, or c.  Plug that value back into a 2 variable equation to solve for a second variable.  Then plug both of those into one of the original equations to find the final variable.