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# How many of each kind of ticket were sold?

There were 305 tickets sold for a basketball game. The activity cardholders' tickets cost \$1.25 and the non-cardholders' tickets cost \$2.50. The total amount of money collected was \$578.75. How many of each kind of ticket were sold?

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William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
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Marked as Best Answer
Let x = number of activity cardholder's tickets sold
Let y = number of non-activity cardholder's tickets sold

We know that x + y = 305     Eq. (1)

We also know that (\$1.25)*x + (\$2.50)*y = \$578.75     Eq. (2)

Solve Eq. 1 for y and substitute into Eq. 2:

y = 305 - x and (\$1.25)x + (\$2.50)(305 - x) = \$578.75

(\$1.25)x + \$762.5 - (\$2.50)x =  \$578.75

(-\$1.25)x = \$578.75 - \$762.5

x = 147 tickets

Therefore, y = 158 tickets

Check

Does (\$1.25)*(147) + (\$2.50)(158) = \$578.75?  Yes!!!

### Comments

thank you so much William, this one was a doozy for me
Kevin F. | Computer Programming and Mathematics TutorComputer Programming and Mathematics Tut...
5.0 5.0 (3 lesson ratings) (3)
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You can solve this by using two equations, each with two unknowns, or variables.

The variables are the number of cardholders tickets (C), and the number of non-cardholders tickets (N).

There were 305 tickets sold, so the total of cardholders plus noncardholders is 305:

N + C = 305

Each cardholders ticket costs 1.25, so the amount from sales of cardholders tickets is 1.25 * C.
Each non-cardholders ticket costs 2.50, so the amount from sales of non-cardholders tickets is 2.50 * N.
The total sales is 578.75:

1.25C + 2.5N = 578.75

You now have two equations in two unknowns. Now, change the first equation to express one of the variables in term of the other (by subtracting C from both sides of the equation:

N + C = 305
N = 305 - C

Now, use substitution to eliminate N from the second equation:

1.25C + 2.5N = 578.75

substitute for N:
1.25C + 2.5(305 - C) = 578.75

distributive property:
1.25C + (2.5 * 305 - 2.5 * C) = 578.75

perform multiplication inside the parentheses
1.25C + (762.5 - 2.5C) = 578.75

eliminate the parentheses
1.25C - 2.5C + 762.5 = 578.75

combine variable terms
-1.25C + 762.5 = 578.75

subtract 762.5 from both sides of the equation
-1.25C = -183.75

divide both sides by -1.25
C = 147

Now that you know how many cardholder tickets were sold, use the original equation to determine the non-cardholder tickets:

N + C = 305

substitute for C:
N + 147= 305

subtract 147 from both sides of the equation:
N = 305 - 147
N = 158

C = 147, N = 158

There were 147 cardholder tickets sold, and 158 non-cardholder tickets sold. Double check by plugging those values into the second equation:

1.25C + 2.5N = 578.75
1.25 * 147 + 2.5 * 158 = 578.75
183.75 + 395 = 578.75
578.75 = 578.75
Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (174 lesson ratings) (174)
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Kevin,

If we let x=number of activity cardholders and y= the number on non-activity cardholders the we were told that:

x+y=305    total number of tickets sold

and:

1.25x+2.5y=578.75    total revenue from ticket sales. This gives 2 equations with 2 unknowns x and y.

If you now multiply both sides of the first equation by 1.25 and subtract it from the second equation we have

1.25y=197.50, or y=158 and since x+y=305 x must be 147

The sales from activity cardholders =1.25*147 = 183.75
The sales from non act.cardholders=2.5*158 = 395 which totals to \$578.75

Regards
Jim
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Kevin;
x=quantity of cardholders' tickets
\$1.25x+\$2.50(305-x)=\$578.75
\$1.25x+(\$762.50-\$2.50x)=\$578.75
\$762.50-\$1.25x=\$578.75
Let's subtract \$762.50 from both sides as we proceed to isolate x...
-\$762.50+\$762.50-\$1.25x=\$578.75-\$762.50
-\$1.25x=-183.75
Let's divide both sides by -\$1.25 as we proceed to isolate x...
(-\$1.25x)/-\$1.25=(-\$183.75)/(-\$1.25)

x=147 cardholders' tickets
305-147=158 non-cardholders' tickets

Let's check our work...
[147(\$1.25)]+[158(\$2.50)]=\$578.75
\$183.75+\$395.00=\$578.75
\$578.75=\$578.75

Please note that I always indicate all information of cents as well as dollars for each monetary figure.  Consistency is very important, and all initial figures of \$1.25, \$2.50 and \$578.75 include such information.  Therefore, all supplemental information must include such too, even if it is 00 cents.

Ralph L. | Algebra I, II, Visual Basic, Beginning C++ tutorAlgebra I, II, Visual Basic, Beginning C...
4.0 4.0 (1 lesson ratings) (1)
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Let C be the number of cardholders.

1.25C + 2.5(305 - C) = 578.75

1.25C + 762.5 - 2.5C = 578.75

-1.25C = 578.75 - 762.5

-1.25C = -183.75

C = 147

Cardholders 147, Non Cardholders 158

Darshan P. | Interactive and Educated Rice University TutorInteractive and Educated Rice University...
0
x=cardholder ticket
y=non cardholder ticket
Do a system of equations:

x+y=305
1.25x+2.50y=578.75

Now multiply the top equation to get rid of one of the variables in order to solve the other:

-1.25x-1.25y=-381.25
1.25x+2.50y=578.75

This leads to:
1.25y=197.5 -> y=1197.5/1.25=158

Plug the y-value back into x+y=305 ->x+158=305 -> x=147

So there were 147 cardholder and 158 non-cardholder tickets sold