A few days ago
Anonymous

Algebra Help pls?

The Problem Goes Like This: The total receipt for 1800 tickets were $1900. If students paid $1.50 per ticket and non students paid $2.50 per ticket, how many student tickets were sold??

I think the question is flawed… if you think you can solve pls tell me how.

tnx

Top 3 Answers
A few days ago
Ozymandius

Favorite Answer

2.5 Y + 1.5 X = 1900

Y = 1800 – x

2.5 (1800 – x) + 1.5 (x) = 1900

4500 – 2.5 x +1.5 x = 1900

4500 – x = 1900

x = 2600 (tickets to students)

y = -800 (tickets to non-students)

The problem is flawed

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A few days ago
green_eyed_lady
Yes, the question is flawed. If the total receipts for 1800 tickets was only $1900.00, the average ticket price would be a little over $1.05 (1.0555555555 repeating), so no combination of $1.50 tickets and $2.50 tickets (totaling 1800 tickets in all) could possibly be $1900.00.
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A few days ago
cidyah
Let there be x students and y non-students

1.50x+2.50y=1900 — (1)

1.50x=1900-2.50y

x=(1900-2.50y)/1.50 — (2)

x+y=1800

y=1800-x

so (1) becomes

1.50x+2.50(1800-x)=1900

1.50x+4500-2.50x=1900

-x+4500=1900

-x=-2600

x=2600 (students)

y=1800-2600=-800 non-students (meaningless answer)

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