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# ticket selling Ron is selling tickets that sell for \$3.00 each and I am selling them for \$6.50 each. Total for 30 tickets was \$125.00, How many did Ron sell?

Ron is selling children's tickets for \$3.00 each.  I am selling adult tickets for \$6.50 each.  We had a total income of \$125.00 from 30 tickets sold.  How many did Ron sell?

### 3 Answers by Expert Tutors

Johnny R. | Exceptional Math and Accounting TutorExceptional Math and Accounting Tutor
4.8 4.8 (162 lesson ratings) (162)
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To solve this problem, we should identify the information that we know.

We know:

• We sold children tickets. We will let the variable C= children's tickets.
• We sold adult tickets. We will let the variable A = adult tickets
• We know we sold 30 tickets

Therefore we will develop an equation for the number of tickets.

C + A = 30 (the total tickets sold)

We also know.

• The children tickets cost \$3.00 Therefore we will let \$3.00 x C (\$3.00C) = amount of money earned from the sale of children tickets.
• The adult tickets cost \$6.50 Therefore we will let \$6.50 x A (\$6.50A) = amount of money earned from the sale of adult tickets.
• We know we earned \$125.00 from the sale of all tickets.

Therefore we will develop an equation for the amount of money earned from the sale of tickets.

\$3.00C + \$6.50A = \$125.00

Now we have 2 equations ( a system of equations)

• C+ A = 30
• 3.00C + 6.50A= 125

We will use the elimination method to solve for one of the variables.

We will solve for A by eliminating the variable C.

We will multiply equation 1 by -3

-3C + -3A = -90

We will add the new equation to equation 2.

• -3C + -3A = -90
• 3C + 6.50A = 125
• 0+ 3.50A = 35 (The result of adding the equations)

3.50A=35

• We will divide both sides by 3.50

3.50A/3.50 = 35/3.50

A= 10 (10 is the number of adult tickets sold.)

We will substitute 10 into our equation for the number of adult tickets in the 1st equation.

C +10 = 30

We will subtract 10 from both sides

C+10-10 = 30 -10

C= 20 (the number of children tickets sold.)

We will substitute these solutions (A=10, C=20) into our 2nd equation to verify that this is a feasible solution.

• \$3.00(20) + \$6.50 (10) = \$125
• \$60 + \$65 = \$125
• \$125 = \$125

The solution satisfies the equationn.

Therefore Ron sold 20 tickets.

J S. | Your key to success in Biology Pre-Med, Nursing, AP, science/math, SATYour key to success in Biology Pre-Med, ...
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1.  First read the problem carefully finding the useful information given.  Highlight or encircle the points you think will help you solve the problem.  For example, you might pick:

Ron is selling children's tickets for \$3.00 each. I am selling adult tickets for \$6.50 each. We had a total income of \$125.00 from 30 tickets sold. How many did Ron sell

2.  Pick symbols to represent ideas.  For example:

## A = number of adult tickets sold

3.  Do you know what  C + A equals?  Look back at the question and see if you can tell.  Finish the equation:

C + A = ?

A hint is to look for the number of tickets sold...

4. Remember that this is not the answer because the question you have to answer is:

How many [tickets] did Ron sell?

Always check that you are answering the question asked.  Here's a hint:

Ron is selling children's tickets for \$3.00 each

So, what symbol do you think will represent the answer the question is looking for,

## C or A?

5.  Look at the problem to determine another question that you could answer in order to help you figure out what C and A are:

Ron is selling children's tickets for \$3.00 each

I am selling adult tickets for \$6.50 each

We had a total income of \$125.00

6.  Put these into an equation:

## \$3.00(What goes here) + \$6.50(What goes here) = the total income

Hint:  What is the total income?

We had a total income of \$125.00 from 30 tickets sold.

7.  You now have two equations you can use to determine C and A.  Solve them by substitution.  You might have to manipulate the equations to do that. (In other words, have only C= something or A = something by moving parts of the equation by subtracting the same thing from both sides of the equation, for example.  (or, multiplying, dividing, or adding...).)

8.  Go back to point 4:  Always check that you are answering the question asked.  What are you looking for?  Be sure to label your answer and clearly mark your answer.  Check if the answer you get makes sense.  If not, go back and verify your equations.  Do they make sense? Good luck.  I hope this helps you have a way to attack solving problems, in addition to helping you solve this one problem.  Let me know.  Thanks.

P.S.  After a while, these steps will become intuitive.  You can simplify the steps, for example jumping right away to one variable.  For example, if:

x + (30-x) = 30

3(x) + 6.5(30-x) = 125

3.x -6.5x +6.50(30) = 125

-3.5x + 195 = 125

Subtract 195 from both sides.  Keep going...

Follow along this way but don't forget to find the answer the question is looking for.  Is that x or 30 - x?  How do you know?  After doing a few problems like this, you will begin to know automatically, or, at least know how to go about finding the answer.

If all of this still looks hard, go back to using boxes instead of letters, like in grade school.  Pull out some apples and oranges, or, some coins or buttons.  Have fun with it.

Dr. J