Algebra Honors (Period 6 & 7)

Cost Income, and Value Problems 3-7
Objective: To organize the facts of a problem in a chart & solve problems involving cost, income, and value
Using a chart to organize the facts of a problem can be a helpful problem solving strategy.
Cost = number of items X price per item
Income = hours worked X wage per hour
Total value = # of items X value per item

Example: Tickets for the senior class play cost $6 for adults and $3 for students. A total of 846 tickets worth $3846 were sold. How many student tickets were sold?
Let x = the number of student tickets sold
Then 846- x = the number of adult tickets sold

The only fact NOT recorded in this chart is that the total cost of the tickets is $3846.
The equation becomes
3x + 6(846 –x) = 3846.
3x + 5076 – 6x = 3826
-3x = -1230
x = 410
Check to make sure what x represented… in this case the number of student tickets so
410 student tickets were sold.


We then turned to Problem 4 on Page 128

A collection of 52 dimes and nickels is worth $4.50. How many nickels are there?
Let d = the number of dimes, so If there are 52 in the collection
52-d must = the number of nickels. So our Chart looks like:

This time we reread the given facts and with the second fact, we realize that we can add the two expressions and set them equal to $4.50
10d + 5(52 - d) = 450
solving this equation, we find that d = 38
Make sure to reread the question… it asked “how many nickels?”
So we need to use 52-d and substitute in 38… 52 -38 = 14
14 nickels is the correct answer


Next we complete # 6
Celia bought 12 apples, ate two and sold the rest at 20 cents more per apple than she paid. Her total profit was $1.00 How much did she sell each apple for?
Let b = the price she bought each of the apples for
We glued in the “yellow-colored” chart here and completed it to look like:

Now, her profit is the difference between what she sold them for and what she bought them for… Therefore the equation becomes
10(b + 20) – 12b = 100
10b + 200 -12b = 100
-2b = -100
b = 50
Re reading the question we realize we solved for what she bought the apples each for. We need to add 20cents to find out what she sold them for

She sold each apple for 70 cents.

The last problem we did from our textbook was # 14 on Page 129

Jo has 37 coins ( nickels, dimes & quarters) for $5.50 She has 4 more quarters than nickels. How many dimes does Jo have?
Let n = the number of nickels
n + 4 = the number of quarters
So if there was a total of 37 the rest must be dimes
37 – [ n + (n+4)] or 37 –(2n +4)
Let’s set up the chart

Now we know that all of them combined equal $5.50
so
5n + 10[37-(2n+4) + 25(n +4) = 550
5n + 370 – 20n – 40 + 25n + 100 = 550
combining all the n’s

10n + 330 +100 = 550
10n = 120
n = 12
Re reading the question, we find we need to see how many dimes so substitute in
37 -[2(12) +4] = 37 – [24 +4] = 37 -28 = 9
Jo had 9 dimes.