Pct 12: Traditional Math:Finding Original Amounts When Amount of Decrease and Increase are Known
Seventh Grade
As I mentioned in the first lesson on discounts and mark-ups, (more generally referred to as decreases and increases) I have taught the unitary approach which was also used to find original amounts and did so some time after the initial teaching of finding discounts and mark-ups. I recommend teaching those methods after the next chapter on functions.
The initial methods were presented as follows:
A discount is the amount by which the regular price (or value) of an item is reduced. The sale price (or final value) =original price (value) – discounted amount.
A mark-up is the amount by which the regular price (or value) of an item is increased. The sale price (or final value) =original price (value) + increased amount.
The amounts of the discount (decrease) and mark-up (increase) are calculated using a method they learned at the beginning of the unit: finding the percent of a number, more generally expressed as finding the part of the whole. This part was then subtracted or added to the whole, which is what the above formulas show.
The students have learned how to do this in one step using the unitary approach, which involved a proportion equation, or translating the problem directly into an equation. The proportion served to not only find the discounted or marked-up values, but what the original amounts (the wholes) prior to taking a discount or mark-up.
They have been working with different types of problems, all variations of the basic part/whole percent equation they learned at the beginning. At the novice level at which most seventh graders operate, the connection to the original percent equation is not obvious. They are really all part/whole problems.
For example: “A television is discounted 20% and the amount of the discount is $240. What is the original price of the television?” Although the problem is an “original amount” problem, and it asks for the whole (original) amount, it is using a different part of the whole: the amount of the discount rather than the final sales price.
The above problem is equivalent to the following: “Madison saves $240 every month. This is 20% of her monthly salary. What is her monthly salary?” The problem essentially translates to “20% of what number is 240?”/ This problem could also be presented as a discount problem using the same numbers, and resulting in the same equation: 0.2x = 240; x = $1,200.
In this section we will work with problems in which the amount of the discount (decrease) or mark-up (increase), is given as well as the percent rates of same. It is important that in providing examples, as well as homework problems, there be a mix of the different types of percentage problems students have been working with. This interleaving process allows students to better identify what the problem is asking for, and the approach to solve it.
Warm-Ups. Some of these problems are part/whole problems stated in terms other than decreases or increases in price.
1. Michelle scored 42 points on a test. This was 75% of the total score. Find the total score. (Hint: 42 is 75% of what number?) Answer: 75/100 = 42/x 0.75/1 = 42/x; 0.75x =42; 42/0.75= 56
2. The discounted price for a fan is $140. This is 70% of its original amount. What was the original amount for the fan? Answer: 0.7x = 140; x = $200; or 70/100 = 140/x; x = $200.
3. 40% of the students in a class are girls. If there are 16 girls in the class, how many boys are there? Answer: Let x = total number of students in the class; 0.4x =16; x = 40
4. 70% of the students in a class are girls. If there are 15 boys in the class how many girls are there? Answer: Let x = total no. of students. Percent of boys in the class are 100% - 70% = 30%. 0.3x = 15; x = 50 students; there are 50 – 15 = 35 girls
5. John has 40 books. Steven has 25% more books than John. How many books does Steven have? Answer: 125/100 =x/40; 1.25/1 = x/40; x = 1.25 ∙ 40 = 50
Finding Original Amounts When Only the Discount Rate and Amount is Known. An example of this type of problem is the following: Mary bought a dress at a discount of 25%. If the discount was $15 find the usual price of the dress.
I first ask if anyone has any idea how to solve it. I do this because it is different than what they’ve seen and chances are that they will do something like this:
This simplifies to 0.75x = 15 and x = 20.
If somebody speaks up and says this is not correct, I’ll ask them to explain. Otherwise, I’ll take the next card:
“So let’s see, the discount was $15, and the original price was $20, so that means the price was reduced by $15, leaving $5. Wait, what? Where is the $5? What happened?”
There is likely to be some discussion which I guide, mostly by asking them to compare this to problems they have solved before.
“What if I had given you a problem that said ‘Mary but a dress discounted 25% and the final sales price was $60. What was the original amount?’ How would you write the proportion?”
It should look like this:
“This is what you’ve been solving. What’s the difference between this problem and the first one?”
At least one person should see that the first is providing the amount of the discount, rather than the final sales price. “So we subtracted 25% from 100%, and that was incorrect. What should we have done?”
The answer I’m searching for is “The discount rate should have been used”, although I might hear “25% instead of 75%” in which case I’ll paraphrase it to the previous sentence, and then write the proportion formula that we use for this type of problem on the board:
“Now let’s try it. What is the discount rate? What is the discount amount? Where do these values go?” and hopefully end up with:
This can also be solved by using a direct translation to an equation. “If we’re finding the amount of a discount, we’re multiplying the rate of the discount by the full price. We know the discount rate; it’s 25% which is what as a decimal?”
Hearing 0.25, I say “We multiply this by the full price. Do we know it? What do we use if we don’t know the price”.
I usually hear x, so I write: 0.25x = ? and ask “Well, what does it equal?” We end up with the same equation as above.
Examples:
1. A television set is discounted 12%. If the discount amount is $132, what was the original amount? We let x = the original amount. Since the amount of the discount is 132, then 0.12x = 132, and x = $1,100.
2. A necklace was discounted by 20% which was $240. What was the original amount of the necklace? Answer: 20/100 = 240/x; 0.2/1 = 240/x; 0.2x = 240; x = $1,200
3. For the above example, what was the final sales price? Answer: The simplest way is to subtract 240 from 1200 to obtain $960. But the proportion or equation method can also be used. 80/100 = x/1200; 0.8/1 = x/1200; x = 0.8 ∙ 1200 = $960.
4. Now I throw one in that gives final sales price: Jenny bought a fan for $140. It was discounted 30%. What was the original price? Answer: Some will try to answer this assuming $140 was the discounted amount. It should be: 70/100 = 140/x; 0.7x = 140; x = $200.
Finding Original Amounts When Only the Mark-up Rate and Amount is Known. Moving on to mark-ups and other types of increases, I pose a problem: “Bicycles are marked up 50% at a bicycle store. If the amount of mark-up is $100 what is the original cost of the bike? How do we solve this?”
Students should make the connection from the previous discussion on discounts that the 50% rate pertains to the amount of the mark-up. I write the proportion on the board:
The proportion for the problem should look like:
And as before, this can be solved using equations directly: 50% of the original amount is $100. “How do we express that?” It is the same equation as above.
Examples (These include discount and other types of problems mixed in):
1. The mark-up on an electric can opener was 30% resulting in a $15 increase above original cost. What was the original cost of the can opener? 30/100 = 15/x 0.3x = 15; x = $50
2. The mark-up on a calculator is 20% and represents an increase of $70. What is the original cost of the calculator? Answer: 0.2/1 = 70/x; 0.2x = 70; x = $350.
3. Emily’s score for a math test was 5% more than her score for a Spanish test. If she scored 84 on her math test, what did she score on the English test? Answer: (This is a final amount problem not an amount of increase problem) (100 +5)/100 = 84/x; 1.05x = 84; x = 80
4. A food blender costs $400. What is the cost if it is discounted 20%? Answer: This is a straightforward “find the final sales price” problem. (100-20)/100 = x/400; 0.8/1 = x/400; x = 0.8 ∙ 400 = $320.
5. Andy’s pay is $300 more than his co-worker’s pay. If this is 20% higher, what is his coworker’s pay? Answer: 20/100 = 300/x; 0.2/1 = 300/x; 0.2x=300; x = $1500
Homework. The homework problems should focus on this lesson, but should also include a mix of problems as illustrated by the examples.