Pct 11 Traditional Math: Finding the Original Amounts When the Discounted or Marked-Up Values are Known
Seventh Grade
In the two previous lessons, two proportion equations were introduced that allow finding both discounted (decreased) and marked-up (increased) values:
The problems so far have been to find the reduced or increased amounts given the original value and the percent decrease (discount) or increase (mark-up). In this lesson the focus is on finding the original amount given the discounted or marked up value and the percent decrease or increase.
The method is simply an extension of the above proportions. The challenge to students is to identify the type of problem (e.g., find discounted amount, or find original amount) and keep track of information so they know where the values are placed in the proportion equation.
The proportion equations above are really just renamed versions of the proportion equation introduced in an earlier lesson:
Also, these problems can be solved by translating directly into equations. For example, the following problem can be solved without using the proportion equation: “John received a 20% raise and now receives $1,200 per week. How much did he make before the raise?” Let x = salary amount prior to the raise; then 1.2x = 1200, and x = $1,000. My experience is that seventh graders have difficulty expressing the increase or decrease directly and need the assistance of the proportion equation.
As in the previous lesson, problem examples will be mixed, so that students may be finding a discounted value or an original amount for example.
Warm-Ups. In this set, some review of working with ratios via tape diagrams or algebra is included. This unit has covered expressing percents in terms of part to
whole, equations, proportions, percents of change and finally to calculating percent decreases and increases (e.g., discounts and mark-ups). Warm-Ups should provide opportunity to practice all the types of problems presented in this unit.
1. The ratio of John’s weight to Peter’s weight is 5:3. Their average weight is 40 lb. Find John’s weight. (Hint: What is their total weight? Use tape diagrams or algebra.) Answer: Let 5x = John’s weight, 3x = Peter’s weight. Total weight would be 40 x 2 = 80. So 5x + 3x = 80, 8x = 80, x = 10. John’s weight = 50 lb, Peter’s weight = 30 lb
2. Anne bought 72 stamps. She bought 27 more stamps then Betty. How many percent more stamps did Anne buy than Betty? Answer: 72-27 = 45: Betty’s stamps. 72/45 = 1.6 = 160% more stamps
3. There are 40 people at a party at 7 PM. By 8 PM, that number has increased by 25%. How many people are at the party at 8 PM? Answer: 1.25 × 40 = 50 OR 125/100 = x/40; 100x = 5000; x = 50
4. Ian has $56. Beatrice has 20% more money than Ian. How much money does Ian have? Answer: x = 1.2 ∙ 56; x = $67.20 OR 120/100 = x/56; 1.2/1 = x/56; x = 1.2 ∙ 56; x = $67.20
5. 85% of what number is 51? Answer: 0.85x = 51; x = 51/0.85; x = 60 OR 85/100 = 51/x; 0.85/1 = 51/x; 0.85x = 51; x = 60
Problem 1 is a review of ratio problems that are solved using either tape diagrams or algebraic equations. The remaining problems focus on various aspects of percents. Problem 2 is percent of change; Problems 3 and 4 is a percent increase problem and Problem 5 requires students to solve using either an equation or the proportion method. This problem also relates directly to this unit, because it is an equation that is at work in finding original amounts.
Finding Original Amounts Using Proportions. I want to relate what we’ve done with finding discounted and marked-up values to what we are doing in this lesson, so I pose a problem that they can do.
“Suppose there is a jacket that sells for $60 and it is discounted 15%. How do I find the sale price?” I will instruct them to do it in their notebooks and go around to see who did it correctly. If it’s possible, I will pick a student who used an algebraic approach and one who used the proportion approach and have both put their worked solutions on the board.
Algebraic approach: x = 0.85 ∙ 60; x = $51.
Proportion approach:
I will take this opportunity to again show the short cut of expressing 85/100 as 0.85/1:
“Now let’s pretend we don’t know that the original price is $60, but we only know that the sales price is $51 and that the discount is 15%. And now let’s set up the proportion again.” I write the proportion formula on the board:
“I wrote “percent decrease” because it’s a general term that includes discounts. It can also mean percent decrease in population or many other things. In this case we are dealing with percent discount. And what does the problem say that is?”
Proceeding in this fashion I now have:
“What does the problem tell us? Do we know the reduced amount or the original amount?”
This is a straightforward question and usually I get an answer right away: We know the reduced amount and don’t know the original amount.
“So where do I put the 51 and where does the variable go?”
We now end up with:
Solving it, we obtain 0.85x = 51, and x = $60.
“What if I said that a dealer bought a CD player and marked it up by 60% and selling it for $320, and I want to know how much the dealer paid for it. That is, the original cost. Can we solve this in similar way?”
Students may or may not have the answer right away, so as a prompt I will ask for the proportion equation for mark-ups. They will know this and if they have forgotten from the last lesson, I tell them to look it up in their notes. I now write it on the board:
Again, I ask questions to get them to put numbers and the variable in their proper places in the equation, ending up with:
Now I proceed to examples which include both discounts/decreases and mark-ups/increases. Students usually see the pattern but need practice identifying the values and placing them properly.
Worked Examples:
1. A fan is on sale for 30% off at $140. What was its original price?
Prompts: “Do we add 30% to 100% or subtract it from 100%. Where does the 140 go—numerator or denominator?” The proportion equation is now:
2. Mr. Tyler received a raise of 10% and is now making $1,408 per week. How much did he make before the raise?”
Same type of prompts as above. The proportion equation is now:
Examples: (These include finding original value as well as reduced or increased value)
3. Juan sold a bicycle at a discount of 15%. The sales price was $340; find the original price. Answer: 0.85/1 = 340/x; 0.85x = 340; x = $400
4. Jake spent 10% more this week than last week. He spent $55 this week. How much did he spend last week? Answer: 1.1/1 = 55/x; 1.1x = 55; x = $50
5. Joan bought a blouse that was 35% off. The original price was $60. What is the sales price? Answer: 0.65/1 = x/60 x = 0.65∙60; x = $49
6. Tyler increased his test score by 10%. He previously scored 80. What is his new score? Answer: 1.1/1 = x/80; x = 1.1∙80 =88.
Finding Original Amounts Using Equations. I start this portion of the lesson by recalling that we learned how to translate into algebra sentences like: 35% of 200 is what number?
“We could use proportions for this, or we could translate it directly as 0.35 ∙ 200 = x. So how would we translate ‘250 is 40% of what number?’”
Students should get 250 = 0.4x; solving, we obtain x = 625.
“We also learned that we can use equations to solve problems like ‘A $100 tire at cost is marked up by 60%. What is the sales price?’ ”
Students should get x = 100 ∙1.6 = $160
“Now let’s use this to solve for original amounts. A radio discounted 25% sells for $120. What was its original price? “If it’s discounted 25% what percent do you pay?”
Hearing 75%, I go on. “So 75% of the original price is $120. How would you translate that?”
I have them write in their notebooks and check their progress. They should have 0.75x = 120; x = $160. For comparison purposes I have them now solve it using proportions; they will get the same equation.
Examples: (In these, they work in their notebooks and I walk around checking and providing hints and guidance)
1. John made $44 an hour after receiving a 10% raise. How much did he make before the raise?
Prompts: “How do we represent 100% as a decimal?” “110% of what amount is 44?” “Can you translate that into an equation?” Answer: 1.1x = 44; x = 44/1.1; x = $40
2. A tire is sold at a discount of 10%. It is sold for $45. Find the original price of the tire. Answer: 0.9x = 45; x = 45/0.9; x = $50
3. Mary bought a dress at a discount of 25%. If the discounted amount was $45, find the original price of the dress. Answer: 0.75x = 45; x = 45/0.75; x = $60
4. If a bicycle sells for $1500 and is discounted 30%, what is the sales price of the bicycle? Answer: x = 0.7 ∙ 1500 = $1,050
Homework. The homework should start off primarily with finding original amounts. Then it should progress to mixed problems that require finding original and discounted or increased amounts. Interspersed in the mix should be problems in which the amount of the increase or decrease is calculated. For example: “A suit sells for $500, and is discounted 30%. What is the amount of the reduction?” Answer: x = 0.3 ∙ 500; x = $150 OR 30/100 = x/500; 0.3/1 = x/500; x = $150