Module 27: Percents Part 3

Morgan Chase

Module 27: Percents Part 3

You may use a calculator throughout this module.

There is one more situation involving percents that often trips people up: working backwards from the result of a percent change to find the original value.

$\text{Amount}=\text{Rate}\cdot\text{Base}$

$A=R\cdot{B}$

Finding the Base After Percent Increase

Suppose a $12\%$ tax is added to a price; what percent of the original is the new amount?

Well, the original number is $100\%$ of itself, so the new amount must be $100\%+12\%=112\%$ of the original.

As a proportion, $\frac{A}{B}=\frac{112}{100}$ . As an equation, $A=1.12\cdot{B}$ .

If a number is increased by a percent, add that percent to $100\%$ and use that result for $R$ .

The most common error in solving this type of problem is applying the percent to the new number instead of the original. For example, consider this question: “After a $12\%$ increase, the new price of a computer is $ $1,120$ . What was the original price?”

People often work this problem by finding $12\%$ of $ $1,120$ and subtracting that away: $12\%$ of $1,120$ is $134.40$ , and $1,120-134.40=985.60$ . It appears that the original price was $ $985.60$ , but if we check this result, we find that the numbers don’t add up. $12\%$ of $985.60$ is $118.272$ , and $985.60+118.272=1,103.872$ , not $1,120$ .

The correct way to think about this is $1,120=1.12\cdot{B}$ . Dividing $1,120$ by $1.12$ gives us the answer $1,000$ , which is clearly correct because we can find that $12\%$ of $1,000$ is $120$ , making the new amount $1,120$ . The original price was $ $1,000$ .

To summarize, we cannot subtract $12\%$ from the new amount; we must instead divide the new amount by $112\%$ .

Exercises

1. A sales tax of $8\%$ is added to the selling price of a lawn tractor, making the total price $ $1,402.92$ . What is the selling price of the lawn tractor without tax?

2. The U.S. population in 2018 was estimated to be $327.2$ million, which represents a $7.6\%$ increase from 2008. What was the U.S. population in 2008?

Finding the Base After Percent DEcrease

Suppose a $12\%$ discount is applied to a price; what percent of the original is the new amount?

As above, the original number is $100\%$ of itself, so the new amount must be $100\%-12\%=88\%$ of the original.

As a proportion, $\frac{A}{B}=\frac{88}{100}$ . As an equation, $A=0.88\cdot{B}$ .

If a number is decreased by a percent, subtract that percent from $100\%$ and use that result for $R$ .

As above, the most common error in solving this type of problem is applying the percent to the new number instead of the original. For example, consider this question: “After a $12\%$ decrease, the new price of a computer is $ $880$ . What was the original price?”

People often work this problem by finding $12\%$ of $880$ and adding it on: $12\%$ of $880$ is $105.60$ , and $880+105.60=985.60$ . It appears that the original price was $ $985.60$ , but if we check this result, we find that the numbers don’t add up. $12\%$ of $985.60$ is $118.272$ , and $985.60-118.272=867.328$ , not $880$ .

The correct way to think about this is $880=0.88\cdot{B}$ . Dividing $880$ by $0.88$ gives us the answer $1,000$ , which is clearly correct because we can find that $12\%$ of $1,000$ is $120$ , making the new amount $880$ . The original price was $ $1,000$ .

To summarize, we cannot add $12\%$ to the new amount; we must instead divide the new amount by $88\%$ .

Exercises

3. A city department’s budget was cut by $5\%$ this year. If this year’s budget is $ $3.04$ million, what was last year’s budget?

4. CCC’s enrollment in Summer 2019 was $9,116$ students, which was a decrease of $2.17\%$ from Summer 2018. What was the enrollment in Summer 2018? (Round to the nearest whole number.)^[1]

5. An educational website claims that by purchasing access for $ $5$ , you’ll save $69\%$ off the standard price. What was the standard price? (Use you best judgment when rounding your answer.)

These enrollment numbers don't match those in Percents Part 2, which makes me wonder how accurate the yearly reports are. Or maybe I inadvertently grabbed data from two different ways that enrollment was being counted. ↵

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