Percent Problems

Learning goals

Solve $p = rw$ for whichever of the three is unknown
Multiply by $1 + r$ to raise and $1 - r$ to cut
Divide by the factor to reverse a percent change
Stack successive changes by multiplying their factors
Show why a rise and equal fall leave $0.96$
Apply $I = Prt$ , with the principal fixed

The percent relationship as an equation

A percent compares a part to a whole. From the Percent lesson, a percent is a count out of $100$ , so a rate like $30\%$ is the decimal $\frac{30}{100} = 0.30$ . Write the percent as this decimal and call it the rate $r$ . Because “of” with a multiplier means multiply, taking $r$ of a whole is exactly $r$ times the whole. That gives the one relationship this whole lesson runs on:

\text{part} = \text{rate} \times \text{whole}, \qquad p = r\,w.

Written as the percent proportion you already know, the same fact is

\frac{p}{w} = \frac{\text{percent}}{100} = r.

Dividing the part by the whole gives the rate; multiplying the rate by the whole gives the part. These are two views of a single equation. Three quantities live inside it, the part $p$ , the rate $r$ , and the whole $w$ , and any percent problem hands you two of them and hides the third. The hidden one is the letter you solve for, and since $p = r\,w$ is linear in every letter, one step of algebra finishes it.

Solving for each of the three quantities

The three classic percent tasks are not three separate methods. They are the one equation $p = r\,w$ solved for its three different letters.

Finding the part

When you know the rate and the whole, the part is a direct multiplication. Convert the percent to the decimal rate first, then multiply.

Worked example 1 Find a part: $15\%$ of $340$

The rate and whole are given, and the part is unknown. Write the percent as a decimal, $15\% = 0.15$ , and substitute into $p = r\,w$ :

p = 0.15 \times 340.

Carry out the multiplication:

p = 51.

So $15\%$ of $340$ is $51$ . As a check, $10\%$ of $340$ is $34$ and $5\%$ is half of that, $17$ , and $34 + 17 = 51$ .

Finding the whole

When you know the part and the rate, the whole sits multiplied by $r$ . Solve $r\,w = p$ by dividing both sides by the rate, which gives $w = \dfrac{p}{r}$ .

Worked example 2 Find the whole from a part

A $45$ dollar deposit is $15\%$ of a bike’s price. What is the full price?

The part is $45$ and the rate is $15\% = 0.15$ ; the whole price $w$ is unknown. Substitute into $p = r\,w$ :

45 = 0.15\,w.

Divide both sides by the rate to isolate $w$ :

w = \frac{45}{0.15} = 300.

The bike costs $300$ dollars. Run it forward to check: $15\%$ of $300$ is $0.15 \times 300 = 45$ , the deposit you were told. Notice the whole is much larger than the part, which is the sign you should divide, not multiply.

Finding the rate

When you know the part and the whole, the rate is what multiplies $w$ to reach $p$ . Solve $r\,w = p$ by dividing by the whole, giving $r = \dfrac{p}{w}$ , then turn that decimal into a percent.

Worked example 3 Find the percent one number is of another

A student answered $34$ of $40$ questions correctly. What percent is that?

The part is $34$ and the whole is $40$ ; the rate is unknown. From $p = r\,w$ ,

34 = r \times 40.

Divide both sides by the whole:

r = \frac{34}{40} = 0.85.

Convert the decimal rate to a percent by multiplying by $100$ , so $r = 85\%$ . The student scored $85\%$ . The whole is the number after “of,” here the $40$ questions, so it goes in the denominator.

Check your understanding

$45$ is $30\%$ of what number?

135

150

13.5

75

Percent change as a single multiplier

A percent change raises or lowers a quantity by a rate of itself. Writing that change as one multiplication is the key move that powers every discount, tax, markup, and interest problem. So that move is worth deriving carefully and then reusing everywhere.

Why a percent change is one multiplication, and undoing it is one division#

Suppose a quantity starts at $w$ and changes by a rate $r$ , written as a decimal, so a $20\%$ rise is $r = 0.20$ . Increasing $w$ by that rate means adding $r\,w$ to $w$ :

\text{new} = w + r\,w.

Both terms on the right carry a factor of $w$ , and the first is $1 \cdot w$ because the whole starting amount is $100\%$ of itself. Factor $w$ out:

\text{new} = (1 + r)\,w.

A single number, the factor $1 + r$ , performs the entire increase. A decrease runs the identical argument with a minus sign, giving $\text{new} = (1 - r)\,w$ : you keep the whole and remove the rate, so the factor is $1 - r$ . Since $r = \frac{p}{100}$ for a percent $p$ , this factor is the same $1 + \frac{p}{100}$ you met before, now packaged as one decimal multiplier.

Because the change is one multiplication, undoing it is one division. If $\text{new} = (1 + r)\,w$ , then dividing both sides by the factor isolates the original amount:

w = \frac{\text{new}}{1 + r}.

This is exactly why you cannot recover the original by subtracting the percent back from the new value. The change multiplied the original, so undoing it must divide by the same factor.

Now put two changes back to back. A change with factor $f_1$ followed by a change with factor $f_2$ multiplies the start by $f_1$ and then by $f_2$ . So the overall factor is the product $f_1 f_2$ . Multiply a $20\%$ rise and a $20\%$ fall:

(1 + 0.20)(1 - 0.20) = (1.20)(0.80) = 0.96,

not $1$ . The two do not cancel, because the fall is taken of the larger, already-raised amount. An equal rise and fall leave you at $96\%$ of where you began, a $4\%$ net loss. Successive changes always combine by multiplying their factors, never by adding their rates.

With the factor in hand, a percent increase or decrease is one multiplication. To grow $w$ by a rate $r$ , multiply by $1 + r$ ; to shrink it, multiply by $1 - r$ . The everyday money problems are exactly these factors:

A discount of rate $r$ multiplies by $1 - r$ , because you pay the part that is left.
Sales tax or a tip at rate $r$ multiplies by $1 + r$ , adding to the bill.
A markup of rate $r$ on cost multiplies by $1 + r$ to set the selling price.
A commission is the part $r \cdot (\text{sales})$ , a straight use of $p = r\,w$ .

Worked example 4 A markup as a single multiplier

A shop buys a bicycle for $180$ dollars and marks it up $35\%$ to set the price. What does it sell for?

A markup is a percent increase on the cost, so the price is $(1 + r)$ times the cost with $r = 0.35$ . The factor is $1 + 0.35 = 1.35$ :

\text{price} = 1.35 \times 180 = 243.

The bicycle sells for $243$ dollars. Reading the factor aloud helps: a $35\%$ markup leaves you paying $135\%$ of the cost, which is why the multiplier is $1.35$ . The two-step route agrees, since $35\%$ of $180$ is $63$ and $180 + 63 = 243$ .

Reversing a percent change

When a problem gives the amount after a change and asks for the amount before it, solve the factor equation for the original. Because the change multiplied by $1 + r$ or $1 - r$ , you divide by that same factor to get back. This is the single most common trap in percent work: subtracting the percent from the new value gives the wrong answer. The reason is that the percent was figured on the original, not on the new amount.

Worked example 5 Undo a tax to find the original price

A phone costs $92$ dollars including $15\%$ sales tax. What was the price before tax?

Tax is a percent increase, so the after-tax total is $(1 + r)$ times the pre-tax price with $r = 0.15$ . Let $w$ be the pre-tax price:

1.15\,w = 92.

The change multiplied by $1.15$ , so undo it by dividing by $1.15$ :

w = \frac{92}{1.15} = 80.

The pre-tax price was $80$ dollars. Check it forward: $15\%$ of $80$ is $12$ , and $80 + 12 = 92$ . The tempting wrong move is to take $15\%$ of $92$ and subtract, but $15\%$ of $92$ is $13.80$ , giving $78.20$ , not $80$ . That fails because the tax was $15\%$ of the smaller original, not $15\%$ of the larger total.

Check your understanding

After a $25\%$ discount, a shirt sells for $45$ dollars. What was the original price, in dollars?

56.25

60

36

33.75

Successive percent changes

As the proof showed, changes stack by multiplying their factors. To apply several in a row, multiply the starting amount by each factor in turn, which is the same as multiplying by the product of the factors. Because the factors multiply, two changes are not the sum of their rates: the second change acts on the amount the first change already produced.

Expanding the product for two increases makes the gap explicit:

(1 + r_1)(1 + r_2) = 1 + r_1 + r_2 + r_1 r_2.

The single overall rate is $r_1 + r_2 + r_1 r_2$ , which beats the naive $r_1 + r_2$ by the cross term $r_1 r_2$ . That extra piece is the change-of-the-change, and it is exactly why a $10\%$ raise on top of a $10\%$ raise is a $21\%$ raise, not $20\%$ . Here the cross term $r_1 r_2 = 0.01$ adds the last percent.

Worked example 6 Two changes in a row

A $200$ dollar item is marked up $20\%$ for the holidays, and then that higher price is cut $20\%$ in a sale. What is the final price, and is it back to $200$ ?

Build a factor for each change and multiply in order. The markup is $1 + 0.20 = 1.20$ and the sale is $1 - 0.20 = 0.80$ :

200 \times 1.20 \times 0.80.

Multiply the factors first to see the net effect:

1.20 \times 0.80 = 0.96, \qquad 200 \times 0.96 = 192.

The final price is $192$ dollars, not $200$ . The equal rise and fall did not cancel: the $20\%$ cut was taken of the raised $240$ , so it removed more than the markup added. The net factor $0.96$ is a $4\%$ decrease from the start.

A 20 percent rise then a 20 percent fall does not return to the start. The rise multiplies by 1.20 and the fall by 0.80, and 1.20 times 0.80 is 0.96, so the final bar falls short of the original by 4 percent.

Check your understanding

A price rises $10\%$ one week and falls $10\%$ the next. Compared with the start, the price is now:

A unchanged B down

1\%

C up

1\%

D down

10\%

Simple interest

Interest is a percent applied to money over time. In simple interest, the interest each year is the same rate of the original principal, and the principal itself never changes. Suppose you invest or borrow a principal $P$ at an annual rate $r$ (a decimal) for $t$ years. Then the interest is the rate times the principal times the number of years:

I = P\,r\,t.

This is $p = r\,w$ applied once per year and added up: each year earns $r \cdot P$ , and over $t$ years that totals $P r t$ , because the principal stays fixed. The total amount is the principal plus the interest,

A = P + I = P + P r t = P(1 + r t).

The factor $1 + r t$ echoes the increase multiplier, except the rate is multiplied by the time first. Because the principal is fixed, simple interest grows in a straight line with $t$ . (Letting each year’s interest join the principal and earn more is compound interest, a separate lesson.)

Worked example 7 Simple interest on a deposit

You deposit $800$ dollars at $5\%$ simple annual interest for $3$ years. How much interest do you earn, and what is the balance?

Identify the pieces: the principal is $P = 800$ , the rate is $r = 5\% = 0.05$ , and the time is $t = 3$ years. Substitute into $I = P r t$ :

I = 800 \times 0.05 \times 3 = 120.

You earn $120$ dollars of interest. Add it to the principal for the balance, or use the factor form $A = P(1 + r t)$ :

A = 800 + 120 = 800(1 + 0.05 \times 3) = 800 \times 1.15 = 920.

The balance is $920$ dollars. Each year adds the same $0.05 \times 800 = 40$ dollars, and three equal years give $120$ . That equal yearly interest is the mark of simple interest: no year earns interest on a previous year’s interest.

Common mistakes

Subtracting to reverse a change instead of dividing. After a $20\%$ discount, the original is the sale price divided by $0.80$ , not the sale price plus $20\%$ of itself. The change multiplied by the factor, so undoing it divides by that factor.
Adding successive percents instead of multiplying factors. A $20\%$ rise then a $20\%$ fall is not a wash; it multiplies by $1.20 \times 0.80 = 0.96$ , a $4\%$ loss. Two changes combine as a product of factors, never a sum of rates.
Confusing “percent of” with “percent more than.” ” $20\%$ of $50$ ” is $0.20 \times 50 = 10$ , but ” $20\%$ more than $50$ ” is $1.20 \times 50 = 60$ . One is the part; the other is the whole plus the part.
Using the wrong base as the whole. The whole is the number after “of,” and a percent change is always measured against the original. Comparing a change to the new value instead of the original is the classic setup error.
Building the multiplier from the wrong piece. A $30\%$ decrease multiplies by $0.70$ , the part that remains, not by $0.30$ , the part removed. Always build the factor from what is left.
Forgetting to convert the percent to a decimal. In $p = r\,w$ the rate is $r = 0.30$ , not $30$ . Plugging in $30$ inflates the answer a hundredfold.

Key takeaways

Every percent problem is the one linear equation $p = r\,w$ (part = rate times whole), with the rate $r$ the percent written as a decimal. Name the unknown letter and solve: $w = \frac{p}{r}$ for the whole, $r = \frac{p}{w}$ for the rate.
A $p\%$ increase multiplies the original by the factor $1 + r$ ; a $p\%$ decrease multiplies by $1 - r$ , where $r = \frac{p}{100}$ . Discounts use $1 - r$ ; tax, tip, and markup use $1 + r$ .
Reverse a change by dividing by its factor: $\text{original} = \dfrac{\text{new}}{1 \pm r}$ . Divide, do not subtract.
Stack successive changes by multiplying their factors, $f_1 f_2$ . Equal rise and fall do not cancel, since $(1.20)(0.80) = 0.96$ , and percents add only when taken of the same base.
Simple interest is $I = P r t$ , with total $A = P(1 + r t)$ . The principal stays fixed, so the interest grows linearly with time.

Percent Problems

Learning goals

The percent relationship as an equation

Solving for each of the three quantities

Finding the part

Finding the whole

Finding the rate

Percent change as a single multiplier

Reversing a percent change

Successive percent changes

Simple interest

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)