Percent Problems

The percent relationship as an equation

A percent compares a part to a whole. For example, $30\%$ of $80$ dollars is $0.30 \times 80 = 24$ dollars: here $24$ is the part, $0.30$ is the rate, and $80$ is the whole. From the Percent lesson, a percent is a count out of $100$, so a rate like $30\%$ is the decimal $\frac{30}{100} = 0.30$. 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:

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

where the percent number is the plain number in front of the percent sign, $30$ for $30\%$.

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. Whichever quantity is missing, substitute the other two and use one multiplication or division to find 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.

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 $w = \dfrac{p}{r}$. That division needs the rate to be more than $0$: a $0\%$ rate could not have produced any part other than $0$, so there would be no single whole to divide back to.

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. That division needs the whole to be more than $0$: a whole of $0$ has nothing to take a part of.

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.

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 $p = r\,w$.

Finding a percent change from two amounts

Sometimes a problem gives you the amount before and after a change and asks for the rate, instead of giving you the rate. Find the size of the change, then divide by the original amount, never by the new amount. Keep the change positive by subtracting the smaller amount from the larger one: for an increase,

and for a decrease, subtract the other way, $r = \dfrac{\text{original} - \text{new}}{\text{original}}$, so $r$ stays the same positive rate you use everywhere else in this lesson. Either way, this is the same $r = \dfrac{p}{w}$ you already know, with the size of the change playing the part and the original amount playing the whole.

Reversing a percent change

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

That division is always safe for an increase, since a rate $r$ is positive and $1 + r$ is never $0$. A decrease undoes the same way, dividing by $1 - r$ instead, but only when $1 - r$ is not $0$, which means the decrease has to be less than $100\%$. A $100\%$ decrease cannot be reversed: it multiplies every original amount by $0$, so it lands on the same new amount, $0$, no matter what the original was, and there is no way to divide back to recover which original amount that $0$ came from. 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. This is the single most common trap in percent work.

Successive percent changes

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 $f_1 f_2$. To apply several changes in a row, multiply the starting amount by each factor in turn, which is the same as multiplying by the product of the factors. That product is not $1$ even when two rates are equal and opposite, because the second change acts on the amount the first change already produced, not on the original amount. Successive changes always combine by multiplying their factors, never by adding their rates.

Two $10\%$ raises in a row multiply by $1.10 \times 1.10 = 1.21$, a $21\%$ raise, not the $20\%$ you would get by just adding the rates. The extra $1\%$ comes from the second raise acting on the already-raised amount, not on the original.

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:

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,

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.)

Just like $p = r\,w$, the equation $I = P\,r\,t$ rearranges to find whichever quantity is missing: divide by the other two to isolate it, for instance $r = \dfrac{I}{Pt}$ to find the rate or $t = \dfrac{I}{Pr}$ to find the time.

Common mistakes