3.4 Percentages, Percent Change, Markup & Profit

Quick Answer: $P\% = \frac{P}{100}$. The no-calculator engine is the decimal (or fractional) multiplier: new = original $\times (1 \pm P/100)$. Percent of a percent multiplies; successive changes multiply; percent change always divides by the old value. Markup lives on cost, margin lives on revenue, and discount lives on the marked price. Official Guide review groups percents with rates and ratios because every one of these is a proportion.

Percent as a Multiplier

Rather than computing a percentage amount and then adding or subtracting it, multiply the base by one factor:

$$\text{New value} = \text{Original value} \times \text{Multiplier}.$$

Fractions beat decimals on the GMAT whenever the percent is a friendly unit fraction ($25\% = \frac{1}{4}$, $12.5\% = \frac{1}{8}$, $33\frac{1}{3}\% = \frac{1}{3}$).

Percent of a Percent

'$20\%$ of $30\%$' is a product of fractions, not a sum:

$$20\% \text{ of } 30\% = \frac{20}{100} \times \frac{30}{100} = 0.06 = 6\%.$$

It is not $50\%$. If a bonus is $20\%$ of a commission that is itself $30\%$ of revenue, the bonus is $6\%$ of revenue.

Chaining 'is a percent of' works the same way. If $X$ is $25\%$ of $Y$ and $Y$ is $40\%$ of $Z$, then

$$X = 0.25 \times 0.40 \times Z = 0.10 Z,$$

so $X$ is $10\%$ of $Z$. Multiply the decimal forms; do not add $25$ and $40$.

Worked example. What is $40\%$ of $25\%$ of $800$? Compute $0.40 \times 0.25 \times 800 = 0.10 \times 800 = 80$. Equivalently, $25\%$ of $800$ is $200$, and $40\%$ of $200$ is $80$. Taking '$40\%$ of $25\%$' as $65\%$ of $800$ produces $520$ and is the standard wrong answer.

Percent of a percent is also how overlapping discounts differ from a single combined discount: a $10\%$ coupon applied to a price that is already $20\%$ off is $0.90 \times 0.80 = 0.72$ of original, which is a $28\%$ total discount, not $30\%$. The extra $2\%$ is exactly the $10\%$ of the $20\%$ that stacking would double-count if you added.

Percent Change and the Asymmetric Base

$$\text{Percent change} = \frac{\text{New} - \text{Old}}{\text{Old}} \times 100\% = \frac{\text{Change}}{\text{Original base}} \times 100\%.$$

The denominator is always the starting value of the quantity whose change you are reporting. Switching the base is the most common GMAT percent error.

Trap. If $A$ is $25\%$ greater than $B$, by what percent is $B$ less than $A$? Not $25\%$.

$$A = 1.25 B = \frac{5}{4} B \implies B = \frac{4}{5} A = 0.80 A,$$

so $B$ is $20\%$ less than $A$. In general, if $A$ is $x\%$ greater than $B$, then $B$ is smaller than $A$ by $\frac{x}{100 + x} \times 100\%$. For $x = 25$, that is $\frac{25}{125} = 20\%$. For $x = 50$, $B$ is $\frac{50}{150} = 33\frac{1}{3}\%$ less than $A$. The two percentages are reciprocals of the two multipliers, not copies of each other.

Going down and then asking about going up is the mirror: if $B$ is $20\%$ less than $A$, then $A$ is $25\%$ greater than $B$, because $\frac{0.20}{0.80} = 0.25$.

Successive Percent Changes

When a value undergoes consecutive percentage changes, the overall multiplier is the product of the individual multipliers:

$$\text{Final} = \text{Initial} \times M_1 \times M_2 \times \cdots \times M_k.$$

Addition of the percents is legal only for successive changes that are each computed on the same original base (rare in word problems). Compound changes use a new base at every step, so they multiply.

A $+20\%$ move followed by a $-20\%$ move is

$$(1.20)(0.80) = 0.96,$$

which is a $4\%$ net decrease, not a wash. In general,

$$\left(1 + \frac{P}{100}\right)\left(1 - \frac{P}{100}\right) = 1 - \left(\frac{P}{100}\right)^2,$$

so an up-and-down of $P\%$ always nets a decrease of $\left(\frac{P}{10}\right)^2$ percent: $10\%$ each way nets $1\%$ down; $20\%$ each way nets $4\%$ down; $50\%$ each way nets $25\%$ down.

Because multiplication commutes, order does not matter for a sequence of ordinary percent changes on one quantity: $+30\%$ then $-10\%$ is $1.30 \times 0.90 = 1.17$, and $-10\%$ then $+30\%$ is the same $1.17$ ($+17\%$ net). Order would matter only if a later step depended on a threshold or a different base (tax vs discount vs tip on different amounts).

Reverse Percents and Discounts

To recover the original value before a percentage change, divide by the multiplier:

$$\text{Original} = \frac{\text{Final}}{\text{Multiplier}}.$$

Worked example. An item sold for $\text{\textdollar}360$ after a $20\%$ discount. What was the original (marked) price?

• Wrong: $\text{\textdollar}360 + 0.20 \times 360 = \text{\textdollar}432$ applies $20\%$ to the sale price, which is the wrong base.
• Correct: original $= \frac{360}{0.80} = 360 \times \frac{5}{4} = \text{\textdollar}450$.

A discount of $P\%$ is a multiplier of $1 - P/100$ on the marked price, not on cost. Two successive discounts of $10\%$ and $20\%$ are $0.90 \times 0.80 = 0.72$ of marked, a $28\%$ single-equivalent discount. They are not a $30\%$ discount.

Profit, Cost, Revenue, Markup, and Margin

GMAT commercial word problems use three dollar amounts and two different percentages of profit.

Markup and margin are not interchangeable. If cost is $\text{\textdollar}80$ and selling price is $\text{\textdollar}100$, profit is $\text{\textdollar}20$:

• markup on cost $= \frac{20}{80} = 25\%$
• margin on revenue $= \frac{20}{100} = 20\%$

The relationship is the same asymmetric-base identity as before: a $25\%$ markup is a $20\%$ margin, because $S = 1.25 C$ implies $C = 0.80 S$. If a problem says 'marked up $40\%$' it means $S_{\text{marked}} = 1.40 C$. If it then says 'discounted $25\%$', the sale price is $1.40 C \times 0.75 = 1.05 C$. Sold at $\text{\textdollar}252$, cost is $\frac{252}{1.05} = \text{\textdollar}240$ and profit is $\text{\textdollar}12$. That two-step (markup multiplier, then discount multiplier) is the standard profit item.

Revenue itself is $S \times Q$ when quantity also changes. A $25\%$ price increase with a $20\%$ drop in units sold produces revenue multiplier $1.25 \times 0.80 = 1.00$ — unchanged revenue. A $20\%$ price increase with a $20\%$ drop in units produces $1.20 \times 0.80 = 0.96$, a $4\%$ revenue decrease. Translate every percent into a multiplier, multiply the multipliers, then read the net percent off the product. That is the entire commercial-math toolkit the Quantitative Reasoning section uses.