3.3 Percent Change, Successive Percent Changes, and Multipliers

3.3 Percent Change, Successive Percent Changes, and Multipliers

Percentage problems on the GMAT Focus Edition frequently conceal algebraic traps beneath simple everyday terminology. Because calculators are prohibited on Quantitative Reasoning, working with percentages through traditional multi-step calculations (e.g., calculating 15% of a number, then adding it back to the original) is dangerously slow and error-prone. Elite scorers view percentages through the lens of single decimal and fractional multipliers. By mastering the multiplier method, you can condense multi-period percentage changes into rapid, single-line computations.

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Translating Prose to Algebraic Equations

Word problems in the Quantitative section require precise grammatical translation into algebraic notation:

Notice the strict role of the baseline in the final two translations: the entity following the word 'than' is always the base (denominator) of the percentage calculation.

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The Single Multiplier Method

Rather than computing the change and adding or subtracting it from the original, multiply the base directly by a single scaling factor:

$\text{New Value} = \text{Original Value} \times \text{Multiplier } (M)$

• For a percent increase of $p%$: $M = 1 + \frac{p}{100}$ Examples: $+15% \implies M = 1.15$; $+40% \implies M = 1.40 = \frac{7}{5}$; $+150% \implies M = 1 + 1.50 = 2.50 = \frac{5}{2}$.

• For a percent decrease of $d%$: $M = 1 - \frac{d}{100}$ Examples: $-20% \implies M = 0.80 = \frac{4}{5}$; $-35% \implies M = 0.65 = \frac{13}{20}$; $-12.5% \implies M = 0.875 = \frac{7}{8}$.

Using fractional forms of multipliers accelerates mental arithmetic: multiplying by $\frac{7}{5}$ is simply dividing by 5 and multiplying by 7, avoiding tedious decimal long multiplication.

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Percent Change and the Base Identification Rule

The universal formula for percent change is:

$\text{Percent Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\% = \frac{\Delta}{\text{Base}} \times 100\%$

The Base Identification Trap: If the price of an asset increases from $$80$ to $$100$, the change is $$20$ and the base is $$80$: $\text{Percent Increase} = \frac{100 - 80}{80} \times 100\% = \frac{20}{80} \times 100\% = +25\%$ If the price later drops from $$100$ back to $$80$, the change is still $$20$, but the base is now $$100$: $\text{Percent Decrease} = \frac{80 - 100}{100} \times 100\% = \frac{-20}{100} \times 100\% = -20\%$

The absolute dollar difference is identical ($$20$), but the percentage changes are unequal ($+25%$ vs. $-20%$) because the reference base changed.

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Successive (Compounded) Percent Changes

A classic GMAT trap tests whether candidates will mistakenly add consecutive percentage changes. Successive percentage changes must never be added; they must be compounded multiplicatively.

Rule for Successive Changes: If a quantity undergoes a series of percentage changes with multipliers $M_1, M_2, \dots, M_k$, the net multiplier $M_{\text{net}}$ is: $M_{\text{net}} = M_1 \times M_2 \times \dots \times M_k$ $\text{Net Percent Change} = (M_{\text{net}} - 1) \times 100\%$

The Symmetric Shift Fallacy ($+x%$ then $-x%$)

Consider a stock price that rises by $20%$ on Monday and falls by $20%$ on Tuesday. What is the net change?

• Multiplier Monday: $M_1 = 1 + 0.20 = 1.20 = \frac{6}{5}$
• Multiplier Tuesday: $M_2 = 1 - 0.20 = 0.80 = \frac{4}{5}$
• Net Multiplier: $M_{\text{net}} = 1.20 \times 0.80 = 0.96 = \frac{24}{25}$
• Net change: $0.96 - 1 = -0.04 = -4%$

The stock did not break even; it suffered a net loss of $4%$. In general, an increase of $x%$ followed by a decrease of $x%$ (in either order) always yields a net decrease of: $\text{Net Loss} = \left(\frac{x}{10}\right)^2\% = \frac{x^2}{100}\%$ For a $30%$ rise and $30%$ drop: $\text{Net Loss} = \frac{30^2}{100}% = 9%$ ($M_{\text{net}} = 1.30 \times 0.70 = 0.91$).

Commutative Property of Multipliers

Because multiplication is commutative ($M_1 \times M_2 = M_2 \times M_1$), the order of successive percentage changes is completely irrelevant to the final outcome:

• A $40%$ markup followed by a $15%$ discount yields $1.40 \times 0.85 = 1.19$ ($+19%$ net).
• A $15%$ discount followed by a $40%$ markup yields $0.85 \times 1.40 = 1.19$ ($+19%$ net).

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Reversing Percentage Changes: The Reciprocal Principle

When a problem states that an item's price was altered and asks what percentage change is required to restore the original price, use the reciprocal of the multiplier.

Let the original multiplier be $M$. To return to the initial value of $1$, the restoring multiplier must be $\frac{1}{M}$: $\text{Restoring Multiplier } M_{\text{restore}} = \frac{1}{M}$

The Fractional Reversal Theorem

Expressing multipliers as fractions reveals a clean, elegant relationship:

• If a price decreases by $\frac{a}{b}$, the remaining fraction is $\frac{b - a}{b}$.
• To restore the original price, multiply by the reciprocal: $\frac{b}{b - a} = 1 + \frac{a}{b - a}$.
• Therefore, the required percentage increase is: $\text{Required Increase} = \frac{a}{b - a} \times 100\%$

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Percentage Points vs. Percent Change

The GMAT frequently tests your ability to distinguish between an absolute change in rate (percentage points) and a relative change in rate (percent change):

• Percentage Points: The simple arithmetic difference between two percentages: $\text{Points} = P_2 - P_1$.
• Percent Change: The relative growth of the percentage itself: $\frac{P_2 - P_1}{P_1} \times 100%$.

Example: An economic report states that a central bank raised its benchmark lending rate from $4%$ to $5%$.

• The interest rate increased by $1$ percentage point ($5% - 4% = 1%$).
• The interest rate experienced a $25%$ increase ($\frac{5 - 4}{4} \times 100% = 25%$).

Conflating these two terms is one of the most reliable ways to fall for a GMAT distractor choice.

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Common GMAT Pitfalls & Cognitive Traps

Trap 1: The Additive Fallacy in Consecutive Discounts

A store advertises "Take 40% off, then take an additional 20% off at the register!" Test-takers who calculate a 60% total discount fall directly into the trap. The actual multiplier is $0.60 \times 0.80 = 0.48$, corresponding to a 52% total discount, not 60%.

Trap 2: Inverted Base in Comparative Phrases

Pay obsessive attention to phrasing:

• "A is 25% greater than B": $B$ is the base $\implies A = 1.25 B$.
• "A is 25% of B": $A = 0.25 B$.
• "By what percent is B less than A?": $A$ is now the base $\implies \frac{A - B}{A} \times 100% = \frac{1.25B - B}{1.25B} = \frac{0.25}{1.25} = 20%$.

Trap 3: Percent of a Percent Confusion

"What is 5% of 20% of 400?" Translate directly to multiplication: $\left(\frac{5}{100}\right) \times \left(\frac{20}{100}\right) \times 400 = \frac{1}{20} \times \frac{1}{5} \times 400 = \frac{1}{100} \times 400 = 4$ Do not confuse 5% with 0.5 or 20% with 0.02.

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Worked Problem Solving Example

Question: Over a three-month period, the inventory of a warehouse changed as follows: in October, inventory increased by $20%$; in November, inventory decreased by $25%$; and in December, inventory increased by $10%$. If the inventory at the end of December was $2,970$ units, how many units were in the warehouse at the beginning of October?

Step-by-step Solution:

1. Express each monthly change as a fractional multiplier:
   • October increase of $20%$: $M_1 = 1 + \frac{1}{5} = \frac{6}{5}$
   • November decrease of $25%$: $M_2 = 1 - \frac{1}{4} = \frac{3}{4}$
   • December increase of $10%$: $M_3 = 1 + \frac{1}{10} = \frac{11}{10}$
2. Calculate the combined net multiplier ($M_{\text{net}}$): $M_{\text{net}} = \frac{6}{5} \times \frac{3}{4} \times \frac{11}{10}$ Cross-cancel 6 and 4 by dividing both by 2: $M_{\text{net}} = \frac{3}{5} \times \frac{3}{2} \times \frac{11}{10} = \frac{99}{100}$
3. Set up the equation for the initial inventory ($I_0$): $I_0 \times \frac{99}{100} = 2,970$
4. Solve for $I_0$ mentally: Notice that $2,970$ is a direct multiple of $99$: $2,970 / 99 = 30$. $I_0 = 30 \times 100 = 3,000$ The warehouse initially held exactly $3,000$ units.