GMAT Quantitative: Percent Change and Successive Percents

Mastering percent change is non-negotiable for GMAT success. These questions test core quantitative reasoning essential for business decisions—analyzing profit margins, sales growth, or investment returns. A clear, methodical approach transforms these common problems from time-consuming traps into reliable points. This guide builds your proficiency from foundational formulas to the layered calculations you'll encounter on test day.

The Core Percent Change Formula and Its Application

Every percent change problem revolves around a single, fundamental formula: Percent Change = $\frac{\text{New Value}-\text{Original Value}}{\text{Original Value}}\times 100\%$. The Original Value is always the baseline or denominator. A positive result indicates an increase; a negative result indicates a decrease.

Consider a classic GMAT word problem: "If a company's revenue was $200,000lastyearand$250,000 this year, what is the percent increase in revenue?"

1. Identify the Original Value ($200,000)andtheNewValue($250,000).
2. Apply the formula: $\frac{250,000-200,000}{200,000}\times 100\%=\frac{50,000}{200,000}\times 100\%=0.25\times 100\%=25\%$.

The common trap is misidentifying the base. If a price decreases from $80to$60, the decrease is calculated from the original $80:$\frac{60 - 80}{80} \times 100\% = -25\%$.OntheGMAT,awronganswermightuse$60 as the base. You must vigilantly ask: "Percent change from what?" The answer is always from the original value.

Successive Percent Changes: The Multiplier Method

Business scenarios often involve layered changes, such as a price increasing by 20% one month and then decreasing by 15% the next. A critical error is simply adding and subtracting the percentages (e.g., 20% - 15% = 5% net increase). This is incorrect because the second percentage change is applied to a new, different base.

The efficient, error-proof method is to use multipliers. Convert each percent change to a decimal multiplier:

• Increase of p%: Multiply by $(1+\frac{p}{100})$.
• Decrease of p%: Multiply by $(1-\frac{p}{100})$.

For the price example: An increase of 20% means multiply by $1.20$. A subsequent decrease of 15% means multiply by $0.85$. The overall multiplier is $1.20\times 0.85=1.02$. This represents a 2% net increase. This method is faster and avoids base-confusion. You can string together any number of changes by multiplying their respective multipliers in sequence.

Solving Reverse Percent Problems

These problems give you the result after a percent change and ask for the original value. For instance: "After a 30% discount, a book costs $28.Whatwasitsoriginalprice?"Theinstincttotake30$28 and add it back is a major pitfall. The 30% discount was applied to the unknown original price, not the final price.

The multiplier method is again your best tool. A 30% discount means you pay 70% of the original price. If $28represents70$0.70 \times \text{Original Price} = 28$.Therefore,OriginalPrice=$\frac{28}{0.70} = 40$. Always translate the percent change into a multiplier for the remaining portion and divide the final value by that multiplier to work backwards efficiently.

Percent Greater Than and Percent Less Than

The GMAT carefully distinguishes between "what percent of" and "what percent greater than." "A is what percent of B?" is calculated as $\frac{A}{B}\times 100\%$. "A is what percent greater than B?" uses the percent change formula: $\frac{A-B}{B}\times 100\%$.

For example: If Product X costs $90andProductYcosts$72, then:

• $90is$125\%$\ast of\ast$72$($\frac{90}{72}=1.25$).
• $90is$25\%$\ast greaterthan\ast$72$($\frac{90-72}{72}=0.25$).

The language is precise. "Greater than" or "less than" signals you must find the difference relative to the base (B). Misreading this is a common source of incorrect answer choices.

Combined Percent Problems with Multiple Variables

Advanced GMAT questions combine percentages with algebra, often involving two or more changing quantities. The key is to assign variables carefully and use the multiplier concept within equations.

Example Scenario: "A store sells only shirts and pants. The number of shirts sold increased by 20% from Year 1 to Year 2, while the number of pants sold decreased by 10%. If the total number of items sold increased by 5%, what fraction of the items sold in Year 1 were shirts?"

Let $S$ = shirts in Year 1, $P$ = pants in Year 1. In Year 2: Shirts = $1.2S$, Pants = $0.9P$. Total in Year 1: $S+P$. Total in Year 2: $1.2S+0.9P$. We know the Year 2 total is 5% greater: $1.2S+0.9P=1.05(S+P)$. Solving: $1.2S+0.9P=1.05S+1.05P$ $0.15S=0.15P$ $S=P$. Therefore, shirts were $\frac{1}{2}$ of the items sold in Year 1.

The process is methodical: define variables, express changes with multipliers, set up an equation based on the given relationship, and solve. Breaking the word problem into these algebraic components is a powerful strategy.

Common Pitfalls

1. Adding Successive Percentages: Remember that successive percent changes are multiplicative, not additive. A 50% increase followed by a 50% decrease does not return you to the start; it results in a 25% overall decrease ($1.5\times 0.5=0.75$).
2. Misidentifying the Base in Reverse Problems: When finding an original value, the percent change applied to the original, not the final number. Always divide the final value by the appropriate multiplier (e.g., divide by 0.7 for a 30% discount).
3. Confusing "Percent Of" with "Percent Greater/Less Than": The word "greater" or "less" means you must calculate the change relative to the base. If asked for the percent by which A exceeds B, use $\frac{A-B}{B}$, not $\frac{A}{B}$.
4. Forgetting the "1" in the Multiplier: A 40% increase means multiplying by $(1+0.4)=1.4$, not by $0.4$. Multiplying by 0.4 would represent a 60% decrease, a critical error.

Summary

• The universal percent change formula is $\frac{\text{New}-\text{Original}}{\text{Original}}\times 100\%$, where the Original Value is the critical base.
• For successive percent changes, convert percentages to decimal multipliers and multiply them sequentially; never add or subtract the percentages directly.
• Solve reverse percent problems by dividing the final value by the multiplier that produced it (e.g., Final Price / 0.7 for a 30% discount).
• Distinguish carefully between "percent of" ($\frac{A}{B}$) and "percent greater/less than" ($\frac{A-B}{B}$).
• Tackle complex combined problems by assigning variables, expressing changes with multipliers, and translating the word problem into a solvable algebraic equation.