What this quiz covers
This quiz focuses on Percents, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.
A retailer raised the price of a jacket by 20 percent and later put the jacket on sale at a discount of 25 percent off the new price. By what percent did the final sale price differ from the jacket's original price?
GMAT Quiz
Practice Percents in GMAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
This quiz focuses on Percents, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A retailer raised the price of a jacket by 20 percent and later put the jacket on sale at a discount of 25 percent off the new price. By what percent did the final sale price differ from the jacket's original price?
Explanation: This problem tests your ability to handle sequential percentage changes, a common GMAT topic that requires careful attention to which value serves as the base for each calculation. Let's work through this step-by-step using a concrete example. Assume the original price is $100. First, the retailer raises the price by 20%: New price = $100 + (20% of $100) = $100 + $20 = $120 Then, the jacket goes on sale at 25% off this new price of $120: Final price = $120 - (25% of $120) = $120 - $30 = $90 Comparing the final price (90)totheoriginalprice(100): The difference is $90 - 100=−10, which represents a 10% decrease from the original price. Answer choice (D) correctly identifies that the final price was 10% lower than the original price. (A) incorrectly suggests a 10% increase, which would require the final price to be $110. (B) claims a 5% decrease, which would mean a final price of $95. (C) suggests a 5% increase, which would result in $105. These wrong answers often trap students who make calculation errors or who mistakenly think the two percentage changes simply cancel each other out partially. Strategy tip: When dealing with sequential percentage changes, never add or subtract the percentages directly. Always calculate each step using the actual dollar amounts (or use a convenient number like 100). The compounding effect of percentage changes on different base amounts is a frequent GMAT trap.
A company's revenue increased by 25% in the first quarter, then decreased by 20% in the second quarter. If the revenue at the end of the second quarter was $300,000, what was the revenue at the beginning of the first quarter?
Explanation: Let x be the initial revenue. After a 25% increase: x × 1.25. After a 20% decrease: x × 1.25 × 0.8 = x × 1.0 = x. Since the final revenue is $300,000, the initial revenue was $300,000. Choice A assumes incorrect order of operations. Choice C results from adding percentages (25% - 20% = 5%) instead of multiplying. Choice D uses only the first increase without considering the decrease.
An employee receives a 12% salary increase in January, followed by a 15% bonus based on the new salary in March. If the employee's total annual compensation (salary plus bonus) is $64,400, what was the original salary in December?
Explanation: Let S be the original salary. After the 12% increase: 1.12S. The 15% bonus is based on the new salary: 0.15 × 1.12S = 0.168S. Total compensation = 1.12S + 0.168S = 1.288S = $64,400. Therefore, S = $64,400 ÷ 1.288 = $50,000. Choice B results from calculating the bonus on the original salary instead of the increased salary. Choice C assumes the bonus is applied before the salary increase. Choice D results from arithmetic errors in the percentage calculations.
A store marks up items by 60% above cost, then offers a 25% discount during a sale. If the sale price of an item is $96, what was the original cost to the store?
Explanation: Let C be the original cost. After markup: C × 1.6. After 25% discount: C × 1.6 × 0.75 = C × 1.2 = $96. Therefore, C = $96 ÷ 1.2 = $80. Choice A results from incorrectly applying 60% markup as simple addition then subtracting 25%. Choice B assumes the markup and discount partially cancel. Choice D results from applying only the discount to the sale price.
The price of a stock increases by x% on Monday and decreases by y% on Tuesday, resulting in a net decrease of 4% over the two days. If x=2y, what is the value of y?
Explanation: Let the initial price be P. After Monday: P(1 + x/100). After Tuesday: P(1 + x/100)(1 - y/100) = 0.96P. This gives us (1 + x/100)(1 - y/100) = 0.96. Substituting x = 2y: (1 + 2y/100)(1 - y/100) = 0.96. Expanding: 1 - y/100 + 2y/100 - 2y²/10000 = 0.96. Simplifying: 1 + y/100 - 2y²/10000 = 0.96, so y/100 - 2y²/10000 = -0.04. Solving: y = 10. Choice A comes from approximating without the quadratic term. Choice C results from incorrect algebraic manipulation. Choice D doubles the correct answer.
The price of an item is reduced by 30%, then increased by 40%, then reduced by 25%. What is the overall percentage change in price?
Explanation: Starting with price P, after the changes: P × 0.7 × 1.4 × 0.75 = P × 0.735. This represents a decrease to 73.5% of the original price, so a decrease of 26.5%. Choice B results from simply adding/subtracting percentages (-30% + 40% - 25% = -15%). Choice C makes the same error but with wrong sign. Choice D has the correct magnitude but wrong direction.
An investment account loses 12% of its value in Year 1, gains 15% in Year 2, and loses 8% in Year 3. If the account has $11,628 at the end of Year 3, what was the initial investment amount?
Explanation: Let x be the initial investment. After 3 years: x × 0.88 × 1.15 × 0.92 = x × 0.93152 = $11,628. Therefore, x = $11,628 ÷ 0.93152 ≈ $12,476, which is closest to $12,500. Choice A underestimates due to rounding errors. Choice C and D overestimate the required initial amount.
The number of bacteria in a culture triples every 4 hours. If there are currently 2,400 bacteria, how many were there 12 hours ago?
Explanation: When you encounter exponential growth or decay problems, the key is determining whether you're moving forward or backward in time, as this affects whether you multiply or divide by the growth factor.
Since the bacteria triple every 4 hours, you need to work backward from the current population of 2,400 bacteria to find how many existed 12 hours ago. Going backward 12 hours means reversing 3 periods of growth (since 12 ÷ 4 = 3 periods).
To reverse tripling, you divide by 3 for each 4-hour period:
Alternatively, you can divide by 33=27 in one step: 2400÷27≈89 bacteria.
Choice A (889) represents a common error of dividing by only 3 instead of 33. Choice B (267) comes from dividing by 32=9, accounting for only 2 periods instead of 3. Choice C (800) results from dividing by 3 just once, representing the population only 4 hours ago.
Remember that exponential problems require careful attention to direction: multiply the growth factor when moving forward in time, divide when moving backward. Always count the number of periods correctly by dividing the total time by the period length.
A population grows at a rate of 8% per year. After how many complete years will the population first exceed 150% of its original size?
Explanation: This is a compound growth problem where you need to find when a population reaches a specific threshold. When you see "grows at X% per year," think exponential growth using the formula P=P0(1+r)t, where P0 is the initial population, r is the growth rate, and t is time in years.
You want to find when the population first exceeds 150% of its original size, meaning P>1.5P0. Setting up the inequality: P0(1.08)t>1.5P0. Dividing both sides by P0 gives us (1.08)t>1.5.
Let's test each year systematically:
Since 1.469<1.5 but 1.586>1.5, the population first exceeds 150% after 6 complete years.
Choice A (5 years) is incorrect because (1.08)5=1.469, which equals only 146.9% of the original population—still below the 150% threshold.
Choice B (7 years) is wrong because while (1.08)7=1.714 does exceed 150%, the population already crossed this threshold in year 6, so 7 years is not the "first" time.
Choice D (8 years) makes the same error as B—it's too late since the threshold was already exceeded in year 6.
Strategy tip: For compound growth problems, always check the year before and after your calculated answer to ensure you're finding the "first" occurrence, especially when the question asks for "complete years."
A city's population decreases by 5% each year due to migration. Additionally, the birth rate exceeds the death rate by 2% per year. What is the net annual percentage change in the city's population?
Explanation: When you encounter population change problems involving multiple factors, you need to understand that percentage changes don't simply add or subtract—they compound multiplicatively.
Let's work through this step-by-step. If the population starts at 100%, after one year it experiences:
The net change is 96.9%−100%=−3.1%, so the population decreases by 3.1% annually.
Answer choice A incorrectly assumes the effects work in the opposite direction, perhaps misinterpreting which factor is positive versus negative. Answer choice B represents the most common trap: simply subtracting the percentages arithmetically (5% - 2% = 3%). This ignores the compounding effect where the 2% increase applies to the already-reduced population, not the original amount. Answer choice C might result from calculation errors or rounding mistakes during the multi-step process.
The key insight is that when you have sequential percentage changes, you must apply them multiplicatively in order. The second percentage change always applies to the result after the first change, not to the original baseline.
Strategy tip: In compound percentage problems, resist the urge to simply add or subtract the percentages. Always multiply the factors: (1±change1)×(1±change2) and calculate the final percentage change from there.