Hard GRE Arithmetic Word Problems Practice Questions

Concept Explanation

Hard GRE Arithmetic Word Problems are complex quantitative reasoning tasks that require students to translate multi-step narrative scenarios into precise mathematical equations involving integers, fractions, percentages, and ratios. These problems often appear in the second quantitative section of the GRE Prep curriculum for students who perform well on the initial section, serving as a primary differentiator for high scores. Unlike basic arithmetic, these questions frequently combine multiple concepts—such as work rates and percentage changes—within a single word-based prompt. Success requires a solid grasp of algebraic modeling and the ability to identify hidden constraints. To improve your speed, you might utilize an AI Question Generator to practice the specific phrasing used by the Educational Testing Service (ETS).

Solved Examples

Review these detailed solutions to understand how to decompose intricate word problems into solvable steps.

1. Example 1: Combined Rates
   Machine A can complete a job in 6 hours, and Machine B can complete the same job in 9 hours. If Machine A starts at 10:00 AM and Machine B joins at 11:00 AM, at what time will the job be finished?
   1. Determine individual rates: Machine A rate = $\frac{1}{6}$ job/hr; Machine B rate = $\frac{1}{9}$ job/hr.
   2. Calculate work done by A alone from 10:00 to 11:00: $1 \times \frac{1}{6} = \frac{1}{6}$.
   3. Find remaining work: $1 - \frac{1}{6} = \frac{5}{6}$.
   4. Set up the combined rate equation: $(\frac{1}{6} + \frac{1}{9})t = \frac{5}{6}$.
   5. Find common denominator: $(\frac{3}{18} + \frac{2}{18})t = \frac{5}{6} \rightarrow \frac{5}{18}t = \frac{5}{6}$.
   6. Solve for $t$: $t = \frac{5}{6} \times \frac{18}{5} = 3$ hours.
   7. Add 3 hours to 11:00 AM: The job finishes at 2:00 PM.
2. Example 2: Percentage Change and Ratios
   In a local election, 40% of the voters were men. If the number of male voters increased by 20% and the number of female voters increased by 10% in the next election, what is the new percentage of male voters among the total?
   1. Assume a total of 100 voters for simplicity. Men = 40; Women = 60.
   2. Calculate new male count: $40 \times 1.20 = 48$.
   3. Calculate new female count: $60 \times 1.10 = 66$.
   4. Calculate new total: $48 + 66 = 114$.
   5. Calculate new percentage of men: $\frac{48}{114} \times 100$.
   6. Simplify: $\frac{48}{114} \approx 42.1\%$.
3. Example 3: Overlapping Sets
   In a group of 120 students, 75 study French, 60 study Spanish, and 30 study both. How many students study neither language?
   1. Use the formula for sets: $\text{Total} = \text{Set A} + \text{Set B} - \text{Both} + \text{Neither}$.
   2. Substitute values: $120 = 75 + 60 - 30 + \text{Neither}$.
   3. Simplify: $120 = 105 + \text{Neither}$.
   4. Solve: $\text{Neither} = 15$.

Practice Questions

Test your skills with these hard GRE arithmetic word problems. Ensure you read every constraint carefully.

1. A water tank is $\frac{2}{5}$ full. After adding 12 liters of water, the tank is $\frac{3}{4}$ full. What is the total capacity of the tank in liters?
2. A merchant buys an item for $80 and marks it up by 50%. During a sale, he offers a discount such that his net profit is 20% of the original cost. What was the discount percentage?
3. Working alone at their respective constant rates, Pump X can fill a pool in 12 hours and Pump Y can fill it in 15 hours. If both pumps start together but Pump X is turned off 4 hours before the pool is full, how many total hours did it take to fill the pool?

4. In a mixture of 80 liters, the ratio of milk to water is 7:3. How many liters of water must be added to make the ratio 2:1?
5. A cyclist travels from City A to City B at an average speed of 15 mph and returns at an average speed of 10 mph. What is the average speed for the entire round trip?
6. Three partners, P, Q, and R, invest in a business. P invests $\frac{1}{3}$ of the total capital, Q invests $\frac{1}{4}$, and R invests the rest. If the total profit is $4,800, how much does R receive?
7. If the price of a commodity increases by 25%, by what percentage must a consumer reduce their consumption so that the total expenditure remains the same?
8. A car's value depreciates by 10% each year. If the car is worth $16,200 after two years, what was its original purchase price?

Answers & Explanations

1. Answer: 34.28 (or $\frac{240}{7}$) liters. Let $C$ be the capacity. $\frac{3}{4}C - \frac{2}{5}C = 12$. Common denominator 20: $\frac{15}{20}C - \frac{8}{20}C = 12 \rightarrow \frac{7}{20}C = 12$. $C = \frac{240}{7}$.
2. Answer: 20%. Original cost = $80. Marked price = $80 \times 1.5 = \$120$. Profit of 20% on cost = $80 \times 0.2 = \$16$. Selling price = $80 + 16 = \$96$. Discount = $120 - 96 = \$24$. Discount % = $\frac{24}{120} \times 100 = 20\%$.
3. Answer: 8.44 hours (or $\frac{76}{9}$). Let the total time be $t$. Pump Y works for $t$ hours, Pump X works for $t-4$ hours. Equation: $\frac{t-4}{12} + \frac{t}{15} = 1$. Multiply by 60: $5(t-4) + 4t = 60 \rightarrow 9t - 20 = 60 \rightarrow 9t = 80$. Total time is $\frac{80}{9}$ hours (approx 8.88 hours). Correction: Ensure the logic follows the timeline correctly; X stops 4 hours before completion.
4. Answer: 4 liters. Initial Milk = $\frac{7}{10} \times 80 = 56$; Initial Water = 24. Let $w$ be added water. $\frac{56}{24+w} = \frac{2}{1} \rightarrow 56 = 48 + 2w \rightarrow 2w = 8 \rightarrow w = 4$.
5. Answer: 12 mph. Average speed = $\frac{ \text{Total Distance}}{ \text{Total Time}}$. Let distance one-way be $D$. Total distance = $2D$. Time 1 = $\frac{D}{15}$, Time 2 = $\frac{D}{10}$. Average = $\frac{2D}{\frac{D}{15} + \frac{D}{10}} = \frac{2}{\frac{2+3}{30}} = \frac{60}{5} = 12$.
6. Answer: $2,000. R's share = $1 - (\frac{1}{3} + \frac{1}{4}) = 1 - \frac{7}{12} = \frac{5}{12}$. Profit for R = $\frac{5}{12} \times 4800 = 2000$.
7. Answer: 20%. Let price be $P$ and consumption be $C$. New price = $1.25P$. New consumption $C'$ must satisfy $1.25P \times C' = P \times C$. $C' = \frac{C}{1.25} = 0.8C$. Reduction is 20%.
8. Answer: $20,000. Let original price be $X$. After 1 year: $0.9X$. After 2 years: $0.9(0.9X) = 0.81X$. $0.81X = 16200 \rightarrow X = \frac{16200}{0.81} = 20000$.

1. A shirt's price is reduced by 20%, then the new price is increased by 25%. What is the net change from the original price?

• A5% decrease
• B0% change
• C5% increase
• D10% increase

Frequently Asked Questions

What makes an arithmetic word problem "hard" on the GRE?

Hard problems typically involve multiple steps, require identifying hidden variables, or combine disparate concepts like ratios and percentage changes. They often use distractor information or phrasing designed to lead students toward common calculation errors.

How do I translate word problems into equations effectively?

Assign variables to the unknown values and look for keywords like "is" (equals), "of" (multiplication), and "per" (division). Drawing a simple diagram or table for rate and work problems can help organize the information visually before writing the equation.

Should I use the on-screen calculator for these problems?

While the calculator is useful for final computations, you should prioritize setting up the algebraic expression first. Many hard problems are designed with numbers that cancel out or simplify easily if the conceptual setup is correct.

How can I improve my speed on multi-step arithmetic problems?

Consistent practice with GRE Practice Questions with Explanations helps you recognize patterns in how ETS structures complex prompts. Learning mental math shortcuts for common percentages and fractions also reduces reliance on the calculator.

What is the most common mistake in rate-related word problems?

The most frequent error is simply averaging the two speeds or rates directly without considering the time spent at each rate. You must always use the harmonic mean or calculate total distance divided by total time for average speed problems.