Hard GRE Algebra Word Problems Practice Questions

Concept Explanation

Hard GRE algebra word problems are complex mathematical scenarios that require translating multi-step narrative descriptions into algebraic equations involving variables, inequalities, or systems of equations. These problems frequently appear on the GRE Prep quantitative reasoning sections and test your ability to model real-world constraints using mathematical logic. Unlike simple computational questions, these problems often hide information or require you to establish relationships between multiple unknown quantities. Success depends on identifying the "fixed" and "variable" components of the story. For example, in a work-rate problem, the combined rate is the sum of individual rates, a concept thoroughly explained in resources like Khan Academy's rational equation modeling. To solve these efficiently, one should assign variables to the specific values the question asks for, set up an equation that reflects the "balance" in the story, and solve for the unknown while being mindful of units and constraints.

Solved Examples

The following examples demonstrate the logic required to dismantle high-difficulty word problems on the GRE.

1. The Mixture Problem: A chemist has 10 liters of a solution that is 10 percent nitric acid. How many liters of a 40 percent nitric acid solution must be added to the 10 liters to create a mixture that is 25 percent nitric acid?
   1. Define the variable: Let $x$ be the liters of 40% solution added.
   2. Set up the equation based on the amount of pure acid: $0.10(10) + 0.40(x) = 0.25(10 + x)$.
   3. Simplify the equation: $1 + 0.4x = 2.5 + 0.25x$.
   4. Isolate $x$: $0.15x = 1.5$.
   5. Solve: $x = \frac{1.5}{0.15} = 10$. The chemist must add 10 liters.
2. The Work-Rate Problem: Machine A can produce 500 widgets in 4 hours, and Machine B can produce the same number of widgets in 6 hours. If Machine A starts at 9:00 AM and Machine B joins at 10:00 AM, at what time will they have produced a total of 1,000 widgets?
   1. Find rates: Rate A = $\frac{500}{4} = 125$ widgets/hr. Rate B = $\frac{500}{6} = \frac{250}{3}$ widgets/hr.
   2. Calculate Machine A's head start: From 9:00 to 10:00, Machine A produces 125 widgets. Remaining widgets needed = $1000 - 125 = 875$.
   3. Set up the combined work equation: $(125 + \frac{250}{3}) \times t = 875$.
   4. Combine rates: $(\frac{375}{3} + \frac{250}{3})t = 875 \rightarrow \frac{625}{3}t = 875$.
   5. Solve for $t$: $t = 875 \times \frac{3}{625} = \frac{7 \times 125 \times 3}{5 \times 125} = \frac{21}{5} = 4.2$ hours.
   6. Convert to time: 4 hours and 12 minutes after 10:00 AM is 2:12 PM.
3. The Distance-Rate Inequality: A cyclist travels uphill at a constant rate of 8 miles per hour and returns downhill at 24 miles per hour. If the total round trip takes less than 2 hours, what is the maximum possible integer distance (in miles) of the one-way trip?
   1. Let $d$ be the distance of the one-way trip.
   2. Time uphill: $\frac{d}{8}$. Time downhill: $\frac{d}{24}$.
   3. Set up the inequality: $\frac{d}{8} + \frac{d}{24} < 2$.
   4. Find a common denominator: $\frac{3d + d}{24} < 2 \rightarrow \frac{4d}{24} < 2$.
   5. Simplify: $\frac{d}{6} < 2 \rightarrow d < 12$.
   6. The maximum integer distance is 11 miles.

Practice Questions

1. A store sells two types of coffee beans: Type A costs $12 per pound and Type B costs $18 per pound. If a 50-pound blend of the two types costs $14.40 per pound, how many pounds of Type A were used?
2. Working alone at their respective constant rates, Pump X can fill a tank in 6 hours and Pump Y can fill the same tank in 9 hours. If Pump X runs alone for 2 hours and then both pumps run together, how many additional hours will it take to fill the tank?
3. A train travels from City P to City Q at an average speed of 60 miles per hour and returns from City Q to City P at an average speed of 90 miles per hour. If the total travel time is 5 hours, what is the distance between City P and City Q?

4. In a group of 120 students, 70 are studying French, 50 are studying Spanish, and 20 are studying both. How many students are studying neither French nor Spanish?
5. A certain investment doubles in value every 7 years. If the investment is worth $10,000 today, how many years will it take for the investment to be worth $160,000?
6. An alloy is 20% copper and 80% tin. How many kilograms of pure copper must be added to 40 kilograms of this alloy to make a new alloy that is 40% copper?
7. If 3 workers can build 2 houses in 9 months, how many months would it take 5 workers to build 10 houses, assuming all workers work at the same constant rate?
8. A car rental agency charges a daily fee of $45 plus $0.25 per mile driven. If a customer's total bill for a 3-day rental was $210, how many miles did the customer drive?
9. The sum of three consecutive integers is 45 more than the smallest integer. What is the value of the largest integer?
10. A rectangular garden has a perimeter of 80 feet. If the length is 10 feet more than the width, what is the area of the garden in square feet?

Answers & Explanations

1. Answer: 30 pounds. Let $x$ be the pounds of Type A. Then $50 - x$ is the pounds of Type B. The equation is $12x + 18(50 - x) = 14.40(50)$. Simplifying: $12x + 900 - 18x = 720$. Thus, $-6x = -180$, so $x = 30$.
2. Answer: 2.4 hours (or 2 hours 24 minutes). Pump X rate is $1/6$ tank/hr; Pump Y rate is $1/9$ tank/hr. In 2 hours, Pump X fills $2/6 = 1/3$ of the tank. Remaining: $2/3$. Combined rate: $1/6 + 1/9 = 3/18 + 2/18 = 5/18$. Time $t = \frac{2/3}{5/18} = \frac{2}{3} \times \frac{18}{5} = \frac{12}{5} = 2.4$ hours.
3. Answer: 180 miles. Let $d$ be the distance. Time 1: $d/60$. Time 2: $d/90$. Equation: $d/60 + d/90 = 5$. Common denominator 180: $3d/180 + 2d/180 = 5 \rightarrow 5d/180 = 5$. Thus $d = 180$.
4. Answer: 20 students. Use the formula $\text{Total} = A + B - (A \cap B) + \text{Neither}$. $120 = 70 + 50 - 20 + \text{Neither}$. $120 = 100 + \text{Neither}$. So, $\text{Neither} = 20$.
5. Answer: 28 years. To get from $10,000 to $160,000, the value must double 4 times ($10k \rightarrow 20k \rightarrow 40k \rightarrow 80k \rightarrow 160k$). Since each doubling takes 7 years, $4 \times 7 = 28$ years.
6. Answer: 13.33 kg (or 40/3 kg). Initial copper: $0.20 \times 40 = 8$ kg. Let $x$ be added copper. Equation: $\frac{8 + x}{40 + x} = 0.40$. $8 + x = 16 + 0.4x$. $0.6x = 8$. $x = 8/0.6 = 80/6 = 40/3$.
7. Answer: 27 months. Use the formula $\frac{W_1}{R_1 \times T_1} = \frac{W_2}{R_2 \times T_2}$. $\frac{2 \text{ houses}}{3 \text{ workers} \times 9 \text{ months}} = \frac{10 \text{ houses}}{5 \text{ workers} \times T_2}$. $\frac{2}{27} = \frac{2}{T_2}$. Thus, $T_2 = 27$.
8. Answer: 300 miles. Total fee = $3 \times 45 = 135$. Mileage cost = $210 - 135 = 75$. Miles $m = 75 / 0.25 = 300$.
9. Answer: 23. Let integers be $n, n+1, n+2$. Sum: $3n + 3 = n + 45$. $2n = 42$, so $n = 21$. Largest is $n + 2 = 23$.
10. Answer: 375 sq ft. $2(L + W) = 80 \rightarrow L + W = 40$. Since $L = W + 10$, then $(W + 10) + W = 40$. $2W = 30$, so $W = 15$. $L = 25$. Area = $25 \times 15 = 375$.

1. A car travels \( x \) miles at 40 mph and then \( 2x \) miles at 60 mph. What is the average speed for the entire trip?

• A48 mph
• B50 mph
• C51.4 mph
• D52.5 mph

Frequently Asked Questions

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

Hard problems typically involve multiple variables, hidden constraints, or require combining different algebraic concepts like rates and inequalities simultaneously. They often use distractor information or require you to solve for a value that is not the primary variable in your equation.

How do I translate text into algebraic equations effectively?

Focus on keywords such as "is" (equals), "of" (multiplication), and "per" (division) while assigning specific variables to unknown quantities. It is helpful to organize data into a table for rate, time, and distance problems to visualize the relationships before writing the equation.

Can I solve these problems using the answer choices?

Yes, "back-solving" by plugging answer choices into the problem constraints is a valid strategy for multiple-choice questions. This is often faster for complex mixture or age problems where setting up the algebra might be prone to errors, though it is less effective for quantitative comparison questions.

Are there specific formulas I should memorize for GRE word problems?

You should be fluent in the distance formula ($d = rt$), the work-rate formula ($W = rt$), and the mixture formula ($C_1V_1 + C_2V_2 = C_fV_f$). Understanding how to manipulate these basic formulas is more important than memorizing niche shortcuts.

How much time should I spend on a hard word problem?

You should aim to spend no more than 2 minutes on any single question. If a word problem is exceptionally dense, it is better to mark it, make an educated guess, and return to it later if time permits, as all questions carry equal weight.