GRE word problems: the method, and 12 worked examples

What counts as a word problem on the GRE?

Any Quant question that describes a situation in sentences and leaves you to build the math. It is not a separate question type. The GRE has four Quant formats, quantitative comparison, multiple choice with one answer, multiple choice with one or more answers, and numeric entry, and a word problem can arrive in any of them.

That is why they are worth studying as a group. ETS describes some Quantitative Reasoning questions as posed in real-life settings and others in purely mathematical settings, and does not publish how many of each a test contains. What we can say from our own bank, which is written to GRE style, is that questions set in a real-life context run through every content area: arithmetic, algebra and data analysis in particular. You will not get through a Quant section without several.

The content never goes beyond high-school math. The difficulty is in reading: working out what is being asked, which numbers matter, and which single relationship ties them together.

How to solve GRE word problems: a four-step method

1. Read the last sentence first. It tells you what is being asked and in what unit: dollars or cents, hours or minutes, how many or how many more. Half of all word problem errors are a correct calculation of the wrong thing.
2. Name the unknown. Write “let $x$be the wholesale cost”, in words, on your scratch paper. If the question asks for something other than $x$, note that too.
3. Translate one sentence at a time. Each sentence gives you one relationship. Do not hold the whole paragraph in your head.
4. Solve, then reread the question. Check the unit and check that you answered what was asked. Then check that the number is plausible: a discount should make a price lower, and two pumps should be faster than one.

Step 3 is mechanical once you know the vocabulary:

The seven families of GRE word problems

Recognizing the family is most of the work, because each one has a single relationship you always start from.

The formulas behind each row are collected on our GRE math formulas cheat sheet.

GRE word problems practice: 12 questions with solutions

Ordered from straightforward to hard, and covering all seven families. Work each one on paper before opening the solution, and name the family before you start. These are original questions written to GRE style and difficulty, not ETS items, and the difficulty labels are our own editorial ratings.

1. Question 1Multiple choice · Easy to medium · Arithmetic

   A car travels $150$ miles in $3$ hours at a constant speed. At this rate, how many miles does it travel in $5$ hours?

   • (A)$200$
   • (B)$225$
   • (C)$250$
   • (D)$300$
   • (E)$450$
   Show the answer and worked solution

   Answer: (C) $250$

   The constant speed is $\frac{150\text{ miles}}{3\text{ hours}}=50$ miles per hour.

   In $5$ hours the car travels $50\times 5=250$ miles.

   Choice (e) $450$ multiplies $150$ by $3$. Choice (d) $300$ uses $6$ hours. Choices (a) and (b) use incorrect rates.

   Do

   Find the unit rate (miles per hour) first, then multiply by the new time.

   Avoid

   Don't multiply the original distance by the new time directly.

2. Question 2Multiple choice · Easy to medium · Arithmetic

   A printer prints $40$ pages per minute. At this constant rate, how many minutes will it take to print $1,800$ pages?

   • (A)$30$
   • (B)$40$
   • (C)$45$
   • (D)$50$
   • (E)$60$
   Show the answer and worked solution

   Answer: (C) $45$

   Time $=\frac{\text{pages}}{\text{rate}}=\frac{1800}{40}=45$ minutes.

   Why the others fail:

   (a) $30$ would require a rate of $60$ pages/min.

   (b) $40$ confuses the rate with the time.

   (d) $50$ would require $900$ pages, not $1800$.

   (e) $60$ would require a rate of $30$ pages/min.

   Do

   Divide total pages by pages-per-minute to get minutes.

   Avoid

   Don't confuse the rate value with the time value.

3. Question 3Multiple choice · Easy to medium · Data Analysis

   The ratio of boys to girls in a class is $3:5$. If there are $40$ students in the class, how many more girls than boys are there?

   • (A)$5$
   • (B)$8$
   • (C)$10$
   • (D)$15$
   • (E)$25$
   Show the answer and worked solution

   Answer: (C) $10$

   The ratio $3:5$ has $3+5=8$ equal parts. With $40$ students, each part is

   $$\frac{40}{8}=5\text{ students}.$$

   Boys $=3\times 5=15$; girls $=5\times 5=25$. The difference is

   $$25-15=10.$$

   Equivalently, the difference is $(5-3)=2$ parts, and $2\times 5=10$.

   Do

   Convert the ratio to a per-part value using the total of $8$ parts.

   Avoid

   Don't read $3$ and $5$ as actual head counts; they are ratio parts, not numbers of students.

4. Question 4Multiple choice · Medium · Arithmetic

   A retailer sells a television for $450, which reflects a markup of 25 percent over the wholesale cost. What is the wholesale cost of the television?

   • (A)$337.50
   • (B)$360.00
   • (C)$375.00
   • (D)$400.00
   • (E)$405.00
   Show the answer and worked solution

   Answer: (B) $360.00

   Let $w$ be the wholesale cost. A 25 percent markup means the selling price is $1.25w$. So $1.25w=450$, giving $w=450/1.25=360$. Choice (a) results from incorrectly taking 25 percent off the selling price instead of dividing by $1.25$: $450-0.25(450)=337.50$, which is not the correct inverse operation. Choices (c), (d), and (e) do not satisfy $1.25w=450$ when checked directly.

   Do

   Set up the equation selling price equals wholesale cost times $1.25$, then solve for wholesale cost by dividing, rather than subtracting a percentage from the selling price.

   Avoid

   Do not subtract 25 percent of the selling price from the selling price; markup percentages are computed on the original (wholesale) cost, not on the final selling price.

5. Question 5Quantitative comparison · Easy to medium · Percents

   A jacket originally priced at $\$80$ is marked up by $25\%$ for the holiday season, and the new price is later discounted by $20\%$. Compare the two quantities.

   Quantity A

   The final price of the jacket

   Quantity B

   $\$80$
   Show the answer and worked solution

   Answer: C. The two quantities are equal.

   Apply the changes in order. After a $25\%$ markup: $80\times 1.25=100$. After a $20\%$ discount on the new price: $100\times 0.80=80$. The final price is $\$80$, equal to Quantity B. The relationship is fully determined because every number is given, so D is wrong. The trap is thinking a $+25\%$ then $-20\%$ must net to a change: here $1.25\times 0.80=1.00$ exactly, so the price returns to its start. Note this equality is special to these particular percentages and base. Answer: C

   Do

   Multiply by the factors $1.25$ and $0.80$ in sequence and compute $1.25\times 0.80=1$.

   Avoid

   Do not add $+25\%$ and $-20\%$ to get a net $+5\%$; percent changes apply to different bases.

6. Question 6Multiple choice · Medium · Rates/ratios/percents

   A store sells notebooks in two package sizes: a pack of 4 notebooks for $10.00 and a pack of 6 notebooks for $13.50. By how many cents is the price per notebook lower for the pack of 6 than for the pack of 4?

   • (A)15
   • (B)20
   • (C)30
   • (D)25
   • (E)35
   Show the answer and worked solution

   Answer: (D) 25

   Price per notebook in the pack of 4: $\frac{10.00}{4}=\$2.50$. Price per notebook in the pack of 6: $\frac{13.50}{6}=\$2.25$. The difference is $\$2.50-\$2.25=\$0.25$, which is 25 cents. Choices (a) and (b) result from arithmetic slips in the division, and choices (c) and (e) overstate the gap.

   Do

   Convert both packages to a common unit, price per notebook, before comparing, then convert the dollar difference to cents as required.

   Avoid

   Don't compare the total prices ($13.50 versus $10.00) directly; the packages contain different numbers of notebooks, so only unit price is meaningful.

7. Question 7Multiple choice · Medium · Data analysis

   A company has four departments with average annual salaries of $50{,}000, $60{,}000, $70{,}000, and $80{,}000, and with 10, 15, 20, and 5 employees, respectively. What is the average annual salary of all employees in the company?

   • (A)$60{,}000
   • (B)$62{,}500
   • (C)$64{,}000
   • (D)$65{,}000
   • (E)$70{,}000
   Show the answer and worked solution

   Answer: (C) $64{,}000

   The total number of employees is $10+15+20+5=50$. The total salary paid is $10(50,000)+15(60,000)+20(70,000)+5(80,000)=500,000+900,000+1,400,000+400,000=3,200,000$. The average salary is $3,200,000/50=64,000$. Choice d, $65{,}000, comes from averaging the four salary figures directly, $(50,000+60,000+70,000+80,000)/4$, without weighting by the number of employees in each department.

   Do

   Compute the weighted average by summing (rate times count) across all groups and dividing by the total count of employees.

   Avoid

   Don't average the four department salaries as if each department had equal weight; the department sizes differ and must be used as weights.

8. Question 8Multiple choice · Medium · Rates, ratios, and percents

   Pump A can fill an empty tank in 6 hours, and Pump B can fill the same tank in 4 hours. If both pumps work together at their constant individual rates, how many hours will it take to fill the tank?

   • (A)2 hours
   • (B)2.4 hours
   • (C)2.5 hours
   • (D)3 hours
   • (E)5 hours
   Show the answer and worked solution

   Answer: (B) 2.4 hours

   Pump A's rate is $\frac{1}{6}$ tank per hour, and Pump B's rate is $\frac{1}{4}$ tank per hour. Working together, their combined rate is $\frac{1}{6}+\frac{1}{4}=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}$ tank per hour. The time to fill one tank is the reciprocal of this rate: $\frac{12}{5}=2.4$ hours.

   Do

   Add the individual rates (tanks per hour) and take the reciprocal of the sum to get the combined time.

   Avoid

   Don't average the two individual times (6 and 4) to get 5 hours; combined work time is found from combined rates, not averaged times.

9. Question 9Multiple choice · Hardest · Algebra

   A solution that is $20\%$ salt by weight is mixed with a solution that is $50\%$ salt by weight. How many liters of the $20\%$ solution must be mixed with $30$ liters of the $50\%$ solution to produce a mixture that is $30\%$ salt by weight?

   • (A)$30$
   • (B)$45$
   • (C)$50$
   • (D)$60$
   • (E)$75$
   Show the answer and worked solution

   Answer: (D) $60$

   Treat 'liters' as proxying for weight so percentages by weight act on volumes directly (a standard GRE convention). Let $x$ be the liters of $20\%$ solution.

   Salt from the $20\%$ solution: $0.20x$.

   Salt from the $50\%$ solution: $0.50\times 30=15$.

   Total mixture: $x+30$ liters, $30\%$ salt, so salt $=0.30(x+30)$.

   Equation: $0.20x+15=0.30(x+30)=0.30x+9$.

   Solve: $15-9=0.30x-0.20x$, so $6=0.10x$, giving $x=60$.

   Check: salt $=0.20(60)+15=12+15=27$; total $=90$ liters; $27/90=30\%$.

   Why the others fail: (A) $30$ assumes equal amounts of each solution, which would give the simple average $35\%$, not $30\%$. (B) $45$ and (C) $50$ come from arithmetic slips on the linear equation. (E) $75$ inverts the alligation ratio (you need more of the weaker solution because the target $30\%$ is closer to $20\%$ than to $50\%$, but not that much more).

   Do

   Set total salt from each solution equal to $30\%$ of the combined volume, then solve a one-variable linear equation.

   Avoid

   Don't average the percents; mixture concentration is a weighted average, not a simple mean.

10. Question 10Multiple choice · Hard · Multi-step word problems

    A tank can be filled by pipe A alone in $6$ hours and by pipe B alone in $9$ hours. A drain at the bottom can empty the full tank in $12$ hours. If all three are open starting from an empty tank, how many hours does it take to fill the tank?

    • (A)$\frac{36}{7}$ hours
    • (B)$\frac{36}{5}$ hours
    • (C)$5$ hours
    • (D)$7$ hours
    • (E)$9$ hours
    Show the answer and worked solution

    Answer: (A) $\frac{36}{7}$ hours

    Convert each to a rate in tanks per hour. Pipe A fills at $\frac{1}{6}$, pipe B at $\frac{1}{9}$, and the drain removes at $\frac{1}{12}$.

    Net rate $=\frac{1}{6}+\frac{1}{9}-\frac{1}{12}$. Using a common denominator of $36$: $\frac{6}{36}+\frac{4}{36}-\frac{3}{36}=\frac{7}{36}$ tanks per hour.

    Time to fill one tank $=\frac{1}{7/36}=\frac{36}{7}$ hours.

    Answer: (A) $\frac{36}{7}$ hours. Choice B ($\frac{36}{5}$) adds the drain rate instead of subtracting it; choices C, D, and E are rounding or addition errors.

    Do

    Add the filling rates and subtract the draining rate, then invert.

    Avoid

    Don't add all three rates; the drain reduces the net fill rate.

11. Question 11Numeric entry · Hard · Arithmetic

    A merchant purchased an item for a certain cost and sold it for $\$120$, earning a profit of 20 percent based on the cost. If the merchant had instead purchased the same item for 10 percent less than the original cost and sold it for the same $\$120$, what would the profit percent have been, based on the new cost?

    Numeric entry: you type the answer, in percent. There are no choices to work back from.

    Show the answer and worked solution

    Answer: 33.3 percent

    Let the original cost be $x$. Since selling at $\$120$ gives a 20 percent profit, $120=1.2x$, so $x=100$. The new cost is 10 percent less: $100-10=90$. Selling at the same $\$120$, the profit is $120-90=30$. The profit percent based on the new cost is $\frac{30}{90}\times 100=33.\overline{3}\%$, which rounds to $33.3\%$. Answer: 33.3

    Do

    Recover the original cost from the given profit percent and selling price, then apply the 10 percent reduction to get the new cost before recomputing profit percent on that new base.

    Avoid

    Do not apply the 10 percent reduction to the selling price instead of the cost, and do not compute the new profit percent using the original cost of 100 as the base instead of the reduced cost of 90.

12. Question 12Quantitative comparison · Hardest · Simple versus compound interest

    A sum of $\$2,000$ is invested for $2$ years. Account A earns $10\%$ simple annual interest, and Account B earns $10\%$ annual interest compounded yearly. Each account starts with $\$2,000$.

    Quantity A

    The balance of Account A after $2$ years

    Quantity B

    The balance of Account B after $2$ years
    Show the answer and worked solution

    Answer: B. Quantity B is greater.

    Both accounts start at $\$2,000$ with a $10\%$ annual rate over $2$ years; the difference is simple versus compound interest.

    Account A (simple interest): interest each year is $10\%$ of the original $\$2,000$, which is $\$200$ per year. Over $2$ years that is $\$400$, so the balance is $2000+400=\$2,400$. Quantity A is $\$2,400$.

    Account B (compound interest): after year one the balance is $2000\times 1.10=\$2,200$. After year two it is $2200\times 1.10=\$2,420$. Quantity B is $\$2,420$.

    Since $\$2,420>\$2,400$, Quantity B is greater. The extra $\$20$ is the second year's interest earned on the first year's $\$200$ of interest, which simple interest does not provide.

    A wrong approach is to assume equal rates and equal principal must yield equal balances; compounding pays interest on accumulated interest, so it pulls ahead whenever the term exceeds one year.

    Do

    Compute simple interest as rate times principal times years, and compound interest by multiplying by $1.10$ each year.

    Avoid

    Do not treat the two accounts as equal; compounding adds interest on prior interest after the first year.

The mistakes that cost points on word problems

• Averaging rates or times. Two pumps that take 6 hours and 4 hours do not take 5 hours together. They take less than either alone. Add the rates, $\frac{1}{6}+\frac{1}{4}=\frac{5}{12}$ of the tank per hour, and invert (question 8).
• Taking a percent of the wrong base. A price that is 25 percent more than cost is $1.25\times \text{cost}$. To get back to the cost you divide by 1.25. You do not subtract 25 percent of the price, and the choices will include the number that mistake produces (question 4).
• Adding and subtracting successive percents. Up 25 percent then down 20 percent is not up 5 percent. Multiply the factors: $1.25\times 0.80=1.00$ (question 5).
• Averaging the averages. When groups differ in size, the overall average is weighted toward the larger group. Compute the total, then divide (question 7).
• Answering a different question.“How many more girls than boys” is not the number of girls. Solving for $x$ when the question asks for $x+y$ is the most common way to lose a question you understood.

Most of these appear in the 12 GRE Quant traps, each with its own example.

Where to practice GRE word problems next

Word problems improve with repetition faster than almost anything else in Quant, because the families repeat. A workable order:

• More of the same, by format. Our free problem solving and numeric entry sets are mostly word problems, each with a worked solution.
• Tables and charts. Data interpretation questions are word problems with the numbers moved into a figure. Practice them in data interpretation.
• Probability. Marbles, committees and dice are word problems with their own five rules, covered in our GRE probability questions.
• The hard end. When these twelve feel routine, move to the hardest GRE math questions.
• Under time. A free full-length mock is where you find out whether the translation holds up at under two minutes a question. Our GRE study plan shows where Quant practice fits week by week.

Frequently asked questions