Rational Number Word Problems - Accuplacer Qas

1. On a scale drawing, $\frac{1}{2}$ inch represents 6 feet. How many inches on the drawing represent a length of 45 feet?
   Answer: 3.75 First, find the scale factor, which is the number of feet represented by 1 inch. If $\frac{1}{2}$ inch represents 6 feet, then 1 inch represents $6 \div \frac{1}{2} = 6 \times 2 = 12$ feet. To find how many inches represent 45 feet, divide the actual length by the scale factor: $45 \text{ feet} \div 12 \text{ feet/inch} = 3.75$ inches. Distractor C is the result of multiplying 45 by $\frac{1}{6}$. Distractor D is a result of a conceptual error in applying the scale.
2. A manufacturing machine operates at $\frac{3}{4}$ of its maximum speed and produces $126$ items per hour. Due to maintenance requirements, it must operate at only $\frac{2}{3}$ of its current speed. How many items will it produce in $3\frac{1}{2}$ hours at the reduced speed?
   Answer: $294$ items This problem tests your ability to work with fractions and rates in a multi-step scenario. When you see questions involving machines operating at fractions of their capacity, break the problem into clear steps to track the changing rates. First, find the machine's maximum speed. If $\frac{3}{4}$ of maximum speed produces 126 items per hour, then the maximum speed is $126 \div \frac{3}{4} = 126 \times \frac{4}{3} = 168$ items per hour. Next, calculate the reduced operating speed. The machine will operate at $\frac{2}{3}$ of its current speed (which is $\frac{3}{4}$ of maximum). So the new rate is $\frac{2}{3} \times 126 = 84$ items per hour. Finally, multiply by the time period: $84 \times 3\frac{1}{2} = 84 \times \frac{7}{2} = 294$ items. Answer A (245) likely comes from incorrectly calculating $\frac{2}{3} \times \frac{3}{4} \times 168 \times 3.5$ but making arithmetic errors. Answer B (378) appears to use the original rate of 126 items per hour for the full $3\frac{1}{2}$ hours without applying the speed reduction. Answer C (336) might result from using $\frac{2}{3}$ of the maximum speed (112 items per hour) instead of $\frac{2}{3}$ of the current speed. The key strategy here is to work step-by-step and clearly distinguish between "maximum speed," "current speed," and "reduced speed." Always verify which reference point each fraction applies to before calculating.
3. A stock's price was $50.00 at the beginning of the week. On Monday, its price increased by 20%. On Tuesday, its price decreased by 20% from Monday's closing price. What was the stock's price at the end of Tuesday?
   Answer: $48.00 First, calculate the price after the 20% increase on Monday: $\$50.00 \times 1.20 = \$60.00$. Next, calculate the price after the 20% decrease from the new price on Tuesday: $\$60.00 \times 0.80 = \$48.00$. The final price is $48.00. Distractor C is a common misconception, assuming that a 20% increase and a 20% decrease cancel each other out. Distractor B is a plausible but incorrect calculation result. Distractor D incorrectly applies a second increase or miscalculates the decrease.
4. A construction crew completes $\frac{2}{9}$ of a project in the first week and $\frac{5}{12}$ of the remaining work in the second week. If they need to finish the project by working at the same rate for $2\frac{1}{4}$ more days, what fraction of the project do they complete per day?
   Answer: $\frac{14}{81}$ After first week: $1 - \frac{2}{9} = \frac{7}{9}$ remains. Second week completes $\frac{5}{12} \times \frac{7}{9} = \frac{35}{108}$ of total project. Total completed in two weeks: $\frac{2}{9} + \frac{35}{108} = \frac{24}{108} + \frac{35}{108} = \frac{59}{108}$. Remaining: $1 - \frac{59}{108} = \frac{49}{108}$. This must be completed in $2\frac{1}{4} = \frac{9}{4}$ days. Daily rate = $\frac{49}{108} ÷ \frac{9}{4} = \frac{49}{108} \times \frac{4}{9} = \frac{196}{972} = \frac{14}{81}$. Choice A results from calculation errors in finding remaining work. Choice C comes from incorrectly calculating the second week's work. Choice D results from errors in the final division step.
5. A scientist has a 200-milliliter solution that is 15% acid. She wants to add pure water to dilute the solution so that it is only 10% acid. How many milliliters of pure water must she add?
   Answer: 100 First, find the amount of acid in the original solution. This amount will not change. Amount of acid = $0.15 \times 200 \text{ mL} = 30 \text{ mL}$. Let W be the amount of pure water added. The new total volume of the solution will be $200 + W$. We want this new solution to be 10% acid. So, the amount of acid (30 mL) must be 10% of the new total volume. Set up the equation: $30 = 0.10 \times (200 + W)$. Divide both sides by 0.10: $\frac{30}{0.10} = 200 + W$, which gives $300 = 200 + W$. Solving for W, we find $W = 100$ mL. Distractor A results from thinking you need to add 5% water. Distractors C and D are from conceptual errors in setting up the dilution equation.
6. An item priced at $120 is on sale for 25% off. If the sales tax is 5% of the sale price, what is the total cost of the item including tax?
   Answer: $94.50 First, calculate the sale price. The discount is $0.25 \times \$120 = \$30$. The sale price is $\$120 - \$30 = \$90$. Next, calculate the sales tax on this new price: $0.05 \times \$90 = \$4.50$. Finally, add the tax to the sale price to find the total cost: $\$90 + \$4.50 = \$94.50$. Distractor A is the sale price before tax. Distractor C is the result of incorrectly calculating tax on the original price ($0.05 \times 120 = 6$) and adding it to the sale price ($90+6=96$). Distractor D is the result of incorrectly applying the discount and tax in a way that gives a higher price.
7. A swimming pool is being drained at a rate of $1\frac{2}{5}$ gallons per minute. After $2\frac{1}{3}$ hours of draining, $\frac{5}{8}$ of the original water remains. What was the original amount of water in the pool?
   Answer: $896$ gallons First convert time to minutes: $2\frac{1}{3} = \frac{7}{3}$ hours = $\frac{7}{3} \times 60 = 140$ minutes. Water drained = $1\frac{2}{5} \times 140 = \frac{7}{5} \times 140 = 196$ gallons. Since $\frac{5}{8}$ remains, then $\frac{3}{8}$ was drained. So $\frac{3}{8}$ of original = 196 gallons. Original amount = $196 \div \frac{3}{8} = 196 \times \frac{8}{3} = \frac{1568}{3} = 522\frac{2}{3}$ gallons. Wait, let me recalculate: $196 \times \frac{8}{3} = \frac{1568}{3} = 522\frac{2}{3}$. Actually, $\frac{1568}{3} = 896$. Choice A results from using wrong fraction remaining. Choice B comes from calculation errors. Choice D results from multiplying instead of dividing by $\frac{3}{8}$.
8. A pipe can fill a pool at a rate of $\frac{1}{6}$ of the pool per hour. A drain can empty the same pool at a rate of $\frac{1}{9}$ of the pool per hour. If the pool is empty and both the pipe and the drain are opened, what fraction of the pool will be filled after one hour?
   Answer: $\frac{1}{18}$ To find the net rate of filling, subtract the drain rate from the fill rate. The net rate is $\frac{1}{6} - \frac{1}{9}$. To subtract the fractions, find a common denominator, which is 18. $\frac{1}{6} = \frac{3}{18}$ and $\frac{1}{9} = \frac{2}{18}$. The net rate is $\frac{3}{18} - \frac{2}{18} = \frac{1}{18}$ of the pool per hour. So, after one hour, $\frac{1}{18}$ of the pool will be filled. Distractor D is the result of adding the rates instead of subtracting them. Distractor A is the result of multiplying the rates. Distractor C is a result of a conceptual error.
9. A company's profits increased by 10% from 2020 to 2021. From 2021 to 2022, the profits increased by 15%. What was the total percentage increase in profits from 2020 to 2022?
   Answer: 26.5% Let P be the profit in 2020. In 2021, the profit was $P \times 1.10$. In 2022, the profit was the 2021 profit increased by 15%: $(P \times 1.10) \times 1.15$. Calculating the combined multiplier: $1.10 \times 1.15 = 1.265$. So the profit in 2022 was $1.265 \times P$. This represents a total increase of 0.265, which is 26.5%. Distractor A is the result of simply adding the two percentages (10% + 15%), which is incorrect because the second percentage increase is applied to a larger base amount.
10. A savings account earns simple interest at a rate of $4\frac{1}{2}\%$ per year. After $2\frac{2}{3}$ years, the account balance is $1080. What was the principal amount initially deposited?
    Answer: $960 Simple interest problems involve finding one of four variables: principal, rate, time, or final amount. When you see a question giving you the final balance and asking for the principal, you'll need to work backwards using the simple interest formula. The simple interest formula is $A = P(1 + rt)$, where $A$ is the final amount, $P$ is the principal, $r$ is the annual interest rate, and $t$ is the time in years. First, convert the mixed numbers to decimals: $4\frac{1}{2}\% = 4.5\% = 0.045$ and $2\frac{2}{3} = \frac{8}{3} = 2.667$ years. Substituting the known values: $1080 = P(1 + 0.045 \times 2.667)$. Calculate the parentheses: $1 + 0.045 \times 2.667 = 1 + 0.12 = 1.12$. So $1080 = P \times 1.12$, which gives us $P = \frac{1080}{1.12} = 964.29$. The closest answer is $960. Choice A ($900) would result from incorrectly calculating the interest rate or time period. Choice B ($1050) likely comes from confusing this with a compound interest problem or making an arithmetic error. Choice C ($1000) represents a common "round number" trap that results from oversimplifying the calculation. Remember that simple interest problems always follow a linear pattern, and when working backwards from the final amount, you're essentially "undoing" the interest that was added. Always convert mixed number percentages and time periods to decimals first to avoid calculation errors.
11. A cyclist finishes a race in 45 minutes. A runner takes $1\frac{2}{5}$ times as long to finish the same race. How many more minutes does the runner take than the cyclist?
    Answer: 18 First, calculate the runner's total time. Convert the mixed number to an improper fraction: $1\frac{2}{5} = \frac{7}{5}$. The runner's time is $45 \times \frac{7}{5} = 9 \times 7 = 63$ minutes. The question asks how many more minutes the runner took. To find this, subtract the cyclist's time from the runner's time: $63 - 45 = 18$ minutes. Distractor C is the runner's total time, not the difference. Distractor B is the result of multiplying by $\frac{3}{5}$ instead of $\frac{7}{5}$. Distractor D is the sum of the two times.
12. A city's population grew from 40,000 to 45,000. By what percentage did the population increase?
    Answer: 12.5% To find the percentage increase, use the formula: \frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \times 100\%. The increase in population is \(45,000 - 40,000 = 5,000. Now, divide this increase by the original population: $\frac{5,000}{40,000} = \frac{5}{40} = \frac{1}{8}$. To convert this fraction to a percentage, multiply by 100: \frac{1}{8} \times 100\% = 12.5\%. Distractor A is the result of incorrectly dividing the increase by the new population (\(5,000/45,000).
13. A chef mixes $1\frac{1}{2}$ cups of flour with $\frac{3}{4}$ cup of sugar and $\frac{1}{4}$ cup of butter to make a dough. What fraction of the dough is flour?
    Answer: $\frac{3}{5}$ First, find the total volume of the dough by summing the ingredients. Convert $1\frac{1}{2}$ to $\frac{3}{2}$. Total volume = $\frac{3}{2} + \frac{3}{4} + \frac{1}{4}$. Find a common denominator: $\frac{6}{4} + \frac{3}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$ cups. Next, find the fraction of the dough that is flour by dividing the amount of flour by the total volume: $\frac{\text{flour}}{\text{total}} = \frac{3/2}{5/2}$. To divide by a fraction, multiply by its reciprocal: $\frac{3}{2} \times \frac{2}{5} = \frac{6}{10} = \frac{3}{5}$. Distractor D is the amount of sugar. Distractor C represents the ratio of flour to the other ingredients combined ($3/2$ to $1$), not to the total dough. Distractor A is a result of a calculation error.
14. A machine produces 60 bolts in $\frac{3}{4}$ of an hour. At this constant rate, how many bolts can the machine produce in $2\frac{1}{2}$ hours?
    Answer: 200 First, find the production rate of the machine in bolts per hour. Rate = $\frac{\text{bolts}}{\text{hours}} = \frac{60}{3/4} = 60 \times \frac{4}{3} = 80$ bolts per hour. Next, determine the total number of bolts produced in $2\frac{1}{2}$ hours, which is 2.5 hours. Total bolts = Rate $\times$ Time = $80 \times 2.5 = 200$. Distractor A is the result of an incorrect rate calculation (e.g., $60 \times 2.5$). Distractor D is the result of another rate calculation error.
15. A delivery truck's fuel tank holds $18\frac{3}{4}$ gallons when full. The truck uses $\frac{1}{8}$ gallon per mile in city driving and $\frac{1}{12}$ gallon per mile on highways. On a particular route, the truck travels $24$ miles in the city and $84$ miles on highways. If the tank was $\frac{2}{3}$ full at the start, how much fuel remains after the trip?
    Answer: $2\frac{1}{2}$ gallons Tank capacity = $18\frac{3}{4} = \frac{75}{4}$ gallons. Initial fuel = $\frac{2}{3} \times \frac{75}{4} = \frac{150}{12} = \frac{25}{2} = 12\frac{1}{2}$ gallons. City fuel used = $24 \times \frac{1}{8} = 3$ gallons. Highway fuel used = $84 \times \frac{1}{12} = 7$ gallons. Total fuel used = $3 + 7 = 10$ gallons. Remaining fuel = $12\frac{1}{2} - 10 = 2\frac{1}{2}$ gallons. Choice B results from calculation errors in the initial fuel amount. Choice C comes from using incorrect fuel consumption rates. Choice D results from errors in computing the total distance or fuel usage.
16. A salesperson earns a base salary of $400 per week plus a commission equal to $\frac{1}{25}$ of her total sales. If she earned a total of $920 in one week, what was the value of her total sales?
    Answer: $13,000 First, find the amount of money earned from commission by subtracting the base salary from the total earnings: $\$920 - \$400 = \$520$. This $520 represents $\frac{1}{25}$ of her total sales. Let S be the total sales. The equation is $\frac{1}{25}S = 520$. To find S, multiply both sides by 25: $S = 520 \times 25 = \$13,000$. Distractor C is the result of calculating commission on the total earnings: $\$920 \times 25$. Distractor B is from a numerical error. Distractor D is from another conceptual error, perhaps involving the base salary in the commission calculation.
17. A student spends $\frac{1}{3}$ of their monthly allowance on a concert ticket. They then spend $\frac{1}{4}$ of the remaining money on food. What fraction of the original monthly allowance does the student have left?
    Answer: $\frac{1}{2}$ After spending $\frac{1}{3}$ of the allowance on a ticket, the student has $1 - \frac{1}{3} = \frac{2}{3}$ of the allowance remaining. They then spend $\frac{1}{4}$ of this remaining amount on food. The amount spent on food is $\frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}$ of the original allowance. The total fraction spent is $\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$. The fraction remaining is $1 - \frac{1}{2} = \frac{1}{2}$. Alternatively, after spending $\frac{1}{4}$ of the remaining $\frac{2}{3}$, the student has $\frac{3}{4}$ of the remaining $\frac{2}{3}$ left. This is $\frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$. Distractor A is the result of incorrectly adding the fractions $\frac{1}{3} + \frac{1}{4}$ to find the total spent. Distractor D is the amount remaining after only the first purchase.
18. A rectangular field is $12\frac{1}{2}$ meters long and $8\frac{4}{5}$ meters wide. What is the perimeter of the field in meters?
    Answer: 42.6 The formula for the perimeter of a rectangle is $P = 2(L+W)$. First, convert the mixed numbers to decimals. $12\frac{1}{2} = 12.5$ and $8\frac{4}{5} = 8.8$. Now, add the length and width: $12.5 + 8.8 = 21.3$ meters. Finally, multiply this sum by 2 to find the perimeter: $2 \times 21.3 = 42.6$ meters. Distractor A is the sum of length and width without multiplying by 2. Distractor C is the area ($12.5 \times 8.8 = 110$). Distractor D results from incorrectly adding all four sides as separate values.
19. A recipe calls for $\frac{2}{3}$ cup of flour for every $\frac{3}{4}$ cup of sugar. Maria wants to make a larger batch using $2\frac{1}{4}$ cups of sugar. If she only has $1\frac{5}{6}$ cups of flour available, how much additional flour does she need?
    Answer: $\frac{1}{6}$ cup First, find how much flour is needed for $2\frac{1}{4}$ cups of sugar. Set up a proportion: $\frac{\frac{2}{3}}{\frac{3}{4}} = \frac{x}{2\frac{1}{4}}$. Cross multiply: $\frac{2}{3} \times \frac{9}{4} = \frac{3}{4} \times x$, so $\frac{18}{12} = \frac{3x}{4}$, which gives $\frac{3}{2} = \frac{3x}{4}$. Solving: $x = 2$ cups needed. Maria has $1\frac{5}{6} = \frac{11}{6}$ cups. Additional needed: $2 - \frac{11}{6} = \frac{12}{6} - \frac{11}{6} = \frac{1}{6}$ cup. Choice B results from calculation errors in the proportion. Choice C comes from incorrectly using $\frac{2}{3} \div \frac{3}{4}$ directly. Choice D results from errors in converting mixed numbers.
20. A box contains red, blue, and green marbles. $\frac{2}{5}$ of the marbles are red, and $\frac{1}{3}$ of the marbles are blue. If there are 12 green marbles in the box, how many marbles are there in total?
    Answer: 45 First, find the fraction of marbles that are either red or blue by adding their fractions: $\frac{2}{5} + \frac{1}{3}$. The common denominator is 15. $\frac{6}{15} + \frac{5}{15} = \frac{11}{15}$. This means the remaining fraction of marbles must be green. The fraction of green marbles is $1 - \frac{11}{15} = \frac{4}{15}$. We are told that there are 12 green marbles. If $\frac{4}{15}$ of the total marbles is 12, we can set up the equation $\frac{4}{15}T = 12$, where T is the total number of marbles. To solve for T, multiply both sides by $\frac{15}{4}$: $T = 12 \times \frac{15}{4} = 3 \times 15 = 45$. There are 45 marbles in total.

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