DAT Quantitative Reasoning Quiz: Word Problems

What this quiz covers

This quiz focuses on Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for DAT Quantitative Reasoning.

How to use this quiz

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.

Question 1

1. 6
2. 8.4 (correct answer)
3. 10
4. 12

Explanation: When you encounter work rate problems, think in terms of rates per unit time. Each pump has a specific rate at which it fills the vat, and when working together, their rates combine. First, find each pump's rate. Pump A fills the vat in 18 minutes, so its rate is $\frac{1}{18}$ vat per minute. Pump B fills the vat in 30 minutes, so its rate is $\frac{1}{30}$ vat per minute. When both pumps work together for 6 minutes, their combined rate is $\frac{1}{18} + \frac{1}{30}$. To add these fractions, find a common denominator of 90: $\frac{5}{90} + \frac{3}{90} = \frac{8}{90} = \frac{4}{45}$ vat per minute. In 6 minutes working together, they fill $6 \times \frac{4}{45} = \frac{24}{45} = \frac{8}{15}$ of the vat. The remaining portion is $1 - \frac{8}{15} = \frac{7}{15}$ of the vat. Now Pump A works alone at rate $\frac{1}{18}$ to finish $\frac{7}{15}$ of the vat. Time needed is $\frac{7/15}{1/18} = \frac{7}{15} \times 18 = \frac{126}{15} = 8.4$ minutes. Choice A (6) likely comes from assuming only the time they worked together matters. Choice C (10) might result from calculation errors with the fractions. Choice D (12) could come from incorrectly calculating the remaining work or using wrong rates. Remember: in combined work problems, always calculate what portion of work is completed first, then determine how long the remaining work takes at the new rate.

Question 2

1. 12 L
2. 16 L
3. 20 L (correct answer)
4. 24 L

Explanation: When you encounter mixture problems involving percentages, you're dealing with weighted averages. The key is tracking the amount of pure substance (acid, in this case) before and after mixing. Let's call the unknown amount of 25% acid solution $x$ liters. Set up an equation based on the fact that the total amount of pure acid before mixing equals the total amount after mixing. Pure acid from 25% solution: $0.25x$ Pure acid from 60% solution: $0.60 \times 15 = 9$ liters Total volume after mixing: $x + 15$ liters Final concentration: 40% The equation becomes: $0.25x + 9 = 0.40(x + 15)$ Solving: $0.25x + 9 = 0.40x + 6$ $9 - 6 = 0.40x - 0.25x$ $3 = 0.15x$ $x = 20$ liters This confirms answer C is correct. Answer A (12 L) would give you $\frac{0.25(12) + 9}{27} = \frac{12}{27} = 44.4\%$ acid - too concentrated. Answer B (16 L) yields $\frac{0.25(16) + 9}{31} = \frac{13}{31} = 41.9\%$ - still too high. Answer D (24 L) produces $\frac{0.25(24) + 9}{39} = \frac{15}{39} = 38.5\%$ - too dilute. For mixture problems, always set up your equation based on the pure substance amounts, not the total volumes. This approach works for any concentration problem, whether it's acids, salt solutions, or medication dosages.

Question 3

1. 45
2. 50
3. 55 (correct answer)
4. 60

Explanation: Age relationship problems require you to set up equations based on different time periods. The key is defining variables for current ages and translating the word relationships into mathematical expressions. Let's define Ivan's current age as $I$ and Mira's current age as $M$. Now we can translate each condition: Ten years ago: Ivan was $(I-10)$ and Mira was $(M-10)$. Since Ivan was three times as old as Mira then: $I-10 = 3(M-10)$, which simplifies to $I-10 = 3M-30$, or $I = 3M-20$. In five years: Ivan will be $(I+5)$ and Mira will be $(M+5)$. Since Ivan will be twice as old as Mira then: $I+5 = 2(M+5)$, which simplifies to $I+5 = 2M+10$, or $I = 2M+5$. Setting our two expressions for $I$ equal: $3M-20 = 2M+5$. Solving: $M = 25$. Therefore, $I = 2(25)+5 = 55$. Let's verify: Ten years ago, Ivan was 45 and Mira was 15, and indeed $45 = 3 \times 15$. In five years, Ivan will be 60 and Mira will be 30, and indeed $60 = 2 \times 30$. Choice A (45) represents Ivan's age ten years ago, not now. Choice B (50) might result from calculation errors in the system of equations. Choice D (60) represents Ivan's age in five years, not his current age. Always define variables for current ages first, then work backward and forward in time. Double-check by substituting your answer back into both original conditions.

Question 4

1. 2.8
2. 3.1 (correct answer)
3. 3.4
4. 3.9

Explanation: When you encounter problems involving currents or winds affecting travel, remember that the current adds to your speed in one direction and subtracts from it in the other. You need to calculate time for each leg separately since the speeds differ. For the downstream journey, the boat's effective speed is $10 + 2 = 12$ mph. Time equals distance divided by speed, so: $\text{Time downstream} = \frac{15 \text{ miles}}{12 \text{ mph}} = 1.25 \text{ hours}$ For the upstream return, the current works against the boat: $10 - 2 = 8$ mph. Therefore: $\text{Time upstream} = \frac{15 \text{ miles}}{8 \text{ mph}} = 1.875 \text{ hours}$ Total round trip time: $1.25 + 1.875 = 3.125$ hours, which rounds to 3.1 hours. Choice A (2.8) likely comes from incorrectly using the boat's still-water speed for both directions, giving $\frac{30}{10} = 3$ hours, then making a calculation error. Choice C (3.4) might result from using an average speed approach, which doesn't work for round trips with different speeds each way. Choice D (3.9) could come from miscalculating the upstream speed as $10 - 2 = 6$ mph instead of 8 mph. The correct answer is B (3.1). Strategy tip: In current/wind problems, always calculate each leg separately using the adjusted speeds. Never average the speeds for round-trip calculations—this is a common trap that leads to incorrect answers.

Question 5

1. 4 h
2. 5 h
3. 6 h (correct answer)
4. 7 h

Explanation: When you encounter work rate problems, you're dealing with the relationship between workers, time, and output. The key insight is that work rate (output per worker per unit time) stays constant, so you can scale up or down based on the number of workers. First, find the rate per assistant. Three assistants prepare 60 kits in 4 hours, so the total work done is $3 \times 4 = 12$ assistant-hours for 60 kits. This means the rate is $\frac{60 \text{ kits}}{12 \text{ assistant-hours}} = 5 \text{ kits per assistant-hour}$. Now apply this rate to the new scenario. You need 150 kits at 5 kits per assistant-hour, which requires $\frac{150}{5} = 30$ assistant-hours total. With 5 assistants working, the time needed is $\frac{30 \text{ assistant-hours}}{5 \text{ assistants}} = 6 \text{ hours}$. Looking at the wrong answers: (A) 4 hours likely comes from assuming the same time regardless of the number of workers or kits. (B) 5 hours might result from incorrectly thinking that since you have 5 assistants, it takes 5 hours. (D) 7 hours could come from various calculation errors, such as incorrectly setting up the proportion. For DAT work rate problems, always establish the rate per individual worker first, then scale to find total worker-time needed, and finally divide by the actual number of workers. This systematic approach prevents the common trap of confusing the number of workers with the time required.

Question 6

1. $\dfrac13$
2. $\dfrac25$
3. $\dfrac23$ (correct answer)
4. $\dfrac34$

Explanation: When you encounter probability questions asking for "at least one" of something, the complement approach is often your most efficient strategy. Instead of calculating all the ways to get one white mask plus all the ways to get two white masks, find the probability that NO white masks are selected, then subtract from 1. To find the probability of selecting no white masks (only blue masks), you need both masks to be blue. The drawer contains 6 blue masks and 4 white masks (10 total). The probability of selecting two blue masks without replacement is: First mask blue: $\frac{6}{10}$ Second mask blue (given first was blue): $\frac{5}{9}$ Combined probability of both blue: $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ Therefore, the probability of at least one white mask is: $1 - \frac{1}{3} = \frac{2}{3}$ Choice A ($\frac{1}{3}$) represents the probability of getting NO white masks—the complement of what we want. Choice B ($\frac{2}{5}$) likely comes from incorrectly using $\frac{4}{10}$ (initial white probability) without accounting for the "without replacement" condition. Choice D ($\frac{3}{4}$) might result from misapplying the white mask ratio or making calculation errors with the complement approach. Study tip: For "at least one" probability questions, always consider the complement method first. Calculate the probability of the opposite outcome (usually "none"), then subtract from 1. This approach typically involves fewer calculations and reduces error risk.

Question 7

1. 10 g
2. 12.5 g (correct answer)
3. 14 g
4. 16 g

Explanation: When you encounter alloy mixture problems, you're dealing with weighted averages and the principle that the total amount of a substance equals the sum of its parts from different sources. Start by identifying what you know: the original alloy has 500 g with 18% silver, so it contains $500 \times 0.18 = 90$ g of pure silver. You're adding $x$ grams of pure silver (100% silver), creating a new alloy weighing $(500 + x)$ grams that should be 20% silver. Set up the equation using the fact that total silver amount equals the desired percentage of the final mixture: $90 + x = 0.20(500 + x)$ Solving: $90 + x = 100 + 0.20x$ $x - 0.20x = 100 - 90$ $0.80x = 10$ $x = 12.5$ So you need 12.5 g of pure silver, making (B) correct. Choice (A) 10 g represents the difference between final and initial silver amounts (100g - 90g), but ignores that adding silver increases the total weight. Choice (C) 14 g might result from incorrectly calculating 20% of the original 500g minus the existing silver percentage. Choice (D) 16 g could come from misapplying the percentage increase to the original silver content. The key strategy for mixture problems is always accounting for how additions change both the numerator (amount of substance) and denominator (total weight) in your percentage calculation. Set up your equation so both sides represent the same quantity.

Question 8

1. 42 mph
2. 43.5 mph (correct answer)
3. 45 mph
4. 47.5 mph

Explanation: When you encounter average speed problems, remember that average speed isn't simply the arithmetic mean of the speeds. Instead, you must use the formula: average speed = total distance ÷ total time. First, calculate the time for each segment. For the first 40 miles at 50 mph: time = distance ÷ speed = 40 ÷ 50 = 0.8 hours. For the next 60 miles at 40 mph: time = 60 ÷ 40 = 1.5 hours. The total distance is 100 miles, and the total time is 0.8 + 1.5 = 2.3 hours. Therefore, average speed = 100 ÷ 2.3 ≈ 43.5 mph. Choice A (42 mph) is too low and likely results from calculation errors in the time conversions. Choice C (45 mph) is the simple arithmetic mean of 50 mph and 40 mph ($\frac{50 + 40}{2} = 45$), which is the most common trap in average speed problems. This would only be correct if equal amounts of time were spent at each speed, not equal distances. Choice D (47.5 mph) appears to be a weighted average that incorrectly emphasizes the higher speed, possibly from misunderstanding how to weight the segments. The key strategy for average speed problems is to resist the temptation to average the speeds directly. Always calculate the actual times spent at each speed, then use total distance divided by total time. Watch for this pattern on quantitative reasoning exams—they frequently test whether you'll fall into the "simple average" trap.

Question 9

1. 65 mL
2. 75 mL
3. 83 mL (correct answer)
4. 90 mL

Explanation: This is a dilution problem where you're adding water to reduce the concentration of active ingredient. The key insight is that while you're adding volume, the absolute amount of active ingredient stays the same. First, calculate the amount of active ingredient in the original solution: $250 \text{ mL} \times 0.12 = 30 \text{ mL}$ of active ingredient. Next, determine what total volume you need to achieve 9% concentration with this same 30 mL of active ingredient. Set up the equation: $\frac{30 \text{ mL active ingredient}}{\text{new total volume}} = 0.09$ Solving: $\text{new total volume} = \frac{30}{0.09} = 333.33 \text{ mL}$ Since you started with 250 mL, the water you must add is: $333.33 - 250 = 83.33 \text{ mL}$, which rounds to 83 mL. Looking at the wrong answers: Choice A (65 mL) would give you a final volume of 315 mL, resulting in a 9.5% concentration—still too strong. Choice B (75 mL) creates a 325 mL solution with 9.2% concentration—closer but still not dilute enough. Choice D (90 mL) gives you 340 mL total volume with 8.8% concentration—this over-dilutes the solution below your target. For dilution problems, always remember that the amount of solute (active ingredient) remains constant while you're changing the total volume. Set up your equation based on this principle, and double-check by calculating the final concentration with your answer.

Question 10

1. 135
2. 180
3. 225 (correct answer)
4. 240

Explanation: Ratio problems test your ability to work with proportional relationships. When you see a ratio like 3:5, think of it as representing parts of a whole rather than actual quantities. The ratio 3:5 means that for every 3 small brackets, there are 5 large brackets. This gives us a total of 3 + 5 = 8 parts in the ratio. Since 360 brackets were ordered total, each part represents $\frac{360}{8} = 45$ brackets. Large brackets make up 5 parts of the ratio, so the number of large brackets is $5 \times 45 = 225$. You can verify this: small brackets would be $3 \times 45 = 135$, and $135 + 225 = 360$ ✓ Looking at the wrong answers: A) 135 is the number of small brackets, not large ones—this represents the classic trap of solving for the wrong part of the ratio. B) 180 would result if you incorrectly calculated each part as $\frac{360}{10} = 36$ (perhaps by adding an extra step), then multiplied by 5. D) 240 comes from miscalculating the total parts as 6 instead of 8, giving $\frac{360}{6} = 60$ per part, then $60 \times 4 = 240$—but this uses the wrong ratio interpretation entirely. The key strategy for ratio problems: always find the total number of parts first, then determine what each part represents. Double-check by ensuring your parts add up to the given total. This systematic approach prevents the common mistake of solving for the wrong component.

Question 11

1. 120 mL
2. 140 mL
3. 160 mL (correct answer)
4. 180 mL

Explanation: When you encounter percentage concentration problems involving evaporation, you need to track how the removal of one component changes the overall composition. The key insight is that while water evaporates, the non-water portion remains constant. Let's call the original volume $V$. Initially, the solution is 70% water, so it contains $0.7V$ mL of water and $0.3V$ mL of other substances. After 40 mL of water evaporates, you have:

• Water remaining: $0.7V - 40$ mL
• Other substances: $0.3V$ mL (unchanged)
• New total volume: $V - 40$ mL

Question 12

1. 150 mg
2. 160 mg
3. 170 mg (correct answer)
4. 180 mg

Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay where a substance decreases by a constant percentage over equal time intervals. The given formula $N=N_0(1-r)^t$ captures this pattern, where $N_0$ is the initial amount, $r$ is the decay rate as a decimal, and $t$ is time. Let's substitute the given values: $N_0 = 250$ mg, $r = 0.12$ (12% as a decimal), and $t = 3$ hours. This gives us: $N = 250(1-0.12)^3 = 250(0.88)^3$ Calculating $(0.88)^3$: $0.88 \times 0.88 = 0.7744$, then $0.7744 \times 0.88 = 0.6815$ Therefore: $N = 250 \times 0.6815 = 170.4$ mg, which rounds to approximately 170 mg. Looking at the wrong answers: Choice A (150 mg) represents too much decay, possibly from using an incorrect decay rate or miscalculating the exponential. Choice B (160 mg) is close but likely results from rounding errors during intermediate calculations. Choice D (180 mg) suggests insufficient decay, perhaps from using simple rather than compound decay or applying the wrong time period. The key strategy here is recognizing that exponential decay is compound—the 12% loss applies to the remaining amount each hour, not the original amount. Always convert percentages to decimals, be careful with your exponential calculations, and remember that "decay rate" goes with $(1-r)$, not just $r$ in the formula.

Question 13

1. 48
2. 168
3. 336 (correct answer)
4. 720

Explanation: This is a permutation problem where you're selecting and arranging objects in a specific order. When you see "arrange in order" or "ordered arrangements," you need to account for the fact that different sequences of the same items count as different arrangements. Since you're selecting 3 cups from 8 different colored cups and arranging them in order, you use the permutation formula: $P(n,r) = \frac{n!}{(n-r)!}$ where n = 8 total cups and r = 3 selected cups. $P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$ You can also think of this step-by-step: you have 8 choices for the first position, 7 remaining choices for the second position, and 6 remaining choices for the third position, giving you $8 \times 7 \times 6 = 336$. Choice A (48) incorrectly uses $8 \times 6$ or might represent some other flawed calculation. Choice B (168) is exactly half of the correct answer, suggesting someone divided by 2 unnecessarily, perhaps confusing this with a combination problem. Choice D (720) represents $6! = 720$, which would be the number of ways to arrange 6 items, not the correct calculation for this problem. Remember: when a problem mentions "ordered arrangements" or "in order," use permutations, not combinations. The sequence matters, so ABC is different from BAC.

Question 14

1. 11.5 in
2. 12.6 in (correct answer)
3. 13.8 in
4. 14.5 in

Explanation: Unit conversion problems on the DAT require careful attention to setting up your conversion factor correctly. When converting between metric and imperial units, you need to ensure you're multiplying and dividing in the right direction. To convert 32 cm to inches using the given conversion factor $1\text{ in} = 2.54\text{ cm}$, set up the calculation so the centimeters cancel out: $32\text{ cm} \times \frac{1\text{ in}}{2.54\text{ cm}} = \frac{32}{2.54}\text{ in} = 12.6\text{ in}$ This confirms that choice B is correct. Looking at the wrong answers: Choice A (11.5 in) results from a calculation error, likely from using an incorrect conversion factor or arithmetic mistake. Choice C (13.8 in) suggests the student may have used an approximate conversion like 1 inch = 2.3 cm instead of the given 2.54 cm. Choice D (14.5 in) indicates a more significant error, possibly from inverting the conversion factor or using 1 inch = 2.2 cm. The most common trap in conversion problems is setting up the fraction incorrectly. Always write your conversion factor so that the units you're converting from appear in the denominator and cancel out. Double-check by ensuring your final answer makes logical sense – since centimeters are smaller than inches, 32 cm should convert to a smaller number when expressed in inches.

Question 15

1. 64 °C
2. 68 °C (correct answer)
3. 72 °C
4. 76 °C

Explanation: This is a multi-step temperature change problem that requires you to work backwards from the final condition. When you see problems involving different rates of change over time, set up the sequence of events chronologically and track the cumulative effect. The autoclave goes through two cooling phases: 4 minutes on the counter (cooling at 3°C/minute), then 2 minutes under the fan (cooling at 8°C/minute) before reaching 40°C. Working backwards from the final temperature: After 2 minutes under the fan, the load cooled $2 \times 8 = 16°C$. So before going under the fan, it was $40 + 16 = 56°C$. Before that, it cooled on the counter for 4 minutes at 3°C/minute, losing $4 \times 3 = 12°C$. Therefore, the starting temperature was $56 + 12 = 68°C$. Answer choice A (64°C) represents the error of adding the temperature drops instead of working backwards: $40 + 16 + 12 = 68°C$, but then mistakenly calculating $68 - 4 = 64°C$. Answer choice C (72°C) likely comes from incorrectly calculating one of the cooling periods, perhaps using $4 \times 4 = 16°C$ instead of $4 \times 3 = 12°C$ for the counter cooling. Answer choice D (76°C) might result from adding an extra 8°C, possibly double-counting one minute of fan cooling. For temperature change problems, always work systematically through each time period and double-check by working forward from your answer to verify you reach the given final condition.

Question 16

1. 78
2. 80
3. 82 (correct answer)
4. 84

Explanation: When you encounter weighted average problems, you're working with components that contribute different percentages to a final result. The key is setting up an equation where each component is multiplied by its weight, and the sum equals your target average. Here, you need to find the final exam score that produces an 80% semester average. Set up the weighted average equation: $0.40(82) + 0.25(74) + 0.35(x) = 80$, where $x$ is the unknown final exam score. Calculate the known components: $32.8 + 18.5 + 0.35x = 80$. This simplifies to $51.3 + 0.35x = 80$. Subtracting 51.3 from both sides gives $0.35x = 28.7$. Dividing by 0.35 yields $x = 82$. Let's examine why the other options don't work. Choice A (78) would give: $32.8 + 18.5 + 0.35(78) = 32.8 + 18.5 + 27.3 = 78.6$, which falls short of 80. Choice B (80) produces: $32.8 + 18.5 + 0.35(80) = 32.8 + 18.5 + 28 = 79.3$, still below the target. Choice D (84) yields: $32.8 + 18.5 + 0.35(84) = 32.8 + 18.5 + 29.4 = 80.7$, which exceeds 80. Only choice C (82) produces exactly 80. Study tip: For weighted average problems, always convert percentages to decimals and verify your answer by substituting back into the original equation. Remember that the final exam score needed depends on how the earlier grades performed relative to the target average.

Question 17

1. 18 min (correct answer)
2. 20 min
3. 22.5 min
4. 24 min

Explanation: When you see "varies inversely" in a quantitative reasoning problem, you're dealing with inverse variation, where two variables have a constant product. As one increases, the other decreases proportionally. For inverse variation, you can write the relationship as $T \times n = k$ (where $k$ is constant), or $T = \frac{k}{n}$. This makes intuitive sense: more workers mean less time needed. First, find the constant $k$ using the given information. With 2 technicians taking 45 minutes: $k = T \times n = 45 \times 2 = 90$ Now solve for 5 technicians: $T = \frac{k}{n} = \frac{90}{5} = 18$ minutes Therefore, A) 18 min is correct. The wrong answers represent common mistakes: B) 20 min might come from incorrectly assuming the relationship is $T = \frac{45 \times 2}{5} = 18$, then rounding or making arithmetic errors. C) 22.5 min could result from setting up a direct proportion instead of inverse: $\frac{45}{2} = \frac{T}{5}$, giving $T = 112.5$, then dividing by 5 instead of multiplying. D) 24 min might come from various calculation errors or misunderstanding the inverse relationship entirely. Strategy tip: For inverse variation problems, always multiply the given values to find your constant first, then divide that constant by the new variable. Watch for keywords like "inversely," "varies inversely," or situations where logically more of one thing means less of another (workers/time, speed/time, etc.).

Question 18

1. $8.6\text{ mph}$
2. $10.3\text{ mph}$ (correct answer)
3. $12\text{ mph}$
4. $14\text{ mph}$

Explanation: When you encounter average speed problems with multiple segments and rest periods, remember that average speed equals total distance divided by total time—not the average of the individual speeds. Let's calculate each component systematically. For the uphill segment: $\text{time} = \frac{6 \text{ miles}}{8 \text{ mph}} = 0.75 \text{ hours} = 45 \text{ minutes}$. The rest period is 10 minutes. For the downhill segment: $\text{time} = \frac{6 \text{ miles}}{24 \text{ mph}} = 0.25 \text{ hours} = 15 \text{ minutes}$. Total distance is $6 + 6 = 12$ miles. Total time is $45 + 10 + 15 = 70$ minutes $= \frac{70}{60} = \frac{7}{6}$ hours. Therefore, average speed $= \frac{12 \text{ miles}}{\frac{7}{6} \text{ hours}} = 12 \times \frac{6}{7} = \frac{72}{7} \approx 10.3 \text{ mph}$, confirming answer B. Choice A ($8.6 \text{ mph}$) likely results from calculation errors or incorrectly weighting the slower uphill speed too heavily. Choice C ($12 \text{ mph}$) is what you'd get if you forgot to include the 10-minute rest period in your total time. Choice D ($14 \text{ mph}$) appears to come from incorrectly averaging the two speeds: $\frac{8 + 24}{2} = 16$, then perhaps adjusting downward, but this approach ignores both the time spent at each speed and the rest period. Always remember: average speed problems require total distance divided by total time. Include all time periods (including rest stops) and avoid the temptation to simply average the individual speeds.

Question 19

1. $9.20
2. $9.60 (correct answer)
3. $8.90
4. $10.20
5. $8.50

Explanation: Let x = mL of Type A and (50-x) = mL of Type B. Set up the equation: 0.02x + 0.04(50-x) = 0.032(50). Solving: 0.02x + 2 - 0.04x = 1.6, so -0.02x = -0.4, thus x = 20 mL of Type A and 30 mL of Type B. Cost = 20($0.15) + 30($0.22) = $3.00 +$6.60 = $9.60.

Question 20

1. 5.25 hours (correct answer)
2. 6.00 hours
3. 5.75 hours
4. 6.25 hours
5. 5.50 hours

Explanation: Sarah's rate = 1/8 of the work per hour, Michael's rate = 1/12 per hour. Combined rate = 1/8 + 1/12 = 3/24 + 2/24 = 5/24 per hour. In 3 hours together, they complete 3 × 5/24 = 15/24 = 5/8 of the work. Remaining work = 1 - 5/8 = 3/8. Time for Sarah to finish alone = (3/8) ÷ (1/8) = 2.25 hours. Total time = 3 + 2.25 = 5.25 hours.