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📝 Distance rate time formula D = rt (12 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 12 questions available

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What is Distance rate time formula D = rt?

Definition:
The distance-rate-time formula $D = rt$ states that distance traveled is equal to the rate (speed) multiplied by the time taken, where $D$ is distance, $r$ is rate (constant speed), and $t$ is time, and this formula applies to uniform motion, allowing us to solve for any of the three variables if the other two are known, using rearranged forms $r = \frac{D}{t}$ and $t = \frac{D}{r}$ .

Working:
To use this formula, ensure that the units for rate and time are consistent (e.g., if rate is in miles per hour, time must be in hours), then plug the known values into the appropriate version; if solving for distance, multiply rate by time; if solving for rate, divide distance by time; and if solving for time, divide distance by rate, and this works for any type of motion as long as speed is constant.

Example:
A car travels at 55 miles per hour for 4 hours; the distance covered is $D = 55 \times 4 = 220$ miles, and if a cyclist covers 30 miles in 2 hours, the rate is $r = \frac{30}{2} = 15$ miles per hour.

Reason:
The distance-rate-time formula is one of the most practical and widely used formulas in everyday life, from calculating travel times to planning logistics, and it forms the basis for understanding motion in physics and algebra, essential for students and professionals alike.

Easy

Medium

Hard

📝 All Distance rate time formula D = rt MCQs

Q1. A cyclist travels at a constant rate of 1818 km/h for 2.52.5 hours. What distance does the cyclist cover?

A.36 km

B.40 km

C.45 km ✅

D.50 km

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Using $D=rt$ , substitute the rate 18 km/h and time 2.5 hours. Thus $D=18(2.5)=45$ km. The key is maintaining consistent units and multiplying rate by elapsed time.

Q2. A train covers 360360 km in 4.54.5 hours at a constant rate. Which expression correctly represents its rate?

360+4.5

360-4.5

360(4.5)

360\div4.5

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Rate measures distance per unit of time, so distance must be divided by time. Therefore, $r=360\div4.5=80$ km/h. Multiplying would produce incorrect units rather than a meaningful rate.

Q3. Two delivery vans leave the same warehouse. Van A travels 6060 km/h for 33 hours, while Van B travels 7575 km/h for 2.42.4 hours. Which conclusion is correct?

A.Van A travels farther by 18 km.

B.Van B travels farther by 18 km.

C.Both travel 180 km. ✅

D.Both travel 225 km.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Van A travels $60(3)=180$ km. Van B travels $75(2.4)=180$ km. Although their rates and travel times differ, the products are equal, showing that different rate-time combinations can produce the same distance.

Q4. A student says, “If a car's speed doubles while the travel time stays unchanged, its distance becomes half as large because speed and time are inversely related.” What is the best evaluation?

A.Correct, because higher speed always means shorter distance.

B.Correct, because distance is divided by rate.

C.Incorrect, because with fixed time, distance doubles when rate doubles. ✅

D.Incorrect, because distance remains unchanged whenever rate changes.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Since $D=rt$ , distance is directly proportional to rate when time is fixed. Doubling the rate therefore doubles the distance. The student's inverse relationship applies when distance is fixed and rate and time vary together.

Q5. A bus normally travels 240240 km at 6060 km/h. Road construction reduces its speed to 4848 km/h. If the distance remains 240240 km, how much additional travel time is required?

A.30 minutes

B.45 minutes

C.60 minutes ✅

D.75 minutes

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: At 60 km/h, the trip takes $240/60=4$ hours. At 48 km/h, it takes $240/48=5$ hours. The difference is 1 hour, or 60 minutes, so the reduced speed adds one hour.

Q6. A boat travels 8484 km downstream in 3.53.5 hours. Its effective downstream rate is constant. A student calculates the rate as 29.429.4 km/h. What error did the student make?

A.The student multiplied distance by time.

B.The student added distance and time.

C.The student divided time by distance.

D.The student used an incorrect division operation. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The correct calculation is $r=D/t=84/3.5=24$ km/h. The value 29.4 results from an incorrect computation rather than from the distance-rate-time relationship. Correct units also provide a useful check on the calculation.

Q7. A taxi charges no distance fee during the first 55 km and then continues at a constant travel rate of 5050 km/h. If the taxi has traveled for 3636 minutes, what total distance has it covered, assuming the stated travel rate applies throughout the trip?

A.25 km

B.30 km ✅

C.35 km

D.40 km

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Convert 36 minutes to $0.6$ hour. Then $D=rt=50(0.6)=30$ km. The pricing condition is irrelevant to physical distance because the question asks how far the taxi travels, not how much the passenger pays.

Q8. A distance-versus-time graph is a straight line passing through $(0,0)$ and $(4,240)$ , where time is measured in hours and distance in kilometers. What does the slope represent?

A.The travel time, 4 hours

B.The distance traveled, 240 km

C.The constant rate, 60 km/h ✅

D.The total distance and time combined

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a distance-versus-time graph, slope is calculated as change in distance divided by change in time. Here the slope is $240/4=60$ km/h, so it represents the constant travel rate.

Q9. A distance-time graph rises from 00 km at 00 hours to 150150 km at 33 hours, then becomes horizontal from 33 to 44 hours. What most reasonably happened during the horizontal segment?

A.The object traveled faster.

B.The object continued at 50 km/h.

C.The object traveled backward.

D.The object stopped moving. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: A horizontal distance-time graph means distance does not change as time passes. Therefore the object covered no additional distance during that interval, indicating that it stopped rather than moving forward or backward.

Q10. A student claims that traveling 100100 km at 5050 km/h and then 100100 km at 100100 km/h gives an overall average rate of 7575 km/h. Which evaluation is most accurate?

A.Correct, because 75 is the average of the two rates.

B.Correct, because both distances are equal.

C.Incorrect, because the average rate is 66.7 km/h. ✅

D.Incorrect, because the average rate is 50 km/h.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The first 100-km segment takes 2 hours and the second takes 1 hour. Total distance is 200 km and total time is 3 hours, so the overall average rate is $200/3\approx66.7$ km/h. Averaging rates directly is inappropriate here.

Q11. A drone travels 9090 km in 1.51.5 hours. Because wind conditions change, its rate during the next trip is $20\%$ lower, while the distance remains 9090 km. What is the new travel time?

A.1.60 hours

B.1.75 hours

C.1.875 hours ✅

D.2.25 hours

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The original rate is $90/1.5=60$ km/h. A $20\%$ reduction gives $60(0.8)=48$ km/h. For the same 90-km distance, the new time is $90/48=1.875$ hours, or 1 hour 52.5 minutes.

Q12. A runner must cover 2020 km in at most 2.52.5 hours. During the first 0.50.5 hour, the runner averages 66 km/h. What minimum constant rate is needed for the remaining 1717 km to finish within the target time?

A.7.5 km/h

B.8 km/h

C.8.5 km/h ✅

D.9 km/h

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: During the first 0.5 hour, the runner covers $6(0.5)=3$ km. The remaining distance is 17 km, and only 2 hours remain. Therefore the required rate is $17/2=8.5$ km/h, so option C is correct.

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