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06 Aug 2026, 03:09
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A delivery cyclist rode from a warehouse through several neighborhoods and then returned to the warehouse. If the entire ride lasted 5 hours, what was the cyclist’s average (arithmetic mean) speed for the entire ride?
(1) The cyclist’s average (arithmetic mean) speed during the first 4 hours was 30 kilometers per hour.
(2) The cyclist’s average (arithmetic mean) speed during the last 4 hours was 35 kilometers per hour.
(1) The cyclist’s average (arithmetic mean) speed during the first 4 hours was 30 kilometers per hour.
(2) The cyclist’s average (arithmetic mean) speed during the last 4 hours was 35 kilometers per hour.
06 Aug 2026, 03:09
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Official Solution:
A delivery cyclist rode from a warehouse through several neighborhoods and then returned to the warehouse. If the entire ride lasted 5 hours, what was the cyclist’s average (arithmetic mean) speed for the entire ride?
The average speed for the entire ride is:
\(\frac{\text{total distance}}{\text{total time}} = \frac{\text{total distance}}{5}\)
So we need the total distance traveled in the 5 hours.
(1) The cyclist’s average (arithmetic mean) speed during the first 4 hours was 30 kilometers per hour.
So the distance traveled during the first 4 hours was \(30 * 4 = 120\) kilometers. But we do not know the distance traveled during the last hour.
Not sufficient.
(2) The cyclist’s average (arithmetic mean) speed during the last 4 hours was 35 kilometers per hour.
So the distance traveled during the last 4 hours was \(35 * 4 = 140\) kilometers. But we do not know the distance traveled during the first hour.
Not sufficient.
(1)+(2) The first 4 hours and the last 4 hours overlap for 3 hours, hours 2 through 4. We do not know the distance traveled during those overlapping hours.
For example, the distances traveled in hours 1 through 5 could be:
30, 30, 30, 30, 50
Then the first 4 hours total 120 kilometers, and the last 4 hours total 140 kilometers. The total distance is 170 kilometers.
But the distances traveled in hours 1 through 5 could also be:
20, 35, 35, 30, 40
Then the first 4 hours total 120 kilometers, and the last 4 hours total 140 kilometers. The total distance is 160 kilometers.
Both cases satisfy both statements, but they give different total distances, so the average speed for the entire ride is not fixed.
Not sufficient.
Answer: E
A delivery cyclist rode from a warehouse through several neighborhoods and then returned to the warehouse. If the entire ride lasted 5 hours, what was the cyclist’s average (arithmetic mean) speed for the entire ride?
The average speed for the entire ride is:
\(\frac{\text{total distance}}{\text{total time}} = \frac{\text{total distance}}{5}\)
So we need the total distance traveled in the 5 hours.
(1) The cyclist’s average (arithmetic mean) speed during the first 4 hours was 30 kilometers per hour.
So the distance traveled during the first 4 hours was \(30 * 4 = 120\) kilometers. But we do not know the distance traveled during the last hour.
Not sufficient.
(2) The cyclist’s average (arithmetic mean) speed during the last 4 hours was 35 kilometers per hour.
So the distance traveled during the last 4 hours was \(35 * 4 = 140\) kilometers. But we do not know the distance traveled during the first hour.
Not sufficient.
(1)+(2) The first 4 hours and the last 4 hours overlap for 3 hours, hours 2 through 4. We do not know the distance traveled during those overlapping hours.
For example, the distances traveled in hours 1 through 5 could be:
30, 30, 30, 30, 50
Then the first 4 hours total 120 kilometers, and the last 4 hours total 140 kilometers. The total distance is 170 kilometers.
But the distances traveled in hours 1 through 5 could also be:
20, 35, 35, 30, 40
Then the first 4 hours total 120 kilometers, and the last 4 hours total 140 kilometers. The total distance is 160 kilometers.
Both cases satisfy both statements, but they give different total distances, so the average speed for the entire ride is not fixed.
Not sufficient.
Answer: E
