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Difficulty: HardRatios, Rates, and Proportions

A cyclist completes a journey consisting of three distinct segments: an uphill segment, a flat segment, and a downhill segment. The ratio of the distances of the uphill, flat, and downhill segments is 2:3:52 : 3 : 5, respectively. The cyclist's average speed on the flat segment is twice her average speed on the uphill segment, and her average speed on the downhill segment is three times her average speed on the uphill segment. If the cyclist's overall average speed for the entire journey is 3030 miles per hour, what is her average speed, in miles per hour, on the flat segment?

Answer: 31 miles per hour

Answer

31
The correct average speed on the flat segment is 31 miles per hour. Setting up segment distances as 2x2x, 3x3x, and 5x5x (total distance 10x10x) and segment speeds as vv, 2v2v, and 3v3v, the segment times are t1=2xvt_1 = \frac{2x}{v}, t2=3x2vt_2 = \frac{3x}{2v}, and t3=5x3vt_3 = \frac{5x}{3v}. The total travel time is T=31x6vT = \frac{31x}{6v}. Dividing total distance 10x10x by total time TT yields an overall average speed of 60v31=30\frac{60v}{31} = 30. Solving for vv gives v=15.5v = 15.5 miles per hour. Thus, the average speed on the flat segment is 2v=312v = 31 miles per hour.

Step-by-Step Solution

1
Define segment distances using ratio multipliers.
Distances are d1=2xd_1 = 2x, d2=3xd_2 = 3x, and d3=5xd_3 = 5x, giving total distance D=10xD = 10x.
The distances of the three segments are in the ratio 2:3:52 : 3 : 5.
2
Express segment speeds relative to the uphill speed vv.
Uphill speed is vv, flat speed is 2v2v, and downhill speed is 3v3v.
The problem states flat speed is twice uphill speed, and downhill speed is three times uphill speed.
3
Calculate the time spent on each segment.
t1=2xvt_1 = \frac{2x}{v}, t2=3x2vt_2 = \frac{3x}{2v}, and t3=5x3vt_3 = \frac{5x}{3v}.
Time equals distance divided by speed (t=dvt = \frac{d}{v}).
4
Calculate total travel time by summing individual segment times.
T=xv(2+32+53)=31x6vT = \frac{x}{v} \left(2 + \frac{3}{2} + \frac{5}{3}\right) = \frac{31x}{6v}.
Combining fractions with a common denominator of 6 gives 12+9+106=316\frac{12 + 9 + 10}{6} = \frac{31}{6}.
5
Relate total distance and total time to the overall average speed.
Average Speed=Total DistanceTotal Time=10x31x6v=60v31=30\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{10x}{\frac{31x}{6v}} = \frac{60v}{31} = 30.
Overall average speed is defined as total distance divided by total time.
6
Solve for vv and calculate the flat segment speed 2v2v.
v=15.5v = 15.5 mph, so flat segment speed =2(15.5)=31= 2(15.5) = 31 mph.
Solving 60v31=30\frac{60v}{31} = 30 yields v=15.5v = 15.5, making 2v=312v = 31.

Key Concept

Weighted Average Speed and Multi-Segment Distance-Rate-Time Ratios
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Topics

Ratios, Rates, and Proportions

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