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Bicyclist's round trip

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Bicyclist's round trip

by bhumika.k.shah » Sun Jan 31, 2010 4:14 am

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A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?
1) 4 4/5
2)5
3)5 1/5
4)5 2/5
5)5 3/5

I am terrible at this topic :(
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by sars72 » Sun Jan 31, 2010 5:25 am
bhumika.k.shah wrote:A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?
1) 4 4/5
2)5
3)5 1/5
4)5 2/5
5)5 3/5

I am terrible at this topic :(
we can directly employ the harmonic mean formula, but i believe the follwing will give you a better understanding of the problem:

average speed = total distance travelled/ total time taken

uphill
speed = 4mph
time = 2 hrs
-> distance = 4*2 = 8 miles

downhill
distance will remain the same = 8 miles
speed = 6mph
-> time = distance/ speed = 8/6 = 4/3 hrs

so, total distance travelled = 8+8 = 16 miles
total time taken = 2 + 4/3 = 10/3 hours

-> average speed = 16/(10/3) = 16*3/10 = 48/10 = 4.8

4.8 = 4 4/5 .. hope that helps (and hope this is the correct answer)
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by sunil.hubli » Sun Jan 31, 2010 5:33 am
Distance = 4*2 = 8 miles
Time taken to come down = 8/6 = 4/3

Now avg speed for complete trip = total distance/total time
=> (8+8)/(2+4/3)=>16*3/10=>8*3/5=>24/5=>4 4/5

Hence answer is A
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by papgust » Sun Jan 31, 2010 5:36 am
This is very simple as you just need to apply the equation, Distance = Speed * Time.

Distance is same. The onward journey takes 2 hours at an avg speed of 4 mph. So, distance b/w A and B is (4*2) = 8 miles.

TOTAL TIME for the round trip is d/S1 + d/S2 [d is same AND S1 and S2 are speeds of onward and return journeys].

Simplifying,

(S1*d + S2*d)/ S1*S2 = d (S1+S2) / S1*S2

Sub values of d, S1, S2.

8 (4+6) / 4*6 = 80 / 24 = 3.33 hrs OR 3 1/3 hrs

Now,
Avg Speed = Total Distance/Total Time

Total Distance = 8+8 = 16 miles
Total Time = 3 1/3 hrs

Avg Speed = 16 / (3 1/3) = 4.8 mph or 4 4/5 mph


Alternate Method:

There is a cool shortcut to calculate Avg speed for round trip. If you find it confusing, just ignore this method. I find it very handy and easy to approach instead of framing those equations.

In this prob, Avg Speeds are 4 mph and 6 mph.

1) Express the avg speeds in the form of ratios (s1:s2)
4:6 = 2:3 (Totally 2+3 OR 5 parts) ==> Assume this simplified ratio as r1:r2

2) Divide the (difference of Total Avg Speeds) by ratio parts.
(6-4) / 5 = 2/5

3) Now, s1 + (r1*2/5)
4 + (2*2/5) = 4 + 4/5 = 24/5 = 4.8 mph OR 4 4/5 mph
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by thephoenix » Sun Jan 31, 2010 7:17 am
let dist=d

for A to B
time=2
dis=d
speed=4

dis=time*speed
===> d=2*4=8miles

now for B to A
d=8
s=6
t=8/6=4/3

abg speed= tot dis/tot time

tot dis=2d=16
tot time=2+4/3=10/3
avg sp=16/10/3=48/10=24/5=4 4/5
A
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by bhumika.k.shah » Sun Jan 31, 2010 10:18 am
Sowree papgust found it too confusing!
Since i aint very good @ math i'll choose picking the normal method of solving this sum!
thanks for the shortcut though :-)
papgust wrote:This is very simple as you just need to apply the equation, Distance = Speed * Time.

Distance is same. The onward journey takes 2 hours at an avg speed of 4 mph. So, distance b/w A and B is (4*2) = 8 miles.

TOTAL TIME for the round trip is d/S1 + d/S2 [d is same AND S1 and S2 are speeds of onward and return journeys].

Simplifying,

(S1*d + S2*d)/ S1*S2 = d (S1+S2) / S1*S2

Sub values of d, S1, S2.

8 (4+6) / 4*6 = 80 / 24 = 3.33 hrs OR 3 1/3 hrs

Now,
Avg Speed = Total Distance/Total Time

Total Distance = 8+8 = 16 miles
Total Time = 3 1/3 hrs

Avg Speed = 16 / (3 1/3) = 4.8 mph or 4 4/5 mph


Alternate Method:

There is a cool shortcut to calculate Avg speed for round trip. If you find it confusing, just ignore this method. I find it very handy and easy to approach instead of framing those equations.

In this prob, Avg Speeds are 4 mph and 6 mph.

1) Express the avg speeds in the form of ratios (s1:s2)
4:6 = 2:3 (Totally 2+3 OR 5 parts) ==> Assume this simplified ratio as r1:r2

2) Divide the (difference of Total Avg Speeds) by ratio parts.
(6-4) / 5 = 2/5

3) Now, s1 + (r1*2/5)
4 + (2*2/5) = 4 + 4/5 = 24/5 = 4.8 mph OR 4 4/5 mph
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by bhumika.k.shah » Sun Jan 31, 2010 10:19 am
Yes for both!
hope that helps (and hope this is the correct answer)
[/quote][/quote]
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by Jeff@TargetTestPrep » Wed Jul 25, 2018 4:25 pm
bhumika.k.shah wrote:A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist's average speed for the round trip, in miles per hour?
1) 4 4/5
2)5
3)5 1/5
4)5 2/5
5)5 3/5
We can use the average rate formula:

average = total distance/total time

average = 2d/(d/4 + d/6)

average = 2d/(3d/12 + 2d/12)

average = 2d/(5d/12) = 24d/5d = 24/5

Alternate Solution:

We see that the distance from town A to town B is 2 x 4 = 8 miles, and thus the round trip distance is 16 miles. We can use the average rate formula:

average = total distance/total time

average = 16/(8/4 + 8/6)

average = 16/(2 + 8/6)

average = 16/(20/6)

average = 16/(10/3)

average = 48/10 = 24/5 = 4 4/5

Answer: A

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