BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Bicyclist's round trip

Expert replies
by bhumika.k.shah » Sun Jan 31, 2010 4:14 am
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 :(
Join the discussion
Source: — Problem Solving |

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)
Join the discussion

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
Join the discussion

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
Join the discussion

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
Join the discussion

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
Join the discussion

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]
Join the discussion

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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
Join the discussion