Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
26 Mar 2026, 00:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (04:06) correct
62%
(03:48)
wrong
based on 13
sessions
History
Date
Time
Result
Not Attempted Yet
Dimitra runs for 1 hour as follows:
• For the first 20 minutes, she runs at 12 km/h
• For the next 20 minutes, she runs at 6 km/h
• For the final 20 minutes, she runs at 20 km/h
At two different points during her run, Dimitra’s average speed since the start is exactly 10 km/h.
What is the distance she runs between those two points?
A) \(\frac{5}{3}\)
B) 2
C)\( \frac{7}{3}\)
D) 3
E) \(\frac{10}{3}\)
• For the first 20 minutes, she runs at 12 km/h
• For the next 20 minutes, she runs at 6 km/h
• For the final 20 minutes, she runs at 20 km/h
At two different points during her run, Dimitra’s average speed since the start is exactly 10 km/h.
What is the distance she runs between those two points?
A) \(\frac{5}{3}\)
B) 2
C)\( \frac{7}{3}\)
D) 3
E) \(\frac{10}{3}\)
26 Mar 2026, 12:13
Kudos
Bookmarks
The first 20 minutes, Dimitra runs at 12 km/hr. The only time this average could go down to 10 km/hr is in the next section of her run, where she runs at 6 km/hr. We can convert the 20 min time periods to hours (20/60 = 1/3 hrs). Let's determine the amount of time she needs to run at 6 km/hr in order to decrease her average to 10 km/hr, denoted by x.
(12*(1/3)+ 6*x) / (1/3 + x) = 10
We get x = 1/6, meaning she first hits an average of 10 km/hr after running for 1/6 of an hour at 6 km/hr.
Next we see that her average speed must equal (12+6)/2 = 9 km/hr at the end of the first two periods, so we need to figure out how long she will need to run at 20 km/hr in order to raise that average back up to 10 km/hr:
(9*(2/3) + (20*x)) / (2/3 + x) = 10
We get x = 1/15, meaning she hits an average of 10 km/hr the second time after running for 1/15 of an hour at 20 km/hr.
Now we need to find the distance between these two times.
|-----------------------|------------A-----------|---B--------------------|
(6*(1/6)) + (20*(1/15)) = 35/15 = 7/3
Answer choice C
(12*(1/3)+ 6*x) / (1/3 + x) = 10
We get x = 1/6, meaning she first hits an average of 10 km/hr after running for 1/6 of an hour at 6 km/hr.
Next we see that her average speed must equal (12+6)/2 = 9 km/hr at the end of the first two periods, so we need to figure out how long she will need to run at 20 km/hr in order to raise that average back up to 10 km/hr:
(9*(2/3) + (20*x)) / (2/3 + x) = 10
We get x = 1/15, meaning she hits an average of 10 km/hr the second time after running for 1/15 of an hour at 20 km/hr.
Now we need to find the distance between these two times.
|-----------------------|------------A-----------|---B--------------------|
(6*(1/6)) + (20*(1/15)) = 35/15 = 7/3
Answer choice C
