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.
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 1408:30 AM PDT
-09:30 AM PDT
Most GMAT test-takers are intimidated by the hardest GMAT Verbal questions. In this session, Target Test Prep GMAT instructor Erika Tyler-John, a 100th percentile GMAT scorer, will show you how top scorers break down challenging Verbal questions..
26 Apr 2026, 11:10
Kudos
Bookmarks
The current of a stream runs at 5 km/h. A motor launch goes 25 km upstream and back again to the starting point in 160 minutes. What is the speed of the motor launch in still water?
A. 15 km/h
B. 18 km/h
C. 22 km/h
D. 20 km/h
E. 25 km/h
A. 15 km/h
B. 18 km/h
C. 22 km/h
D. 20 km/h
E. 25 km/h
26 Apr 2026, 11:10
Kudos
Bookmarks
Official Solution:
The current of a stream runs at 5 km/h. A motor launch goes 25 km upstream and back again to the starting point in 160 minutes. What is the speed of the motor launch in still water?
A. 15 km/h
B. 18 km/h
C. 22 km/h
D. 20 km/h
E. 25 km/h
Let the speed in still water be \(v\) km/h.
• When the boat goes upstream, the current opposes the boat’s motion and reduces its effective speed relative to the shore.
• When the boat goes downstream, the current helps the boat and increases its effective speed relative to the shore.
So:
• Upstream speed = \(v - 5\)
• Downstream speed = \(v + 5\)
Total time is 160 minutes = \(\frac{160}{60} = \frac{8}{3}\) hours, so:
\(\frac{25}{v - 5} + \frac{25}{v + 5} = \frac{8}{3}\)
\(\frac{25((v + 5) + (v - 5))}{v^2 - 25} = \frac{8}{3}\)
\(4v^2 - 75v - 100 = 0\)
\((4v + 5)(v - 20) = 0\)
\(v = 20\) discard \(v = -\frac{5}{4}\) since speed cannot be negative.
Answer: D
The current of a stream runs at 5 km/h. A motor launch goes 25 km upstream and back again to the starting point in 160 minutes. What is the speed of the motor launch in still water?
A. 15 km/h
B. 18 km/h
C. 22 km/h
D. 20 km/h
E. 25 km/h
Let the speed in still water be \(v\) km/h.
• When the boat goes upstream, the current opposes the boat’s motion and reduces its effective speed relative to the shore.
• When the boat goes downstream, the current helps the boat and increases its effective speed relative to the shore.
So:
• Upstream speed = \(v - 5\)
• Downstream speed = \(v + 5\)
Total time is 160 minutes = \(\frac{160}{60} = \frac{8}{3}\) hours, so:
\(\frac{25}{v - 5} + \frac{25}{v + 5} = \frac{8}{3}\)
\(\frac{25((v + 5) + (v - 5))}{v^2 - 25} = \frac{8}{3}\)
\(4v^2 - 75v - 100 = 0\)
\((4v + 5)(v - 20) = 0\)
\(v = 20\) discard \(v = -\frac{5}{4}\) since speed cannot be negative.
Answer: D
