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Difficulty:
45%
(medium)
Question Stats:
80% (03:00) correct
20%
(02:38)
wrong
based on 138
sessions
History
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A traveler journeys from Marseille to Geneva, covering 180 km by ferry, 360 km by train, and 30 km by horse carriage. The entire journey takes 15 hours. The speed of the train is three times the speed of the horse carriage and 3/2 times the speed of the ferry. What is the speed of the train?
A. 36 km/h
B. 40 km/h
C. 48 km/h
D. 54 km/h
E. 60 km/h
A. 36 km/h
B. 40 km/h
C. 48 km/h
D. 54 km/h
E. 60 km/h
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M47-35
14 Feb 2026, 14:30
Kudos
Bookmarks
Bunuel
GMAT Club Official Solution:
Let the train speed be t km/h.
Then horse carriage speed = t/3, and ferry speed = 2/3 * t.
Total time:
180/((2/3) * t) + 360/t + 30/(t/3) = 15
270/t + 360/t + 90/t = 15
t = 48
Answer: C.
General Discussion
06 Feb 2026, 11:36
Kudos
Bookmarks
This is a classic distance-speed-time problem with linked speeds, and the key is expressing everything in terms of one variable.
Step 1: Define variables using the speed relationships.
Let the train's speed = T km/h.
We're told:
- Train speed = 3 × horse carriage speed → horse carriage speed = T/3
- Train speed = (3/2) × ferry speed → ferry speed = 2T/3
Step 2: Write the time equation.
Time = Distance ÷ Speed, and total time = 15 hours.
Ferry time: 180 ÷ (2T/3) = 180 × 3/(2T) = 270/T
Train time: 360 ÷ T = 360/T
Horse carriage time: 30 ÷ (T/3) = 30 × 3/T = 90/T
Step 3: Solve.
270/T + 360/T + 90/T = 15
720/T = 15
T = 720/15 = 48 km/h
Step 4: Quick sanity check.
Ferry speed = 2(48)/3 = 32 km/h → time = 180/32 = 5.625 hrs
Train speed = 48 km/h → time = 360/48 = 7.5 hrs
Horse carriage speed = 48/3 = 16 km/h → time = 30/16 = 1.875 hrs
Total = 5.625 + 7.5 + 1.875 = 15 hours ✓
Answer: C (48 km/h)
Common trap: Students often get the speed ratios backwards — reading "train is 3/2 times the ferry" as ferry = (3/2) × train instead of train = (3/2) × ferry. Always re-read ratio statements carefully and ask yourself "which one is faster?"
Takeaway: When speeds are given as ratios of each other, pick the variable that sits at the center of all the relationships (here, the train). It keeps every fraction clean and saves you time.
Step 1: Define variables using the speed relationships.
Let the train's speed = T km/h.
We're told:
- Train speed = 3 × horse carriage speed → horse carriage speed = T/3
- Train speed = (3/2) × ferry speed → ferry speed = 2T/3
Step 2: Write the time equation.
Time = Distance ÷ Speed, and total time = 15 hours.
Ferry time: 180 ÷ (2T/3) = 180 × 3/(2T) = 270/T
Train time: 360 ÷ T = 360/T
Horse carriage time: 30 ÷ (T/3) = 30 × 3/T = 90/T
Step 3: Solve.
270/T + 360/T + 90/T = 15
720/T = 15
T = 720/15 = 48 km/h
Step 4: Quick sanity check.
Ferry speed = 2(48)/3 = 32 km/h → time = 180/32 = 5.625 hrs
Train speed = 48 km/h → time = 360/48 = 7.5 hrs
Horse carriage speed = 48/3 = 16 km/h → time = 30/16 = 1.875 hrs
Total = 5.625 + 7.5 + 1.875 = 15 hours ✓
Answer: C (48 km/h)
Common trap: Students often get the speed ratios backwards — reading "train is 3/2 times the ferry" as ferry = (3/2) × train instead of train = (3/2) × ferry. Always re-read ratio statements carefully and ask yourself "which one is faster?"
Takeaway: When speeds are given as ratios of each other, pick the variable that sits at the center of all the relationships (here, the train). It keeps every fraction clean and saves you time.
