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20 Feb 2026, 00:40
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
53% (02:34) correct
47%
(02:29)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
Airplane Cruising Speed (miles per hour)
A: 480
B: 600
C: 650
D: 520
E: 550
If Vernon and Tracey are two on one of the five airplanes above, which airplane is Vernon on?
(1) The airplane Vernon is on would take less than 17 seconds to travel three miles at its cruising speed.
(2) The cruising speed of the airplane Tracey is on is 20 percent lower than that of the airplane Vernon is on.
A: 480
B: 600
C: 650
D: 520
E: 550
If Vernon and Tracey are two on one of the five airplanes above, which airplane is Vernon on?
(1) The airplane Vernon is on would take less than 17 seconds to travel three miles at its cruising speed.
(2) The cruising speed of the airplane Tracey is on is 20 percent lower than that of the airplane Vernon is on.
20 Feb 2026, 01:05
Kudos
Bookmarks
We have 5 airplanes with speeds: 480, 520, 550, 600, 650 mph.
Statement 1: Vernon's airplane takes less than 17 seconds to travel 3 miles.
Speed = Distance ÷ Time
Let's find the minimum speed needed:
• Time = 17 seconds = 17/3600 hours
• Speed > 3 ÷ (17/3600) = 3 × 3600/17 ≈ 635.3 mph
Now check which airplane exceeds 635.3 mph:
• 480 mph — No
• 520 mph — No
• 550 mph — No
• 600 mph — No
• 650 mph — Yes!
Only ONE airplane (650 mph) works. Vernon must be on airplane C.
Statement 1 is SUFFICIENT.
---
Statement 2: Tracey's speed is 20% lower than Vernon's speed.
This means: Tracey's speed = 0.8 × Vernon's speed
Let's check which pairs satisfy this:
• Case 1: Vernon on 650 mph → Tracey on 650 × 0.8 = 520 mph ✓
• Case 2: Vernon on 600 mph → Tracey on 600 × 0.8 = 480 mph ✓
Two valid scenarios exist! Vernon could be on 650 OR 600.
Statement 2 is NOT SUFFICIENT.
Statement 1: Vernon's airplane takes less than 17 seconds to travel 3 miles.
Speed = Distance ÷ Time
Let's find the minimum speed needed:
• Time = 17 seconds = 17/3600 hours
• Speed > 3 ÷ (17/3600) = 3 × 3600/17 ≈ 635.3 mph
Now check which airplane exceeds 635.3 mph:
• 480 mph — No
• 520 mph — No
• 550 mph — No
• 600 mph — No
• 650 mph — Yes!
Only ONE airplane (650 mph) works. Vernon must be on airplane C.
Statement 1 is SUFFICIENT.
---
Statement 2: Tracey's speed is 20% lower than Vernon's speed.
This means: Tracey's speed = 0.8 × Vernon's speed
Let's check which pairs satisfy this:
• Case 1: Vernon on 650 mph → Tracey on 650 × 0.8 = 520 mph ✓
• Case 2: Vernon on 600 mph → Tracey on 600 × 0.8 = 480 mph ✓
Two valid scenarios exist! Vernon could be on 650 OR 600.
Statement 2 is NOT SUFFICIENT.
kevincan
20 Feb 2026, 11:50
Kudos
Bookmarks
The key concept here is Data Sufficiency with a unit conversion trap. Most people get this wrong not because the math is hard, but because they don't convert seconds to hours before applying the speed formula — and then they either eliminate the wrong airplane or incorrectly evaluate Statement 2.
Setup: Five airplanes with cruising speeds: A=480, B=600, C=650, D=520, E=550 mph.
Statement (1): Vernon's airplane takes less than 17 seconds to travel 3 miles.
This is a Speed = Distance ÷ Time problem, but the trap is the units mismatch — speed is in mph, but time is given in seconds.
Convert 17 seconds to hours: 17 seconds = 17/3600 hours.
For an airplane to travel 3 miles in less than 17 seconds:
Speed > 3 ÷ (17/3600) = 3 × (3600/17) ≈ 635.3 mph
Now check the five speeds against 635.3 mph:
- A (480): No
- B (600): No
- C (650): Yes ✓
- D (520): No
- E (550): No
Only airplane C exceeds 635.3 mph. Vernon must be on airplane C.
Statement (1) is SUFFICIENT.
Statement (2): Tracey's cruising speed is 20% lower than Vernon's.
"20% lower" means Tracey's speed = 0.8 × Vernon's speed. Check which pairs from our list satisfy this:
- 600 × 0.8 = 480 ✓ → Vernon on B, Tracey on A
- 650 × 0.8 = 520 ✓ → Vernon on C, Tracey on D
Two valid pairs exist. We cannot pin down which plane Vernon is on.
Statement (2) is NOT SUFFICIENT.
Answer: A
Common trap: Almost everyone loses Statement (1) because they compute "3 miles ÷ 17 seconds" without converting to mph — you get ~0.18, which means nothing when your speeds are in the hundreds. Always check your units in Distance-Rate-Time problems before plugging in. If speed is in mph, time must be in hours.
Takeaway: In DS Distance-Rate-Time problems, convert all units to the same system before comparing — a seconds-vs-hours mismatch will silently eliminate the right answer.
Setup: Five airplanes with cruising speeds: A=480, B=600, C=650, D=520, E=550 mph.
Statement (1): Vernon's airplane takes less than 17 seconds to travel 3 miles.
This is a Speed = Distance ÷ Time problem, but the trap is the units mismatch — speed is in mph, but time is given in seconds.
Convert 17 seconds to hours: 17 seconds = 17/3600 hours.
For an airplane to travel 3 miles in less than 17 seconds:
Speed > 3 ÷ (17/3600) = 3 × (3600/17) ≈ 635.3 mph
Now check the five speeds against 635.3 mph:
- A (480): No
- B (600): No
- C (650): Yes ✓
- D (520): No
- E (550): No
Only airplane C exceeds 635.3 mph. Vernon must be on airplane C.
Statement (1) is SUFFICIENT.
Statement (2): Tracey's cruising speed is 20% lower than Vernon's.
"20% lower" means Tracey's speed = 0.8 × Vernon's speed. Check which pairs from our list satisfy this:
- 600 × 0.8 = 480 ✓ → Vernon on B, Tracey on A
- 650 × 0.8 = 520 ✓ → Vernon on C, Tracey on D
Two valid pairs exist. We cannot pin down which plane Vernon is on.
Statement (2) is NOT SUFFICIENT.
Answer: A
Common trap: Almost everyone loses Statement (1) because they compute "3 miles ÷ 17 seconds" without converting to mph — you get ~0.18, which means nothing when your speeds are in the hundreds. Always check your units in Distance-Rate-Time problems before plugging in. If speed is in mph, time must be in hours.
Takeaway: In DS Distance-Rate-Time problems, convert all units to the same system before comparing — a seconds-vs-hours mismatch will silently eliminate the right answer.
