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Nora and Felix are riding electric scooters at constant speeds along the same bike path from the same starting plaza to Central Station. Felix leaves first and by the time Nora leaves, Felix is already 14 kilometers ahead. If Nora rides faster than Felix, how long after Nora leaves will Nora be 1 kilometer ahead of Felix?
(1) Nora catches up to Felix 7 minutes after she leaves the plaza.
(2) Nora arrives at Central Station 12 minutes before Felix does.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) Nora catches up to Felix 7 minutes after she leaves the plaza.
(2) Nora arrives at Central Station 12 minutes before Felix does.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
M47-70
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14 Mar 2026, 03:15
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GMAT Club Official Solution:
Nora and Felix are riding electric scooters at constant speeds along the same bike path from the same starting plaza to Central Station. Felix leaves first and by the time Nora leaves, Felix is already 14 kilometers ahead. If Nora rides faster than Felix, how long after Nora leaves will Nora be 1 kilometer ahead of Felix?
(1) Nora catches up to Felix 7 minutes after she leaves the plaza.
Felix is 14 km ahead when Nora leaves. If Nora catches him 7 minutes later, then Nora gains 14 km on Felix in 7 minutes. To gain the additional 1 km needed to be 1 kilometer ahead of Felix, she needs an additional 7/14 = 0.5 minutes. Thus, the total time after Nora leaves before she is 1 kilometer ahead of Felix is 7 + 0.5 = 7.5 minutes. Sufficient.
(2) Nora arrives at Central Station 12 minutes before Felix does.
Knowing only that Nora arrives 12 minutes before Felix does does not determine how long it takes her to become 1 km ahead, since that depends on the total trip distance and the two speeds. Not sufficient.
Answer: A.
Nora and Felix are riding electric scooters at constant speeds along the same bike path from the same starting plaza to Central Station. Felix leaves first and by the time Nora leaves, Felix is already 14 kilometers ahead. If Nora rides faster than Felix, how long after Nora leaves will Nora be 1 kilometer ahead of Felix?
(1) Nora catches up to Felix 7 minutes after she leaves the plaza.
Felix is 14 km ahead when Nora leaves. If Nora catches him 7 minutes later, then Nora gains 14 km on Felix in 7 minutes. To gain the additional 1 km needed to be 1 kilometer ahead of Felix, she needs an additional 7/14 = 0.5 minutes. Thus, the total time after Nora leaves before she is 1 kilometer ahead of Felix is 7 + 0.5 = 7.5 minutes. Sufficient.
(2) Nora arrives at Central Station 12 minutes before Felix does.
Knowing only that Nora arrives 12 minutes before Felix does does not determine how long it takes her to become 1 km ahead, since that depends on the total trip distance and the two speeds. Not sufficient.
Answer: A.
General Discussion
06 Mar 2026, 23:16
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Statement (1) alone is sufficient because it gives the time for Nora to close the 14 km gap, which is 7 minutes. The relative speed is then 14 km / 7 min = 2 km/min. To be 1 km ahead, she needs to gain an additional 15 km, which takes 15 km / 2 km/min = 7.5 minutes.
Statement (2) alone is insufficient because it relates the arrival times but does not provide enough information to determine the relative speed uniquely; multiple combinations of speeds and distances satisfy the condition.
A
Statement (2) alone is insufficient because it relates the arrival times but does not provide enough information to determine the relative speed uniquely; multiple combinations of speeds and distances satisfy the condition.
A
07 Mar 2026, 01:00
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Key Concept: Relative Rate and Gap Problems in Data Sufficiency
This is a classic relative speed question dressed up in DS clothing — and the trap is in Statement 2, which feels sufficient but isn't.
Setting up the framework first:
When Nora leaves, Felix is already 14 km ahead. Since Nora is faster, she's gaining on Felix at a rate of (N − F) km/min, where N and F are their speeds. The question asks: how long until Nora is 1 km ahead of Felix? That means she needs to erase the 14 km gap AND gain an additional 1 km — a total relative gain of 15 km. So the answer = 15 ÷ (N − F). We just need to know the relative speed.
Step 1 — Test Statement 1 alone:
"Nora catches up to Felix 7 minutes after she leaves."
Catching up means she closes the full 14 km gap. So: (N − F) × 7 = 14 → N − F = 2 km/min.
Now we can answer the question: time to be 1 km ahead = 15 ÷ 2 = 7.5 minutes.
Statement 1 is SUFFICIENT.
Step 2 — Test Statement 2 alone:
"Nora arrives at Central Station 12 minutes before Felix does."
Let D = distance from plaza to Central Station. Nora's travel time = D/N. From when Nora leaves, Felix has (D − 14) km remaining, so his time = (D − 14)/F. The condition gives: (D − 14)/F − D/N = 12. This is one equation with three unknowns (D, N, F). You can construct multiple scenarios with different relative speeds that all satisfy this constraint — try D = 100 with one set of speeds, then D = 200 with another. The answer to "how long until Nora is 1 km ahead" changes across scenarios.
Statement 2 is INSUFFICIENT.
Answer: A — Statement 1 alone is sufficient.
Common trap: Statement 2 feels satisfying because "12 minutes earlier" sounds like concrete timing information. But it ties together the total distance, individual speeds, and the head-start — it doesn't isolate the relative speed. On DS, always ask: does this statement uniquely determine what I'm solving for? A time-difference at the destination doesn't pin down how fast Nora is closing in before that point.
Takeaway: In relative speed DS problems, you need the relative rate (speed difference) directly or a way to calculate it in isolation — arrival time differences almost never give you that cleanly.
This is a classic relative speed question dressed up in DS clothing — and the trap is in Statement 2, which feels sufficient but isn't.
Setting up the framework first:
When Nora leaves, Felix is already 14 km ahead. Since Nora is faster, she's gaining on Felix at a rate of (N − F) km/min, where N and F are their speeds. The question asks: how long until Nora is 1 km ahead of Felix? That means she needs to erase the 14 km gap AND gain an additional 1 km — a total relative gain of 15 km. So the answer = 15 ÷ (N − F). We just need to know the relative speed.
Step 1 — Test Statement 1 alone:
"Nora catches up to Felix 7 minutes after she leaves."
Catching up means she closes the full 14 km gap. So: (N − F) × 7 = 14 → N − F = 2 km/min.
Now we can answer the question: time to be 1 km ahead = 15 ÷ 2 = 7.5 minutes.
Statement 1 is SUFFICIENT.
Step 2 — Test Statement 2 alone:
"Nora arrives at Central Station 12 minutes before Felix does."
Let D = distance from plaza to Central Station. Nora's travel time = D/N. From when Nora leaves, Felix has (D − 14) km remaining, so his time = (D − 14)/F. The condition gives: (D − 14)/F − D/N = 12. This is one equation with three unknowns (D, N, F). You can construct multiple scenarios with different relative speeds that all satisfy this constraint — try D = 100 with one set of speeds, then D = 200 with another. The answer to "how long until Nora is 1 km ahead" changes across scenarios.
Statement 2 is INSUFFICIENT.
Answer: A — Statement 1 alone is sufficient.
Common trap: Statement 2 feels satisfying because "12 minutes earlier" sounds like concrete timing information. But it ties together the total distance, individual speeds, and the head-start — it doesn't isolate the relative speed. On DS, always ask: does this statement uniquely determine what I'm solving for? A time-difference at the destination doesn't pin down how fast Nora is closing in before that point.
Takeaway: In relative speed DS problems, you need the relative rate (speed difference) directly or a way to calculate it in isolation — arrival time differences almost never give you that cleanly.
