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Difficulty:
85%
(hard)
Question Stats:
30% (01:30) correct
70%
(01:41)
wrong
based on 67
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Lucas drove from Warehouse Orion to Warehouse Vega and then returned to Warehouse Orion along the same highway. Was Lucas’s average (arithmetic mean) speed for the entire round trip at least 90 kilometers per hour?
(1) The distance from Warehouse Orion to Warehouse Vega is 210 kilometers.
(2) Lucas’s average (arithmetic mean) speed from Warehouse Orion to Warehouse Vega was 45 kilometers per hour.
(1) The distance from Warehouse Orion to Warehouse Vega is 210 kilometers.
(2) Lucas’s average (arithmetic mean) speed from Warehouse Orion to Warehouse Vega was 45 kilometers per hour.
M47-65
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13 Mar 2026, 04:00
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GMAT Club Official Solution:
Lucas drove from Warehouse Orion to Warehouse Vega and then returned to Warehouse Orion along the same highway. Was Lucas’s average (arithmetic mean) speed for the entire round trip at least 90 kilometers per hour?
(1) The distance from Warehouse Orion to Warehouse Vega is 210 kilometers.
Knowing only the one-way distance does not determine the total driving time for the round trip, so the overall average speed cannot be determined. Not sufficient.
(2) Lucas’s average (arithmetic mean) speed from Warehouse Orion to Warehouse Vega was 45 kilometers per hour.
Let the one-way distance be d kilometers, and let the return speed be r kilometers per hour.
Total distance for the round trip is 2d.
Total time for the round trip is d/45 + d/r.
So the overall average speed is
2d/(d/45 + d/r) = 2/(1/45 + 1/r).
So, the question becomes:
Is 2/(1/45 + 1/r) ≥ 90?
Is 2/90 ≥ 1/45 + 1/r?
Is 0 ≥ 1/r?
Is r ≤ 0?
But r is a speed for the return trip, so r must be greater than 0. Therefore, the answer to the question is No. Sufficient.
Answer: B.
Lucas drove from Warehouse Orion to Warehouse Vega and then returned to Warehouse Orion along the same highway. Was Lucas’s average (arithmetic mean) speed for the entire round trip at least 90 kilometers per hour?
(1) The distance from Warehouse Orion to Warehouse Vega is 210 kilometers.
Knowing only the one-way distance does not determine the total driving time for the round trip, so the overall average speed cannot be determined. Not sufficient.
(2) Lucas’s average (arithmetic mean) speed from Warehouse Orion to Warehouse Vega was 45 kilometers per hour.
Let the one-way distance be d kilometers, and let the return speed be r kilometers per hour.
Total distance for the round trip is 2d.
Total time for the round trip is d/45 + d/r.
So the overall average speed is
2d/(d/45 + d/r) = 2/(1/45 + 1/r).
So, the question becomes:
Is 2/(1/45 + 1/r) ≥ 90?
Is 2/90 ≥ 1/45 + 1/r?
Is 0 ≥ 1/r?
Is r ≤ 0?
But r is a speed for the return trip, so r must be greater than 0. Therefore, the answer to the question is No. Sufficient.
Answer: B.
General Discussion
05 Mar 2026, 08:20
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Average speed asked.
Whenever its a question of avg speed, the two most important categories are:
1) What is the speed
2) How long does it travel WITH that speed
These two factors determine the avg speed
Statement 1:
210 kilimeters
Well alright we do have the distance but we dont have anything regarding time whatsovever.
INSUFFICIENT
Statement 2:
Now, we have the avg speed itself!
For equal distances formula is actually independent of distance itself:
2v1v2/v1+v2
Even if one had to derive and intuitively solve
Avg speed = total dist/total time
= 2D/(D/v1+D/v2) The distance gets cancelled through as going and coming back is the same distance. It will be independent as long as distance are some multiple of some rational number.
Therefore 2v1v2/v1+v2 = 2*45*v2/45+v2
= 90v2/45+v2
Now is the actual fun part.
The above expression CANNOT be greater than 90 itself.
90(v2/45+v2) is basically 90*[value smaller than 1]
Hence SUFFICIENT.
Answer: Option B
Whenever its a question of avg speed, the two most important categories are:
1) What is the speed
2) How long does it travel WITH that speed
These two factors determine the avg speed
Statement 1:
210 kilimeters
Well alright we do have the distance but we dont have anything regarding time whatsovever.
INSUFFICIENT
Statement 2:
Now, we have the avg speed itself!
For equal distances formula is actually independent of distance itself:
2v1v2/v1+v2
Even if one had to derive and intuitively solve
Avg speed = total dist/total time
= 2D/(D/v1+D/v2) The distance gets cancelled through as going and coming back is the same distance. It will be independent as long as distance are some multiple of some rational number.
Therefore 2v1v2/v1+v2 = 2*45*v2/45+v2
= 90v2/45+v2
Now is the actual fun part.
The above expression CANNOT be greater than 90 itself.
90(v2/45+v2) is basically 90*[value smaller than 1]
Hence SUFFICIENT.
Answer: Option B
Bunuel
