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06 Oct 2025, 01:13
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
65% (01:47) correct
35%
(01:37)
wrong
based on 170
sessions
History
Date
Time
Result
Not Attempted Yet
Greg travels from city A to city B along route K and returns along route L. Greg's average speed during the onward journey is 40kmph. What is his average speed for the entire trip?
(1) Greg's average speed for the return journey is 3/4ths that for the onward journey.
(2) Route K is 2/5ths longer than route L.
Source: 4gmat
(1) Greg's average speed for the return journey is 3/4ths that for the onward journey.
(2) Route K is 2/5ths longer than route L.
Source: 4gmat
06 Oct 2025, 02:34
Kudos
Bookmarks
Let’s visualize with a line diagram:
City A -----K-----> City B
City A <----L------ City B
Onward journey (K): speed = 40 kmph
Return journey (L): speed unknown.
We want: overall average speed => (total distance) / (total time).
---------------------------------------------------------------------------
**(1)** Return speed = (3/4) × 40 = 30 kmph.
But route lengths K and L not known (same or different). Cannot compute exact overall average speed.
Not sufficient.
---------------------------------------------------------------------------
(2) K = (1 + 2/5)L = 7/5 L.
So ratio of distances is known.
But return speed unknown. Cannot compute overall average speed.
Not sufficient.
---------------------------------------------------------------------------
(1) + (2):
Let L = d → K = 7/5 d.
Total distance = K + L = (7/5)d + d = (12/5)d.
Time = K/40 + L/30 = (7/5 d)/40 + d/30
= (7d/200) + (d/30).
Take LCM 600: (21d/600 + 20d/600) = 41d/600.
Average speed = total distance / total time
= (12/5 d) ÷ (41d/600)
= (12/5) × (600/41)
= 720/41 ≈ 17.56 kmph.
Unique value.
---------------------------------------------------------------------------
(1) alone → not sufficient
(2) alone → not sufficient
Together → sufficient
Answer: C
City A -----K-----> City B
City A <----L------ City B
Onward journey (K): speed = 40 kmph
Return journey (L): speed unknown.
We want: overall average speed => (total distance) / (total time).
---------------------------------------------------------------------------
**(1)** Return speed = (3/4) × 40 = 30 kmph.
But route lengths K and L not known (same or different). Cannot compute exact overall average speed.
Not sufficient.
---------------------------------------------------------------------------
(2) K = (1 + 2/5)L = 7/5 L.
So ratio of distances is known.
But return speed unknown. Cannot compute overall average speed.
Not sufficient.
---------------------------------------------------------------------------
(1) + (2):
Let L = d → K = 7/5 d.
Total distance = K + L = (7/5)d + d = (12/5)d.
Time = K/40 + L/30 = (7/5 d)/40 + d/30
= (7d/200) + (d/30).
Take LCM 600: (21d/600 + 20d/600) = 41d/600.
Average speed = total distance / total time
= (12/5 d) ÷ (41d/600)
= (12/5) × (600/41)
= 720/41 ≈ 17.56 kmph.
Unique value.
---------------------------------------------------------------------------
(1) alone → not sufficient
(2) alone → not sufficient
Together → sufficient
Answer: C
06 Oct 2025, 11:17
Kudos
Bookmarks
Bunuel
Greg travels from city A to city B via route K, and returns back to city A from route L.
Greg’s average speed from A to B is 40 kmph.
Average speed of entire journey = ?
Average speed = Total distance travelled / Total time taken.
Statement 1:
(1) Greg's average speed for the return journey is 3/4ths that for the onward journey.
return journey speed = (3/4)*40 = 30 Kmph.
we don’t know the distance or time taken for each travel. Hence, insufficient.
Statement 2:
(2) Route K is 2/5ths longer than route L.
let the distance of route K = (7/5) of route L
route K : Route L = 7x : 5x
Hence, insufficient.
combining statements 1 and statements 2, we get
Average speed = (7x + 5x) / (7x/40) + ( 5x/40)
Hence, Sufficient
Option C
