Total distance = x
75%x/50 = T1
25%x/S = T2
T1+T2 = 75%x/50 + 25%x/S = (12Sx + 200x)/800S
X/(T1+T2) = 40
x / (12Sx + 200x)/800S = 40
800Sx = 480Sx + 8000x
320Sx = 8000x
cancel x from both sides
S = 8000/320 = 25
OA?
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Average Speed
Source: Beat The GMAT — Problem Solving |
Here is another method: Pluggin In number
Let total distance = 100
75/50 = T1
25/S = T2
T1 + T2 = (75S + 1250)/50S
Avg. Speed = Total Distance/ Total time
40 = 100 /(75S + 1250)/50S
40 = 5000S / (75S +1250)
3000S + 50000 = 5000S
2000S = 50000
S = 25
Hope this helps.
Let total distance = 100
75/50 = T1
25/S = T2
T1 + T2 = (75S + 1250)/50S
Avg. Speed = Total Distance/ Total time
40 = 100 /(75S + 1250)/50S
40 = 5000S / (75S +1250)
3000S + 50000 = 5000S
2000S = 50000
S = 25
Hope this helps.
No rest for the Wicked....
Yup, it is C....
I am surprised nobody made the mistake of taking average speed to be average speed over distance (it is actually over time)......I made that mistake when i first solved this sum
I am surprised nobody made the mistake of taking average speed to be average speed over distance (it is actually over time)......I made that mistake when i first solved this sum

















