Mischa drove from A to B, a distance of 100 miles, at an average speed of 50 miles per hour. SO time taken to go from A to B = 100/50 hrs = 2hrs
Average speed for going to B and coming back from B to A = 2uv/(u+v)
where u = speed from A to B = 50
v = speed from B to A = m
Average Speed = 100m/(50+m)
Total time taken = total distance/ avg speed = 200/(100m/(50+m)) = (100+2m)/m hrs
STAT1
Average speed = (50+m)/2
= 100m/(50+m)
=> 2500 + m^2 + 100m = 200m
=> m^2 - 100m + 2500 = 0
=> m = 50
So, SUFFICIENT
STAT2
If Mischa's avg speed from B to A was 20% slower => New Avg speed = 2*50*(0.8m)/(50+0.8m)
= 80m/(50+0.8m) = 400m/(250+4m)
Total time would be 12.5% greater => new total time = old total time *1.125 = 1.125 *(100+2m)/m ...(1)
New total time = total distance / new avg speed = 200/ (400m/250+4m))
= (250+4m)/2m = 1.125*(100+2m)/m (from ...(1))
=> (250+4m) = 2.25*(100+2m)
=> 25 = 0.5m
=> m = 50
So, SUFFICIENT
So, Answer will be D
Hope it helps!
MonSama
Mischa drove from A to B, a distance of 100 miles, at an average speed of 50 miles per hour, and then back from B to A along the same route at an average speed of m miles per hour. What is the value of m?
(1) Mischa’s average speed for the entire round trip, excluding any time he spent at point B, was \(\frac{m+50}{2}\) miles per hour.
(2) If Mischa’s average speed from B to A had been 20% slower, the total time for the entire round trip, excluding any time Mischa spent at point B, would have been 12.5% greater.