All entries for Tuesday 25 October 2011
October 25, 2011
Below is a statistics question I was given in subject studies, followed by how I approached it and my solution.
(9) a) Yesterday I drove 100 miles. My average speed for the first 50 miles was 40mph; my average speed for the last 50 miles was 60mph. What was my average speed for the whole trip?
The temptation here is to immediately find the average of the 2 speeds, however we are spending 2 different amounts of time travelling at these speeds, and so this approach does not work.
Instead we must find how long we have spent travelling the two sections of the trip, then hence the trip as a whole and divide total distance by total time, thusly:
T= 50/40 + 50/60 = 25/12 hours. D= 100 miles
S= D/T .'. S= 100 / (25/12) = 48mph.
b) Here's a related question: Quinten and Xavier have a cycle race. Quinten travels at a constant 15mph. Xavier travels at 10mph uphill, 15mph on level ground and 20mph downhill. The course is evenly divided between uphill, level and downhill sections. Does Quinten win, does Xavier win or do they cross the finishing line at the same time?
We do not need to know exact timings for the course of the race. It is enough to know that the course is evenly divided by distance (not by time).
If we use our s=d/t triangle, Quinten's time taken is d/15.
Xavier's time taken will be 1/3 (d/10 + d/15 + d/20) = 13d/180
d/15 < 13d/180 for all d, therefore Quinten is quicker than Xavier.
This can be seen because Xavier spends longer going slower than Quinten than going quicker, despite the distances being equal, and the speeds being equidistant from Quinten's average.
If we wanted to give a general formula for the overall average speed of a journey broken up into k sections of equal distance, where we travel an average of s1 mph in section 1, s2 mph in section to up to sk mph in section k, we could write this as:
t= Σ[d/(k*sa)] Where the summation is between a=1 and a=k. Hence as s= d/t, ST= d/tT (T= total).