Anonymous Wrote:M travelled X miles/hr the 1st y miles & 1.25X for the last 40-y miles. The time M took to travel 40 miles was what % of the time taken her if travelled at an average of X miles for the entire trip?
A: X=48
B: y=20
you can't use t and t + 1, because you don't know that there is a one-hour difference between the times. (that '1' would have to be a literal '1': i.e., it would have to stand for 1 hour.)
rhetorical question: are you in the 9-session class? if so, did you mistakenly assume that the 't' and 't + 1' from the rate problem in session 5 apply to
all time differentials? they don't: the '1' in that problem specifically comes from the one-hour time difference between the two parts.
(1) only:
insufficient, because you don't know the lengths of time for which M traveled at the two different speeds.
consider the extreme cases:
if y is 0.0001 or so, then M traveled at 1.25x = 60 mph for virtually the whole trip. therefore, her time will be very close to 80% of what it would have been at x = 48 mph.
if y is 39.999 or so, then M traveled at 48 mph for virtually the whole trip. therefore, her time will be very close to 100% of what it would have been at x = 48 mph.
(2) only:
if y = 20, then M traveled 20 miles at 1.25x mph and the other 20 miles at x mph.
solve d = rt for time: t = d/r. therefore, the total time is 20/(1.25x) + 20/x = 16/x + 20/x = 36/x hours.
the time she'd have taken at a constant speed of x mph is 48/x hours.
we know that the percentage difference can be calculated:
(36/x) / (48/x) = 36/48 = 75%
(notice that there's no need to actually calculate this, unless you don't otherwise see that the x's cancel)
sufficient
ans = b