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Graham walked to school at an average speed of 3 miles an hour and jogged back along the same route at 5 miles an hour. If his total traveling time was 1 hour, what was the total number of miles in the round trip? a) 3 b) 3 and 1/8 c) 3 and 3/4 d)4 e) 5 please help me with the answer and explanation (show me how to answer it!!!) SAT IS IN LESS THAN A MONTH!!! XD
Wrong. It's, distance = rate * time (d = rt)
total time (1 hour) = t1 + t2
t1 = time spent walking to school t2 = time spent jogging home from school
t1 = d1/r1, r1 = 3 mph t2 = d2/r2, r2 = 5 mph
t1 + t2 = d1/3 + d2/5 = 1 hour d1 = d2 (The same route was traveled, so both distances are equal. d = length (or distance) of route)
(1/3 + 1/5)d = 8/15d = 1 hour
d = 15/8 = length of route one way 2d = 30/8 = total miles traveled in round trip = 3.75 miles
distance = rate * time (d = rt)
total time (1 hour) = t1 + t2
t1 = time spent walking to school t2 = time spent jogging home from school
t1 = d1/r1, r1 = 3 mph t2 = d2/r2, r2 = 5 mph
t1 + t2 = d1/3 + d2/5 = 1 hour d1 = d2 (The same route was traveled, so both distances are equal. d = length (or distance) of route)
(1/3 + 1/5)d = 8/15d = 1 hour
d = 15/8 = length of route one way 2d = 30/8 = total miles traveled in round trip = 3.75 miles
Just for anyone who finds this later on, I want to point out that the answer from mamak is really, really wrong.
really. Notice that the question says 3 miles per hour one way and 5 miles per hour the other way, each being the same distance. Yet mamak's answer assumes that going different the same distance at different speeds will take the same amount of time. That would kind of defeat the purpose of speed yeah?
The correct answer is 3 and 3/4.
The total time for one way is 3 you divide that by half. The other half of the way he did half of 5. Add 1 1/2 (half of 3 miles per hour) plus 2 1/2 (half of the 5 miles per hour). Equals 4 miles
thx for the answer but a little late XD no worries I didnt get anything similar!
ya thanks i sort of figured that out later while studying physics.
Thanks!!
