On his way from suburb A to surburb B, the bus driver kept to the speed of 44km/h. When he arrived at B, he found out he needed to hurry to be back in A on time. So he drove the bus at a constant speed of 66km/h on the way home along the same route. What speed did he need to keep constant during the whole trip so that it would take the same amount of time.
The answer might be 55, but my tutor says it might not be, so I'm confused.
2007-06-29 18:54:01 · 6 answers · asked by Elder Price 2 in Science & Mathematics ➔ Mathematics
Let d = distance from A to B,
d / 44 = time to travel from A to B, and
d / 66 = time to travel from B to A.
average speed for round trip
= (total distance) / (total time )
= 2d / [ d / 44 + d / 66 ]
The least common denominator of this complex fraction is 132. Multiply every term by 132.
= 264d / [ 3d + 2d]
= 264d / [5d]
= 52.8
Answer: 52.8 m/h
If 2 parts of a trip are equal in distance and traveled at different speeds (x and y), then the average speed for the whole trip is the harmonic mean of the speeds: 2 / [ 1/x + 1/y ].
In this problem, 2 / [1/44 + 1/66] = 52.8
2007-06-29 19:20:05 · answer #1 · answered by mathjoe 3 · 0⤊ 0⤋
No, those are not the answer.
You have an unknown distance d between the suburbs, and are told that he drove the distance at two different speeds. You have to find the total time, and knowing that you will find a speed that could have been used in a round-trip to take the same amount of time, but it will NOT be the average of the two speeds. Just don't worry about d, keep it unknown. It turns out it will cancel out.
A to B:
time = distance / speed = d/44
B to A:
time = d/66 [if the denominator is larger, obviously it's a shorter time]
So the total time for round trip: d/44 + d/66 . Take the cross-products:
= (66d + 44d)/(66*44) = 110*d/2904 .
Looking at this total time, you want to find the speed needed to cover a round trip, a distance of 2d, in the SAME total time.
distance = speed * time
2d = x * (110*d/2904)
2 = x (110/2904)
therefore x = 2 * 2904/110 = 52.8 km/h . [an answer that's true no matter what d is. Test it out with any d of your choice.]
There's a similar kind of problem where you're told the bus-driver usually makes an average speed but today was too slow, and has to speed up on the trip back to make a total time. If the speed is 1/2 the usual speed or less, then no matter how fast he manages to make the return journey he still can't complete the round trip in the usual amount of time, he will always be late since all the time already elapsed on the trip out.
2007-06-30 02:02:52 · answer #2 · answered by PIERRE S 4 · 0⤊ 0⤋
I would guess 50. 44+66=100 The average of the two is 50. That's my answer
2007-06-30 01:57:25 · answer #3 · answered by Little Miss confused 4 · 0⤊ 2⤋
well 44 + 66 is 110, which divided by 2 (for the average) is 55, so I think you're right :)
2007-06-30 02:17:41 · answer #4 · answered by Anonymous · 0⤊ 1⤋
I am almost certain that it is 55km/h. Your tutor could just be saying that so you could think about it more and think of other answers.
I got 55 because it is the middle number. I couldn't think of anything else it could possible be.
2007-06-30 02:01:35 · answer #5 · answered by The Wise One 2 · 0⤊ 1⤋
its 55.
44+66=110
110/2=55
2007-06-30 02:02:13 · answer #6 · answered by kagome in blue 3 · 0⤊ 1⤋