SOLUTION: a private plane traveled from seattle to a rugged wilderness at an average of 132 mph. On the return trip, the average speed was 231 mph. If the total traveling time was 6 hours, h

Question 1157210: a private plane traveled from seattle to a rugged wilderness at an average of 132 mph. On the return trip, the average speed was 231 mph. If the total traveling time was 6 hours, how far is Seattle from the wideness?
Found 4 solutions by josgarithmetic, josmiceli, greenestamps, MathTherapy:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
Both travel time sum to 6 hours.

Solve for d.

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Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = the distance between Seattle & the wilderness
Let = the time for the 1st trip in hrs

and

Seattle is 504 mi from wilderness
----------------------
check:

error due to rounding off
OK

Answer by greenestamps(13215)   (Show Source): You can put this solution on YOUR website!

The distances are the same, and the ratio of the speeds is 132:231 = 4:7. That means the ratio of times is 7:4.

So 4/11 of the 6 hours was spent at the higher speed. The distance is then

Or of course you could calculate the distance as 7/11 of the total 6 hours at 132mph.

Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!

a private plane traveled from seattle to a rugged wilderness at an average of 132 mph. On the return trip, the average speed was 231 mph. If the total traveling time was 6 hours, how far is Seattle from the wideness?

Let distance be D
Then we get the following TIME equation:
7D + 4D = 5,544 ------ Multiplying by LCD, 924
11D = 5,544
Distance, or
NOTHING else to this!!
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