Question 71973: A consultant traveled 4 hours to attend a meeting. The return trip took only 3 hours because the speed was 7 miles per hour faster. What was the consultant's speed each way?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Use the equation:
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D = R*T
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Where D represents distance, R represents rate or speed, and T represents Time.
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Let's say the unknown Rate while going to the conference is represented by S (standing
for speed). And the problem tells us that it takes 4 hours to get there. Substituting
these values into the equation tells us that the distance (D) is equal to S times 4 hours
or in more conventional form our first equation becomes:
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D = 4*S
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For the trip back home the rate is 7 miles per hour faster which can be written as (S + 7).
And this trip only requires 3 hours of driving time. Substituting these values into the
distance equation results in a second equation of:
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D = (S + 7)* 3
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and after doing the multiplication this simplifies to:
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D = 3S + 21
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But the two distances (going and coming back) are the same. Therefore, we can set the
right side of these two distance equations equal to get:
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4S = 3S + 21
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Subtracting 3S from both sides to get rid of the 3S on the right side gives us:
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S = 21
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So the consultant drives at a rate of 21 miles per hour while going to the meeting.
[And since he drives for 4 hours, he travels 4*21 = 84 miles to get there.]
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On the way back the consultant drives 7 miles per hour faster or 21+7 = 28 miles per hour.
[Since he drives back in 3 hours, the distance he travels is 28 * 3 = 84 miles back.]
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Note that the problem checks out. I hope he's not driving on the Interstate Highway
system because at these rates he's a traffic hazard (unless he's driving in a blizzard)!
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Hope this explanation helps you to understand how to view word problems involving
the factors of distance, rate, and time.
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