SOLUTION: Normally it takes John 2.5 hours to drive from his house to his office. Due to yesterday's snowy weather he was driving 10 mph slower than usually, and it took 3 hours. How many mi

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Question 1134427: Normally it takes John 2.5 hours to drive from his house to his office. Due to yesterday's snowy weather he was driving 10 mph slower than usually, and it took 3 hours. How many miles is the office away from John's house? Thank you for solving this problem! :)
Found 4 solutions by ankor@dixie-net.com, josgarithmetic, MathTherapy, greenestamps:
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
Normally it takes John 2.5 hours to drive from his house to his office.
Due to yesterday's snowy weather he was driving 10 mph slower than usually, and it took 3 hours.
How many miles is the office away from John's house?
:
let s = speed normally
then
(s-10) = speed in the snow
the distance of both trips are the same,
write a distance question: dist = speed * time
3(s-10) = 2.5s
3s - 30 = 2.5s
3s - 2.5s = 30
.5s = 30/.5
s = 60 mph
:
Find the distance
2.5 * 60 = 150 mi
confirm the dist using the slower speed
3 * 50 = 150 mi

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
                    SPEED         TIME        DISTANCE

NORMALLY            r             2.5          d

SNOWY WEATHER       r-10          3            d

Question asks for d.




-






Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Normally it takes John 2.5 hours to drive from his house to his office. Due to yesterday's snowy weather he was driving 10 mph slower than usually, and it took 3 hours. How many miles is the office away from John's house? Thank you for solving this problem! :)
Let distance be D
Then we get the following SPEED equation:
6D = 5D + 150 ----- Multiplying by LCD, 15
6D - 5D = 150
Distance, or
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


You have received, so far, three responses to your question, showing similar algebraic solutions. Here is another quite different method that I like to use on this kind of problem.

The ratio of times required is 2.5:3, or 5:6.
Since the distances are the same, that means the ratio of the speeds is 6:5.
If the ratio of the two speeds is 6:5 and the difference between the two speeds is 10mph, then the two speeds are 50mph and 60mph.

The distance is then rate times time: 2.5 hours at 60mph = 150 miles; or 3 hours at 50mph = 150 miles.

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