SOLUTION: Jay traveled from his house to his office at an average speed of 25mph. by traveling 5mph faster, he took 30 minutes less to return home. How far is Jay's office from his house

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Question 1203653: Jay traveled from his house to his office at an average speed of 25mph. by traveling 5mph faster, he took 30 minutes less to return home. How far is Jay's office from his house
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
Jay traveled from his house to his office at an average speed of 25mph.
by traveling 5mph faster, he took 30 minutes less to return home.
How far is Jay's office from his house
~~~~~~~~~~~~~~~~~~~~~~ 

Let d be the distance from the house to the office, in miles.

Traveling at 25 mph, Jay spends d%2F25  hours.

Traveling back at the speed 25 mph + 5 mph = 30 mph, he spends d%2F30  hours.


The difference is 30 minutes, or  1%2F2 of an hour;

so we write this "time equation"

    d%2F25 - d%2F30 = 1%2F2  of an hour.


To solve it, multiply all the terms by 150.  you will get then

    6d - 5d = %281%2F2%29%2A150

       d    =   75.


ANSWER.  The distance from the house to the office is 75 miles.


CHECK.   Time to the office  75%2F25 = 3 hours.

         Time returning back is  75%2F30 = 2.5 hours.

         The difference is  3 - 2.5 = 0.5 of an hour, correct.

Solved.



Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
                         SPEED         TIME hours    DISTANCE each way, miles

House to Office            25           L/25           L

Office to House            25+5=30      L/30           L

DIFFERENCE                              1/2

highlight_green%28L%2F25-L%2F30=1%2F2%29
l.c.d., 2*5*5*3

%282%2A3%2A5%2A5%29%28L%2F25-L%2F30%29=%282%2A3%2A5%2A5%29%2F2
6L-5L=75
highlight%28L=75%29

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


You have received two responses showing a standard algebraic method for solving the problem.

Here is a solution using a very different method.

The two speeds are 25mph and 30mph, a ratio of 5:6.

Since the distances are the same, the ratio of times at the two speeds is 6:5.

The time at the higher speed is 1/2 hour less than the time at the lower speed.

x = hours at lower speed
x-1/2 = hours at higher speed

The ratio is 6:5

x%2F%28x-%281%2F2%29%29=6%2F5
6%28x-%281%2F2%29%29=5x
6x-3=5x
x=3

The time spent at the lower speed was 3 hours. Since the lower speed was 25mph, the distance was 3*25 = 75 miles.

ANSWER: 75 miles