Question 955979: Michael drove to work on Monday at 45 mph and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late. How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
Found 4 solutions by lwsshak3, ikleyn, n2, josgarithmetic:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Michael drove to work on Monday at 45 mph and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late. How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
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min=1/60 hr
d=distance
s=speed
let x=0n-time travel time
distance/speed=travel time
d/45=x-(1/60)
d/40=x+(1/60)
45(x-(1/60)=40(x+(1/60)
45x-45/60=40x+40/60
5x=95/60
x=95/300 hr or 19 min (on-time travel time)
d=45(x-(1/60)=45(95/300-5/300)=45(90/300)=13.5 mi
d/s=95/300-5/60
13.5/s=95/300-25/300=70/300
s=13.5/(70/300)≈57.86 mph
How far does Michael live from work? 13.5 mi
What speed does he need to drive to arrive 5 minutes early? 57.86 mph
Answer by ikleyn(53937)  (Show Source):
You can put this solution on YOUR website! .
Michael drove to work on Monday at 45 mph and arrived one minute early.
The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late.
How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
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The solution by @lwsshar3 is incorrect: it has an arithmetic error, which ruins everything and leads to wrong answer.
I don't like the way how @lwsshar3 construct his solution.
Actually, this problem is one of standard (and beautiful) problems on Travel and Distance,
and it has a standard solution, which is equally beautiful as the problem itself is.
So, anyone involved with this or a similar task. should know its standard method of solution.
It is why I came to help.
Let 'd' be the distance from home to work (the same as the returning distance), in miles.
The time Michael spent on Monday was  hours (the distance divided by the average speed).
The time Michael spent on Tuesday was  hours (the distance divided by the average speed).
These two time intervals differ in 1 + 1 = 2 minutes.
So, we write this "time equation"
- =  .
To solve, multiply both sides by 40*45. You will get
45d - 40d = 
5d = 60
d = 60/5 = 12.
So. the distance from home to work is 12 miles.
Thus first question is answered.
To answer second question, calculate the travel time on Monday.
It is  of a hour, which is the same as  of an hour, or 16 minutes.
+----------------------------------------------------------+
| So, if Michaels leaves his home at the same time |
| every day (as the problem states), |
| he has 16+1 = 17 minutes to get the office on time. |
+----------------------------------------------------------+
To get the office 5 minutes earlier the control time, the travel time should be 17-5 = 12 minutes, or  of an hour.
Hence, to get the office 5 minutes earlier the control time, his average speed should be  = 60 miles per hour.
It is the answer to the second question.
At this point, the problem is solved completely.
I think you agree with me that this beautiful Travel and Distance problem deserves a perfect solution.
Answer by n2(91) (Show Source):
You can put this solution on YOUR website! .
Michael drove to work on Monday at 45 mph and arrived one minute early.
The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late.
How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let 'd' be the distance from home to work (the same as the returning distance), in miles.
The time Michael spent on Monday was hours (the distance divided by the average speed).
The time Michael spent on Tuesday was hours (the distance divided by the average speed).
These two time intervals differ in 1 + 1 = 2 minutes.
So, we write this "time equation"
- = .
To solve, multiply both sides by 40*45. You will get
45d - 40d =
5d = 60
d = 60/5 = 12.
So. the distance from home to work is 12 miles.
Thus first question is answered.
To answer second question, calculate the travel time on Monday.
It is of a hour, which is the same as of an hour, or 16 minutes.
+----------------------------------------------------------+
| So, if Michaels leaves his home at the same time |
| every day (as the problem states), |
| he has 16+1 = 17 minutes to get the office on time. |
+----------------------------------------------------------+
To get the office 5 minutes earlier the control time, the travel time should be 17-5 = 12 minutes, or of an hour.
Hence, to get the office 5 minutes earlier the control time, his average speed should be = 60 miles per hour.
It is the answer to the second question.
At this point, the problem is solved completely.
Answer by josgarithmetic(39835) (Show Source):
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