Donna K. answered • 04/04/15
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Hi Grant!
So... This is an "average speed" problem. When we hear the word "average," we often think of the method for taking the average of two or more numbers. For example, the average of 10 and 50 would be found by adding up 10 + 50 and then dividing by 2, to get 10+50=60, divided by 2 = 30. BUT ... this problem is about average SPEED, and so you cannot just add 15 and 12 and then divide by two. Why not? Well, let's think about it ... What is a rate of speed? A rate is equal to how far you travel in a certain amount of time. Every rate problem then has 3 elements: "R" which is the speed (or rate) and "D" for Distance and "T" for time. When we say "mph" it means Miles Per Hour. "Miles" is a Distance, "per" means "divide," and an "hour" is a unit of Time. SO ... "R=D/T" is the formula for determining rate. Distance would be the amount of miles traveled going at a certain rate for a certain amount of time, so D=RxT (Distance = Rate x Time) And lastly, Time can be derived if one knows the rate or speed and how far someone drove at that speed, then we would know how long the whole trip took, or Time = Distance divided by Rate or "T=D/R." So to get an average speed for an entire trip, R=D/T, you need to know the ENTIRE distance divided by the ENTIRE time!
OK, now back to the problem at hand!
This is a tricky one! Often in problems such as these, there are two things that are the SAME (in respect to time, OR distance, OR rate.) In this case, the thing that is the same is DISTANCE because the miles from Fritz's house to his work is the SAME as the miles going back home after work. The distance is the same BUT the rates are different and the times therefore would be different as well. So let's just say that his work is 60 miles away (I made this up - the problem does not tell us how far his work is, so I just chose a distance that was evenly divisible by 12 and 15 to make my math easier. It does not matter which distance you choose as long as it is the same for each direction - to work and home from work. You could even just use "D" but using a number here makes it easier to understand.)
Ok, so Fritz usually goes at a speed of 15mph to work. That means the first hour he would have gotten 15 miles closer to work, the second hour he would be 30 miles closer, or halfway, and then the 3rd hour 45 miles, and finally after the 4th hour, he would be 60 miles away from home and arriving at work. So it would take him 4 hours to get to work going 15 miles per hour. Going home at 15mph would then take him another 4 hours for a total of 8 hours round-trip. He will have traveled then exactly 120 miles in 8 hours - 4 hours in the AM and 4 hours in the PM. (I know, that's a long commute! Those were not very realistic numbers I chose, but oh well!) But THIS morning, the problem says, Fritz only could go 12 mph. So how long did it take him to get to work? Well, according to our formula for time, t=D/r, it would be t=60/12, or t=5. So it took him FIVE hours to get to work this morning, whereas it usually takes him only 4 hours! His average time for the whole trip is usually 8 hours to go 120 miles. That would give us a rate of r=D/t or: r=120/8 or 15 mph. (Reduce 120 over 8 to 15 over 1.) Since normally his trip is 8 hours total and he has already used up 5 hours in the morning commute, he only has 3 hours left to match his average daily commute time of 8 hours.
Although he only now has 3 hours left out of the 8, he still has the exact same 60 miles to travel in that reduced time of 3 hours. That means he needs to go at a rate of r=D/t or r=60mi/3hrs or r=60/3 = 20. So his rate in the evening must be 20 mph in order to make up for his slow commute in the morning at 12 mph. And that is your ANSWER - 20 mph!
NOTE: Notice that 20 + 12 = 32 and 32 divided by 2 =16, or the average of the two speeds, which is NOT the CORRECT answer of 20 mph, the average speed of the whole trip. So just remember... to get an average RATE, you need to find the entire distance and the entire time and divide, r=D/t! (The reason that 16 is not the average is because he spends more time going the slower speed than he does the faster speed, so it would need to be a "weighted average." They could have asked the question in a different way: If Fritz spent 3 hours going to work at 20 mph and 4 hours going to work at 15 mph, how far is Fritz's round-trip to work and back home? You would then have D=D equation (because the distance there is the same as the distance back). Since Distance = Rate times Time then you would have this equation: RxT=RxT. Then you could just fill in the two pairs of numbers and get 3 x 20 = 4 x 15 or 60 = 60. Either way, his distance to work and back home = 120.
Hope that helps you to see how to set up any Distance/Rate/Time type of problem!
Kind regards,
Donna K :-)
So... This is an "average speed" problem. When we hear the word "average," we often think of the method for taking the average of two or more numbers. For example, the average of 10 and 50 would be found by adding up 10 + 50 and then dividing by 2, to get 10+50=60, divided by 2 = 30. BUT ... this problem is about average SPEED, and so you cannot just add 15 and 12 and then divide by two. Why not? Well, let's think about it ... What is a rate of speed? A rate is equal to how far you travel in a certain amount of time. Every rate problem then has 3 elements: "R" which is the speed (or rate) and "D" for Distance and "T" for time. When we say "mph" it means Miles Per Hour. "Miles" is a Distance, "per" means "divide," and an "hour" is a unit of Time. SO ... "R=D/T" is the formula for determining rate. Distance would be the amount of miles traveled going at a certain rate for a certain amount of time, so D=RxT (Distance = Rate x Time) And lastly, Time can be derived if one knows the rate or speed and how far someone drove at that speed, then we would know how long the whole trip took, or Time = Distance divided by Rate or "T=D/R." So to get an average speed for an entire trip, R=D/T, you need to know the ENTIRE distance divided by the ENTIRE time!
OK, now back to the problem at hand!
This is a tricky one! Often in problems such as these, there are two things that are the SAME (in respect to time, OR distance, OR rate.) In this case, the thing that is the same is DISTANCE because the miles from Fritz's house to his work is the SAME as the miles going back home after work. The distance is the same BUT the rates are different and the times therefore would be different as well. So let's just say that his work is 60 miles away (I made this up - the problem does not tell us how far his work is, so I just chose a distance that was evenly divisible by 12 and 15 to make my math easier. It does not matter which distance you choose as long as it is the same for each direction - to work and home from work. You could even just use "D" but using a number here makes it easier to understand.)
Ok, so Fritz usually goes at a speed of 15mph to work. That means the first hour he would have gotten 15 miles closer to work, the second hour he would be 30 miles closer, or halfway, and then the 3rd hour 45 miles, and finally after the 4th hour, he would be 60 miles away from home and arriving at work. So it would take him 4 hours to get to work going 15 miles per hour. Going home at 15mph would then take him another 4 hours for a total of 8 hours round-trip. He will have traveled then exactly 120 miles in 8 hours - 4 hours in the AM and 4 hours in the PM. (I know, that's a long commute! Those were not very realistic numbers I chose, but oh well!) But THIS morning, the problem says, Fritz only could go 12 mph. So how long did it take him to get to work? Well, according to our formula for time, t=D/r, it would be t=60/12, or t=5. So it took him FIVE hours to get to work this morning, whereas it usually takes him only 4 hours! His average time for the whole trip is usually 8 hours to go 120 miles. That would give us a rate of r=D/t or: r=120/8 or 15 mph. (Reduce 120 over 8 to 15 over 1.) Since normally his trip is 8 hours total and he has already used up 5 hours in the morning commute, he only has 3 hours left to match his average daily commute time of 8 hours.
Although he only now has 3 hours left out of the 8, he still has the exact same 60 miles to travel in that reduced time of 3 hours. That means he needs to go at a rate of r=D/t or r=60mi/3hrs or r=60/3 = 20. So his rate in the evening must be 20 mph in order to make up for his slow commute in the morning at 12 mph. And that is your ANSWER - 20 mph!
NOTE: Notice that 20 + 12 = 32 and 32 divided by 2 =16, or the average of the two speeds, which is NOT the CORRECT answer of 20 mph, the average speed of the whole trip. So just remember... to get an average RATE, you need to find the entire distance and the entire time and divide, r=D/t! (The reason that 16 is not the average is because he spends more time going the slower speed than he does the faster speed, so it would need to be a "weighted average." They could have asked the question in a different way: If Fritz spent 3 hours going to work at 20 mph and 4 hours going to work at 15 mph, how far is Fritz's round-trip to work and back home? You would then have D=D equation (because the distance there is the same as the distance back). Since Distance = Rate times Time then you would have this equation: RxT=RxT. Then you could just fill in the two pairs of numbers and get 3 x 20 = 4 x 15 or 60 = 60. Either way, his distance to work and back home = 120.
Hope that helps you to see how to set up any Distance/Rate/Time type of problem!
Kind regards,
Donna K :-)
