Answer :
If Terrance trains 6 miles in half an hour, that means he trains 12 miles in a whole hour. You can think of it as a fraction, [tex] \frac{6mi}{1/2hr} [/tex] and multiply the top and bottom of the fraction to get the unit rate [tex] \frac{12mi}{hr} [/tex] or [tex] 12\frac{mi}{hr} [/tex]. Jesse runs 2 miles every 15 minutes. 15 minutes is a quarter of an hour, so 2 miles every quarter hour -> [tex] \frac{2mi}{1/4hr} [/tex]. Multiply the top and bottom by 4 to get the unit rate [tex] \frac{8mi}{hr} [/tex] or [tex]8 \frac{mi}{hr} [/tex].
For the next question, use a proportion. We know that in an hour, Terrance will run 12 miles, but we want to know how long it takes for him to run 50 miles.
[tex] \frac{12mi}{1\ hr} = \frac{50mi}{x\ hrs} [/tex]
Cross multiply:
[tex]12x=50\\x= \frac{50}{12}= \frac{25}{6} = \frac{24}{6} + \frac{1}{6} =4 \frac{1}{6} [/tex]
So it takes Terrance 4 1/6 hours to run 50 miles. A sixth of an hour is ten minutes, so that becomes 4 hours and 10 minutes.
In an hour, Jesse runs 8 miles. How long will it take for him to run 50?
[tex] \frac{8mi}{1hr} = \frac{50mi}{xhr} [/tex]
Cross multiply:
[tex]8x=50\\ x= \frac{50}{8} = \frac{25}{4} = \frac{24}{4}+ \frac{1}{4} =6 \frac{1}{4} [/tex]
It takes Jesse 6 1/4 hours to run 50 miles. A quarter of an hour is 15 minutes, so that becomes 6 hours and 15 minutes.
Sandra runs 8 miles in 45 minutes. If Lee runs 8 miles in an hour, that means Lee takes longer to run the same distance. Therefore, Sandra is faster than Lee. Terrance runs 12 miles in an hour. To compare his speed to Sandra's, we have to do some mathemagics:
[tex]Terrence's\ speed:\ \frac{12mi}{1hr} = \frac{12mi}{60min}\ |divide\ top\ and\ bottom\ by\ 4\\ \frac{3mi}{15min}\ | multiply\ top\ and\ bottom\ by\ three\\ \frac{9mi}{45min} [/tex]
This might not have been the way your teacher showed you, but it's the way that makes sense to me. The new rate we have is really the same rate as before, we just made it look different so that the denominator would be the same as Sandra's rate. Now we can clearly see who is faster -- Sandra runs 8 miles in 45 minutes, Terrance runs 9 miles in 45 minutes, so Sandra is slower than Terrance.
For the next question, use a proportion. We know that in an hour, Terrance will run 12 miles, but we want to know how long it takes for him to run 50 miles.
[tex] \frac{12mi}{1\ hr} = \frac{50mi}{x\ hrs} [/tex]
Cross multiply:
[tex]12x=50\\x= \frac{50}{12}= \frac{25}{6} = \frac{24}{6} + \frac{1}{6} =4 \frac{1}{6} [/tex]
So it takes Terrance 4 1/6 hours to run 50 miles. A sixth of an hour is ten minutes, so that becomes 4 hours and 10 minutes.
In an hour, Jesse runs 8 miles. How long will it take for him to run 50?
[tex] \frac{8mi}{1hr} = \frac{50mi}{xhr} [/tex]
Cross multiply:
[tex]8x=50\\ x= \frac{50}{8} = \frac{25}{4} = \frac{24}{4}+ \frac{1}{4} =6 \frac{1}{4} [/tex]
It takes Jesse 6 1/4 hours to run 50 miles. A quarter of an hour is 15 minutes, so that becomes 6 hours and 15 minutes.
Sandra runs 8 miles in 45 minutes. If Lee runs 8 miles in an hour, that means Lee takes longer to run the same distance. Therefore, Sandra is faster than Lee. Terrance runs 12 miles in an hour. To compare his speed to Sandra's, we have to do some mathemagics:
[tex]Terrence's\ speed:\ \frac{12mi}{1hr} = \frac{12mi}{60min}\ |divide\ top\ and\ bottom\ by\ 4\\ \frac{3mi}{15min}\ | multiply\ top\ and\ bottom\ by\ three\\ \frac{9mi}{45min} [/tex]
This might not have been the way your teacher showed you, but it's the way that makes sense to me. The new rate we have is really the same rate as before, we just made it look different so that the denominator would be the same as Sandra's rate. Now we can clearly see who is faster -- Sandra runs 8 miles in 45 minutes, Terrance runs 9 miles in 45 minutes, so Sandra is slower than Terrance.
