Terry and Phyllis went to the park to train for a long distance run. Terry reached earlier and started her run 10 minutes befor Phyllis arrived and joined in. Terry finished 5 minutes beforw Phyllis. They both ran the same distance. At the point when Terry finished, Phyllis had completed 5/6 of the joirney and still had 1 mile to go. Find each of their speeds. Who was faster? Give the answers in miles per hour.
step one:
5/6 left one mile to go
so
one mile is 1/6 of the course
so
It is 6 miles long
-----------------------------------------------------
now times
say Terry started at noon and finished at t minutes after so ran t min
P started at 10 past and finished at t + 5 so ran t -5 min
at t, P had run 5 miles and T had run 6 miles
if ST is speed of T
so 6 miles = ST * t where ST is speed of T
5 miles = SP * (t-10)
6 miles = SP * (t-5)
=======
three equations, three unknowns
5 = SP t - 10 SP
6 = SP t -5 SP
------------------- subtract
-1 = -5 SP
so speed of P = 1/5 mile/min (times 60 = 12 miles/hour)
go back and get t and ST
Grade 6 ????
Yes it’s grade 6
To solve this problem, let's assign variables to the unknowns:
Let T represent Terry's speed (in miles per hour),
Let P represent Phyllis's speed (in miles per hour).
We know that both Terry and Phyllis ran the same distance, so their time ratios will be equal.
Now, let's break down the information given to us:
- Terry started running 10 minutes before Phyllis. Therefore, when Phyllis started her run, Terry had already been running for 10 minutes. We need to convert these 10 minutes into hours:
10 minutes = 10/60 = 1/6 of an hour.
- Terry finished her run 5 minutes before Phyllis. So, when Terry finished her run, Phyllis had only 5 minutes left to complete her journey. We also need to convert these 5 minutes into hours:
5 minutes = 5/60 = 1/12 of an hour.
Now, we can set up an equation to represent the time ratio:
(Terry's time + 1/6) / (Phyllis's time + 1/12) = 1
Since both Terry and Phyllis ran the same distance, we can also set up an equation for their speeds:
Terry's speed = distance / Terry's time
Phyllis's speed = distance / Phyllis's time
We know that at the point when Terry finished, Phyllis had completed 5/6 of the journey and still had 1 mile to go. So, we can set up another equation:
(5/6) * distance = 1
Now we have three equations:
(Terry's speed) / (Phyllis's speed) = (Phyllis's time + 1/12) / (Terry's time + 1/6)
Terry's speed = distance / Terry's time
Phyllis's speed = distance / Phyllis's time
(5/6) * distance = 1
To solve this system of equations, we need to further simplify the equations and then solve for the variables.
First, we can simplify the equation (5/6) * distance = 1 by multiplying both sides by (6/5):
distance = 6/5
Now, we can substitute the value of distance in the Terry's speed equation:
Terry's speed = (6/5) / Terry's time
Then, we can substitute the value of Terry's speed in the time ratio equation:
(6/5) / Terry's time / (Phyllis's speed) = (Phyllis's time + 1/12) / (Terry's time + 1/6)
Now, we can simplify this equation further:
(6/5) / Terry's time / (Phyllis's speed) = (Phyllis's time + 1/12) * (Terry's time + 1/6)^-1
This equation allows us to find the relationship between Terry's speed and Phyllis's speed. By solving this equation, we can determine their speeds and identify who was faster. However, finding the exact numerical values for their speeds requires more information or additional equations.