Question

Ryan drives 10mi/h slower than Maya, And it takes ryan 1 hour longer to travel 300 miles. How long does it take Maya to make a trip

Answer 1

I'll let amistre get this :p

Answer 2

well, you need two equations and two variables to be able to solve this

Answer 3

one variable is the time it takes Maya to take the trip, and one variable is the time it takes ryan to take the trip

Answer 4

I got 300(x)= 300(x-1) + 10x(x-1)

Answer 5

well, actually, let's make the two variables the speed of each person - then we can easily solve for how long it takes maya to drive 300 miles.

So:
1. R = M - 10
2. 300/R - 300/M = 1

Answer 6

i think i have an answer.

Answer 7

what did you get?

Answer 8

i got 60 but it was kind of a pain the way i did it. maybe an easier way.

Answer 9

i used maya's rate as r, ryan's as r - 10 and then
\[\frac{300}{r}=\frac{300}{r-10}+1\]

Answer 10

that is ryan's time is one hour more than maya's.

Answer 11

right. yeah that makes sense

Answer 12

sorry but i do not understand

Answer 13

i made a mistake somewhere i am trying again.

Answer 14

first of all the answer: ryan travels at 50 mpr and maya at 60. it is easy to check:
\[d=rt\]
\[t=\frac{d}{r}\]
if maya travels 300 miles at 60 miles an hour she makes the trip in \[\frac{300}{60}=5\] hours
while ryan takes \[\frac{300}{50}=6\] hours or one hour longer. so this is the correct answer.

Answer 15

O i understand thank you!

Answer 16

well i got the correct answer, but i am trying to find the right algebra to solve it for you.

Answer 17

whew finally.
ok here goes. ready?

Answer 18

yes!

Answer 19

ok we use \[t=\frac{d}{r}\]
and we know that ryans rate is r-10 and his maya's time is one less than ryan's. Maya's time is \[\frac{300}{r}\] and ryan's is \[\frac{300}{r-10}\]
and \[\frac{300}{r}-1=\frac{300}{r-10}\]

Answer 20

that says maya's time is one less than ryans. to solve for r, easiest to subtract first on the left hand side:
\[\frac{300-r}{r}=\frac{300}{r-10}\]
now cross multiply:
\[(r-10)(300-r)=300r\]
mutiply out

\[300r-3000-r^2+10r=300r\]
subtract 300r
\[-3000 -r^2+10r=0\]

Answer 21

damn

Answer 22

thought i had \[(r-60)(r+50)=0\] so r = 60 but i seem to have made another mistake somewhere. i will try to find it.

Answer 23

oh lord. i put the one on the wrong side!
should have been \[\frac{300}{r}+1=\frac{300}{r-10}\]

Answer 24

maya's time plus one gives ryans time. ok now.
\[\frac{300+r}{r}=\frac{300}{r-10}\]
\[(300+r)(r-10)=300r\]
\[300r -3000+r^2-10r=300r\]
\[r^2-10r-300=0\]
\[(r-60)(r+50)=0\]
\[r=60\] or \[r=-50\]
which makes no sense in this problem so
\[r=60\]
sorry it took a while.

Answer 25

its okay and thanks again

Answer 26

sorry i made a mistake originally. hope it was not too confusing.

Answer 27

it wasnt thanks