Equations and problem solving

[Angiejane51]

I hate this! Please help me!
Question:
Ellen and Kate raced on their bicycles to the library after school. they both left school at 3:00 p.m. and bicycled along the same path. Ellen rode at speed of 12mph and Kate rode at 9mph Ellen got to the library 15 minutes before Kate what time did Ellen get to the library?

 

[Denis]

http://www.gmat-mba-prep.com/gmat-cours ... Speed.html

 

[soroban]

Hello, Angiejane51!

Ellen and Kate raced on their bicycles to the library after school.
They both left school at 3:00 p.m. and bicycled along the same path.
Ellen rode at speed of 12 mph and Kate rode at 9 mph.
Ellen got to the library 15 minutes before Kate.
What time did Ellen get to the library?

Let \(\displaystyle D\) = distance to the library.

Ellen rode \(\displaystyle D\) miles at 12 mph.
. . \(\displaystyle \text{This took her: }\:\frac{D}{12}\text{ hours.}\)

Kate road \(\displaystyle D\) miles at 9 mph.
. . \(\displaystyle \text{This took her: }\:\frac{D}{9}\text{ hours.}\)

\(\displaystyle \text{Ellen's time is 15 minutes (}\tfrac{1}{4}\text{ hour) less than Kate's time.}\)

\(\displaystyle \text{We have: }\;\frac{D}{12} \;=\;\frac{D}{9} - \frac{1}{4}\)

\(\displaystyle \text{Multiply by 36: }\;3D \;=\;4D - 9 \quad\Rightarrow\quad D \;=\;9\text{ miles.}\)


\(\displaystyle \text{Hence, Ellen rode 9 miles at 12 mph.}\)
. . \(\displaystyle \text{This took her: }\:\frac{9}{12} \:=\:\frac{3}{4}\text{ of an hour} \:=\:45\text{ minutes}\)


Therefore, Ellen reached the library at 3:45 p.m.