Clueless: Tim leaves at 9:30 by bike, going 12 mph....

helpmyson

New member
Joined
Jul 9, 2007
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2
Tim leaves Exodus at 9:30am traveling to Portsmouth by bicycle at 12mph. Sara leaves Portsmouth at 10:00am traveling to Exodus by bicycle at 16mph. The distance between Exodus and Portsmouth is 20miles. What time do they meet?

Please show the formula on how to figure this mess out. Three of us has worked on it and can not figure it out. HELP!!!
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,480
Sir/Madam, not being able to solve "12t + 16t - 8 = 20" means that
you are unfamiliar (not retarded!) with the basics of Algebra;
unfortunately, it also means that these basics need to be taught to
you in order that you understand HOW to solve an equation; this
means classroom environment, which this site IS NOT.

However, the problem you posted can be "reasoned out":

> Tim leaves Exodus at 9:30am traveling to Portsmouth by bicycle at 12mph.
> Sara leaves Portsmouth at 10:00am traveling to Exodus by bicycle at 16mph.
> The distance between Exodus and Portsmouth is 20miles.

So when Sara leaves, Tim has already travelled 6 miles: 12mph for 1/2 hour;
so there is 14 miles left, right?

Now, before they meet, this 14 miles will be "covered" at a combined speed of
28mph, right?

14 miles at 28mph takes 1/2 hour, right?

So they'll meet 1/2 hour after 10.00am, right?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, helpmyson!

Denis is absolutely right . . . it can be solved without Algebra.
. . But I suspect an algebraic solution is expected.


Tim leaves Exodus at 9:30am traveling to Portsmouth by bicycle at 12mph.
Sara leaves Portsmouth at 10:00am traveling to Exodus by bicycle at 16mph.
The distance between Exodus and Portsmouth is 20 miles.
What time do they meet?

We are expected to know: \(\displaystyle \:\text{Distance} \;=\;\text{Speed }\times\text{ Time}\)

Tim has a half-hour headstart.. Then, starting at 10:00,
. . they both travel for \(\displaystyle x\) hours and meet.

Tim travels for \(\displaystyle x\,+\,\frac{1}{2}\) hours at \(\displaystyle 12\) mph.
. . His distance is: \(\displaystyle \,12\left(x\,+\,\frac{1}{2}\right)\) miles.

Sara travels for \(\displaystyle x\) hours at \(\displaystyle 16\) mph.
. . Her distance is: \(\displaystyle \,16x\)miles.

Since the total distance is 20 miles: \(\displaystyle \:12\left(x\,+\,\frac{1}{2}\right)\,+\,16x \;=\;20\)

Solve for \(\displaystyle x:\;\;12x\,+\,6\,+\,16x\;=\;20\;\;\Rightarrow\;\;28x \,=\,14\quad\Rightarrow\quad x \,=\,\frac{1}{2}\)

They meet a half-hour after 10 o'clock . . . \(\displaystyle \fbox{\text{10:30 am}}\)

 

helpmyson

New member
Joined
Jul 9, 2007
Messages
2
Thank you. I appreciate your help.
 
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