# Distance (Word Problem)

• May 10th 2011, 03:57 AM
Beastkun
Distance (Word Problem)
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...

"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"

I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
• May 10th 2011, 04:12 AM
earboth
Quote:

Originally Posted by Beastkun
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...

"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"

I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!

1. Let d denote the covered distance and t the elapsed time. Then you know:

$\displaystyle \dfrac dt = 50$
and
$\displaystyle \dfrac{d}{t-3}=60$

2. Solve for d.
Spoiler:
You should come out with d = 900
• May 10th 2011, 04:38 AM
Beastkun
I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.

Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
• May 10th 2011, 04:47 AM
mr fantastic
Quote:

Originally Posted by Beastkun
I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.

Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.

Substitute d = 50t from the first equation into the second equation and solve for t and hence d.
• May 10th 2011, 04:52 AM
jgv115
What prove it is doing is simultaneous equations.

If we rearrange the first equation $\displaystyle \frac{d}{t} = 50$ to make d the subject we get: $\displaystyle d= 50t$ we can now sub this into the second equation $\displaystyle \frac{d}{t-3} = 60$ which gets us $\displaystyle t=18$.

Now we can sub it back into any equation to find d!
• May 10th 2011, 05:03 AM
Beastkun
So, would it be something along the lines of 50t=60(t-3)? No.. that can't be right.

Are you saying that it should become 50t/t-3=60?
• May 10th 2011, 05:04 AM
jgv115
Yes, now solve for t
• May 10th 2011, 05:20 AM
Soroban
Hello, Beastkun!

Let me give it a try . . .

Quote:

Driving from Dallas towards Memphis, JoJo averages 50 mph.
She figured that if she had averaged 60 mph,
the driving time would have decreased by 3 hours.
How far did she drive?

$\displaystyle \text{The formula is: }\:D \:=\:R\!\cdot\!T$

$\displaystyle \text{Use this variation: }\:T \:=\:\frac{D}{R}$

$\displaystyle \text{She drove }x\text{ miles at 50 mph.}$
. . $\displaystyle \text{This took: }\:T_1 \:=\:\frac{x}{50}\text{ hours.}$

$\displaystyle \text{She drove }x\text{ miles at 50 mph.}$
. . $\displaystyle \text{This took: }\:T_2 \:=\:\frac{x}{60}\text{ hours.}$

$\displaystyle \text{We are told that }T_2\text{ is 3 hour less than }T_1.$

$\displaystyle \text{There is our equation! }\;\;\frac{x}{60} \:=\:\frac{x}{50} - 3$

• May 10th 2011, 05:22 AM
Beastkun
Hrmmm so, 50t=60(t-3)..

50t=60t-180
-10t=-180
t=18

THEN

I plug T into the first equation d=50t, so d=50*18

900!

Thanks guys, I really appreciate your help!