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Two bikers leave the Harley-Davidson shop, traveling on the same road and in the same direction. One biker drives 5 mph faster than the other. The faster biker stops after 1 1/2 h but the slower biker stops after 1 h. If the two bikers are 32 1/2 miles apart, what was the speed of each biker?

I am learning Algebra at 38 years old, just because I never got it in high school!
 
afreemanny said:
Two bikers leave the Harley-Davidson shop, traveling on the same road and in the same direction. One biker drives 5 mph faster than the other. The faster biker stops after 1 1/2 h but the slower biker stops after 1 h. If the two bikers are 32 1/2 miles apart, what was the speed of each biker?

** Warning ** This LOOKS W-A-A-A-A-Y more scary than it is. Just go one step at a time, slowly and easily. understand each step.

Make definitions for stuff you need.

B = Speed of Faster Biker
K = Speed of Slower Biker

Distance = Rate * Time <== I just wrote this because I thought I might need it.

"One biker drives 5 mph faster than the other."

B = K + 5 mph

Now we have a problem. Your problem statement did not state that they left at the same time. We'll have to assume that.

A little thinking is required. The faster biker drove for more time. Oh, then he must be way beyond the other.

Write the equations.

Distance<sub>1</sub> = Rate<sub>1</sub>*Time<sub>1</sub>
Distance<sub>2</sub> = Rate<sub>2</sub>*Time<sub>2</sub>

Sustitute what we know or have defined.

Distance<sub>1</sub> = B*Time<sub>1</sub>
Distance<sub>2</sub> = K*Time<sub>2</sub>

"The faster biker stops after 1 1/2 h but the slower biker stops after 1 h."

Time<sub>1</sub> = 1.5 hour
Time<sub>2</sub> = 1.0 hour

Distance<sub>1</sub> = B*1.5 hr
Distance<sub>2</sub> = K*1.0 hr

"the two bikers are 32 1/2 miles apart"

Distance<sub>1</sub> = Distance<sub>2</sub> + 32.5

Distance<sub>2</sub> + 32.5 mi = B*1.5 hr
Distance<sub>2</sub> = K*1.0 hr

B = K + 5 mph

Distance<sub>2</sub> + 32.5 mi = (K + 5 mph)*1.5 hr
Distance<sub>2</sub> = K*1.0 hr

Whew! That was a lot to keep track. Just one step at a time. Don't try to do too much at once.

Old addage - Solve for what you DON'T want. The problem statement wants the speeds.

Distance<sub>2</sub> = (K + 5 mph)*1.5 hr - 32.5 mi
Distance<sub>2</sub> = K*1.0 hr

Transitive Property of Equals.

(K + 5 mph)*1.5 hr - 32.5 mi = K*1.0 hr

Solve for 'K'

1.5*K*hr + 7.5*mi - 32.5*mi = K*hr
0.5*K*hr = 25 mi
K*hr = 50*mi
K = 50 mph <== If your units don't come out right, you did something wrong.

B = K + 5 mph = 55 mph

Great. Now, let's test it.

55 mph * 1.5 hr = 82.5 mi
55 mph * 1.0 hr = 50.0 mi
82.5 mi - 50.0 mi = 32.5 mi

Looks good.
 
Hello, afreemanny!

I always respect and enjoy tkhunny's explanations.
I used a different approach, though. [Is anyone surprised?]

Two bikers leave the Harley-Davidson shop, traveling on the same road and in the same direction.
One biker drives 5 mph faster than the other.
The faster biker stops after 1 1/2 h but the slower biker stops after 1 h.
If the two bikers are 32 1/2 miles apart, what was the speed of each biker?
Of course, we will use: . Distance .= .Speed x Time

Let x = speed of the slower Hawg.
. . In one hour, it went: . (1)(x) .= .x miles.

Then x + 5 = speed of the faster Harley.
. . In 1.5 hours, it went: . (3/2)(x + 5) miles.

When they stopped, they were 32.5 miles apart.
. . So their distances <u>differ</u> by 65/2 miles.

And there is our equation! . (3/2)(x + 5) - x . = . 65/2 . **


You could probably use a walk-through on the rest of it, right?

Multiply through by 2 (to eliminate the fractions):

. . . . . .3(x + 5) - 2x . = . 65

Then: . 3x + 15 - 2x . = . 65

. . . . . . . . . . x + 15 . = . 65

. . . . . . . . . . . . . . x . = . 50

Therefore: . the slower biker went 50 mph,
. . . . . . . . . . the faster biker went 55 mph.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

The slower bike moved for 1 hours at 50 mph.
. . . Its distance is 50 miles.

The faster biked moved for 1.5 hurs at 55 mph.
. . . Its distance is 82.5 miles.

And, sure enough, they are 32.5 miles apart!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
Which way should we subtract the distances?
. . As TK pointed out, the faster bike rode longer, so it has the larger distance.
So, it is "Faster minus Slower".
 
love baby steps!

Thanks very much for your help. I had it almost right until I transposed a number. :oops:
 
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