# Grade 8 Algebra 1, Course 2 - Acid Solutions

• May 24th 2007, 04:21 PM
Harryhit4
Grade 8 Algebra 1, Course 2 - Acid Solutions
THIS IS VERY URGENT

This isn't actually an acid solution, just something like it. I would GREATLY appreciate it if you can help me understand this

A man invested $1200 for one year, part at 4% and part at 6%. He earned 58$ from his investments. How much did he invest at each rate?

The way I learned it was doing something like this

1...|2.........|New
.04 |.06......|.1
x....|1200-x.|1200
.04x|72-.06x|120

-.02x = 48

I know I'm doing something wrong there - The chart may look screwed up, but it was multiply down and add across.
EDIT:
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I'd also like some help on understanding this:

Jake can run downhill at a rate 2 times faster than he can run uphill. If the distance each way is 5km, and the total time for the trip is 30 minutes, what is his rate downhill?
• May 24th 2007, 07:40 PM
Soroban
Hello, Harryhit4!

Quote:

A man invested $1200 for one year, part at 4% and part at 6%. He earned$58 from his investments.
How much did he invest at each rate?

Your chart is not quite correct . . .

He invested $x$ dollars at 4%.
. . His interest is: . $0.04x$ dollars.

He invested the other $1200-x$ dollars at 6%.
. . His interest is: . $0.06(1200-x)$ dollars.

His total interest is \$58.

There is our equation: . $0.04x + 0.06(1200 - x) \:=\:58$

Quote:

Jake can run downhill at a rate 2 times faster than he can run uphill.
If the distance each way is 5km, and the total time for the trip is 30 minutes,
what is his rate downhill?

We will use: . $\text{Distance }= \text{ Speed } \times \text{ Time}\quad\Rightarrow\quad T \:=\:\frac{D}{S}$

Let $U$ = Jake's uphill speed (in km per hour).
Then $2U$ = Jake's downhill speed.

Running 5km uphill at $U$ kph, his time is: . $\frac{5}{U}$ hours.

Running 5km downhill at $2U$ kph, his time is . $\frac{5}{2U}$ hours.

His total time is $\frac{1}{2}$ hour (30 minutes).

There is our equation: . $\frac{5}{U} + \frac{5}{2U} \:=\:\frac{1}{2}$

• May 25th 2007, 11:34 AM
Harryhit4
Thanks
Sadly, I didn't get enough time to check your answer before going to bed. But, for the first question, I just figured out during the test [A different problem, though] that I can use substitution as an alternative

x + y = 1200
.04x + .06y = 58

The problem was that I kept thinking .04x + .06y was 1200, not 58.

As for the second one, that type of problem wasn't on the test, so I was saved. I would have been completely wrong [Although I know how to do regular rate/time/distance problems]

That helped me understand a lot, thanks!