Help With These 2 Interest Problems!

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Hanajima

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Aug 27, 2009
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Could someone please help me solve these two interest problems.
I tried to solve them using an example problem I got from class, but I don't seem to be getting the right answer.

Problem #1

Eva invested a certain amount of money at 7% interest and $1600 more than that amount at 8%. Her total yearly interest was $815. How much did she invest at each rate?

At 7%?

At 8%?

Problem #2

Use an algebraic approach to solve the problem.
A total of $4000 was invested, part of it at 8% interest and the remainder at 11%. If the total yearly interest amounted to $350, how much was invested at each rate?

At 8%?

At 11%?

Work and/or and explanation of how you got to your answer would be appreciated!
Thank you so much!
 
Problem #1

Eva invested a certain amount of money at 7% interest and $1600 more than that amount at 8%. Her total yearly interest was $815. How much did she invest at each rate?

You don't indicate whether you may use 2 variables or only one. I'll get you started using one variable.

ALWAYS start by naming things and write it down so you can refer back to it.

Let x represent the amount of money invested at 7%. (I wrote it down)
Therefore, the amount invested at 8% = x+1600. (I wrote this down, also.)

Now, we have to figure out how to write an equation that represents the problem. We know that we calculate the amount of interest earned with the formula...

Principal * rate of interest * time = Amount of interest.
In this case we have two amounts of principal invested and we know that if we add these together we get the total amount of interest earned.

Now let's put it all together.

Amount of interest from 7% investment for one year = .07x
Amount of interest from 8% investment for one year = .08(x+1600)
Total amount of interest earned is 815.

You take it from there.
 
Hanajima said:
Could someone please help me solve these two interest problems.
I tried to solve them using an example problem I got from class, but I don't seem to be getting the right answer.

Problem #1

Eva invested a certain amount of money at 7% interest and $1600 more than that amount at 8%. Her total yearly interest was $815. How much did she invest at each rate?

At 7%?

At 8%?

Problem #2

Use an algebraic approach to solve the problem.
A total of $4000 was invested, part of it at 8% interest and the remainder at 11%. If the total yearly interest amounted to $350, how much was invested at each rate?

At 8%?

At 11%?

Work and/or and explanation of how you got to your answer would be appreciated!
Thank you so much!
Click to expand...

You haven't shown us ANY of your thoughts and/or effort.

I'll see if I can get you started on problem 1...and then you show us what you can do.

Let x = number of dollars invested at 7% interest

Now, the problem tells us that $1600 MORE than that amount was invested at 8% interest. $1600 MORE than x would be (x + 1600)...so,
x + 1600 = number of dollars invested at 8% interest

You'll also need the formula for simple interest. If P dollars is invested at an interest rate of r% for t years, then the simple interest I earned is this: I = P * (r/100)*t

if you invest x dollars at 7% interest for 1 year, the interest earned (using the above formula) would be I = x * (7/100) * 1, or 0.07x

If you invest (x + 1600) dollars at 8% for 1 year, the interest earned would be I = (1600 + x)*(8/100)*1, or 0.08(1600 + x)

If the total amount of interest earned in 1 year is $815, we have this:

815 is equal to the interest for 1 year on x dollars invested at 7% PLUS the interest for 1 year on (1600 + x) dollars invested at 8%

815 = 0.07x + 0.08(1600 + x)

Please show us what you would do to complete this problem.

Then, see if you can apply the same sort of reasoning on problem #2.
 
Sorry..I was doing this on scraps and I forgot to scan them before throwing them away...

From 0.07x + 0.08(x+1600) =815

I distributed the 0.08 then combined like terms to get a new equation of

0.15x + 128 = 815

I then subtracted 128 from both sides to get:

0.15x = 687

I then divided 687 by .15
which made x = 4580

Which is the amount invested at 7%.
To find the amount invested at 8% I simply added 1600 to 4580 to get 6180, which is the amount invested at 8%
 
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