Question 1162273: Juanita invested $13,000 for one year, part at 9 percent annual interest and the rest at 7 percent annual interest. How much must she invest at 9 percent to get $1,000 in interest in a year?
Jonathon invested $8,400, part at 12 percent annual interest and the remainder at 8 percent annual interest. In one year, he earned three times as much interest from the 12 percent investment as he did from the 8 percent investment. How much interest did Jonathon earn at 8 percent interest?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
Let x be the amount invested at 9%.
Then the amount invested at 7% is the rest (13000-x) dollars.
An equation for the total interest is
0.09x + 0.07*(13000-x) = 1000 dollars.
From the equation
x =  = 4500.
ANSWER. $4500 invested at 9% and the rest, 13000-4500 = 8500 dollars invested at 7%.
Solved.
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It is a typical and standard problem on investment.
To see many other similar solved problems on investment, look into the lesson
- Using systems of equations to solve problems on investment
in this site.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
I use a non-standard, non-algebraic method to solve "mixture" problems like this.
I will show a typical algebraic setup for each problem and let you finish the solution by those methods.
Then I will show how I like to solve these problems informally, using logical reasoning and some simple arithmetic.
first problem....
Let x be the amount invested at 9%; then the amount invested at 7% is 13000-x.
The total interest is to be $1000:
Solve using basic algebra... although the decimals make the arithmetic tedious.
My method compares the actual amount of interest to the amounts that would have been earned at each rate separately to determine how the money needs to be split between the two investments.
All $13,000 at 7% would yield $910 interest.
The actual amount of interest is to be $1000.
All at 9% would yield $1170 interest.
Use simple arithmetic to find that the actual interest is 9/26 of the way from $910 to $1170:
$910 to $1170 is a difference of $260
$910 to $1000 is a difference 0f $90
The actual interest is 90/260 = 9/26 of the way from $910 to $1170.
That means 9/26 of the total needs to be invested at the higher rate.
9/26 of $13,000 is $4500.
ANSWER: $4500 at 9%; the other $8500 at 7%.
CHECK: .09(4500)+.07(8500) = 405+595 = 1000
second problem....
An algebraic setup:
x = amount invested at 8%
8400-x = amount invested at 12%
The interest from the 12% investment was 3 times the interest from the 8% investment:
Again solve using basic algebra.
The problem is posed in a way that a little logical reasoning can lead to a quick solution.
Specifically, since the 12% rate is 1.5 times the 8% rate, and since the interest from the 12% investment was 3 times the interest from the 8% investment, we can immediately conclude that the amount invested at 12% was 2 times the amount invested at 8%.
Dividing the total of $8400 into the ratio 2:1, we see that the amount invested at 8% was $2800.
And then the amount of interest from the 8% investment was .08($2800) = $224.
CHECK: The interest from ($8400-$2800 = $5600) at 12% is twice the interest from the 8% investment: .12($5600) = $672 = 3*$224.
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