Question 1133346: Two investments totaling $57,500 produce an annual income of $4245 . One investment yields 8% per year, while the other yields 7% per year. How much is invested at each rate?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The traditional algebraic approach: 7% of amount x, plus 8% of amount ($57,500-x), equals $4245:
Solve using basic algebra, although the decimals make for some ugly computations.
I leave it to you to finish the problem by that method.
An alternative method, based on finding where the actual interest of 4245 lies between the two amounts if all $57,500 had been invested in either one of the two accounts....
$57,500 at 7% = 4025
$57,500 at 8% = 4600
Find where the actual amount of interest lies between those two amounts:
The difference between those two amounts is $575.
The difference between the actual amount of interest and the amount of interest at 7% is $4245-$4025 = $220.
So the amount of actual interest is 220/575 of the way from 7% to 8%; that means 220/575 of the total was invested at the higher rate of 8%. Since the total amount was $57,500, the amount invested at 8% was $22,000; that means the amount invested at 7% was $57,500-$22,000 = $35,500.
ANSWER: $35,500 at 7%; $22,000 at 8%
Of course that is the answer you should get by solving the problem by the algebraic method.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
7% of x plus 8% of (57500 -x) is equal to the total annual income of $4245.
It is your equation
0.07*x + 0.08*(57500-x) = 4245
Simplify and solve for x:
x =  = one click in Excel = 35500.
Answer. $35500 was invested at 7% and the rest, 57500 - 35500 = 22000 dollars was invested at 8%.
CHECK. 0.07*35500 + 0.08*22000 = 4245 dollars. ! Correct !
Solved.
------------------
It is a standard and typical problem on investments.
If you need more details, or if you want to see other similar problems solved by different methods, 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.
|
|
|
