SOLUTION: A woman invests $39,000, part at 8% and the rest at 9 1\2% annual interest. If the 9 1/2% investment provides $642.50 more income than the 8% investment, how much is invested at ea

Algebra.Com
Question 1121634: A woman invests $39,000, part at 8% and the rest at 9 1\2% annual interest. If the 9 1/2% investment provides $642.50 more income than the 8% investment, how much is invested at each rate?
Found 3 solutions by Boreal, solver91311, Theo:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
x at 8%
39000-x at 9.5%
interest is .08x and 0.095(39000-x)
we know that .08x+642.50=3705-.095x
0.175x=3062.50
x=$17,500@8%=$1400
39000-x=$21,500@9.5%=$2042.50
$17,500 at 8%
$21,500 at 9.5%

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Let represent the amount invested at 9.5%. Then is the amount invested at 8%. The amount of interest earned on the 9.5% investment is then and the amount invested at 8% is . These two amounts differ by $642.50, so:



Solve for , then calculate


John

My calculator said it, I believe it, that settles it


Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
let x = part invested at 8% = .08.
let y = part invested at 9.5% = .095.

total investment is 39000, therefore x + y = 39000.

you are given that part invested at .095 is 642.5 more than part invested at .08.

this means .095y = .08x + 642.5

solve for y to get y = (.08x + 642.5) / .095

in the equation of x + y = 39000, replace y with that to get:

x + (.08x + 642.5) / .095 = 39000

multiply both sides of this equation by .095 to get:

.095x + (.08x + 642.5) = 39000 * .095

simplify to get:

.095x + .08x + 642.5 = 3705

subtract 642.5 from both sides of this equation and combine like terms to get:

.175x = 3062.5

divide both sides of this equation by .175 and solve for x to get x = 17500

since x + y = 39000, then y must be equal to 21500 because 17500 + 21500 = 39000.

you have:

x = 17500
y = 21500

your solution is that the amount invested at 8% is equal to 17500 and the amount invested at 9.5% is equal to 21500.

8% of 17500 = .08 * 17500 = 1400
9.5% of 21500 = .095 * 21500 = 2042.5

income invested at 9.5% minus income invested at 8% = 642.5.

requirements of the problem are satisfied, therefore the solution is confirmed to be good.














RELATED QUESTIONS

A woman invests $33,000, part at 8% and the rest at 9 1/2% annual interest. If the 9 1/2% (answered by Boreal)
A woman invests $39,000, part at 8% and the rest at 9.5% annual interest. If the 9.5%... (answered by ikleyn)
ms.berger has two investments. one has 8% annual interest rat, and the other investment... (answered by ankor@dixie-net.com)
Cyril invested part of 185 000 at 7% and the rest at 9. If the interest from the 9%... (answered by checkley77)
I have two questions #1-After inheriting some money, a woman wants to invest enough to (answered by checkley75)
Larry Mitchell invested part of his $30,000 advance at 5% annual simple interest and the... (answered by greenestamps)
A woman invest $10 000;part at 9% annual interests, and the rest at 14%. In each case the (answered by greenestamps,josgarithmetic)
Egbert invested $5,000 for one year, part at 9% annual interest and the rest at 12%... (answered by mananth)
A man invests part of 17,000 pesos at 8% and the rest at 9%. If the annual income... (answered by checkley79)