SOLUTION: PLease show all steps and work show that i may have an idea as to how to solve the rest. Thanks in advance. I ahve no idea how to even begin this type of problem.
Sam's annual
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: PLease show all steps and work show that i may have an idea as to how to solve the rest. Thanks in advance. I ahve no idea how to even begin this type of problem.
Sam's annual
Log On
Question 433111: PLease show all steps and work show that i may have an idea as to how to solve the rest. Thanks in advance. I ahve no idea how to even begin this type of problem.
Sam's annual interest from an investment is $225. If sam had invested $600 more at an annual interest rate that was %1 less, the annual interest would have been $234. What are the amount and the rate of his actual investment. Found 2 solutions by stanbon, ankor@dixie-net.com:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Sam's annual interest from an investment is $225. If sam had invested $600 more at an annual interest rate that was %1 less, the annual interest would have been $234. What are the amount and the rate of his actual investment.
-------------------------------
Original interest: Pr*1 = 225
-------------
New investment: (P+600)*(r-0.01)*1 = 234
----
Pr-0.01P+600r-6 = 234
---
Substitute for Pr to get:
---
225-0.01P+600r-6 = 234
------
Then:
600r-0.01P = 15
=======
That's about as for as you can go with the
information you have.
Cheers,
Stan H.
You can put this solution on YOUR website! Sam's annual interest from an investment is $225.
If sam had invested $600 more at an annual interest rate that was %1 less, the
annual interest would have been $234.
What are the amount and the rate of his actual investment.
:
Let p = original amt invested
Then
(p+600) = hypothetical amt invested
:
Let r = rate of interest originally, in decimal form
then
(r-.01) = hypothetical interest rate
:
The actual equation:
rp = 225
p = , us the for substitution
;
The hypothetical equation
(r-.01)*(p+600) = 234
FOIL
rp + 600r - .01p - 6 = 234
rp + 600r - .01p = 234 + 6
Replace rp with 225
225 + 600r - .01p = 240
600r - .01p = 240 - 225
600r - .01p = 15
replace p with
600r - .01() = 15
600r - ) = 15
multiply by r
600r^2 - 2.25 = 15r
A quadratic equation
600r^2 - 15r - 2.25 = 0
Using the quadratic formula, we get a positive solution of
r = .075 or 7.5% original interest rate
:
Find p
p = 225/.075
p = $3000 is the original amt invested
:
:
See if that satisfies the hypothetical equation
.065 * 3600 = 234
(