SOLUTION: Sam's annual interest from an investment iss 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

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Sam's annual interest from an investment iss 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       Log On


   



Question 113310: Sam's annual interest from an investment iss 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 rate of his actual investment?

Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) About Me  (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 rate of his actual investment?
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Interest = Prt; in your problem t=1
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EQUATIONS:
225 = Pr
234 = (P+600)(r-1)
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234 = Pr -P + 600r -600
234 = 225 -P + 600r -600
609 = -P + 600r
P -600r+609 = 0
But r = 225/P:
P -[600*225/P]+609 = 0
Multiply thru by P to get:
P^2 +609P - (600*225) = 0
Use the Quadratic Formula to solve for P:
P = [-609 +- sqrt(609^2 -4*(-600*225)]/2
P = [-609 + 954.4]/2
P = $172.70
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r = 225/P = 0.768
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Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the amount of Sam's investment in dollars, and I = the interest rate.
We are told that the annual interest earned on the x dollars at I% is $225.
This can be expressed as:
1) x%2A%28I%29+=+225 Rewrite this in terms of x.
1a)x+=+225%2FI
If the investment had been $600 more (x+600) and the interest rate 1% less (I-0.01), the earnings would have been $234, so we can set up the second equation in x and I as follows:
2) %28x%2B600%29%28I-0.01%29+=+234
Now we have two equations in two unknowns (x and I) which can be solved.
Substitute the x from equation 1a) into equation 2) and solve for I, the interest rate.
2a) %28%28225%2FI%29%2B600%29%28I-0.01%29+=+234 Simplify.
%28%28225%2B600I%29%2FI%29%28I-0.01%29+=+234 Multiply both sides by I
%28225%2B600I%29%28I-0.01%29+=+234%28I%29 Perform the indicated multiplication on the left side.
225I%2B600I%5E2-2.25-6I+=+234I Combine like-terms.
600I%5E2-15I-2.25+=+0 Use the quadratic formula to solve for I:
I+=+%28-%28-15%29%2B-sqrt%28%28-15%29%5E2-4%28600%29%282.25%29%29%29%2F4%28600%29
I+=+0.075 or I+=+-0.05 Discard the negative solution, interest rates are positive.
So the interest rate would be I = 7.5%
The investment, x, would be:
x+=+225%2FI
x+=+225%2F0.075
x+=+3000
The investment would be $3,000
Check:
($3,000)(0.075) = $225 and...
($3,000+$600)(0.075-0.01) = ($3,600)(0.065) = $234