SOLUTION: Using I=PRT. Shawndra has $66,000 to invest. She decides to invest part at 6.5%and the rest at 4.5%. She figured she would make a total of $3,720 on her investments. How much did S

Algebra ->  Test -> SOLUTION: Using I=PRT. Shawndra has $66,000 to invest. She decides to invest part at 6.5%and the rest at 4.5%. She figured she would make a total of $3,720 on her investments. How much did S      Log On


   



Question 1080081: Using I=PRT. Shawndra has $66,000 to invest. She decides to invest part at 6.5%and the rest at 4.5%. She figured she would make a total of $3,720 on her investments. How much did Shawndra invest at each rate?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+I+=+P%2Ar%2At+
+r%5B1%5D+=+.065+
+r%5B2%5D+=+.045+
Let the amount invest @ 6.5% = +P%5B1%5D+
The amount invested @ 4.5% = +P%5B2%5D+=+66000+-+P%5B1%5D+
Let the interest earned @ 6.5% = +I%5B1%5D+
The interest earned @ 4.5% = +I%5B2%5D+=+3720+-+I%5B1%5D+
-----------------------------------------------------
(1)
+I%5B1%5D+=+P%5B1%5D%2Ar%5B1%5D%2At+
(2)
+I%5B2%5D+=+P%5B2%5D%2Ar%5B2%5D%2At+
-------------------------
I'll assume that +t+=+1+ ( one year )
(1)
+I%5B1%5D+=+P%5B1%5D%2A.065%2A1+
(2)
+3720+-+I%5B1%5D+=+%28+66000+-+P%5B1%5D+%29%2A.045%2A1+
-----------------------------------------
Substitute (1) into (2)
+3720+-+P%5B1%5D%2A.065+=+%28+66000+-+P%5B1%5D+%29%2A.045%2A1+
+3720+-+.065P%5B1%5D+=+2970+-+.045P%5B1%5D+
+.02P%5B1%5D+=+750+
+P%5B1%5D+=+37500+
and
+P%5B2%5D+=+66000+-+P%5B1%5D+
+P%5B2%5D+=+66000+-+37500+
+P%5B2%5D+=+28500+
---------------------------
She invested $37,500 @ 6.5%
She invested $28,500 @ 4.5%
----------------------------
check:
+I%5B1%5D+=+P%5B1%5D%2A.065%2A1+
+I%5B1%5D+=+37500%2A.065%2A1+
+I%5B1%5D+=+2437.5+
and
+3720+-+I%5B1%5D+=+P%5B2%5D%2A.045%2A1+
+3720+-+I%5B1%5D+=+28500%2A.045%2A1+
+3720+-+I%5B1%5D+=+1282.5+
+I%5B1%5D+=+2437.5+
OK
Get a 2nd opinion if needed