Question 899353: The annual interest on a $15,000 investment exceeds the amount of interest earned by a $1000 investment by $410. The $15,000 is invested at a .4% higher rate of interest than the $1000 investment. What is the rate of interest for each investment?
Your help is appreciated! I can't seem to figure it out!
Found 2 solutions by mananth, MathTherapy: Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let interest rate on $1000 be x
15000 @ (x+4)%
interest earned on 15000 less interest earned on 1000 = 410
15000*(x+4)% -1000*x%=410
multiply by 100
15000(x+4)-1000x= 41000
15000x+60000-1000x=41000
there appears to be a typo in the problem
Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
The annual interest on a $15,000 investment exceeds the amount of interest earned by a $1000 investment by $410. The $15,000 is invested at a .4% higher rate of interest than the $1000 investment. What is the rate of interest for each investment?
Your help is appreciated! I can't seem to figure it out!
Let interest rate on $15,000 investment be F
Since $15,000 investment's interest rate was higher than the $1,000 investment's interest rate by .4%,
then interest rate on $1,000 investment = F - .004
Therefore, interest earned on $15,000 investment = 15,000F
Interest earned on $1,000 investment = 1,000(F - .004), or 1,000F - 4
Since the interest earned on the $15,000 investment exceeds the interest earned on the $1,000 investment
by $410, then we can say that:
15,000F = 1,000F - 4 + 410
15,000F - 1,000F = 406
14,000F = 406
F, or interest rate on $15,000 investment = , or , or %
Interest rate on $1,000 investment = 0.029 - .004, or 0.025, or %
You can do the check!!
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