SOLUTION: The annual interest rate on a $14000 investment exceeds the interest on a $7000 investment by $672. The $14000 is invested at a 0.3% higher rate of interest than the $7000 . what
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Question 853703: The annual interest rate on a $14000 investment exceeds the interest on a $7000 investment by $672. The $14000 is invested at a 0.3% higher rate of interest than the $7000 . what is the interest rate if each investment?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
The annual interest on a $14000 investment exceeds the interest on a $7000 investment by $672. The $14000 is invested at a 0.3% higher rate of interest than the $7000 . what is the interest rate if each investment?
Let interest rate of $14000 be x%
and Interest rate for $7000 be y %
According to the first condition
14000 *x% - 7000*y% = 672
multiply by 100
14000x-7000y = 67200
/700
20x-10y=96.................(1)
II condition
x%=y%+0.3%
x=y+0.3
x-y=0.3..................................(2)
solve (1) & (2)
20 x -10 y = 96 .............1
Total value
1 x -1 y = 0.30 .............2
Eliminate y
multiply (1)by 1
Multiply (2) by -10
20.00 x -10.00 y = 96.00
-10.00 x + 10.00 y = -3.00
Add the two equations
10.00 x = 93.00
/ 10.00
x = 9.30
plug value of x in (1)
20.00 x -10.00 y = 96.00
186.00 -10.00 y = 96.00
-10.00 y = 96.00 -186.00
-10.00 y = -90.00
y = 9.00
x= 9.30 Interest rate on $14000
y= 9.00 Interest rate on $7000
m.ananth@hotmail.ca
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