SOLUTION: the annual interest on a 9000 investment exceeds the interest earned on a 3000 investment by 438. the 9000 is invested at 0.7% higher rate of interest than the 3000. what is the in
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Question 414328: the annual interest on a 9000 investment exceeds the interest earned on a 3000 investment by 438. the 9000 is invested at 0.7% higher rate of interest than the 3000. what is the interest of each
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
interest on 3000 investment = x
interest on 9000 investment = x + 438
interest rate of 3000 investment = y
interest rate of 9000 investment = y + .007
you convert interest rate percent to interest rate by dividing by 100%.
you get .7% = .7%/100% = .007
investment times interest rate = interest.
you get:
9000 * (y + .007) = x + 438
3000 * y = x
you can replace x in the first equation with the value of x from the second equation.
this reduces the first equation to one unknown which can be solved for.
you get:
9000 * (y + .007) = (3000 * y) + 438
all you have to do now is solve for y.
simplify the equation to get:
9000 * y + .007 * 9000 = 3000*y + 438
simplify further to get:
9000 * y + 63 = 3000 * y + 438
subtract 63 from both sides of the equation and subtract 3000 * y from both sides of the equation to get:
9000 * y - 3000 * y = 438 - 63
simplify to get:
6000 * y = 375
divide both sides of the equation by 6000 to get:
y = .0625
that's the interest rate for the 3000 investment.
3000 * .0625 = 187.5
that's the interest for the 3000 investment which equal to x.
the interest rate on the 9000 investment is equal to .0625 + .007 = .0695
9000 * .0695 = 625.5
that's the interest for the 9000 investment which should be equal to x + 438
x = 187.5
x + 438 = 187.5 + 438 = 625.5
the requirements of the problem are satisfied so the answer is correct.
the answer is:
the interest rate of the 3000 investment is .0625
the interest rate of the 9000 investment is .0695
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