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) About Me  (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