Word problem on interest/investment

Question

Thirty-five thousand in bonds, part paying 5% interest, part 4.5% and the remainder 3.5%. How much is invested at the 4.5% rate, if the total annual income is $1510 and that on the 5% bond equals on that 3.5 bonds?

Answer 1

There are three pools of money - n invested at 5%, m invested at 4.5%, and q invested at 3.5%. The interest on any one of these pools is I = R x P where I is the interest, R is the interest rate, and P is the principle.

n + m + q = 35,000 (given) and n = q (given), so 2n + m = 35000

since m is the desired quantity, solve the above for n: n = (35000 - m)/2

The total interest is the sum of the interest of the individual amounts invested:

I = 1510 = n(0.05) + m(0.045) + n(0.035) = 0.085 n + 0.045 m replace n with the equation found above:

1510 = (0.085)(35000 - m)/2 + (0.045)m

solve for m:

multiply both sides of the equation by 2 to eliminate the fraction:

3020 = (0.085)(35000) - (0.085)m + (0.090)m

Group like terms and reduce:

3020 - 2975 = (0.005)m

Solve for m:

m = 45/(0.005) = 9000

Answer: $9000 are invested at 4.5%

Answer 2

Let x dollars invested at 4.5% interest rate. Then (35000 - x)/2 invested at 3.5% and 5%.

Then we have equation: 0.045x + 0.035(35000 - x)/2 + 0.05(35000 - x)/2 = 1510;

then let multiply this equation by 2000 we get 90x + 35(35000 - x) + 50(35000 - x) = 3020000;

90x - 35x - 50x = 3020000 - 85·35000; 5x = 45000; x = 9000.

Answer: $9000 invested with 4.5% rate