Let the amount invested at 9% be $y

therefore amount invested at 12% becomes 2y and at 10% becomes y +500

Total Interest in year = 1555

9% of y + 12 % of (2y) + 10% of (y +500) = 1555

0.09y + 0.24 y + 0.1(y+500) = 1555

0.33y + 0.1y + 50 = 1555

0.43 y = 1555 – 50

y = 1505/0.43 = 3500

Money intested at 9% = y = $3500

money invested at 12% = 2y = 2 x 3500 = $7000

money invested at 10% = y + 500 = 3500 + 500 = $4000

Sol2:

Let the rate of interest on $30,000 be x %

therefore the interest on 48000 becomes x + 2 %

Total interest for a month = 600

[x% of 30000 x 1/12] + [(x+2)% of 48000 x 1/12] = 600

300x / 12 + (x+2) x 40 = 600

25x + 40x + 80 = 600

65x = 520

x = 8

She pays 8% p.a on $30000 and 10% p.a. on $48000

#2… ($30K * x) + ($48K * 1.02x) = $6K

Solve for x to get the interest rate on the cheaper loan, multiply that answer by 1.02 to get the interest rate on the more expensive loan.