The interest rate of a $16,000 investment is 7.5% and the interest rate
of $6000 investment is 8%
Compound interest is the interest levied on the interest. The annual interest formula is I = P r t, where
1. P is the money invested
2. r is the interest rate in decimal
3. t is the time
The annual interest rate on $16,000 investment exceeds the interest
earned on $6000 investment by $830 the $16,000 investment at
0.5% higher rate of interest than the $11,000
Assume that $16,000 investment is I1 , the annual interest
Assume that $6000 investment is I2 , the annual interest
The annual interest rate on a $16,000 investment exceeds the interest
earned on $6000 investment by $830
I1 - I2 = 830
I1 = 16,000 (r1)(1)
I2 = 6000 (r2)(1)
Substitute them in the equation above
16,000 (r1)(1) - 6000(r1)(1) = 830
16,000 (r2) - 6000 (r2) = 830⇒ (1)
The $16,000 investment at 0.6% higher rate of interest then the $11,000
r1 = r2 + 0.005
Substitute equation (2) in equation (1)
16,000 ( r2 + 0.006) - 6000(r2) = 830
16,000r2 + 80 - 6000r2 = 830
Subtract 80 from both sides and add like terms in L.H.S
10000r2 = 750
r2 = 0.075
- Substitute the value of in equation (2)
∴ = 0.075 + 0.005 = 0.080
- Change them to per cent by multiplying them by 100%
The interest rate of $16,000 investment = 0.075 × 100% = 7.5%
The interest rate of $6000 investment = 0.08 × 100% = 8%
The interest rate of $16,000 investment is7.5 % and the interest
rate of $6000 investment is 8%
To know more about interest follow
https://brainly.com/question/24924853
#SPJ1
