Question

The annual interest rate on $16,000 investment exceed the interest earned on $11,000 investment by $496 the $16,000 investment at 0.6% higher rate of interest then the $11,000 what is the interest rate of each investment

Answer 1

The interest rate of $16,000 investment is 8.6% and the interest rate

of $11,000 investment is 8%

Step-by-step explanation:

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 exceed the interest

earned on $11,000 investment by $496 the $16,000 investment at

0.6% higher rate of interest then the $11,000

Assume that $16,000 investment is $P_{1}$ , the annual interest

rat is $r_{1}$ and the interest is $I_{1}$

Assume that $11,000 investment is $P_{2}$ , the annual interest

rat is $r_{2}$ and the interest is $I_{2}$

∵ The annual interest rate on $16,000 investment exceed the interest

   earned on $11,000 investment by $496

∴ $I_{1}$ - $I_{2}$ = 496

∵ $I_{1}$ = 16,000 ($r_{1}$)(1)

∵ $I_{2}$ = 11,000 ($r_{2}$)(1)

- Substitute them in the equation above

∴ 16,000 ($r_{1}$)(1) - 11,000 ($r_{2}$)(1) = 496

∴ 16,000 ($r_{1}$) - 11,000 ($r_{2}$) = 496 ⇒ (1)

∵ The $16,000 investment at 0.6% higher rate of interest then

   the $11,000

∴ $r_{1}$ = $r_{2}$ + (0.6/100)

∴ $r_{1}$ = $r_{2}$ + 0.006 ⇒ (2)

- Substitute equation (2) in equation (1)

∴ 16,000 ($r_{2}$ + 0.006) - 11,000 ($r_{2}$) = 496

∴ 16,000 $r_{2}$ + 96 - 11,000 $r_{2}$ = 496

- Subtract 96 from both sides and add like terms in L.H.S

∴ 5,000 $r_{2}$ = 400

- Divide both sides by 5,000

∴ $r_{2}$ = 0.08

- Substitute the value of $r_{2}$ in equation (2)

∴ $r_{1}$ = 0.08 + 0.006 = 0.086

- Change them to percent by multiply them by 100%

∴ The interest rate of $16,000 investment = 0.086 × 100% = 8.6%

∴ The interest rate of $11,000 investment = 0.08 × 100% = 8%

The interest rate of $16,000 investment is 8.6% and the interest

rate of $11,000 investment is 8%

Learn more:

You can learn more about interest in brainly.com/question/11149751

#LearnwithBrainly