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Mathematics 16 Online
nightmare54362:

Chanelle deposits 7,500 into the bank. She does not withdraw or deposit money for 6 years. She earns 6% interest during that time. A. how much interest will she have earned at the end of 6 years? B. What will the balance be when she is finally able to withdraw her money?

NerdyLizard:

One second. :D

NerdyLizard:

Ok, so, every year she is gaining interest of 0.06

NerdyLizard:

That means that at the end she should earn 6*0.06.

NerdyLizard:

Aka 0.36. 0.36 will allow you to find your second answer. :D

NerdyLizard:

@nightmare54362

Florisalreadytaken:

simple interest formula (just googled it) \[ A=P(1+it) \] where A is the final amount, P is the principal, i is the interest rate and t is the time (in years) in our case we have to find A, we are given P=$7500 ; i=6% or 0.06 ; t=6 years thus, we just solve this: \[ A=7500(1+0.06 \times 6) \] \[ A=7500 \times 1.36 \] \[ A=10200 \] usually, i would do it the long way around (every year they get their 6%): 1st year: \( 7500 \times 1.06 = $ 7950 \) 2nd year: \( 7950 \times 1.06 = $ 8427 \) 3rd year: \( 8427 \times 1.06 =$8932.62 \) 4th year: \( 8932.62 \times 1.06 = $ 9468.5772 \) 5th year: \( 9468.5772 \times 1.06 = $ 10036.69 \) 6th year: \( 10036.69 \times 1.06 = \color{lightskyblue}{$10638.8914} \) i am not sure if this is the same, but the logic seems right

mhanifa:

@florisalreadytaken wrote:
1st year: \( 7500 \times 1.06 = $ 7950 \) 2nd year: \( 7950 \times 1.06 = $ 8427 \) 3rd year: \( 8427 \times 1.06 =$8932.62 \) 4th year: \( 8932.62 \times 1.06 = $ 9468.5772 \) 5th year: \( 9468.5772 \times 1.06 = $ 10036.69 \) 6th year: \( 10036.69 \times 1.06 = \color{lightskyblue}{$10638.8914} \) i am not sure if this is the same, but the logic seems right
This part is compounded interest

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