Question

Please Help!!! :)

Ned currently has an account balance of $3,634.51. He opened the account 13 years ago with a deposit of $2,564.65. If the interest compounds twice a year, what is the interest rate on the account?

A. 4.7%

B. 5.4%

C. 2.7%

D. 1.4%



Leroy opened a savings account 18 years ago with a deposit of $2,651.43. The account has an interest rate of 2.8% compounded twice a year. How much interest has Leroy earned?


A. $4,373.68

B. $2,725.67

C. $4,358.69

D. $1,722.25

Answer 1

I can help! Let's start with 1.
Your formula is r = n(ln(A/P)/(nt)) R being the interest rate of what your trying to find. So A= 3,634.51 P = 2,564.65 N is the number of times compounded per year.N= 2 and t is the number of years t= 13. Also while doing so use PEMDAS Let me know if you need any more help!

Answer 2

Compound interest formula is :

`A=p(1+(r)/(n) )^(nt)`

1.

A = 3634.51

p = 2564.65

r = ?

n = 2

t = 13

Putting the values in formula, we get

`3634.51=2564.65(1+(r)/(2) )^(2*13)`

=>
`3634.51=2564.65(1+(r)/(2) )^(26)`

=>
`(3634.51)/(2564.65)=((2+r)/(2) )^(26)`

=>
`1.417=((2+r)/(2) )^(26)`

We get
`\sqrt[26]{1.417}=(2+r)/(2)`

We get r = 0.02699 and r = -4.02699 (neglecting the negative value)

We get r = 0.027 (rounded)

And in percentage, it is 2.7%.

So, option C is the answer.

2.

p = 2651.43

r = 2.8% or 0.028

n = 2

t = 18

Putting the values in formula we get;

`A=2651.43(1+(0.028)/(2) )^(2*18)`

=>
`A=2651.43(1.014)^(36)`

A = $4373.67

So, interest earned =
`4373.67-2651.43=1722.24` dollars

Hence, option D. $1,722.25 is the answer.