Question

Anthony is going to invest in an account paying an interest rate of 2.7% compounded quarterly. How much would Anthony need to invest, to the nearest hundred dollars, for the value of the account to reach $130 in 10 years?

Answer 1

Answer: 95810

Step-by-step explanation:

Compound interest formula

A=240000\hspace{35px}t=20\hspace{35px}r=0.046\hspace{35px}n=12

A=240000t=20r=0.046n=12

Given values

240000=

240000=

\,\,P\left(1+\frac{0.046}{12}\right)^{12(20)}

P(1+

12

0.046

​

)

12(20)

Plug in values

240000=

240000=

\,\,P(1.003833333)^{240}

P(1.003833333)

240

Simplify

240000=

240000=

\,\,P(2.504880829)

P(2.504880829)

Evaluate in calculator

\frac{240000}{2.504880829}=

2.504880829

240000

​

=

\,\,\frac{P(2.504880829)}{2.504880829}

2.504880829

P(2.504880829)

​

Divide by 2.504880829

95812.9413769=

95812.9413769=

\,\,P

P

Use calculator

P\approx

P≈

\,\,95810

95810 .

Answer 2

$99

Explanation:

Given that :

Rate (r) = 2.7% = 0.027 Compounded quarterly

Final amount (A) = $130

Period (t) = 10 years

Principal = invested amount (p) =?

n = number of compounding tines per period = 12/3 = 4

Using the formula :

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

130 = p(1 + 0.027/4)^(4*10)

130 = p(1 + 0.00675)^40

130 = p(1.00675)^40

130 = p(1.3087766)

P = 130 / 1.3087766

P = 99.329402

P = $99