Question

Jocelyn invested $56,000 in an account paying an interest rate of 7 3 8 7 8 3 ​ % compounded continuously. Oliver invested $56,000 in an account paying an interest rate of 7 1 4 7 4 1 ​ % compounded monthly. After 13 years, how much more money would Jocelyn have in her account than Oliver, to the nearest dollar?

Answer 1

$2761

Explanation:

Rate 1: 7/38%=7+3/8=

7.375%→0.07375

Rate 2: 7/14%=7+1/4=

7.25%→0.0725

​

Compounded Continuously:

A=Pe^{rt}

P=56000 r=0.07375 t=13

Given values

A=56000e^0.07375(13)

Plug in

A=56000e^{0.95875}

Multiply

A=146072.298

Use calculator (with e button)

Compound interest formula

P=56000 r=0.0725 t=13 n=12

Given values

A=56000\(1+{0.0725}{12})^12(13)

Plug in values

A=56000(1.0060417)^{156}

Simplify

A=143310.8391

Use calculator

How much more money Jocelyn has:

146072.298−143310.8391

2761.4589

$2761

Answer 2

$4,205

Explanation:

Jocelyn

A = p × e^rt

Where,

P = principal = $56000

e = Napier's number = 2.7283

r = interest rate = 7 3/8% = 0.07375

t = time = 13 years

A = 56,000 × 2.7283^(0.07375×13)

= 56,000 × 2.7283^0.95875

= 56,000 × 2.6177

= 146,591.2

Approximately

A = $146,591 to the nearest dollar

Oliver

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

= 56,000(1 + 0.0725/12)^12*13

= 56,000(1 + 0.0060)^156

= 56,000(1.0060)^156

= 56,000(2.5426)

= 142,385.6

Approximately

A = $142,386 to the nearest dollar

Jocelyn will have $4,205 more than Oliver