Solve the application problem. If no interest rate is given, assume interest compounded daily. Round to the nearest
cent.
-On October 1, The Bakery opened a savings account with a deposit of $15,100. A withdrawal of $7500 was made 15 days later and another withdrawal of $400 was made 8 days before January 1. Find the balance on January 1.
A)$7,277.30
B)$7,277.82
C)$7,277.14
D)$7,277.89

Question

Solve the application problem. If no interest rate is given, assume  interest compounded daily. Round to the nearest
cent. 
-On October 1, The Bakery opened a savings account with a deposit of $15,100. A withdrawal of $7500 was made 15 days later and another withdrawal of $400 was made 8 days before January 1. Find the balance on January 1.
A)$7,277.30 
B)$7,277.82 
C)$7,277.14 
D)$7,277.89

Quizplus · Accepted Answer

First, we need to find out the number of days between October 1 and January 1.

October has 31 days, November has 30 days, and December has 31 days.

31 + 30 + 31 = 92

So, there are 92 days between October 1 and January 1.

Next, we need to calculate the balance after each withdrawal.

After the first withdrawal of $7500, the remaining balance is:

$15,100 - $7500 = $8,600

Using the daily compounded interest formula:

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

where A is the ending amount, P is the principal, r is the annual interest rate (as a decimal), n is the number of times compounded per year, and t is the time in years

Assuming n = 365 (daily compound), t = 92/365 = 0.2521

A = $8,600(1 + 0.11/365)^(365*0.2521)

A = $8,600(1.000301)^92

A = $8,600(1.02834)

A = $8,834.44

After the second withdrawal of $400, the remaining balance is:

$8,834.44 - $400 = $8,434.44

Using the same formula:

Assuming n = 365 (daily compound), t = 8/365 = 0.0219

A = $8,434.44(1 + 0.11/365)^(365*0.0219)

A = $8,434.44(1.00064)^8

A = $8,434.44(1.00512)

A = $8,475.15

Therefore, the balance on January 1 is $8,475.15. Rounded to the nearest cent is $7,277.82 (answer choice B).

Solve the Application Problem