Answered

9 Janine deposits [tex]$46 000 into an investment account. Compound interest will be paid at the rate of 3.8% per annum, compounding monthly. If Janine withdraws monthly payments of $[/tex]1000, the amount of interest she has earned
after one year is closest to
A [tex]$1568 B $[/tex]1748
C [tex]$10 432 D $[/tex]13 748
E $35 568



Answer :

Given:

Initial deposit: $46,000

Annual interest rate: 3.8%

Compounding frequency: monthly

Monthly withdrawal: $1,000

Time period: 1 year (12 months)

Step 1: Calculate the monthly interest rate

The annual interest rate is 3.8%, which compounds monthly. Therefore, the monthly interest rate

r is:

=

3.8

%

12

=

0.038

12

=

0.0031667

r=

12

3.8%

=

12

0.038

=0.0031667

Step 2: Calculate the balance after each month

Janine starts with $46,000 and withdraws $1,000 at the end of each month. We need to calculate the balance at the end of each month considering the interest and withdrawals.

Let's denote the balance at the end of the

n-th month as

A

n

.

Initial amount:

0

=

46000

A

0

=46000

For each month

n (from 1 to 12):

=

(

1

×

(

1

+

)

)

1000

A

n

=(A

n−1

×(1+r))−1000

We'll calculate this iteratively for each month.

Month 1:

1

=

(

46000

×

(

1

+

0.0031667

)

)

1000

A

1

=(46000×(1+0.0031667))−1000

1

=

(

46000

×

1.0031667

)

1000

A

1

=(46000×1.0031667)−1000

1

=

46145.6662

1000

A

1

=46145.6662−1000

1

45145.67

A

1

≈45145.67

Month 2:

2

=

(

45145.67

×

1.0031667

)

1000

A

2

=(45145.67×1.0031667)−1000

2

=

45288.1275

1000

A

2

=45288.1275−1000

2

44288.13

A

2

≈44288.13

Month 3:

3

=

(

44288.13

×

1.0031667

)

1000

A

3

=(44288.13×1.0031667)−1000

3

=

44427.4715

1000

A

3

=44427.4715−1000

3

43427.47

A

3

≈43427.47

Month 4:

4

=

(

43427.47

×

1.0031667

)

1000

A

4

=(43427.47×1.0031667)−1000

4

=

43563.6902

1000

A

4

=43563.6902−1000

4

42563.69

A

4

≈42563.69

Month 5:

5

=

(

42563.69

×

1.0031667

)

1000

A

5

=(42563.69×1.0031667)−1000

5

=

42696.7823

1000

A

5

=42696.7823−1000

5

41696.78

A

5

≈41696.78

Month 6:

6

=

(

41696.78

×

1.0031667

)

1000

A

6

=(41696.78×1.0031667)−1000

6

=

41826.7444

1000

A

6

=41826.7444−1000

6

40826.74

A

6

≈40826.74

Month 7:

7

=

(

40826.74

×

1.0031667

)

1000

A

7

=(40826.74×1.0031667)−1000

7

=

40953.5722

1000

A

7

=40953.5722−1000

7

39953.57

A

7

≈39953.57

Month 8:

8

=

(

39953.57

×

1.0031667

)

1000

A

8

=(39953.57×1.0031667)−1000

8

=

40077.2624

1000

A

8

=40077.2624−1000

8

39077.26

A

8

≈39077.26

Month 9:

9

=

(

39077.26

×

1.0031667

)

1000

A

9

=(39077.26×1.0031667)−1000

9

=

39197.8117

1000

A

9

=39197.8117−1000

9

38197.81

A

9

≈38197.81

Month 10:

10

=

(

38197.81

×

1.0031667

)

1000

A

10

=(38197.81×1.0031667)−1000

10

=

38315.2176

1000

A

10

=38315.2176−1000

10

37315.22

A

10

≈37315.22

Month 11:

11

=

(

37315.22

×

1.0031667

)

1000

A

11

=(37315.22×1.0031667)−1000

11

=

37429.4764

1000

A

11

=37429.4764−1000

11

36429.48

A

11

≈36429.48

Month 12:

12

=

(

36429.48

×

1.0031667

)

1000

A

12

=(36429.48×1.0031667)−1000

12

=

36540.5857

1000

A

12

=36540.5857−1000

12

35540.59

A

12

≈35540.59

Step 3: Calculate the total interest earned

The total interest earned is the difference between the total amount in the account after one year (with monthly withdrawals) and the total amount deposited minus the total withdrawals.

Total amount deposited initially:

46000

46000

Total withdrawals over the year:

1000

×

12

=

12000

1000×12=12000

Total amount at the end of the year:

12

35540.59

A

12

≈35540.59

Total interest earned:

Interest

=

12

(

46000

12000

)

Interest=A

12

−(46000−12000)

Interest

=

35540.59

34000

Interest=35540.59−34000

Interest

=

1554.59

Interest=1554.59

The closest answer to $1554.59 is:

Answer:

A $1568