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## Ryan was 8 when his parents invested \$4000 in a certificate of deposit that pays 6%. if ryan leaves the account alone until it reaches \$10,000, how old will he be? (assume that the interest is not compounded annually.)

Ryan was 8 when his parents invested \$4000 in a certificate of deposit that pays 6%. if ryan leaves the account alone until it reaches \$10,000, how old will he be? (assume that the interest is not compounded annually.)

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## Jason corporation has invested in a machine that cost \$75,000, that has a useful life of fifteen years, and that has no salvage value at the end of its useful life. the machine is being depreciated by the straight-line method, based on its useful life. it will have a payback period of six years. given these data, the simple rate of return on the machine is closest to: (ignore income taxes in this problem.)

We can solve
this problem by first calculating the annual net cash inflow. This can be
solved by remembering that:

Payback period
= Initial investment / Annual net
cash inflow

6 years = \$75,000
/ Annual net
cash inflow

Therefore,
Annual net
cash inflow = \$12,500

Next, we
calculate for the cost. The cost we will consider here is the depreciation
value of the machine.
Annual depreciation
= \$75,000 / 15 years = \$5,000

Therefore the annual net operating income is:
Annual net operating income = \$12,500 – \$5,000 = \$7,500

Simple rate of
return is calculated by:
Simple rate of
return = Annual net operating income / Initial
investment
Simple rate of
return = \$7,500 /
\$75,000 = 0.1 = 10%

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## Mrs. Ming invested an amount of money in two accounts for one year. She invested some at 8% interest and the rest at 6% interest. Her total amount invested was \$1,500. At the end of the year, she had earned \$106.40 in interest. How much had Mrs. Ming invested in the account paying 6%?A.\$117B.\$680C.\$760D.\$820

Mrs. Ming invested an amount of money in two accounts for one year. She invested some at 8% interest and the rest at 6% interest. Her total amount invested was \$1,500. At the end of the year, she had earned \$106.40 in interest. How much had Mrs. Ming invested in the account paying 6%?A.\$117B.\$680C.\$760D.\$820

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## Calculating the return on investment using financial leverage. suppose Dave invested only 20,000 of his own money and borrowed 180,000 interest-free from his rich father. what was his return on investment?

The correct option is B) \$680.

Step-by-step explanation:

Consider the provided information.

Her total amount invested was \$1,500. At the end of the year, she had earned \$106.40 in interest.

Let she invested x amount with 8% interest rate.

Total amount she invested was \$1,500, thus the amount she invested with 6% interest rate was 1500-x.

Total interest she earn was \$106.40

Write this information into mathematical form.      Hence, she invested \$820 with 8% interest rate.

The amount she invested with 6% interest rate was 1500-x.

Substitute the value of x in above.

1500-820=680

Hence, she invested \$680 with 6% interest rate.

The correct option is B) \$680.

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## Things did not go quite as planned. you invested \$20,000, part of it in a fund that paid 12% annual interest. however, the rest of the money suffered a 5% loss. if the total annual income from both investments was \$1890, how much was invested at each rate?

Initial investment on Jan 1, 2013 = (500 shares)*(\$24 per share) = \$12,000

Dividend collected at the end of 2013 = \$2.50*500 = \$1,250
Dividend collected at the end of 2014 = \$4*500 = \$2,000
Dividend collected at the end of 2015 = \$3*500 = \$1,500
Mony received from sellng the 500 shares at the end of 2015 = \$20*500 = \$10,000

Total returns at the end of 2015 = 1,250+2,000+1,500+10,000 = \$14,750
Net gains = 14750 – 12000 = \$2,750
Duration = 3 years
Realized total rate of return = 2750/12000 = 0.2292 = 22.9%

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## Mrs. Ming invested an amount of money in two accounts for one year. She invested some at 8% interest and the rest at 6% interest. Her total amount invested was \$1,500. At the end of the year, she had earned \$106.40 in interest. How much had Mrs. Ming invested in the account paying 6%? \$117 \$680 \$760 \$820

The correct option is B) \$680.

Step-by-step explanation:

Consider the provided information.

Her total amount invested was \$1,500. At the end of the year, she had earned \$106.40 in interest.

Let she invested x amount with 8% interest rate.

Total amount she invested was \$1,500, thus the amount she invested with 6% interest rate was 1500-x.

Total interest she earn was \$106.40

Write this information into mathematical form.      Hence, she invested \$820 with 8% interest rate.

The amount she invested with 6% interest rate was 1500-x.

Substitute the value of x in above.

1500-820=680

Hence, she invested \$680 with 6% interest rate.

The correct option is B) \$680.

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## A principal of \$3800 is invested at 6.75% interest, compounded annually. How much will the investment be worth after 6 years?

The formula of the present value of annuity ordinary is
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv present value 1200000
PMT semiannual payment 75000
R interest rate 0.0635
K compounded semiannual 2
N time?
1200000=75000[(1-(1+0.0635/2)^(-2n))÷(0.0635/2)]
Solve for n
1,200,000÷75,000=[(1-(1+0.0635/2)^(-2n))÷(0.0635/2)]

16=[(1-(1+0.0635/2)^(-2n))÷(0.0635/2)]

16×(0.0635÷2)=(1-(1+0.0635/2)^(-2n))

0.508=(1-(1+0.0635/2)^(-2n))

0.508−1=-(1+0.0635/2)^(-2n)

−0.492=-(1+0.0635/2)^(-2n)

0.492=(1+0.0635/2)^(-2n)

-2n=log(0.492)÷log(1+0.0635÷2)

N=-[log(0.492)÷log(1+0.0635÷2)]÷2

N=11.35 years

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## A principal of \$2600 is invested at 7.5% interest, compounded annually. How much will the investment be worth after 8 years?

The formula of the present value of annuity ordinary is
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv present value 1200000
PMT semiannual payment 75000
R interest rate 0.0635
K compounded semiannual 2
N time?
1200000=75000[(1-(1+0.0635/2)^(-2n))÷(0.0635/2)]
Solve for n
1,200,000÷75,000=[(1-(1+0.0635/2)^(-2n))÷(0.0635/2)]

16=[(1-(1+0.0635/2)^(-2n))÷(0.0635/2)]

16×(0.0635÷2)=(1-(1+0.0635/2)^(-2n))

0.508=(1-(1+0.0635/2)^(-2n))

0.508−1=-(1+0.0635/2)^(-2n)

−0.492=-(1+0.0635/2)^(-2n)

0.492=(1+0.0635/2)^(-2n)

-2n=log(0.492)÷log(1+0.0635÷2)

N=-[log(0.492)÷log(1+0.0635÷2)]÷2

N=11.35 years

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## A total of \$6000 is invested: part at 5% and the remainder at 9%. How much is invested at each rate if the annual interest is \$530?

A total of \$6000 is invested: part at 5% and the remainder at 9%. How much is invested at each rate if the annual interest is \$530?

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## An executive invests \$20,000, some at 9% and some at 10% annual interest. If he recieves an annual return of 1,980,how much is invested at each rate?

An executive invests \$20,000, some at 9% and some at 10% annual interest. If he recieves an annual return of 1,980,how much is invested at each rate?

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