Financial Algebra Cengage

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 46 Problem 1 Answer

Given: A corporation has a market capitalization of $24,000,000,000 with 250M outstanding shares.

To calculate the price per share. We will have to divide the market capitalization with outstanding shares.

We have the market capitalization as $24,000,000,000

The number of outstanding shares is 250,000,000

Price per share= Market capitalization/ Number of outstanding shares

​=$24,000,000,000/250,000,000

=96

​The price per share is obtained to be $96.

Page 46 Problem 2 Answer

Given: Qual Comm, Inc. instituted a4−for−1 split in November.  After the split, Elena owned 12,800 shares.

To find the number of shares owned before the split.

We will use the fact that 4−for−1split means the number of share after the split gets four times.

There is 4−for−1 split.

Hence the  number of share after the split gets four times.

Therefore the number of shares before the split will be:

Number of shares after the split /4

=12800/4

=3200​

The number of shares owned by Elena before the split is 3200.

Page 48 Problem 3 Answer

Given: A major drugstore chain whose stocks are traded on the New York Stock Exchange was considering a 2-for-5 reverse split.

The pre-split market cap was1.71B

To find: The post-split market cap

Solution: The market cap pre and post split remains the same.

Splitting and reverse splitting only affects the number of shares and the value per share, it does not change the total market capital

A major drugstore chain whose stocks are traded on the New York Stock Exchange was considering a 2-for-5 reverse split.

If the pre-split market cap was1.71B, then the post-split market cap will also be1.71B

Page 48 Problem 4 Answer

Given: Gabriella owned1,045shares of Hollow Corporation at a price of $62.79. The stock split3−for−2

To find: Financial effect of the split for Gabriella

Solution: We will find the following: Post split number of sharesPost split value per shareCash settlement for fractional shares

Post-split number of shares=a/b×Pre-split number of shares

=3/2×1045

=1567.5

Post-split share price=b/a×Pre-split share price

=2/3×62.79

=41.86

Fractional shares are not traded, so the corporation paid him the market value of 0.5 shares.

Fractional part×Market price=0.5×41.86=20.93

Gabriella received$20.93 in cash and 1567 shares worth$41.86 each.

Page 49 Problem 5 Answer

We have the statement: Perception is strong and sight is weak.

In strategy, it is important to see distant things as if they were close and to take a distanced view of close things.

The given statement implies that the perception of the value per stock is strong and might strongly influence any decision-making of an investor, while the investors don’t tend to notice the true reason of a stock split .

This is way it is important to see distant things as if they are close, such that an investor can see the true reason by stock splits and can take them into account for any decision-making.

A split does not have any meaningful benefit to the investor as the main reason of the split is the perception.

The perception of the value per stock is strong and might strongly influence any decision making of an investor.

Page 49 Problem 6 Answer

Given:  In February, Robbins and Myers, Inc. executed a 2-for-1split. Janine had 470 shares before the split.

Each share was worth $69.48.To find the number of shares she hold after the split.

We will use the fact that 2 for one split means the number of share gets doubled.

Now we have that Janine held 470 shares before  the split.

It was 2 for one split therefore the number of shares get doubled.

Hence we have the number of shares as: 470×2=940

The number of shares that Janine held after the splits is 940.

Page 49 Problem 7 Answer

Given:  In February, Robbins and Myers, Inc. executed a 2-for-1 split. Janine had 470 shares before the split.

Each share was worth $69.48.To find the post split price per share. We will use the worth of each share.

Now we have that there was 2−for one split,this will imply that price per share will get halved.

Now each share was originally $69.48

Hence after the split we will have: 69.48/2=$34.74

The post-split price per share is $34.74.

Page 49 Problem 8 Answer

Given:  In February, Robbins and Myers, Inc. executed a2−for−1 split. Janine had 470 shares before the split.

Each share was worth $69.48.To show that the split was monetary non-event for Janine.

We will have to show that the  Total worth before split=Total worth after split

Now we know that before the split Janine had 470 shares worth $69.48 each.

Hence the total worth before the split was: 470×69.48=$32,655.60

After the split, she had 940 shares worth $34.74. each.

Hence the total worth after the split was: 940× 34.74=$32,655.60

Therefore, Total worth before split=Total worth after split

We have shown that the split was a monetary non-event for Janine.

Page 49 Problem 9 Answer

Given:  In February, Robbins and Myers, Inc. executed a 2-for-1 split. Janine had 470 shares before the split.

Each share was worth $69.48.To find the number of shares she hold after the split.

We will use the fact that 2 for one split means the number of share gets doubled.

Now we have that Janine held 470 shares before  the split.

It was 2 for one split therefore the number of shares get doubled.

Hence we have the number of shares as: 470×2=940

The number of shares that Janine held after the splits is 940.

Page 49 Problem 10 Answer

Given:  In February, Robbins and Myers, Inc. executed a 2-for-1 split. Janine had 470

shares before the split. Each share was worth $69.48.To find the post split price per share.We will use the worth of each share.

Now we have that there was 2−for one split,this will imply that price per share will get halved.

Now each share was originally $69.48

Hence after the split we will have: 69.48/2

=$34.74

The post-split price per share is $34.74.

Page 49 Problem 11 Answer

Given:  In February, Robbins and Myers, Inc. executed a 2-for-1 split. Janine had 470

shares before the split. Each share was worth $69.48.To show that the split was monetary non-event for Janine.

We will have to show that the  Total worth before split=Total worth after split

Now we know that before the split Janine had 470 shares worth $69.48 each.

Hence the total worth before the split was: 470×69.48=$32,655.60

After the split she had 940 shares worth $34.74. each.

Hence the total worth after the split was: 940× 34.74=$32,655.60

Therefore,  Total worth before split=Total worth after split

We have shown that the split was a monetary non-event for Janine.

Page 49 Problem 12 Answer

Given: Vilma owns 750 shares of Aeropostale. On August 22, the corporation instituted a 3−for−2 stock split.

Before the split, each share was worth $34.89.To find the number of shares that Vilma hold after the split.

We will use the fact that three for two stock split means we will have to multiply the shares with 3/2.

Now we know that there is 3−for−2 share split.

Before the split the number of shares were 750.

Hence after the split it will be multiplies by 3/2 that is:3/2×750=1125

The number of shares that Vilma hold after the split is 1125.

Page 49 Problem 13 Answer

Given: Vilma owns 750 shares of Aeropostale. On August 22, the corporation instituted a 3−for−2  stock split.

Before the split, each share was worth $34.89.To find the post-price per split after the split. We will use the worth of each share.

There is a three for two split share.Hence the price will get two third of the price before the split.

Now each share was originally $34.89

Hence after the split it will be:2/3×34.89=23.26

The post-price per share after the split is $23.26.

Page 49 Problem 14 Answer

Given: Vilma owns 750 shares of Aeropostale. On August 22, the corporation instituted a 3−for−2

stock split. Before the split, each share was worth $34.89.To show that the split was a monetary non-event for Vilma.

We will have to show that the  Total worth before split=Total worth after split

Now we know that before the split there were 750 shares worth $34.89 per share

Hence the total worth before the split was: 750×34.89=$26167.50

After the split there were 1125 shares worth $23.26

Hence  the total worth after the split was: 1125×23.26=$26167.50

Therefore Total worth before split=Total worth after split

We have shown that  the split was a monetary non-event for Vilma.

Page 49 Exercise 1 Answer

Given:  Versant Corporation executed a 1−for−10 reverse split on August 22.

At the time, the corporation had 35,608,800 shares outstanding and the pre-split price per share was $0.41

.To find the number of shares that were outstanding after the split.

We will use the fact that one for ten stock split means we will have to multiply the shares with 1/10

Now we know that there is1−for−10 share split.

Before the split the number of shares were 35,608,800 shares Hence after the split it will be multiplies by1/10

Hence we get: 1/10×35608800=3,560,880

The number of shares that were outstanding after the split were 3,560,880.

Page 49 Exercise 2 Answer

Given:  Versant Corporation executed a 1−for−10 reverse split on August 22.

At the time, the corporation had 35,608,800 shares outstanding and the pre-split price per share was $0.41.

To find the post-price per share after the split.We will use the worth of each share.

There is a one for ten split share. Hence the price will get ten times  the price before the split.

Now each share was originally $0.41

Hence after the split it will be: 10×0.41=$4.10

The post price per share after the split was $4.10. 

Page 49 Exercise 3 Answer 

Given: Versant Corporation executed a 1−for−10 reverse split on August 22.

At the time, the corporation had 35,608,800 shares outstanding and the pre-split price per share was $0.41.

To show that this split was a monetary non-event for the corporation.

We will have to show that the  Total worth before split=Total worth after split

Now we know that before the split there were35,608,800 shares worth $0.41.

Hence the total worth before the split was: 35608800×0.41=$14,599,608

After the split there were3,560,880 shares worth $4.10.

Hence  the total worth after the split was: 3560880×4.10= $14,599,608

Therefore Total worth before split=Total worth after split

We have shown that the split was a monetary non-event for the corporation.

Page 50 Exercise 4 Answer

Given: Jon noticed that most traditional splits are in the form of x−for−1

He says that in those cases, all you need do is multiply the number of shares held by x

and divide the price per share by x to get the post-split numbers.

To verify Jon’s method works to determine the post-split price and shares outstanding for Hansen Natural Corporation which executed a 4−for−1 split on July 10 with 22,676,800 outstanding shares and a market price of $203.80 per share before the split.

We will have to find the number of shares after the split.

Now we have four for one split.

This implies that the number of shares will become four times.

Now before the split there were 22,676,800 shares.

Hence after the split there will be: 22,676,800×4=90,707,200 shares Now the price will get one fourth of the before the split.

Hence the price per share after the split will be:

1/4×203.80=$50.95

Hence the number of shares were multiplied by x=4

whereas the price was divided by it after the split.

We have verified that Jon’s method works to determine the post-split price and shares outstanding for Hansen Natural Corporation which executed a 4−for−1 split on July 10 with 22,676,800 outstanding shares and a market price of $203.80 per share before the split.

Page 50 Exercise 5 Answer

Given:  Jon also noticed that every traditional split ratio can be written in the form x−for−1.

Examine how the  3−for−2  traditional split can be expressed as 1.5−for1.

Given split:5−for−4 To express the given split ratios as x−for−1.

We will use one of the given example in the question.

5−for−4

5/4=x/1

⇒5=4x

⇒x=1.25

Hence  1.25−for−1

The traditional split ratio 5−for−4 can be expressed as 1.25−for−1.

Page 50 Exercise 6 Answer

Given:  Jon also noticed that every traditional split ratio can be written in the form x−for−1.

Examine how the  3−for−2  traditional split can be expressed as 1.5−for−1 .

Given split:6−for−5

To express the given split ratios as x−for−1. We will use one of the given example in the question.

6-for-5

6/5=x/1

=>6=5x

=>x=1.2

Hence 1.2 – for 1

The traditional split ratio6−for−5 can be expressed as 1.2−for−1.

Page 50 Exercise 7 Answer

Given:  Jon also noticed that every traditional split ratio can be written in the form x−for−1.

Examine how the  3−for−2  traditional split can be expressed as 1.5−for−1 .Given split: 5−for−2

To express the given split ratios as x−for−1. We will use one of the given example in the question.

5-for-2

5/2=x/1

=> 5=2x

=>x=2.5

Hence 2.5 -for- 1

The traditional split ratio5−for−2 can be expressed as 2.5−for−1.

Given:  Jon also noticed that every traditional split ratio can be written in the form x−for−1.

Examine how the 3−for−2 traditional split can be expressed as 1.5−for−1.

Given split:8−for−5

To express the given split ratios as x−for−1.

We will use one of the given example in the question.

8-for-5

8/5=x/1

=>8=5x

=>x=1.6

Hence 1.6 -for-1

The traditional split ratio8−for−5 can be expressed as 1.6−for−1.

Page 50 Exercise 8 Answer

Given: Monarch Financial Holdings, Inc. executed a 6−for−5 traditional split on October 5.

Before the split there were approximately 4,800,000 shares outstanding, each at a share price of $18.00.

To determine the post-split share price and the number of shares.We will use the worth of each share.

Now we know that there is6−for−5 share split.

Before the split the number of shares were 4,800,000 shares Hence after the split it will be multiplies by 6/5 Hence we get: 6/5×4800000=5,760,000

There is a6−for−5  split share.Hence the price will get 5/6  the price before the split.

Now each share was originally$18.00

Hence after the split it will be : 5/6×18=$15.00

The post-split share price is $15.00 The number of shares outstanding is 5,760,000.

Page 50 Exercise 9 Answer

Given: Monarch Financial Holdings, Inc. executed a 6−for−5 traditional split on October 5.

Before the split there were approximately 4,800,000 shares outstanding, each at a share price of $18.00.To compare the results from part a. with that obtained by using Jon’s method.

Jon’s method says that 6−for−5 is the same as 1.2−for−1. We will find the answer using 1.2-for-1 and then compare.

Now we know that there is1.2−for−1 split Before the split the number of shares were 4,800,000 shares

Hence after the split it will be multiplies by1.2

Hence we get:

1.2×4800000=5,760,000

There is a 1.2−for−1 split. Hence the price will get divided by 1.2

Now  each share was originally$18.00

Hence after the split it will be: 18/1.2=15.00

Hence we get the same answers.

We obtain the same answers by using both the methods of that in part a and by John’s method.

Page 50 Exercise 10 Answer

Given: On June 19  California Pizza Kitchen, Inc. instituted a 3−for−2 split.

At that time Krista owned 205 shares of that stock. The price per share was $33.99.

After the split, Krista received a check for a fractional part of a share.To find the amount of the check.We will use the fact that 3−for−2

split means we will have to multiply two third for the original price.

Now we know that there is3−for−2 share split.

Before the split the number of shares were205 Hence after the split it will be multiplies by 3/2

Hence we get 3/2×205=307shares. there is 3 -for-2-share split

Hence the price will get two third of the original price Hence after the split it will be:2/3×33.99=22.66

Now the fractional part .5 is exchanged for the monetary.

Hence the amount of the check was $11.33.

Krista received a check of $11.33.

Page 50 Exercise 11 Answer

Given spreadsheet:

Write a spreadsheet formula that will calculate the post-split number of outstanding shares in C3.

We will use the formula:  Post−split number of shares =a/b× Pre−split number of shares

In general for any a−for−b split the formula which is applied is:  Post−split number of shares =a/b× Pre−split number of shares

Hence applying that here we get: the formula on cell C₃=(B₂/C₂)×B₃

The formula that calculates the post-split number of outstanding shares in C₃ is: (B₂/C₂)×B₃

Page 50 Exercise 12 Answer

Given spreadsheet:

Write the spreadsheet formula that will calculate the post-split price per share in C4.

We will use the formula:  Post−split number of shares =b/a× Pre−split number of shares

In general, for any a−for−b split, the formula which is can be applied is: Post−split share price =b/a× Pre−split share price

Hence applying that here we get: the formula of cell C₄=(C₂/B₂)×B₄

The spreadsheet formula that will calculate the post-split price per share in C₄ is: (C₂/B₂)×B₄

Page 50 Exercise 13 Answer

Given spreadsheet:

To write the pre-split market cap formula in cell B9 and the post-split market cap formula in C9.

We will use the formula Market cap = outstanding shares price per share

Now we know that that market cap is given by: Market cap = outstanding shares per share

Hence we get:

Formula on B₅=B₃×B₄

Formula on C₅=C₃×C₄

The pre-split market cap formula in cell B5 is B3×B4 and the post-split cap formula for cell C5 is C3×C4.

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.9 The Stock Market

Financial Algebra 1st Edition Chapter 4 Consumer Credit

Page 175 Problem 1 Answer

Given: Heather’s guitar costs x and she could savey.

To find: express algebraically the number of months it would take Heather to save for the guitar.

We will use the unitary method to solve the problem and get the result.

Since, we know that for a month y dollar is saved.

So, we can say that for n months, we get the saving as n y.

As, the price of guitar costs x

dollars. So, we get the number of months as

x=ny

⇒n=x/y

Thus, we can say that x/y number of months it would take Heather to save for the guitar when Heather’s wants to buy a guitar.

Page 176 Problem 2 Answer

Given: Heather’s wants to buy a guitar.To find: Express the finance charge algebraically.

Total money will be given by the sum of total monthly payment along with down payment.

Since, we know that the monthly payment is w.

So, we get the down payment value as

Also, we know that finance charge, will be found by subtracting purchase price from the total amount paid.

So, we will get the value as

12×w−p+0.2p=12w−0.8p.

Thus, we can say that when Heather’s wants to buy a guitar then the finance charge algebraically will be 12w−0.8p.

Page 177 Problem 3 Answer

Given: The Whittendale family purchases a new refrigerator on a no-interest for-one-year plan.

To find: express their last month’s payment algebraically.In order to get the finance charge, we will subtract purchase price from the total amount paid.

When we have the down payment as zero dollars and we will pay the money at zero interest for x dollars per month then we get the last month’s payment algebraically as

1385−11×x=1385−11x.

Thus, we can say that 1385−11x

will be the last month’s payment algebraically when the Whittendale family purchases a new refrigerator on a no-interest for-one-year plan.

Page 178 Problem 4 Answer

Given:  Craig wants to purchase a boat that costs$1420.

To find:Does he have enough for the down payment.

We will get the down payment as the product of the purchase price along with rate that is being given.

Since, we know that down payment is the product of the purchase price along with rate that is being given.

So, we get the value of down payment as

​=20%×1420

=0.20×1420

=284

Hence, there is a shortage of $284−$250=$34.

Thus, we can say that $34 is shortage for the down payment when Craig wants to purchase a boat that costs$1420.

Page 178 Problem 5 Answer

Given: Jean bought a$1980 snow thrower on the installment plan.

To find: How much is the down payment.

We will get the down payment as the product of the purchase price along with rate that is being given.

Since, we know that down payment is the product of the purchase price along with rate that is being given.

So, we get the value of down payment as

=10%×1980

=0.10×1980

=198

​Thus, we can say that$198 is the down payment when Jean bought a $1980 snow thrower on the installment plan.

Page 178 Problem 6 Answer

Given: Jean bought a snow thrower on the installment plan.

To find: What is the total amount of the monthly payments.

We will get the value of total amount of monthly payment will be a product of number of months along with monthly payment.

Since, we know that value of total amount of monthly payment will be a product of number of months along with monthly payment.

So, we get the value as18×116=2088.

Thus, we can say that $2088 is the total amount of the monthly payments when Jean bought a snow thrower on the installment plan.

Page 178 Problem 7 Answer

Given: Jean bought a snow thrower on the installment plan.

To find: How much did Jean pay for the snow thrower on the installment plan.

Total money will be given by the sum of total monthly payment along with down payment.

Since, we know that total money will be given by the sum of total monthly payment along with down payment.

So, Jean pay for the snow thrower will be198+2,088=2,286.

Thus, we can say that $2286 did Jean pay for the snow thrower on the installment plan.

Page 178 Problem 8 Answer

Given: Jean bought a snow thrower on the installment plan.

To find: What is the finance charge.

In order to get the finance charge, we will subtract purchase price from the total amount paid.

Since, we know that finance charge, will be found by subtracting purchase price from the total amount paid.

So, we will get the value as

2,286−1,980=306.

Thus, we can say that $306 will be the finance charge when Jean bought a snow thrower on the installment plan.

Page 178 Problem 9 Answer

Given: Linda bought a washer and dryer from Millpage Laundry Supplies.

To find: Express her down payment algebraically.

We will get the down payment as the product of the purchase price along with rate that is being given.

Since, we know that down payment is the product of the purchase price along with rate that is being given.

So, we get the value of down payment as15/100×y=0.15y.

Thus, we can say that$0.15y is the down payment when Linda bought a washer and dryer from Millpage Laundry Supplies.

Page 178 Problem 10 Answer

Given:  Linda bought a washer and dryer from Millpage Laundry Supplies.

To find: How many monthly payments must Linda make.

Payment is done monthly which means that we have to give the money for twelve months to complete one year.

Since, we know that x dollars are paid for a year, Thus, we can say that payment is done monthly which means that we have to give the money for twelve months to complete one year.

Thus, we can say that 12 monthly payments must Linda make when Linda bought a washer and dryer from Millpage Laundry Supplies.

Page 178 Problem 11 Answer

Given: Linda bought a washer and dryer from Millpage Laundry Supplies.

To find: Express the total amount of the monthly payments algebraically.

We will get the value of total amount of monthly payment will be a product of number of months along with monthly payment.

Since, we know that value of total amount of monthly payment will be a product of number of months along with monthly payment.

So, we get the value as12×x=12x.

Thus, we can say that12x is the total amount of the monthly payments when Linda bought a washer and dryer from Millpage Laundry Supplies.

Page 178 Problem 12 Answer

Given: Linda bought a washer and dryer from Millpage Laundry Supplies.

To find: Express the total amount Linda pays for the washer and dryer on the installment plan algebraically.

Total money will be given by the sum of total monthly payment along with down payment.

Since, we know that total money will be given by the sum of total monthly payment along with down payment.

So, the total amount Linda pays for the washer and dryer on the installment plan algebraically 0.15×y+12×x=0.15y+12x.

Thus, we can say that $0.15y+12x is the total amount Linda pays for the washer and dryer on the installment plan algebraically.

Page 178 Problem 13 Answer

Given: Linda bought a washer and dryer from Millpage Laundry Supplies.

To find: Express the finance charge algebraically.

Total money will be given by the sum of total monthly payment along with down payment.

Since, we know that finance charge, will be found by subtracting purchase price from the total amount paid.

So, we will get the value as(0.15y+12x)−y=12x−0.85y.

Thus, we can say that $12x−0.85y will be the finance charge when Linda bought a washer and dryer from Millpage Laundry Supplies.

Page 178 Problem 14 Answer

Given: Zeke bought a bobsled on the installment plan.

To find: How much interest will he pay.

Total money will be given by the sum of total monthly payment along with down payment.

Since, we know that value of total amount of monthly payment will be a product of number of months along with monthly payment.

So, we get the value as​

=24×93.50

=2244

Also, we know that total money will be given by the sum of total monthly payment along with down payment. So, the total amount will be

​2244+450

=2694

​Since, we know that finance charge, will be found by subtracting purchase price from the total amount paid.

So, we will get the value as

=2694−2300

=394

​Thus, we can say that $394 interest will be paid when Zeke bought a bobsled on the installment plan.

Page 178 Problem 15 Answer

Given: Gary is buying a computer on the installment plan.

To find: What is the finance charge.

We will get finance charge when we subtract purchase price from the total amount paid.

Since, we know that value of total amount of monthly payment will be a product of number of months along with monthly payment.

So, we get the value as

​=30×48.25

=1447.50

​Also, we know that total money will be given by the sum of total monthly payment along with down payment. So, the total amount will be

​=1447.50+150

=1597.50

Since, we know that finance charge, will be found by subtracting purchase price from the total amount paid.

So, we will get the value as​

=1597.50−1250

=347.50

​Thus, we can say that $347.50 is the finance charge when Gary is buying a computer on the installment plan.

Page 178 Exercise 1 Answer

Given: Mazzeo’s Appliance Store requires a down payment of 1/3 and Norton’s Depot requires a 30% down payment.

To find: Which store’s down payment rate is lower.We will compare the values to get the result.

Let us take the price of the product as x.

When Mazzeo’s Appliance Store give us the value for down payment as x/3=0.33x.

Now, for the Norton’s Depot, we get the down-payment as 30%x=0.3x.

Thus, we can say that Norton’s Depot store’s down payment rate is lower.

Page 178 Exercise 2 Answer

We are given : Purchase price=m dollars

Down payment rate​=20%

=20/100

=0.20.

Monthly payment=x dollars Number of months =24.

We have to express the finance charge algebraically.

In the question, first we will find the down payment .

Then, find the total amount of monthly payments .

Calculate the total cost and then, find the finance charge.

Firstly, we will find the down payment.

As, down payment is the product of the down payment rate and the purchase price.

Therefore,Down payment=Down payment rate×purchase price

Down payment=0.20×m

Down payment=0.20m .

​Now, we will find the total amount of monthly payments .

As, the total amount of monthly payments is the product of the number of months and the monthly payment.

Therefore,Total amount of monthly payments=Number of months×Monthly payment

Total amount of monthly payments=24×x

Total amount of monthly payments=24x.​

We will find the total cost .

As, the total cost is the sum of the down payment and the total amount of monthly payments .

Therefore,Total cost=Down payment+Total amount of monthly payments

Total cost=0.20m+24x.

Now, we will find the finance charge .

As, the finance charge is the difference between the total cost and the purchase price .

Therefore,Finance charge=Total cost−Purchase price

Finance charge=0.20m+24x−(m)

Finance charge=24x+0.20−1

Finance charge=24x+(0.20−1)m

Finance charge=24x−0.8 .

The finance charge is 24x-0.80.

Page 178 Exercise 3 Answer

We are given : A spreadsheet for installment purchase calculations.

We have to write a spreadsheet formula to compute the down payment in cellC2.

As, we have to find the down payment in cellC2.

So, we will be using A2andB2, that is of the same row and same column .

The down payment is the product of the down payment rate and the purchase price.

Therefore, C2=A2∗B2.

The spreadsheet formula to compute the down payment in C2is : C2=A2∗B2

Page 178 Exercise 4 Answer

We are given :

A spreadsheet for installment purchase calculations.

We have to write a spreadsheet formula to compute the time in months in cellF2.

We have to compute the time in months in cell F2.

Time in months=F2.

The time in years for the same row (purchase) is given in cell E2.

Time in years=E2.

Since there are12 months in a year, the time in months is the time in years should be multiplied by 12​​

Time in months = Time in years ×12

F2=E2×12 .

​The spreadsheet formula to compute the time in months in cell F2 is :F2=E2×12.

Page 178 Exercise 5 Answer

We are given :

A spreadsheet for installment purchase calculations.

We have to write the spreadsheet formula to compute the finance charge in cell H2 .

We have to find the finance charge .

We will subtract the purchase price of the item from the total amount paid on installment.

Total amount is the sum of down payment and the total of monthly payments.

And, the finance charge is the difference between the total cost and the purchase price .

Finance charge is given by :

H2=(C2+G2)−A2 .

The spreadsheet formula to compute the finance charge in cell H2 is :H2=(C2+G2)−A2 .

Page 178 Exercise 6 Answer

We are given :

A spreadsheet for installment purchase calculations.

We have to use our answers to a−d

to fill in the missing entries f−v.

As, down payment is the product of the down payment rate and the purchase price.

Therefore,

(1)Down payment=1200×0.20

Down payment=$240 .

(2)Down payment=1750×0.10

Down payment=$175 .​

(3)Down payment=1340×0.15

Down payment=$201.

(4)Down payment=980×0.10

Down payment=$98 .​

The time in months is the time in years multiples by 12 , as there are 12months in a year.

Therefore, (5)Time in months = Time in years ×12

Time in months =1 ×12

Time in months =12.​

(6)Time in months =2×12

Time in months =24 .

(7)Time in months=1.5×12

Time in months=18 .​

(8)Time in months =0.5×12

Time in months =6.​

Now, we will find the total amount of monthly payments.

(9)Total amount of monthly payments=12×97.01

Total amount of monthly payments=$1164.12​

(10)Total amount of monthly payments=24×71.12

Total amount of monthly payments=$1706.88

​(11)Total amount of monthly payments =18×77.23

Total amount of monthly payments =$1390.14

​(12)Total amount of monthly payments =6×165.51

Total amount of monthly payments =$993.06 .​

Now, we will find the finance charge .

Finance charge = Total cost − Purchase price

Finance charge= Total amount of monthly payments + Down payment − Purchase price .

​(13)Finance charge =1164.12+240−1200

Finance charge =$204.12

​(14)Finance charge =1706.88+175−1750

Finance charge =$131.88

​(15)Finance charge =1390.14+201−1340

Finance charge =$251.14

(16)Finance charge =993.06+98−980

Finance charge =$111.06.​

The missing entries are  :

(1)$240

(2)$175

(3)$201

(4)$98

(5)12

(6)24

(7)18

(8)6

(9)$1,164.12

(10)$1,706.88

(11)$1,390.14

(12)$993.06

(13)$204.12

(14)$131.88

(15)$251.14

(16)$111.06.

​Page 179 Exercise 7 Answer

 We are given :

Purchase price=$1700

Down payment=$0

Monthly payment=$201

Number of months=9 .

We have to find the sumo of the monthly payments and determine the fee charged for the layaway plan .

We will find the total amount of monthly payments .

As, the total amount of monthly payments is the product of the number of months and the monthly payment.

Therefore, Total amount of monthly payments =9×201

Total amount of monthly payments =$1809 .

Now, we will find the total cost .

As, the total cost is the sum of the down payment and the total amount of monthly payments .

Therefore,

Total cost =1809+0

Total cost =$1809 .

Now, we will find the finance charge .

As, the finance charge is the difference between the total cost and the purchase price .

Therefore,

Finance charge =1809−1700

Finance charge =$109 .

​The sum of the monthly payments is$1809.

The fee charged for the layaway plan was :$109 .

Page 179 Exercise 8 Answer

We are given :

Purchase price=$4345

Down payment=$0

Monthly payment=$15

Number of months=11 .

We have to find the sum of the monthly payments .

We will find the total amount of monthly payments .

As, the total amount of monthly payments is the product of the number of months and the monthly payment.

Therefore, Total amount of monthly payments =11×15

Total amount of monthly payments =$165 .

​The sum of the monthly payments is$165 .

Page 179 Exercise 9 Answer

We are given :

Purchase price=$4345

Down payment=$0

Monthly payment=$15

Number of months=11.

We have to find the amount Chris must pay in the last month of the plan .

We will find the total amount of monthly payments .

As, the total amount of monthly payments is the product of the number of months and the monthly payment.

Therefore, Total amount of monthly payments =11×15

Total amount of monthly payments =$165 .

​Now, we will find the total cost .

As, the total cost is the sum of the down payment and the total amount of monthly payments .

Therefore,

Total cost =165+0

Total cost =$165.

The remaining payment is then the difference between the purchase price and the total cost and, there is no interest in this case.

Therefore, only the remaining amount of the purchase price should still have to be paid.

Therefore,Remaining amount =4345−165

Remaining amount =$4180.

​Chris must pay$4180

in the last month of the plan.

Page 179 Exercise 10 Answer

We are given :

Fee charged for layaway plan=$109 in the previous exercise.

We have to determine the difference between the layaway plan in the previous exercise and the deferred payment plan.

The layaway price in the prior exercise was$109, however no fee was levied in this exercise’s deferred payment plan.

Furthermore, with a layaway plan, the buyer receives the product once it has been paid in full, whereas with a deferred payment plan, the consumer receives the merchandise at the time of purchase (not after it has been completely paid).

The difference between the layaway plan in and the deferred payment plan is that there is no fee in the deferred payment plan, while the layaway plan contains a fee.

Page 179 Exercise 11 Answer

We are given :

Purchase price=x dollars

Down payment=d dollars

Monthly payment=m dollars

Number of months=23.

In this case, we will not consider the last month as it will have a different payment amount .

We have to express the amount of the last payment algebraically.

We will find the total amount of monthly payments .

As, the total amount of monthly payments is the product of the number of months and the monthly payment.

Therefore,Total amount of monthly payments =23×m

Total amount of monthly payments =23m .

​Now, we will find the total cost .

As, the total cost is the sum of the down payment and the total amount of monthly payments .

Therefore,Total cost​=23m+d .

​The remaining payment is then the difference between the purchase price and the total cost and, there is no interest in this case .

Therefore, only the remaining amount of the purchase price should still have to be paid .

Therefore,Remaining amount​=x−(23m+d)

​Remaining amount​=x−23m−d .​

​The amount of the last payment is x−23m−d dollars .

Page 180 Exercise 12 Answer

We are given :

Purchase price=$2100

Down payment rate​=10%

=10/100

=0.10.

​We have to find the dollar value of the down payment.

We will find the down payment.

As, down payment is the product of the down payment rate and the purchase price.

Therefore,Down payment=$2,100×10%

Down payment=$2,100×0.10

Down payment=$210.

​The dollar value of the down payment is$210.

Page 180 Exercise 13 Answer

We are given :

Monthly payments=$75

Purchase price=$2100

Down payment rate=10%

Number of months=6 .

We have to find the cost of the rent ,if Sharon decides not to buy the HDTV after the six months .

We will find the down payment .Then, we will find the total amount of monthly payments .We will find the total cost .

Firstly, we will find the down payment .

As, down payment is the product of the down payment rate and the purchase price.

Therefore,Down payment=$2,100×10%

Down payment=$2,100×0.10

Down payment=$210 .

​Now, we will find the total amount of monthly payments .

As, the total amount of monthly payments is the product of the number of months and the monthly payment.

Therefore,Total amount of monthly payments =6×75

Total amount of monthly payments = $450 .

​Now, we will find the total cost .

As, the total cost is the sum of the down payment and the total amount of monthly payments .

Therefore,Total cost =$210+$450

Total cost=$660 .

​The cost of rent was$660 .

Page 180 Exercise 14 Answer

We are given : Regular selling price=$670

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment=$58 .

We have to find the discount .

We will find the discount .

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67 .​

The discount is $67 .

Page 180 Exercise 15 Answer

We are given :

Regular selling price=$670​

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment=$58.

We have to find the sale price .

First, we will find the discount and then, we will find the sale price .

Firstly, we will find the discount amount .

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67 .

Now, we will find the sale price .

Sale price=Regular selling price−Discount amount

Sale price=670−67

Sale price=$603 .​

The sale price is$603 .

Page 180 Exercise 16 Answer

We are given :

Regular selling price=$670

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment= $58.

We have to find the sales tax.

Firstly, we will find the discount amount .

Then, we will find the sale price . We will find the sales tax.

Firstly, we will find the discount amount.

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67 .

Now, we will find the sale price .

Sale price=Regular selling price−Discount amount

Sale price=670−67

Sale price=$603.

​We will find the sales tax.

As,Sales tax=Tax rate×Sale price

Sales tax =8%×603

Sales tax =0.08×603

Sales tax =$48.24 .​

The sales tax is$48.24.

Page 180 Exercise 17 Answer

We are given :

Regular selling price=$670

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment=$58.

We have to find the total cost of the guitar.

Firstly, we will find the discount amount .

Then, we will find the sale price . We will find the sales tax . We will find the total cost.

Firstly, we will find the discount amount.

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67 .

Now, we will find the sale price .

Sale price=Regular selling price−Discount amount Sale price=670−67

Sale price= $603.

​We will find the sales tax.

As,Sales tax=Tax rate×Sale price

Sales tax =8%×603

Sales tax =0.08×603

Sales tax =$48.24.

​Now, we will find the total cost.

As,Total cost=Sales price+Sales tax

Total cost=603+48.24

Total cost= $651.24.

​The total cost of the guitar is$651.24.

Page 180 Exercise 18 Answer

We are given :

Regular selling price=$670

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment= $58.

We have to find the down payment.

Firstly, we will find the discount amount.

Then, we will find the sale price .

We will find the sales tax .

We will find the total cost .

We will find the down payment .

Firstly, we will find the discount amount .

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67 .

​Now, we will find the sale price .

Sale price=Regular selling price−Discount amount

Sale price=670−67

Sale price=$603.

We will find the sales tax.

As,Sales tax=Tax rate×Sale price

Sales tax =8%×603

Sales tax =0.08×603

Sales tax =$48.24.​

Now, we will find the total cost .

As,Total cost=Sales price+Sales tax

Total cost=603+48.24

Total cost=$651.24.

We will find the down payment.

As,Down payment=Total cost×Down payment percentage

Down payment =651.24×15

Down payment =651.24×0.15

Down payment =$97.69.​

The down payment is$97.69.

Page 180 Exercise 18 Answer

We are given :

Monthly payment=$58.

We have one year, therefore,

Number of months=12.

We have to find the total of the monthly payments.

We will find the total amount of monthly payments.

Total amount of monthly payments=Number of months×Monthly payment

Total amount of monthly payments=12×58

Total amount of monthly payments=$696.​

$696is the total of the monthly payments.

Page 180 Exercise 19 Answer

We are given : Regular selling price=$670

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment=$58

Number of months=12, as, we have one year .

We have to find the total Lillian paid for the guitar on the installment plan .

Firstly, we will find the discount amount .

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67.

Now, we will find the sale price .

Sale price=Regular selling price−Discount amount

Sale price=670−67

Sale price=$603 .

We will find the sales tax.

As,Sales tax=Tax rate×Sale price

Sales tax =8%×603

Sales tax =0.08×603

Sales tax =$48.24.

​Now, we will find the total cost.

As,Total cost=Sales price+Sales tax

Total cost=603+48.24

Total cost=$651.24.​

We will find the down payment.

As,Down payment=Total cost×Down payment percentage

Down payment =651.24×15

Down payment =651.24×0.15

Down payment =$97.69.​

We will find the total amount of monthly payments.

Total amount of monthly payments=Number of months×Monthly payment

Total amount of monthly payments =12×58

Total amount of monthly payments =$696.​

The total Lillian paid is the sum of the down payment and the total of the monthly payments.

Therefore,

$97.69+$696=$793.69 .

The total Lillian paid for the guitar on the installment plan is $793.69.

Page 180 Exercise 20 Answer

We are given :

Regular selling price=$670

Discount rate=10%

Tax rate=8%

Down payment rate=15%

Monthly payment=$58

Number of months=12.

We have to find the finance charge.

Firstly, we will find the discount amount.

As,Discount=Discount rate×Regular selling price

Discount=10%×670

Discount=0.10×670

Discount=$67 .

Now, we will find the sale price.

Sale price=Regular selling price−Discount amount

Sale price=670−67

Sale price=$603.​

We will find the sales tax.

As,Sales tax=Tax rate×Sale price

Sales tax =8%×603

Sales tax =0.08×603

Sales tax =$48.24.

​Now, we will find the total cost.

As,Total cost=Sales price+Sales tax

Total cost=603+48.24

Total cost=$651.24.

​We will find the down payment.

As,Down payment=Total cost×Down payment percentage

Down payment =651.24×15

Down payment =651.24×0.15

Down payment =$97.69.

We will find the total amount of monthly payments.

Total amount of monthly payments=Number of months×Monthly payment

Total amount of monthly payments =12×58

Total amount of monthly payments =$696.

The total Lillian paid is the sum of the down payment and the total of the monthly payments.

Therefore, $ 97.69+$696=$793.69.

Now, we will find the finance charge.

Finance charge=Total paid−Total cost

Finance charge =$793.69−$651.24

Finance charge =$142.45.​

The finance charge is$142.45.

Page 180 Exercise 21 Answer

We are given: The inequalities give information on the credit scores.

Let x represent your credit score.

x>700,  your credit score is excellent.

680<x<700, your credit score is good.

620<x<680, your credit score should be watched carefully.

580<x<620,  your credit score is low.

x<580, your credit score is poor.

If Mary Ann’s credit score is low, but she receives 40 points for paying off some delinquent debts, we have to explain whether it is possible that her credit rating is now good.

If Mary’s credit score is low, it may be as high as 619.

So her new credit score will be619+40=659 once we acquire40

points. It should be closely monitored.

However, she needs a credit score of 681 to be considered good.

We can conclude that  Mary Ann’s credit score will not be good even after a rise by 40 points.

If Mary Ann’s credit score is low, but she receives 40 points for paying off some delinquent debts, it is not possible that her credit rating is now good.

No. Mary Ann’s credit score will not be good even after a rise by 40 points.

Page 180 Exercise 22 Answer

We are given :

Credit line=$8000

Previous balance=$567.91

Payment=$1200

Total purchases=$986.79

Finance charge=$10.00.

We have to find the available credit.

Firstly, we will find the current balance.

As, the current balance is the previous balance decreased by the payment and purchases and finance charge.

Therefore,​Current balance =567.91−1200−986.79−10

Current balance =−$1,628.88 .

Now, we will find the available credit.

As, the available credit is the sum of the credit line and the current balance.

Therefore,Availabe credit =8000−1628.88

Availabe credit =$6,371.12.​

The available credit is $6,371.12.

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 167 Problem 1 Answer

To find: Interview a bank representative about trust accounts.

Find out what the abbreviations POD, ATF, and ITF mean.

Prepare questions about FDIC insurance limits and beneficiaries.

Ask for any brochures they offer about trust accounts.

Prepare a report or a poster on trust accounts to present to the class.

FDIC or Federal deposit insurance corporation which gives insurance to the saving account of the people in U.S. Some questions about FDIC insurance limits and beneficiaries –

Q 1) Where can I get to know about insurance rules and limits? Ans- Using electronic deposit insurance estimator.

Q 2) What is the limit of FDIC beneficiaries? Ans- 250000 is the limit up to which you and your family in total can have deposits in your account.

Q 3) what type of accounts are eligible for FDIC insurance?

Trust accounts report Trust account is a savings account deposited in the name of a trustee who controls it during his lifetime, after which the balance is payable to a pre-nominated beneficiary There are two types of Trust Accounts revocable and irrevocable, the extent of FDIC insurance depends on which out of these 2, our account is.

A revocable trust is that, which assigns beneficiaries, receiving the funds of trust after the death of the owner.

But the trust can also be terminated or there can be a change of terms by the owner.

A revocable trust may become an irrevocable trust once the owner dies because effectively there is no longer an option to cancel or change the trust. In a revocable trust, the owner is considered the depositor and the account are only eligible for 250,000 of FDIC insurance coverage.

However, if a bank fails after a trust owner has died but before the trust has been paid out, the beneficiaries are considered the depositors and each is entitled to 250,000 in coverage.

In an irrevocable trust, the beneficiaries are considered to be the depositors even before the owner dies, as long as there are no conditions the beneficiaries have to meet to remain eligible for trust proceeds.

There are conditions the beneficiary has to meet in order to be eligible for trust proceeds.

Hence the answer is  POD− Payable on death, ITF − In trust for, ATF = As Trustee For

Page 167 Problem 2 Answer

To find:  Compare and contrast the accounts offered by the same bank.

What are the benefits of each? What are the drawbacks of each?

Who might be better served by each type of checking account?

Explain which account might be best for your financial situation.

When you start looking online at the different types of checking accounts, there are many different types such as:

Free checking account: No monthly maintenance fee, Possible overdrafts fees, Possible transactions/services fees

Online checking account: Checking accounts available online, many online checking accounts have no monthly maintenance fees, often have lower fees.

Second Chance checking account: Checking accounts for people who had there account terminated due to not covering overdrafts;

Low fees, Low minimum balances, Restrict use of certain services

Student checking account: Checking accounts intended for students; Low fees, Low minimum balances

Senior checking account: Checking accounts intended for a certain age group; Breaks in monthly fees.

Joint checking account: At least two people are authorized to make transactions and write checks; Convenient for married couples

Business checking account: Designed to handle a large amount of transactions

Interest-bearing checking account: You receive interest on your money (at a decent rate; even though it will still be quite low)

Money market checking account: A type of interest bearing checking account, higher interest rate, possibly a higher minimum balance, possibly a limited number of monthly transactions

In your case, a student checking account is most likely the most interesting, as you are still a student.

Hence the answer could vary

Page 167 Problem 3 Answer

To find; While the law states that free checking accounts cannot have minimum balances or per-check fees, there are other fees and penalties that are allowable.

Research the allowable fees and penalties on checking accounts. Make a list and explain the purpose and cost of each.

There are many different fees and penalties on checking accounts.

You can find them easily using a search engine and searching for “checking account allowable fees” or something similar.

The most common fees/penalties on checking accounts are:

Monthly maintenance fee: Fee that has to be paid each month

ATM fees: Fee paid for every transaction at an ATM

Overdraft fees

Abandoned account fee: applies when the account is unused for three to five years

Account closed early fee: when you close an account within 90 or 180 days of opening it.

Reconciliation fee: when there is a discrepancy between your records and the bank’s records (tends to be a fee per hour).

Check printing fee: Charge for checks

Counter check fee: Offered when you run out of checks, but only a limited number of counter checks are free.

Hence the answer could vary

Page 167 Problem 4 Answer

To find: What are the penalties for withdrawing money from a CD before it is due?

What are the minimum balances for different types of accounts? What are the fees for insufficient funds?

What are the different types of checking accounts they offer? What are the fees and requirements for these accounts?

What are the hours of service? Think of other questions to ask. Prepare the findings in a report.

When you start looking online at the different types of checking accounts, there are many different types such as:

Free checking account: No monthly maintenance fee, Possible overdrafts fees, Possible transactions/services fees

Online checking account: Checking accounts available online, many online checking accounts have no monthly maintenance fees, often have lower fees.

Second Chance checking account: Checking accounts for people who had there account terminated due to not covering overdrafts;

Low fees, Low minimum balances, Restrict use of certain services

Student checking account: Checking accounts intended for students; Low fees, Low minimum balances

Senior checking account: Checking accounts intended for a certain age group; Breaks in monthly fees.

Joint checking account: At least two people are authorized to make transactions and write checks; Convenient for married couples

Business checking account: Designed to handle a large amount of transactions

Interest bearing checking account: You receive interest on your money (at a decent rate; even though it will still be quite low)

Money market checking account: A type of interest bearing checking account, higher interest rate, possibly a higher minimum balance, possibly a limited number of monthly transactions

The hours of service tend to vary strongly from bank to bank.

However, generally, the opening hours will range from about 9 A.M. to 5 P.M. with a possible break at noon.

Hence the answer could vary

Page 167 Problem 5 Answer

To find:  Interest rates have historically fluctuated with the economy.

Go online and/or use the library to find interest rates over the past 50 years. Make a graph to display the information.

The following table contains the average interest rates for a 6 Month CD.

Year	Percent	Year	Percent	Year	Percent	Year	Percent	Year	Percent
1965	4.43	1975	6.89	1985	8.25	1995	5.98	2005	3.72
1966	5.63	1976	5.62	1986	6.51	1996	5.47	2006	5.23
1967	5.21	1977	5.92	1987	7	1997	5.72	2007	5.23
1968	6	1978	8.61	1988	7.9	1998	5.44	2008	3.14
1969	7.89	1979	11.44	1989	9.08	1999	5.46	2009	0.88
1970	7.66	1980	12.99	1990	8.17	2000	6.58	2010	0.44
1971	5.22	1981	15.77	1991	5.91	2001	3.64	2011	0.42
1972	5.02	1982	12.57	1992	3.76	2002	1.81	2012	0.44
1973	8.31	1983	9.27	1993	3.28	2003	1.17	2013	0.2
1974	9.98	1984	10.68	1994	4.96	2004	1.74	2014	0.13

Time plot

Year is on the horizontal axis and the interest rate is on the vertical axis

We connect points corresponding to consecutive years be a straight line.

Hence the graph is

Page 167 Problem 6 Answer

we have to Research how experts memorize long sequences of digits.

Visit a few local businesses to see if they would be willing to donate a prize for the contest. Ask the school newspaper to cover the contest.

Emcee the contest in class.

The decimal representation of the number e is:

2.718281828459045235360287471352662497757247093699959574966….

The decimal representative of e contains no repeating pattern (as $e$ is an irrational number).

There are many different tips on how to memorize a long sequence of digits. A few of these tips could be:

Create associations (such as a birthday, anniversary, number of jersey of favorite football player, etc.)

Break the long sequence into smaller sequences

Check for patterns

When learning the number, speak the words out at loud (which helps you remember better).

Repetition: often repeat the memorized number with some time frame left in between different repetitions (such as an hour)

Visualize the shape of the digits on a keypad

Convert the numbers to words/images

Hence the answer could vary

Page 167 Problem 7 Answer

To find: Some employers allow employees to have money deducted from their accounts and automatically placed into a savings account.

Interview three adults working in different professions.

Ask them about employer-sponsored savings plans. Prepare a report on the findings.

A employer-sponsored savings plan is used by employees who elect to save a portion of their paycheck in an investment account.

This type of plan is set up by the employers, who often sponsor the plan as well and these plans are often based around retirement.

There are annual contribution limits, but also tax benefits accompanying this type of plan.

It could be possible to withdraw funds early (with some additional fees) or to bottom against your account balance.

The employees are fully responsible in what their money is invested in and the employees can take their employer-sponsored savings plan with them when they change work (change their employer).

Hence the answer could vary

Page 167 Problem 8 Answer

To find:  Visit a local bank. Get brochures they offer about their services. If the brochures are two-sided, take two of each so you can cut them out and paste them onto a poster board.

Pick several services to highlight. Cut out the portions of the brochures that explain each service.

Give each service an original, short title, and print out your title.

Organize the titles and descriptions of the banking services onto a poster board.

The answers to this questions will vary very strongly.

You will need to pick up brochures at your local bank about their services.

Per service, you will need to pick a title. However, this could be very easy as the brochures tend to use titles to separate the different (most important) services.

If a title has not been specified, then either use the name of the service as title or use a (very) short description of the service.

Organize the titles in some manner. For example, organize the titles from cheapest to most-expensive service.

Hence the answer could vary

Page 167 Problem 9 Answer

To find: The Rule of 72 is a method for quickly estimating how many years it will take principal to double, assuming the interest was compounded.

Go to the library and/or use the Internet to research the Rule of 72 beyond what was presented in Lesson 3-7.

Prepare some examples to illustrate the rule. Discuss the history and the use of the rule. Display your research on a poster board.

The Rule of 72 is used to determine the time it takes for an investment to double.

To use the rule of 72 , you require the knowledge of the annual rate of return (or annual interest rate).

The rule of 72 states that the product of the rate of return and the number of years (time) should be roughly equal to 72 .

On equivalently, the time it takes to double your investment is 72 divided by the annual interest rate

Although, most people attribute the rule of 72 to Albert Einstein.

The Rule of 72 is used to determine the time it takes for an investment to double.

Page 168 Problem 10 Answer

Given: We have been given the following data regarding the occurrences of counterfeit money.

To find: We have to draw the graph for both sets of columns and show them on the same axes

Solution: The graph for the data will be as follows:

The Secret Service drastically reduced the occurrences of counterfeit money since the Civil War but the problem still exists.The graph showing the catching of counterfeit bills in the U.S. is as follows:

Page 168 Problem 11 Answer

Given: The graph showing the catching of counterfeit bills in the U.S. is as follows:

To find: If the pattern of catching counterfeit bills before and after circulation follows the same pattern of increases and decreases.

Solution: In the year2000 and2005, the dollars found before and after circulation both increased from the year before but at a different rate.

In the year2001 and 2004, compared to their respective previous years the dollars found after circulation increased but dollars found before circulation decreased the year 2002, both the dollars found before and after circulation decreased but at a different rate.

Similarly, in the year2003, dollars found after circulation decreased whereas dollars found before circulation increased.

Hence we can see that the pattern of catching counterfeit bills before and after circulation do not follow the same pattern of increases and decreases

On examining the graph drawn in the previous part of the question we can observe that the graph showing the catching of counterfeit bills before and after they enter circulation are not similar nor are they completely opposite.

It is a mixture of increases and decreases and both the graph does not show any relation between them.

Thus, the pattern of catching counterfeit bills before and after circulation does not follow the same pattern of increases and decreases.

Page 169 Problem 12 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 13 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 14 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 15 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 16 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 17 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 18 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 19 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 20 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Problem 21 Answer

we have to Go to www.cengage.com/school/math/financialalgebra and download a blank check register. Complete all of the necessary information in the check register

Hence the answer is

Page 169 Exercise 1 Answer

we have to Use the check register from Exercise 1. It is now one month later, and the checking account statement has arrived. Does the account balance?

The ending balance is given on the last line of the statement which is $2,495.91.

Hence the answer is $2,495.91.

Page 169 Exercise 2 Answer

we have to Use the check register from Exercise 1. It is now one month later, and the checking account statement has arrived. Does the account balance?

In question 1(g) there is deposit made on 12/22 which is not recorded so that is the deposits outstanding and it is $120

Hence the answer is $120

Page 169 Exercise 3 Answer

we have to Use the check register from Exercise 1. It is now one month later, and the checking account statement has arrived. Does the account balance?

All the checks are mentioned in the statement except the one written on dec 20 and the one written on dec 29 for the amounts 1250 and 150.8 respectively hence total checks outstanding is $1,250+$150.8=$1,400.80

Hence the answer is $1,400.80

Page 169 Exercise 4 Answer

we have to Use the check register from Exercise 1. It is now one month later, and the checking account statement has arrived.

Does the account balance?

Revised statement balance = Ending balance – Checks outstanding + deposits outstanding

$2,495.91+$1,400.80−$120=$1,215.11

Hence the answer is $1,215.11

Page 169 Exercise 5 Answer

we have to Use the check register from Exercise 1. It is now one month later, and the checking account statement has arrived.

Does the account balance?

The balance from the checkbook is given in exercise lj at the bottom right corner of the checkbook register, which contains $1,215.11.

Hence the answer is $1,215.11.

Page 170 Exercise 6 Answer

Given: Ruth has a savings account at a bank .To find: her balance in five months if her account balance never reaches 1,500?

Interest will be the difference of amount and the principle value.

Since, we know that interest will be the difference of amount and the principle value.

So, we get the balance before the withdrawal is$1,722−$400=$1,322.

Hence, the balance in five months will be 1322−(3.50)⋅5=1304.50.

Thus, we can say that her balance in five months if her account balance never reaches 1,500,will be $1304.50.

Page 170 Exercise 7 Answer

Given: Nine months ago Alexa deposited 7,000 in a three-year CD.

To find her balance after the withdrawal.Interest will be the difference of amount and the principle value.

On performing the algebraic operations, we get the value as$7,000+$224.16−$1,000−$250=$5,974.16.

Thus, we can say that when Nine months ago Alexa deposited 7,000 in a three-year CD then her balance after the withdrawal will be $5,974.16.

Page 170 Exercise 8 Answer

Given: Ralph deposited 910  in an account that pays 5.2% simple interest, for3.5 years.

To find: How much interest did the account earn Simple interest is given as the product of principal amount with rate and the time for the interest.

Since, we know that simple interest is given as the product of principal amount with rate and the time for the interest.

So. we get the value as

​$910×5.2%×31/2

=$910×0.052×3.5

=$165.62​

Thus, we can say that the interest did the account earn will be $165.62.

Page 170 Exercise 9 Answer

Given:  Ralph deposited 910 in an account that pays 5.2% simple interest.

To find: What is the ending balance.Interest will be the difference of amount and the principle value.

On performing the algebraic operations, we get the value as

$910+$165.62=$1,075.62

Thus, we can say that when Ralph deposited 910 in an account that pays 5.2% simple interest then the ending balance will be $1075.62.

Page 170 Exercise 10 Answer

Given:  Ralph deposited 910 in an account that pays 5.2% simple interest.

To find: How much interest did the account earn the first year?

Simple interest is given as the product of principal amount with rate and the time for the interest.

Since, we know that simple interest is given as the product of principal amount with rate and the time for the interest.

So. we get the value as

​$910×5.2%×1

=$910×0.052×1

=$47.32

Thus, we can say that $47.32 interest did the account earn the first year.

Page 170 Exercise 11 Answer

Given: Ralph deposited 910  in an account that pays 5.2% simple interest.

To find: How much interest did the account earn the third year.

Simple interest is given as the product of principal amount with rate and the time for the interest.

Since, we know that simple interest is given as the product of principal amount with rate and the time for the interest.

So. we get the value as

​$910×5.2%×1

$910×0.052×1

=$47.32

​Thus, we can say that $47.32 interest did the account earn the third year.

Page 170 Exercise 12 Answer

Given: Matt has two single accounts at Midtown Bank.To find: sum of Matt’s balances.

Interest will be the difference of amount and the principle value.

On performing the algebraic operations, we get the value as

74112.09+77,239.01=151,350.10

Thus, we can say that $151,350.10 is the sum of Matt’s balances when Matt has two single accounts at Midtown Bank.

Page 170 Exercise 13 Answer

Given: Matt has two single accounts at Midtown Bank.To find: Is all of Matt’s money insured by the FDIC?

We will insure the total value is less than or greater than 250,000 or not.

Since, we know that $151,350.10 is the sum of Matt’s balances which is less than 250,000.

Thus, we can say that all of Matt’s money insured by the FDIC.

Thus, we can say that Matt’s money insured by the FDIC when Matt has two single accounts at Midtown Bank.

Page 170 Exercise 14 Answer

Given: Rhonda deposits 5,600  in a savings account that pays4.5%.

To find: How much interest does the account earn in the first six months. Interest will be the difference of amount and the principle value.

On using the compound interest formula, we can get the value as

A=5600(1+0.045/2)¹

=5726

Hence, we get the interest as 5726−5600=126.

Thus, we can say that 126$ interest does the account earn in the first six months.

Page 170 Exercise 15 Answer

Given: Rhonda deposits 5,600 in a savings account that pays4.5%

To find:  What is the ending balance after six months.We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value as

A=5600(1+0.045/2)¹

=5726​

Thus, we can say that $5726 is the ending balance after six months.

Page 170 Exercise 16 Answer

Given: Rhonda deposits 5,600 in a savings account.

To find: How much interest does the account earn in the second six months

We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value as

A1=5600(1+0.045/2)¹

=5726

A2=5600(1+0.045/2)²

=5854.84

Hence, interest does the account earn will be

A2−A1

=5854.84−5726

=128.84

​Thus, we can say that interest does the account earn in the second six months will be $128.84.

Page 170 Exercise 17 Answer

Given: Rhonda deposits 5,600 in a savings account.To find: What is the balance after one year.

We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value as

A=5600(1+0.045/2)²

=5854.84

Thus, we can say that the balance after one year will be $5854.84.

Page 170 Exercise 18 Answer

Given: Rhonda deposits 5,600 in a savings account.To find: How much interest does the account earn the first year.

Interest will be the difference of amount and the principle value.

On using the compound interest formula, we can get the value as

A=5600(1+0.045/2)²

=5854.84

Hence, we get the interest as5854.84−5600=254.84.

Thus, we can say that $254.84 interest does the account earn the first year.

Page 170 Exercise 19 Answer

Given: Nick deposited 3,000 in a three-year CD account.To find: What is the ending balance?

We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value as

A=3000(1+0.0408/52)^52×3

=3390.46​

Thus, we can say that $3390.46 is the ending balance when Nick deposited 3,000 in a three-year CD account.

Page 170 Exercise 20 Answer

Given: three years compounded daily.

To find: How much more would 10,000 earn in three years compounded daily at 4.33%?

We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value for daily as

​A=10000(1+0.0433/365)^365×3

=11,387.06

On using the compound interest formula, we can get the value for semiannually as

A=10000(1+0.0433/2)^2×3

=11,371.37

Hence, we get the interest as $11,387.06−$11,371.37=$15.69.

Thus, we can say that$15.69 more would $10,000 earn in three years compounded daily at 4.33%, than compounded semiannually at 4.33%.

Page 171 Exercise 21 Answer

Given: Austin deposits 2,250 into a one-year CD.To find: What is the ending balance after the year.

We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value as

A=2250(1+0.053/365)^365×1

=2372.46​

Thus, we can say that $2372.46 is the ending balance after the year.

Page 171 Exercise 22 Answer

Given: Austin deposits 2,250 into a one-year CD .To find: How much interest did the account earn during the year.

We will use the formula of compound interest to get the result.

On using the compound interest formula, we can get the value as

A=2250(1+0.053/365)^365×1

=2372.46

Hence, we get the interest as 2372.46-2250=122.46.

Thus, we can say that$122.46 interest did the account earn during the year.

Page 171 Exercise 23 Answer

Given: Austin deposits 2,250 into a one-year CD.To find: What is the annual percentage yield.

We will use the formula of annual percentage yield to get the result.

On using the annual percentage yield formula, we can get the value as

APY=(1+0.053/365)³⁶⁵−1

=0.0544

=5.44%

Thus, we can say that 5,44% the annual percentage yield when Austin deposits 2,250 into a one-year CD.

Page 171 Exercise 24 Answer

Given:  interest earned compounded continuously.

To find: interest earned on a 25,000 deposit .

Interest will be the difference of amount and the principle value.

On using the compounded continuously interest formula, we can get the value as

A=25000e0.047×2.5

=28117.04​

Hence, we get the interest as

I=A−P

=28117.04−25000

=3117.04

​Thus, we can say that $3117.04 interest earned on a 25,000  deposit for21/2years .

Page 171 Exercise 25 Answer

Given: You want to put a $40,000 down payment on a store front for a new business that you plan on opening in 5 years.

To find: Identify which of the three cases below applies.

We will try to estimate the values based on the investment done.

Since, we can see that we have a monthly investment.

So. we can say that investment will be present value of a periodic deposit investment.

Thus, we can say that investment of a 40,000 down payment on a store front for a new business that you plan on opening in five years then the investment will be present value of a periodic deposit investment.

Page 171 Exercise 26 Answer

Given: You deposit $3,000 into an account that yields 3.2% interest compounded semiannually.

To find: Identify which of the three cases below applies.

We will try to estimate the values based on the investment done.

Since, we can see that there is single deposit in an account.

So, we get investment as future value of a single deposit investment.

Thus, we can say that when deposit of $3,000 into an account that yields 3.22% interest compounded semiannually then we can say that investment will be future value of a single deposit investment.

Page 171 Exercise 27 Answer

Given: You want to save for a new car that you will buy when you graduate college in 4 years.

To find: How much will you be able to afford.

We will try to estimate the values based on the investment done.

As, we know that we have a deposit in every quarter. So, we get investment that will be a future value of a periodic deposit investment.

Thus, we can say that when saving is done for a new car that you will buy when you graduate college in four years then the future value of a periodic deposit investment is being done.

Page 171 Exercise 28 Answer

Given: Santos deposited 1,800 in an account that yields 2.7% interest, compounded semiannually.

To find: How much is in the account after 54 months.

Interest will be the difference of amount and the principle value.

On using the compound interest formula, we can get the value as

A=1800(1+0.027/2)^54÷6

=2030.89

​Thus, we can say that$2030.89 is in the account after 54  months when Santos deposited 1,800  in an account that yields 2.7% interest, compounded semiannually.

Page 171 Exercise 29 Answer

Given: Stephanie signed up for a direct deposit transfer into her savings account from her checking account.

To find: How much will be in the account at the end of 6.5 years.

We will use the formula of Future value of a investment to get the result.

On using the formula for Future value of a investment, we get the value as

B=P((1+r/n)^nt−1)/r/n

=150((1+0.026/12)^12×6.5−1)/0.026/12

≈12,731.79​

Thus, we can say that $12,731.79 be in the account at the end of 6.5 years when  Stephanie signed up for a direct deposit transfer into her savings account from her checking account.

Page 171 Exercise 30 Answer

Given: Jazmine needs 30,000 to pay off a loan at the end of 5 years.

To find: How much must she deposit monthly into a savings account.

We will use the formula of present value of a investment to get the result.

On using the formula for present value of a investment, we get the value as

P=B×r/n(1+r/n)^nt−1

=30,000×0.03/12(1+0.03/12)^12×5−1

≈464.06

​Thus. we can say that $464.06 must she deposit monthly into a savings account that yields 3% interest, compounded monthly.

Page 171 Exercise 31 Answer

Given: Tom wants to have50,000 saved sometime in the future.

To find: How much must he deposit every month into an account .We will use the formula of present value of a investment to get the result.

On using the formula for present value of a investment, we get the value as

P=B×r/n(1+r/n)^nt−1

=50,000×0.028/12(1+0.028/12)^12×x/12−1

=50,000×0.028/12(1+0.028/12)^x−1

that give us the graph as

interest, compounded monthly that gives us the graph as

 

 

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 162 Problem 1 Answer

To find: How many years would it take for $10,000 to grow to $20,000 in the same account?

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

10000=20000

(1+0.5/1)^1(t)

1=2/(1.5)t/(1.5)t

=2log(1.5)t

=log 2 t log(1.5)=log2

t=log2/log1.5

t=1.70​

Hence the answer is 1.7 years

Page 163 Problem 2 Answer

To find: How does the equation from Example 2 change if the interest is compounded weekly?

Use the formula for the present value of a single deposit investment. Let B=100,r=0.038,t=10, and n=52.

Substitute.

P=B(1+r/n)ⁿ​1

P=100,000

(1+0.038/52)⁵²(10)

Calculate.P=68,395.632

Ritika must deposit approximately$68,395632.

Hence the answer is $68,395632.

Page 164 Problem 3 Answer

To find: Write the formula to find the present value of an x-dollar balance that is reached by periodic investments made semiannually for y years at an interest rate of r.

PV=X×(1−1

(1+(r/2))^yx2)/(r/2)

Periodic payment (PMT) = X

Time (n) =yx²

Interest rate (r) =r/2

PV=PMT×(1−1/(1+r)/n)/r

=X×(1−1(1+(r/2))^yx2)/(r/2) on simplifying

=X⋅(1−1(1+(r/2))^2y)/r/2

Hence the present value is X⋅(1−1(1+(r/2))/2y)/r/2

Page 165 Problem 4 Answer

To find: How might those words apply to what has been outlined in this lesson?

In order to start setting financial goals, you need to take into account what you can currently afford to save (thus determine how much money you have left over after you have paid all bills and bought all necessary items such as food and clothes).

This is necessary, because else you will start making debts to pay for your food or bills, which will not aid your overall financial situation.

Moreover, it is also useful to look ahead at the future value savings, such that you know what you will earn from these future value savings in the future and also such that you can determine what you will do with this money in the future.

Hence the answer is In order to start setting financial goals, you need to take into account what you can currently afford to save.

Page 165 Problem 5 Answer

To find: Complete the table to find the single deposit investment amounts.

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

=1,000(1+0.04/1)^1×3

=1,000(1.04)³≈889.00

​2)P=B(1+r/n)^nt

=2,500(1+0.03/2)^2×5

=2,500(1.015)10

≈2,154.17

3)P=B(1+r/n)^nt

=10,000(1+0.05/4)^4×10

=10,000/(1.0125)⁴⁰

≈6,084.13

​4)P=B(1+r/n)^nt

=50,000(1+0.0275/12)^12×8

≈40,136.04

​Hence the answers are

(1) $889.00

(2) $2,154.17

(3) $6,084.13

(4) $40,136.04

Page 165 Problem 6 Answer

To find: Complete the table to find the periodic deposit investment amounts.

a)P=B×r/n(1+r/n)^nt−1

=50,000×0.02/1(1+0.02/1)^1×8−1

≈5,825.49

b)P=B×r/n(1+r/n)^nt−1

=25,000×0.015/2(1+0.015/2)^2×4−1

≈3,043.89

c)P=B×r/n/(1+r/n)^nt−1

=10,000×0.05/4(1+0.05/4)^4×10−1

≈194.21

d)P=B×r/n(1+r/n)^nt−1

=1,000,000×0.04/12(1+0.04/12)^12×20−1

≈2,726.47​

Hence the answer is

(1) $5,825.49

(2) $3,043.89

(3) $194.21

(4) $2,726.47

Page 165 Problem 7 Answer

To find; how much must he deposit each year in order to reach his goal?

P=B×r/n/(1+r/n)^nt−1

=50,000×0.042/1/(1+0.042/1)^1×7-1

≈6,292.16

Hence the answer is $6292.16

Page 165 Problem 8 Answer

To find; How much must he deposit now into an account that yields 2.75% interest, compounded monthly, so he can be assured of reaching his goal?

P=B×r/n/(1+r/n)^nt−1

=80,000/(1+0.0275/12)^12×20

≈46,185.05

Hence the answer is $46185.04

Page 165 Problem 9 Answer

To find: She sets a goal of saving $10,000 by the time she is 24 years old. How much must she deposit each month?

P=B×r/n/(1+r/n)^nt−1

=10,000×0.044/12/(1+0.044/12)^12/3-1

≈260.36

Hence the answer is $260.36

Page 165 Problem 10 Answer

To find:  Suni needs to repay her school loan in 4 years. How much must she semiannually deposit into an account that pays 3.9% interest, compounded semiannually, to have $100,000 to repay the loan?

P=B×r/n/(1+r/n)^nt−1

=10,000×0.039/2/(1+0.039/2)^2×4-1

≈11671.58

Hence the answer is $11671.58

Page 165 Problem 11 Answer

To find:  Rich needs $50,000 for a down payment on a home in 5 years. How much must he deposit into an account that pays 6% interest, compounded quarterly, in order to meet his goal?

P=B×r/n/(1+r/n)^nt−1

=50,000/(1+0.06/4)4×5

≈37123.52

Hence the answer is $37123.52

Page 165 Problem 12 Answer

To find:  Marcy wants to have $75,000 saved sometime in the future. How much must she deposit into an account that pays 3.1% interest, compounded monthly? Use a graphing calculator to graph the present value function.

P=B×r/n/(1+r/n)^nt−1

=75,000/(1+0.031/12)¹²xr/12

75,000/(1+0.031/12)^x

Hence the answer is 75,000/(1+0.031/12)^x  where x represent the number of months

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 157 Problem 1 Answer

Given: Periodic deposit amount(P)=$5000

Annual interest rate expressed as a decimal(r)=0.045

Number of times interest is compounded annually(n)=1

To find: The account balance(B) if the length of investment in years(t) is 21 years

Solution: We will find the account balance using the formula for future value:

B=P((1+r/n)^nt−1)/r/n

We will substitute the values in the formula to find the account balance

B=P((1+r/n)^nt−1)/r/n

=5000((1+0.045/1)^(1)(21)−1)/0.045/1

=5000((1.045)21−1)/0.045

≈168915.68

The account balance will be$168,915.68 if Rich and Laura decide to hold off retirement for an extra year.

Page 158 Problem 2 Answer

Given: From Example1, Example2 and check your understanding question after example1

we have the following details regarding periodic investment by Rich and Laura :

Total amount deposited in 21

years=5000×21=$105,000

Balance in the account after 21 years=$168,915.68

Interest they will earn after 20 years=$56,857.11

To find: How much more interest would Rich and Laura earn by retiring after 21 years?

Solution: We will first find the interest earned if they retire after 21 Years Then we will compare it with the interest earned if they retire after 20 years.

We will first find the interest Rich and Laura will earn if they retire after 21 year.

Interest= Account balance− amount deposited

=168,915.68−105,000

=$63,915.68

The interest they will earn if they retire after 20 years is $56,857.11. So they will earn $7058.57 more in interest.

Rich and Laura would earn$7058.57 interest more by retiring after 21 years.

Page 158 Problem 3 Answer

Given: Linda and Rob open an online savings account that has a 3.6% annual interest rate compounded monthly.

If they deposit $1,200 every month, $173,022.87  will be in the account after 10 years

To find: If opening an account at a higher interest rate for fewer years would have assured Linda and Rob at least the same final balance

Solution: We will calculate the future value using the formula:

B=P((1+r/n)^nt−1)/r/n

We can see from the formula that the value of the deposit will increase if the rate of interest is higher.

So, if the rate of interest is higher we will achieve the amount they will receive at the current interest rate in fewer years.

Thus, opening an account at a higher interest rate for fewer years would give Linda and Rob at least the same final balance.

Yes, opening an account at a higher interest rate for fewer years would have assured Linda and Rob at least the same final balance.

Page 158 Problem 4 Answer

Given Example 1

To do: During the computation of the numerator, is the 1 being subtracted from the 20. Explain your reasoning

Firstly we have the solution as

B=5,000((1+0.045/1)¹(20)−1)/0.045/1

where 1 is subtracted from the before computation but not the exponent means that (1+0.045/1)1(20) is computed first and then one is subtracted from the value not from 20

Therefore, No, using the order of operations (1+0.045/1)1(20) is computed first and then 1 is subtracted from the value not from 20

Page 159 Problem 5 Answer

Given:p= Principle=$1,000

r= Interest rate=5%=0.05

n=Number or periods per year=1

t=Time expressed in years=4 years

Find:How much money will be in that account at the end of 4 years.

(Given)

Then, By using the formula:

A=P(1+r/n)^n×t

=1,000(1+0.05/1)^1×4

≈1,215.51

​Hence, The required answer is: $1,215.51

Page 159 Problem 6 Answer

Given:p= Principle=$500

r= Interest rate=3%=0.03

n= Number or periods per year=2

t= Time expressed in years=2 years

Find: How much is in the account at the end of two years.

p= Principle =$500

r= Interest rate =3%=0.03

n= Number or periods per year =2

t= Time expressed in years =2 years

​(Given)

By using the formula:

A=P(1+r/n)^n×t

=500(1+0.03/2)^2×2

=500(1.015)⁴

≈530.68​

Hence, the account at the end of two years is:$530.68

Page 159 Problem 7 Answer

Given:p= Principle=$10,000

r= Interest rate=4%=0.04

n= Number or periods per year=4

t= Time expressed in years=3 years

Find: How much does Derrick receive from his grandparents on his 18th birthday.

According to question,

A=P(1+r/n)^n×t

=10,000(1+0.04/4)^2×3

=10,000(1.01)⁶

≈10,615.20

​Hence, The required answer is: $10,615.20

Page 159 Problem 8 Answer

Given: Barbara wants to restore her ′66 Mustang in 4 years. She puts $200 into an account every month that pays 4.5% interest, compounded monthly.

Find: How much is in the account after 4 years.

According to question,

B=P((1+r/n)^nt−1)/r/n

=200((1+0.045/12)^12×4−1)/0.045/12

≈10,496.77

​Hence, the account after 4 years is:$10,496.77

Page 159 Problem 9 Answer

Given: Robbie opens an account at a local bank by depositing $100.

The account pays 2.4% interest, compounded weekly. He deposits $100 every week for three years.

Find: How much is in the account after three years.

According to question,

B=P((1+r/n)^nt−1)/r/n

=100((1+0.024/52)^52×3−1)/0.024/52

≈16,171.46

​Hence, the account after three years is:$16,171.46

Page 159 Problem 10 Answer

Given: Robbie opens an account at a local bank by depositing $100.

The account pays 2.4% interest, compounded weekly. He deposits $100 every week for three years.

Find:Write the future value function if x represents the number of weeks.

According to question,

B=P((1+r/n)^nt−1)/r/n

=100((1+0.024/52)^52×x/52−1)/0.024/52

=100((1+0.024/52)x−1)/0.024/52 the future value function if x represents the number of weeks is:=100((1+0.024/52)x−1)/0.024/52

Page 159 Problem 11 Answer

Given: Robbie opens an account at a local bank by depositing $100.

The account pays 2.4% interest, compounded weekly. He deposits $100 every week for three years.

Find:Use a graphing calculator to graph the future value function.

Then, According to question,

Therefore, The graph is:

Page 159 Problem 12 Answer

Given: Robbie opens an account at a local bank by depositing $100.

The account pays 2.4% interest, compounded weekly. He deposits $100 every week for three years.

Find:  Using the graph, what is the approximate balance after 2 years.

According to question,

The approximate balance after 2 years is $10,651.

Hence, The graph of solution is:

Page 159 Exercise 1 Answer

Given: Suppose $600 is deposited into an account every quarter.

The account earns 5% interest, compounded quarterly.

Find: What is the future value of the account after 5 years.

According to question,

B=P((1+r/n)^nt−1)/r/n

=600((1+0.05/4)^4×5−1)/0.05/4

≈13,537.79

​Hence,  the future value of the account after 5 years is:$13,537.79

Page 159 Exercise 2 Answer

Given: Suppose $600 is deposited into an account every quarter. The account earns 5% interest, compounded quarterly.

Find: Write the future value function if x represents the number of quarters.

According to question,

B=P((1+r/n)^nt−1)/r/n

=600((1+0.05/4)4×x/4−1)/0.05/4

=600((1+0.0125)x−1)/0.0125

=48,000((1.0125)x−1)

Hence, the future value function if x represents the number of quarters is:

=48,000((1.0125)x−1)

Page 159 Exercise 3 Answer

Given: Suppose $600 is deposited into an account every quarter. The account earns 5% interest, compounded quarterly.

Find:  Use a graphing calculator to graph the future value function.

According to question,

Hence, The graph is:

Page 159 Exercise 4 Answer

Given: Suppose $600 is deposited into an account every quarter. The account earns 5% interest, compounded quarterly

Find: Using the graph, what is the approximate balance after 3 years.

According to question,

The approximate balance after 3 years is $7,716.

Then, The graph is:

Hence, The graph is:

Page 160 Exercise 5 Answer

To find: Suppose that you invest $2,000 at a 1% annual interest rate.

Use your calculator to input different values for t in the compound interest formula.

What whole number value of t will yield an amount closest to twice the initial deposit?

B=P(1+r/n)^n×t

40000=20000(1+0.06/1)^1×t

40000=20000(1.06)^t

2=(1.06)^t

ln2=ln(1.06)^t

ln2=tln(1.06)

ln2/ln1.06=t

​Evaluate: t=ln2/ln1.01≈70

Thus we note that the amount will be twice the initial deposit after roughly 70 years.

Hence the answer is 70 years

Page 160 Exercise 6 Answer

To find: Suppose that you invest $4,000 at a 2% annual interest rate.

Use your calculator to input different values for t in the compound interest formula.

What whole number value of t will yield an amount closest to twice the initial deposit?

B=P(1+r/n)^n×t

40000=20000(1+0.06/1)^1×t

40000=20000(1.06)^t

2=(1.06)^t

ln2=ln(1.06)^t

ln2=tln(1.06)

ln2/ln1.06=t

​Evaluate: t=ln2/ln1.02≈35

Thus we note that the amount will be twice the initial deposit after roughly 35 years.

Page 160 Exercise 7 Answer

To find; Suppose that you invest $20,000 at a 6% annual interest rate.

Use your calculator to input different values for t in the compound interest formula.

What whole number value of t will yield an amount closest to twice the initial deposit?

B=P(1+r/n)^n×t

40000=20000(1+0.06/1)^1×t

40000=20000(1.06)^t

2=(1.06)^t

ln2=ln(1.06)^t

ln2=tln(1.06)

ln2/ln1.06=t

​Evaluate: t=ln2/ln1.06≈12

Thus we note that the amount will be twice the initial deposit after roughly 12 years.

Hence we note that the amount will be twice the initial deposit after roughly 12 years.

Page 160 Exercise 8 Answer

To find: Albert Einstein noticed a very interesting pattern when an initial deposit doubled.

In each of the three examples above, multiply the value of t that you determined times the percentage amount.

For example, in a. multiply t by 1. What do you notice?

First example of part (a)

Result part (a):

t=70

r=1%

​We need to multiply the value of t by the percentage amount of r.

t×r=70×1=70

Second example of part (b)

Result part (b):

t=35

r=2%

We need to multiply the value of t by the percentage amount of r.

t×r=35×2=70

Third example of part (c)

Result part (c):

t=12

r=6%

​We need to multiply the value of t by the percentage amount of r.

t×r=12×6=72

We notice that the product in all three examples is equal to 70 or 72 , thus the product is always close to 72 .

Hence We notice that the product in all three examples is equal to 70 or 72 , thus the product is always close to 72 .

Page 160 Exercise 9 Answer

To find: Einstein called this the Rule of 72 because for any initial deposit and for any interest percentage, 72 (percentage) will give you the approximate number of years it will take for the initial deposit to double in value.

Einstein also said that “If people really understood the Rule of 72 they would never put their money in banks.”

Suppose that a 10-year-old has $500 to invest.

She puts it in her savings account that has a 1.75% annual interest rate.

How old will she be when the money doubles?

Given:r= Interest rate =1.75%

By Einstein’s rule of 72 , we know that the product of the time $t$ and the interest rate percentage $r$ needs to be equal to 72 when we double the money.

t×r=72

In this case: r=1.75

t×1.75=72

Divide each side by 1.75:

t=72

1.75

≈41

This then means that the money doubles in roughly 41 years.

Since she is currently 10 years old, the money will double when she is roughly 10+41=51 years.

Hence the answer is 51 years old

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 151 Problem 1 Answer

We will use your calculator to find the value of g(x) for each of the increasing values of x in the table.

The values will be:

x	g(x)=−5x+1
100	−499
1000	−4999
90000	−449,999
900,000	−4,499,999
8,000,000	−39,999,999
50,000,000	−249,999,999

As x approaches infinity, the value of g(x) decreases without. Therefore,g(x) has no limit.

As the values of x increase towards infinity, the values of g(x)=−5x+1 keeps on decreasing.

Page 151 Problem 2 Answer

Given: f(x)=1/x

To find: lim f(x)
_x→∞

Solution: Set up a table with increasing values of x.

x	f(x)=1x
100	0.01
1,000	0.001
90,000	0.00001111111
900,000	0.00000111111
8,000,000	0.000000125
50,000,000	0.00000002
2,000,000,000	0.0000000005

We will first set up a table with increasing values of x

The pattern in the table shows that as x approaches infinity,f(x) approaches 0.

It keeps getting closer to 0; it, never reaches 0.

You can say, “The limit of f(x), as x approaches infinity, is 0,”

If f(x)=1/x, then lim _x→∞=0

Page 152 Problem 3 Answer

Given:f(x)=1x

To find: lim
             _x→∞

Solution: The value of the function will always be 1 for any value of x

So, the limit of f(x), as x approaches infinity is 1

Iff(x)=1x, then lim  = 1
                          _x→∞

Page 152 Problem 4 Answer

Given: f(x)=(1+0.05/x)^x

To find: lim (1+0.05/x)^x
            x→∞

Solution: We will create a table to find values of the function for v

The table showing various values of the function is; x f(x) to five decimal places

x	f(x) to five decimal places
100	1.05126
1,000	1.05127
90,000	1.05127
900,000	1.05127
8,000,000	1.05127
50,000,000	1.05127
2,000,000,000	1.05127

The pattern in the table shows that as x approaches infinity,f(x) approaches a number around1.05127…

Using a table we get, Lim (1+0.05/x)^x
                                   x→∞

=1.05127 rounded to five decimal places.

Page 153 Problem 5 Answer

Given: e=2.718281828

π=3.141592654

To find: Difference between e π and π e

Solution

e π=23.1406926328

Π e=22.4591577184

Difference=e π−πe

=23.1406926328−22.4591577184

=0.6815349144

=0.681  (rounded to the nearest thousandth)

The difference between eπ and πe  rounded to the nearest thousandth is0.681

Page 153 Problem 6 Answer

Given: Craig deposits $5,000  at 5.12% interest compounded continuously for four years

To find: Ending balance to the nearest cent

Solution: Balance= Deposit amount×e years

=5000×(2.7182818)4

=$272,990.7

If Craig deposits$5,000 at 5.12% interest compounded continuously for four years, then the balance in the account at the end would be$272,990.7

(rounded off to nearest cent)

Page 154 Problem 7 Answer

According to question,”Infinite” is a concept that is hard to understand for people, because we cannot observe infinity and thus we need our imagination to be able to take into account the meaning of “infinite”.

This then implies that the question of “infinite” profoundly moves the spirit of a human.

Hence, The required asnwer is: “Infinite” is a concept that is hard to understand for people

Page 154 Problem 8 Answer

Given:p= Principle =$2,000r= Interest rate =4%=0.04t= Time expressed in years =3 years

Find the interest if it is computed using simple interest.

According to question,

I=prt

=2000×0.04×3

=240

​Hence, The interest if it is computed using simple interest is:$240

Page 154 Problem 9 Answer

Given:​

p= Principle =$2,000

r= Interest rate =4%=0.04

t= Time expressed in years =3 years

​Find the interest if it is compounded annually.

p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3years

Then by using the formula,

B=p(1+r/n)^n×t

=2000(1+0.04/1)^1×3

=2000(1.04)³

≈2,249.73

Then, The interest is the balance decreased by the principle:

I=B−p

=2,249.73−2,000

=249.73

​Hence, The interest if it is compounded annually is:$249.73

Page 154 Problem 10 Answer

Given:p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3 years

n=Number of periods per year=2

Find the interest if it is compounded semiannually.

p= Principle =$2,000

r= Interest rate =4%=0.04

t= Time expressed in years =3 years

n= Number of periods per year =2

​(Given)

Then, by using the formula:

B=p(1+r/n)^n×t

=2000(1+0.04/2)^2×3

=2000(1.02)⁶

≈2,252.32

​And then according to question,

I=B−p

=2,252.32−2,000

=252.32

​Hence, the interest if it is compounded semiannually is:$252.32

Page 154 Problem 11 Answer

Given:p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3years

n=Number of periods per year=4

Find the interest if it is compounded quarterly.

p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3 years

n=Number of periods per year=4

Then, by using the formula:

B=p(1+r/n)^n×t

=2000(1+0.04/4)^4×3

=2000(1.01)12

≈2,253.65

And then according to question,The interest is the balance decreased by the principle:

I=B−p

=2,253.65−2,000

=253.65

​Hence, The interest if it is compounded quarterly is:$253.65

Page 154 Problem 12 Answer

Given:p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3 years

n=Number of periods per year=12

Find the interest if it is compounded monthly.

p= Principle =$2,000

r= Interest rate =4%=0.04

t= Time expressed in years =3 years

n= Number of periods per year =12

​(Given)

Then, by using the formula:

B=p(1+r/n)^n×t

=2000(1+0.04/12)^12×3

≈2,254.54

​And then according to question,The interest is the balance decreased by the principle:

I=B−p

=2,254.54−2,000

=254.54

Hence,the interest if it is compounded monthly is:$254.54

Page 154 Problem 13 Answer

Given:p=Principle=$2,000

r= Interest rate=4%=0.04

t= Time expressed in years=3 years

n= Number of periods per year=365

Find the interest if it is compounded daily.

p= Principle =$2,000

r= Interest rate =4%=0.04

t= Time expressed in years =3 years

n= Number of periods per year =365

​(Given)

Then, By using the formula:

B=p(1+r/n)^n×t

=2000(1+0.04/365)^365×3

≈2254.98

And then according to question,The interest is the balance decreased by the principle:

I=B−p

=2,254.98−2,000

=254.98

​Hence, the interest if it is compounded daily is:$254.98

Page 154 Problem 14 Answer

Given:p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3 years

n=Number of periods per year=365×24=8760

Find: the interest if it is compounded hourly..

According to question,

By using the formula:

B=p(1+r/n)^n×t

=2000(1+0.04/8760)^8760×3

≈2254.99

Then, As we know that:

The interest is the balance decreased by the principle:

I=B−p

=2,254.99−2,000

=254.99

​Hence, The interest if it is compounded hourly is:$254.99

Page 154 Problem 15 Answer

Given:p=Principle=$2,000

r=Interest rate=4%=0.04

t=Time expressed in years=3 years

n=Number of periods per year=365×24×60=525,600

Find: the interest if it is compounded every minute.

According to question,

B=p(1+r/n)^n×t

=2000(1+0.04/525,600)^525,600×3

≈2254.99

Then, As we know that:

I=B−p

=2,254.99−2,000

=254.99

​Hence,  the interest if it is compounded every minute is:$254.99

Page 154 Problem 16 Answer

Given:p= Principle =$2,000r= Interest rate =4%=0.04t= Time expressed in years =3 years

Find: the interest if it is compounded continuously.

According to question,

B=per×t

=2000e0.04×3

=2000e0.12

≈2254.99

​Then, As we know that,

The interest is the balance decreased by the principle:

I=B−p

=2,254.99−2,000

=254.99

​Hence, The interest if it is compounded continuously is:$254.99

Page 154 Problem 17 Answer

According to question,

Result of first part:$240

(simple interest)

Result pf second part:$254.99

(compounded continuously)

Then, the difference of interest between compounded continuously and simple interest.

$254.99−$240=$14.99

Hence, The difference in interest between simple interest and interest compounded continuously is:$14.99

Page 154 Exercise 1 Answer

Given:p=Principle=$50,000r=Interestrate=4\dfrac{1}{8}%=4.125%=0.04125t=Timeexpressedinyears=6years

Find: the interest earned .

According to question,

By using the formula:

B=per×t

=50000e0.04125×6

=50000e0.2475

≈64,040.97

​Then,The interest is the balance decreased by the principle:

I=B−p

=64,040.97−50,000

=14,040.97

​Hence, The interest is:$14,040.97

Page 154 Exercise 2 Answer

Given: p= Principle=$9,000

r=Interest rate=4.1%=0.041

t=Time expressed in years=2 years

n=Number of periods per year=12

Find: How much interest does State Bank pay.

According to question,

By using the formula:

B=p(1+r/n)^n×t

=9000(1+0.041/12)^12×2

≈9,767.74

Then,The interest is the balance decreased by the principle:

I=B−p

=9,767.74−9,000

=767.74

​Hence, interest does State Bank pay is:$767.74

Page 154 Exercise 3 Answer

Given:p= Principle =$9,000r= Interest rate =4.01%=0.0401t= Time expressed in years =2 years

Find: How much interest does Kings Savings pay.

According to question,

By using the formula:

B=per×t

=9000e0.0401×2

=9000e0.0802

≈9,751.53

​Then,The interest is the balance decreased by the principle:

I=B−p

=9,751.53−9,000

=751.53

​Hence, The interest does Kings Savings pay is:$751.53

Page 154 Exercise 4 Answer

According to question,

Result of first part: $767.74

Result of second part: $751.53

Then the result of part first is higher and thus State Bank earns more interest.

The difference of the interests:

$767.74−$751.53=$16.21

Therefore, The State Bank will result in$16.21 more interest.

Hence, The State Bank will result in$16.21 । more interest.

Page 154 Exercise 5 Answer

As we know that,There are many factors that could affect Whitney’s choice, such as:

Location bank Opening hours of the bank Other services offered by the bank

Hence, The answer is above explanation.

Page 155 Exercise 6 Answer

Given:p= Principle =$1,000r= Interest rate =16%=0.16t= Time expressed in years =5 years

Find: what is the ending balance.

According to question,

By using the formula,

B=per×t

=1000e0.16×5

=1000e0.8

≈2,225.54​

Hence, The ending balance is:$2,225.54

Page 155 Exercise 7 Answer        

Given:p= Principle =$1,000r= Interest rate =5%=0.05t= Time expressed in years =5 years

Find: What is the ending balance,

According to question,

B=per×t

=1000e0.05×5

=1000e0.25

≈1,284.03​

Hence, The ending balance is: $1,284.03

Page 155 Exercise 8 Answer        

According to question,

Firstly2225.54 is the ending bal. At a 16% interest rate for a 5-year period.

And then 1284.02 is the ending bal. At a 5% interest rate for a 5-year period.

2225.54−1284.02=941.52

Hence, The difference between the two ending balances is: 941.52

Page 155 Exercise 9 Answer   

Given:p= Principle =$30,000r= Interest rate =4\ddrac{1}{2}%=4.5%=0.045t= Time expressed in years =0.5 years (6 months)

Find: interest earned.

According to question, By using the formula:

B=per×t

=30,000e0.045×0.5

=30,000e0.0225

≈30,682.65

​Then, The interest is the balance decreased by the principle:

I=B−p

=30,682.65−30,000

=682.65

Hence, The interest is: $682.65

Page 155 Exercise 10 Answer    

Given:p= Principle=$4,000

r=Interest rate=3.8%=0.038

t=Time expressed in years=3years

n=Number of periods per year=4

Find: How much interest does Option 1 pay.

According to question,

B=p(1+r/n)^n×t

=4000(1+0.038/4)^4×3

≈4,480.60

Then, The interest is the balance decreased by the principle:

I=B−p

=4,480.60−4,000

=480.60​

Hence, The interest is: $480.60

Page 155 Exercise 11 Answer        

Given:p= Principle =$4,000r= Interest rate =3.5%=0.035t= Time expressed in years =3 years

Find: interest does Option 2 pay

According to question,

B=per×t

=4000e0.035×3

=4000e0.105

≈4442.84

Then, The interest is the balance decreased by the principle:

I=B−p

=4,442.84−4,000

=442.84

​Hence, The interest is:$442.84

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 145 Problem 1 Answer

Given: Rico deposits $800 at 3.87% interest, compounded quarterly.

We have to find his ending balance after one year.

We will use above formula.

Here,

P=800

r=.0387

n=12

⇒FV=800∗1.00322512

⇒FV=$831.515

​Therefore, His ending balance after one year is$831.515

Page 146 Problem 2 Answer

Given:  Kate deposits $2,350 in an account that earns interest at a rate of 3.1%, compounded monthly.

We have to find her ending balance after five years.

We will use above formula.

Here,​t=5

P=2,350

r=.031

n=12

⇒FV=2350∗1.002560

⇒FV=$2,743.45

​Therefore, Her ending balance after five years is$2,74345

Page 147 Problem 3 Answer

Given; Barbara deposits 3,000 in a one-year CD at 4.1 % interest, compounded daily.

To find: What is the APY to the nearest hundredth of a percent?

The APY using the compound interest formula and the simple interest formula.

​B=p(1+r/n)^nt

B=3,000(1+0.041/365)^365×1

B=3125

I=3125−2,000

=125

I=prt

=I/pt

=125/3,000×1

=0.041%

​r= 4.1 %

The annual percentage yield is 4.10 %

APY can also be found by using the formula APY (1+r/n)ⁿ⁻¹

APY=(1+r/n)ⁿ⁻¹

APY=(1+0.041/365)³⁶⁵⁻¹

APY≈0.041=4.1%

​The annual percentage yield is 4.1 %, which is the same as the previous answer

Hence The APY to the nearest hundredth of a percent is found to be 4.10 %

Page 145 Problem 4 Answer

Given; One is an account that pays 2.78 % compounded daily. The other account pays 3.25 % compounded quarterly.

To find: Would either of these accounts provide Nancy with a better return than her current account? If so, which account?

To find which account earns the most money compute the ratio of A/P for each account then compare them.

For the second account: r=2.78% and the compound rate is continuous, therefore:

A/P=e^2.78%≈1.028

For the Second account:r=3.25% and n=4  therefore:A/P

=(1+3.25%/4)⁴ ≈1.032

1.032>1.028

The second account earns more money because its ratio of Future Value to Initial Value higher than the other account.

Hence we conclude that The second account earns more money because its ratio of Future Value to Initial Value higher than the other account.

Page 146 Problem 5 Answer

Given;  The ending balance after k years of an account that starts with a balance of 2,000 $ and earns interest at a rate of3.5 %, compounded daily.

To find: Write an algebraic expression

we have given

P=2000,

r=0.035,

t=k

A(t)=P⋅er⋅(t)

A(t)=2000⋅e0.035⋅(t)

A(k)=2000⋅e0.035⋅(k)

Hence an algebraic expression for the ending balance after k years of an account that starts with a balance of 2,000 and earns interest at a rate of 3.5 %, compounded daily.

will be given by A(k)=2000⋅e^0.035⋅(k)

Page 147 Problem 6 Answer

Given ; Consider an amount x deposited into a CD at 2.4 %  interest compounded daily, and the same amount deposited into a CD at the same rate that compounds monthly.

To find : Explain why, after 1 year, the balance on a CD that compounds daily is greater than the CD that compounded monthly.

Find the APY using the compound interest formula and the simple interest formula.

​B=p(1+r/n)^nt

B=x(1+0.024/365)^365×1

Subtract the principal from the new balance.

I = x(1+0.024/365)^365×1

=x(1+0.024/365)^365×1

−x/x×1

APY can also be found by using the formula APY=(1+r/n)ⁿ⁻¹  is the interest rate and n is the number of times interest is compounded per year.

1-Year CD with both of them at the same 2.4 % APR interest rate, and compare how much interest you have at the end of the year. Both credit interest monthly.

Since MoBank compounds monthly, you are getting 2.4 5/12 =0.002 every month.

so it much less that the daily compounded interest Since MoBank compounds monthly, we are getting 2.4 % /12= 0.002 every month.

so it much less that the daily compounded interest

Page 148 Problem 7 Answer

According to question,

The words imply that if you want to earn a lot of money through interest, then you need to start with a lot of money.

As we know that, This means that you will not get rich from bank interest alone, because the interest rates are often quite small and you can earn much more from other business investments.

However, business investments contain a lot more risk compared to interest.

Hence, The required answer is:

You will not get rich from bank interest alone .

Page 148 Problem 8 Answer

Given: p=4,000 Rate percent

Find: What is his balance, to the nearest cent, at the end of 10 years.

According to question,

B=p(1+r/n)ⁿt

=4000(1+0.05/2)^2×10

=4000(1+0.025)20

=4000(1.025)20

≈6554.47

​Hence, Balance, to the nearest cent, at the end of 10 years is:$6,554.47

Page 148 Problem 9 Answer

Given: On Olga′s 16 th birthday, her uncle invested $2,000 in an account that was locked into a 4.75% interest rate, compounded monthly.

Find:How much will Olga have in the account when she turns 18? Round to the nearest cent.

According to question,

B=p(1+r/n)ⁿt

=2000(1+0.0475/12)^12×2

=2000(1+0.0475/12)²⁴

≈2198.91

​Hence, The required answer is: $2,198.91

Page 148 Problem 10 Answer

Given: Samantha deposits $1,500 into the Park Street Bank. The account pays 4.12% annual interest, compounded daily. To the nearest cent,

Find: To the nearest cent, how much is in the account at the end of three non-leap years

According to question,

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

=1500(1+0.0412/365)^365×3

≈1697.33

​Hence, The required answer is: $1,697.33

Page 148 Problem 11 Answer

Given: Joanne deposits $4,300 into a one−year CD at a rate of 4.3%, compounded daily.

Find: What is her ending balance after the year.

According to question,

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

=4300(1+0.043/365)^365×1

≈4488.92

​Hence, The required answer is: $4,488.92

Page 148 Problem 12 Answer

Given: Joanne deposits $4,300 into a one−year CD at a rate of 4.3%,

Find:How much interest does she earn.

According to question,

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

=4300(1+0.043/365)^365×1

≈4488.92

Then,The interest is the difference between the new balance and the principle:

I=B−p

=4488.92−4300

=188.92

​Hence, The required answer is: $188.92

Page 148 Problem 13 Answer

Given: Joanne deposits $4,300 into a one−year CD at a rate of 4.3%, compounded daily.

Find:What is her annual percentage yield to the nearest hundredth of a percent.

According to question,

By usiing the annual percentage yield formula to determine the APY:

APY=(1+r/n)ⁿ⁻¹

=(1+0.043/365)³⁶⁵⁻¹

≈1.0439−1

=0.0439

=4.39%

​Hence, Annual percentage yield to the nearest hundredth of a percent is:=4.39%

Page 148 Problem 14 Answer

Given: Mike deposits $5,000 in a three−year CD account that yields 3.5% interest, compounded weekly.

Find: What is his ending balance at the end of three years.

According to question,

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

=5000(1+0.035/52)^52×3

≈5553.36

​Hence, The required answer is: $5,553.36

Page 148 Exercise 1 Answer

Given: Mike and Julie receive $20,000 in gifts from friends and relatives for their wedding.

Find: Will their money double in fourteen years

​p= Principal=$20,000

r=Interest rate=4.75%=0.0475

t=Time expresses in years=14 years

(Given) Then, By using the formula,

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

=20000(1+0.0475/365)^365×14

≈38888.12

Hence, the balance in 14 years is$38,888.12, which is not the double of the principle of$20,000  and thus the money was not doubled in 14 years..

Hence, The required answer is: No

Page 148 Exercise 2 Answer

Given:​

p= Principle =$20,000

r= Interest rate =4.75%=0.0475

t= Time expressed in years =15 years

​Find: Will their money double in fifteen years.

According to question,

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

=20000(1+0.0475/365)^365×15

≈40779.76

Then, the balance in 15 years is$40,779.76, which is more than the double of the principle of$20,000 and thus the money was doubled in 15 years.

Hence, The required answer is: Yes.

Page 149 Exercise 3 Answer

Given:p= Principle =$5,000r= Interest rate =6%=0.06t= Time expressed in years =1 year

Find: the simple interest for a one-year CD for $5,000 at a 6% interest rate.

Given:p= Principle =$5,000r= Interest rate =6%=0.06t= Time expressed in years =1 year

(Given) Then,by using the formula:

I=prt

=5000×0.06×1

=300

​Hence, The simple interest is:$300

Page 149 Exercise 4 Answer

Given: CD for $5,000 at an interest rate of 6%, compounded annually.

Find: the interest for a one-year.

According to question,

B=p(1+r/n)^n t

=5000(1+0.06/1)^1×1

=5000(1.06)

=5300

​Then, The interest is the difference between the balance and the principle:

I=B−p

=5300−5000

=300

​Hence, The interest is:=300

Page 149 Exercise 5 Answer

Then, We note that the results of part (a) and (b) are both$300, which implies that the simple interest formula and the compound interest formula result in the same interest when the time is l year and the compounding occurs annually.

Hence, Both the result is same.

Page 149 Exercise 6 Answer

Given:CD for $5,000 at an interest rate of 6%.

Find: Simple interest.

According to question,

I=prt

=5000×0.06×3

=900

​Hence, The interest is:$900

Page 149 Exercise 7 Answer

Given:$5,000 at an interest rate of 6%, compounded annually.

Find: the interest for a three-year.

According to question,

B=p(1+r/n)^n t

=5000(1+0.06/1)^1×3

=5000(1.06)3

=5955.08

Then,The interest is the difference between the balance and the principle:

I=B−p

=5955.08−5000

=955.08

​Hence, The interest is:$955.08

Page 149 Exercise 8 Answer

According to question,

We note that the result of part (e) is higher than the result of part (d), which implies that the compound interest is higher than the simple interest when the time is longer than 1 year and when the compounding is annually.

Therefore, compound in part (e) is higher.

Page 149 Exercise 9 Answer

Given: CD for $5,000 at an interest rate of 4%.

Find: The simple interest.

According to question,

I=prt

=5000×0.04×6

=1200

Hence, The simple interest is:$1,200

Page 149 Exercise 10 Answer

Given:$5,000 at an interest rate of 4%, compounded annually

Find The interest.

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

=5000(1+0.04/1)^1×6

=5000(1.04)6

≈6326.60

Then, The interest is the difference between the balance and the principle:

I=B−p

=6326.60−5000

=1326.60

​Hence, The interest is:

$1,326.60

Page 149 Exercise 11 Answer

According to question,

We note that the result of part (h) is higher than the result of part (g), which implies that the compound interest is higher than the simple interest when the time is longer than 1 year and when the compounding is annual.

Hence, Compound interest in part (h) is higher.

Page 149 Exercise 12 Answer

According to question,

No, compound interest is not the same as sample interest when compounding annually, specifically when the time is more than 1 year.

This is due to the fact that the interest will earn more interest in the future when using compound interest .

Hence, The required answer is:No

Page 149 Exercise 13 Answer

Given: Rodney invests a sum of money, P, into an account that earns interest at a rate of r, compounded yearly.

Gerald invests half that amount into an account that pays twice Rodney′s  interest rate

​Find: Which of the accounts will have the higher ending balance after one year.

According to the question, Firstly we can calculate for Rodney:

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

 =P(1+r/1)^1×1

=P(1+r)

=P+Pr

​Then we can calculate for Gerald:

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

=P/2(1+2r/1)^1×1

=P/2(1+2r)

=P/2+Pr

Then comparison is:

Rodney has a higher ending balance.

Hence, Rodney has a higher ending balance.

Page 149 Exercise 14 Answer

Given: Island Bank is advertising a special 6.55% APR for CDs. Manny takes out a one−year CD for $40,000. The interest is compounded daily

Find the annual percentage yield for Manny’s account to the nearest hundredth of a percent.

According to question,

APY=(1+r/n)ⁿ⁻¹

=(1+0.0655/365)³⁶⁵⁻¹

≈1.0677−1

=0.0677

=6.77%

​Hence, the annual percentage yield for Manny’s account to the nearest hundredth of a percent is=6.77%

Page 149 Exercise 15 Answer

Given: An elite private college receives large donations from successful alumni.

The account that holds these donations has $955,000,000 currently.

Find:  How much would the account earn in one year of simple interest at a rate of 5.33%.

According to question,

I=prt

=955,000,000×0.0533×1

=50,901,500

​Hence, The simple interest is: $50,901,500

Page 149 Exercise 16 Answer

Given: An elite private college receives large donations from successful alumni.

The account that holds these donations has $955,000,000 currently.

Find: How much would the account earn in one year at 5.33% if the interest was compounded daily.

According to question,

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

=955,000,000(1+0.0533/365)^365×1

≈1,007,278,530.91

​Then,The interest is the difference between the balance and the principle:

I=B−p

=1,007,278,530.91−955,000,000

=52,278,530.91

​Hence, The required answer is: $52,278,530.91

Page 149 Exercise 17 Answer

Given: An elite private college receives large donations from successful alumni. The account that holds these donations has $955,000,000 currently.

Find: How much

According to question,

Result of first part is:$50,901,500 (simple interest)

Result of second part:$52,278,530.91 (compounded daily)

The differences of the interests:

$52,278,530.91−$50,901,500=$1,377,030.91

Hence, As we note that the compounded daily earns$1,377,030.91 more than the simple interest.

Hence,$1,377,030.91 more interest is earned by compounded daily as compared to simple interest.

Page 149 Exercise 18 Answer

Given: An elite private college receives large donations from successful alumni.

The account that holds these donations has $955,000,000 currently.

Find: If the money is used to pay full scholarships, and the price of tuition is $61,000 per year to attend, how many more students can receive full four-year scholarships if the interest was compounded daily rather than using simple interest.

According to question,Result of first part is:$50,901,500 (simple interest)

Result of second part is:$52,278,530.91 (compounded daily)

The differences of the interests:

$52,278,530.91−$50,901,500=$1,377,030.91

Hence, As we have $1,377,030.91 more when using daily compounding.

Then according to question,Every student receives a tuition of $61,000per year, which is thus$61,000×4=$244,000 per full four-year scholarship.

Then, The number of additional students that can then receive a full four-year scholarship are then the additional interest divided by$244,000.

$1,377,030.91

$244,000≈5.6436≈5

Therefore, 5 more students can receive a full four-year scholarship.

Hence,5 more students can receive a full four-year scholarship.

• Chapter 1 Solving Linear Equations
• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.9 The Stock Market
• Chapter 2 Solving Linear Inequalities
• • Cengage Financial Algebra 1st Edition Chapter 2 Assessment Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.1 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.2 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.3 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.4 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.5 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.6 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.7 Modeling a Business
  • Cengage Financial Algebra 1st Edition Chapter 2 Exercise 2.8 Modeling a Business

• Chapter 3 Graphing Linear Functions

• • Cengage Financial Algebra 1st Edition Chapter 3 Assessment Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.1 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.2 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.3 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.4 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.5 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.6 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.7 Banking Services
  • Cengage Financial Algebra 1st Edition Chapter 3 Exercise 3.8 Banking Services

• Chapter 4 Writing Linear Functions

• • Cengage Financial Algebra 1st Edition Chapter 4 Assessment Consumer Credit
  • Cengage Financial Algebra 1st Edition Chapter 4 Exercise 4.1 Consumer Credit
  • Cengage Financial Algebra 1st Edition Chapter 4 Exercise 4.2 Consumer Credit
  • Cengage Financial Algebra 1st Edition Chapter 4 Exercise 4.3 Consumer Credit
  • Cengage Financial Algebra 1st Edition Chapter 4 Exercise 4.4 Consumer Credit
  • Cengage Financial Algebra 1st Edition Chapter 4 Exercise 4.5 Consumer Credit
  • Cengage Financial Algebra 1st Edition Chapter 4 Exercise 4.6 Consumer Credit

• Chapter 5 Solving Systems of Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.1 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.2 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.3 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.4 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.5 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.6 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.7 Automobile Ownership
  • Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership

• Chapter 6 Exponential Functions and Sequences
• Chapter 7 Polynomial Equations and Factoring
• Chapter 8 Graphing Quadratic Functions

Financial Algebra 1st Edition Chapter 1 The Stock Market

Page 53 Problem 1 Answer

Given: Number of shares = x  Price per share=$y

Annual dividend per share=$d

To find: The percent yield

Solution: Write the yield as a fraction.

Then convert the fraction to a decimal. Finally write the decimal as a percent.

Yield=Annual dividend per share Current price of one share

=d/y                                                [in fraction]

=d/y×100%                               [in percentage]

If we bought x shares of a stock for $y  per share and the annual dividend per share is $d then yield is 100 d/y%

Page 53 Problem 2 Answer

Given: For one share of Skroy Corporation:

Annual Dividend=$1.55

Today’s close=$x

Net change=+0.40

To find: The yield at yesterday’s close

Solution: First, find the closing of the

Yesterday’s close=Today’s close−Net change

=x−0.40

Yield=Dividend

Yesterday′ close

=1.55/x−0.40          [in fraction]

=1.55/x−0.40×100%    [in percentage]

When one share of Skro y Corporation stock pays an annual dividend of$1.55 and today’s Skro y closed at x dollars with a net change of +0.40, then the yield at yesterday’s close will be 100×1.55/x−0.40%

Page 54 Problem 3 Answer

Given: A corporation was paying a $2.10 annual dividend. The stock underwent an a3−for−2 split

To find: The new annual dividend per share

Solution: New annual dividend per share=b/a× Pre Split annual dividend per share

=2/3×2.10

=$1.40 (rounded off to two decimals)

A corporation was paying a$2.10 annual dividend and the stock underwent a a3−for−2 split so the post-split annual dividend per share is $1.40

Page 53 Problem 4 Answer

Given: Details of bond held by Adam Principal amount(P)=$1000

Rate of interest (R)=5.7% per year i.e.0.057

To find Total interest received if the time period (T) of holding bonds is 11years.

Solution: Interest=PRT

=1000×0.057×11

=627

If Adam holds the bond from Example 6 for 11 years, he will receive $627 in total interest.

Page 55 Problem 5 Answer

It is given: I believe non-dividend stocks aren’t much more than baseball cards.

They are worth what you can convince someone to pay for them.

The analogy of non-dividend stocks and baseball cards represents the similarity of two things in which value is determined by the demand, so they can be worthless in one moment and in another very expensive.

If we talk in terms of baseball cards, their value might vary a lot.

The card of some barely known baseball player is worth very little. It has only its intrinsic value.

However, if he becomes very famous, the value of the card increases drastically.

Some card-lovers are ready to pay huge amounts just to get that particular card.

And if people buy cards for their pleasure, that is completely fine.

But, if some person buys the card, or the stock share, with the hope to sell later for a higher price, we can notice the similarity.

Gaining profit from both the investments is dependent on the random value that the buyers will decide to give.

Page 55 Problem 6 Answer

Given: Years ago, Home Depot had an annual dividend of $0.90.

If you owned 4,000 shares of Home Depot. To find how much was received annually in dividends.

We will use the formula  Income dividends =Number of shares × Dividend p

Now the number of shares is given to be 4000 Dividend per share is $0.90

Hence Income dividends = Number of shares × Dividend per share

​=4000×0.90

=3600

​The amount that was received annually in dividends was $3,600.

Page 55 Problem 7 Answer

Given: you own r shares of a stock with an annual dividend of p dollars.

To express the  amount of your quarterly dividends algebraically.

We will use the formula:  Income dividends quarterly = Number of shares × Dividend per share quarterly

We have the number of shares as r.

Dividend per share annually is $p

Hence we get:​

Income dividends quarterly = Number of shares × Dividend per share quarterly

​=r×p/4

=rp/4

The amount of the quarterly dividends can be expressed as rp/4  dollars.

Page 55 Problem 8 Answer

Given: Mike owned 3000 shares of Merck Corporation and received a quarterly dividend check for $1140.

To find the annual dividend for one share of Merck.

We will use the formula: Annual divided per share = Income dividend annually ÷ Number of shares

Now we have the number of shares as 3000

Income dividend quarterly is $1,140

Therefore we get;​

Income Dividend annually =4× Income Dividend quarterly

​=4×$1,140

=$4560

and hence Annual divided per share = Income dividend annually ÷ Number of shares

​=$4560÷3000

=$1.52

The annual dividend for one share of Merck was $1.52.

Page 55 Problem 9 Answer

Given: The Walt Disney Company paid a $0.35 annual dividend on a day it closed at a price of $33.86 per share.

To find the annual dividend for 500 shares.

We will use the formula  Income dividends = Number of shares × Dividend per share

Now we have the number of shares as 500

Dividend per share is given to be $0.35

Therefore we get:​

Income dividends = Number of shares × Dividend per share

​=500×0.35

=175

The annual dividend for 500 shares was $175.

Page 55 Problem 10 Answer

It is given that :

Number of shares:500

Dividend per share annually:$0.35

Current price of per share:$33.86.

We have to find the quarterly dividend for500 shares.

We will use the fact that there are four quarters in a year, so the dividend per share annually is annually dividend per share divided by 4.

Firstly, we will find the dividend per share quarterly.

Dividend per share quarterly = Dividend per share annually /4

Dividend per share quarterly=$0.35/4

Dividend per share quarterly=$0.0875.

Now, we will find the quarterly dividend.​​

Income dividends quarterly = Number of shares × Dividend per share quarterly​

Income dividends quarterly =500×0.0875

Income dividends quarterly =43.75.​​

We get $43.75.

The quarterly dividend for 500 shares was$43.75.

Page 55 Problem 11 Answer

It is given that : Number of shares:500

Dividend per share annually:$0.35

Current price of per share:$33.86.

We have to express yield as a fraction.

As, the yield is the annual divided per share divided by the current price of one share.

Therefore, Yield= Annual dividend per share

Current price of one share

Yield=0.35/33.86

Yield=35/3386.​

We get the yield as a fraction:=35/3386.

Page 55 Problem 12 Answer

Number of shares:500

Dividend per share annually:$0.35

Current price of per share:$33.86.

We have to find the yield to the nearest tenth of a percent.

As, the yield is the annual divided per share divided by the current price of one share.

Therefore, Yield= Annual dividend per share

Current price of one share

Yield=0.35/33.86

Yield=35/3386

Yield≈0.010

Yield=1.0% .​

The yield to the nearest tenth of a percent is 1.0%.

Page 55 Problem 13 Answer

We are given : A spreadsheet in which price per share and annual dividend are given.

The Black Oyster Corporation is going out of business. All of the corporate assets are being sold.

The money raised will be split by the stockholders.

We have to find among stockholders, the common or preferred, who receive money first.

Among the common and preferred stockholders, the preferred stockholders will receive money first, as the name “preferred” indicates that they will be preferred first than any common stockholder.

Also, preferred stockholders receive their dividends before the common stockholders and they also receive a set amount of dividends, which doesn’t vary too often, while the common stockholders will only receive dividends when the board of directions issues the dividends.

Among stockholders, the common or preferred, money will be received first by the preferred stockholders.

Page 56 Exercise 1 Answer

We are given: A table which gives the last price and the annual dividend for 15 corporations.

We have to compute the yield for each corporation and round answers to the nearest tenth of a percent.

We will be finding the yield for each corporation.

3 M Co Yield= Annual dividend per share

Current price of one share

Yield=2.00/76.90

Yield≈0.026

Yield=2.6% .

Alcoa. Inc​Yield=0.68/41.57

Yield≈0.016

Yield=1.6% .

American Express Co​Yield =0.72/46.15

Yield≈0.016

Yield=1.6% .​

American International Group, Inc

Yield =0.80/34.91

Yield≈0.023

Yield=2.3% .

American International Group, Inc

Yield =0.80/34.91

Yield ≈0.023

Yield =2.3%.

AT &T

Yield =1.60/39.51

Yield≈0.040

Yield=4.0% .​

Bank of America Corp​​​

Yield =2.56/33.87​

Yield≈0.076

Yield=7.6% .​

Boeing Co Yield =1.60/82.13

Yield ≈0.019

Yield =1.9% .

Caterpillar, Inc.

Yield =1.44/83.19

Yield≈0.017

Yield=1.7% .

Chevron Corp.

Yield =2.60/100.42

Yield≈0.026

Yield=2.6% .

Citigroup, Inc.

Yield =1.28/21.60

Yield≈0.059

Yield=5.9% .​

Coca-Cola Co/The

Yield =1.52/57.44

Yield≈0.026

Yield=2.6% .​

El Du Pont de Nemours & Co

Yield =1.64/47.63

Yield≈0.034

Yield=3.4% .​

Exxon Mobil Corp

Yield =1.60/90.43

Yield≈0.018

Yield=1.8% .​

By computing the yield for each corporation, we get :

3 M Co:2.6%

Alcoa, Inc:1.6%

American Express Co:1.6%

American International Group, Inc:2.3%

AT & T:4.0%

Bank of America Corp:7.6%

Boeing Co:1.9%

Caterpillar, Inc.:1.7%

Chevron Corp:2.6%

Citigroup, Inc.:5.9%

Coca-Cola Co/The:2.6%

El Du Pont de Nemours & Co :3.4%

Exxon Mobil Corp:1.8%.

Page 56 Exercise 2 Answer

We are given : Annual dividend per share=$1.6

Friday closing=$44

Net change=$0.35.

We have to find the yield that was on Friday and then round answers to the nearest tenth of a percent.

As, the yield is the annual divided per share divided by the current price of one share.

Therefore, yield on friday=dividend on friday price on friday

yield on friday=1.6/44

yield on friday=0.036

yield on friday=36% .​

On Friday, the yield was:36%.

Page 56 Exercise 3 Answer

We are given : Annual dividend per share=$1.6

Friday closing=$44

Net change=$0.35.

We have to find the price at which Revreg closed on Thursday and round answers to the nearest tenth of a percent.

We will be using the formula: Thursday closing = Friday closing − net change.

The Thursday closing is given by : thursday closing = Friday closing − net change

thursday closing =$44−$0.35

thursday closing =$43.65.​

$43.65 is the price at which Revreg closed on Thursday.

Page 56 Exercise 4 Answer

We are given: Annual dividend per share=$1.6

Friday closing=$44

Net change=$0.35.

We have to find the yield at Thursday’s close.

We will be using the formula :

Thursday closing = Friday closing − net change.

Firstly, we will find the Thursday closing.

Thursday closing = Friday closing − net change

Thursday closing =$44−$0.35

Thursday closing =$43.65.

Now, we will find the yield at Thursday’s close.

yield on Thursday =dividend on Thursday price on Thursday

yield on Thursday =1.6/43.65

yield on Thursday =0.0366

yield on Thursday =3.66%

yield on thursday =3.7% .

3.7% is the yield at Thursday’s close.

Page 56 Exercise 5 Answer

We are given : Annual dividend per share=$1.60

Friday closing=$44

Friday’s net change=$0.35

Thursday’s net change=$1.22.

We have to find the price at which Revreg closed on Wednesday and round answers to the nearest tenth of a percent.

We will be using the formula: Wednesday closing = thursday closing −thursday′s net change.

Firstly, we will find the Thursday closing.

thursday closing = Friday closing −friday′s net change

thursday closing =$44−$0.35

thursday closing =$43.65 .

We will find the Wednesday closing .

wednesday closing = thursday closing − thursday′s net change

wednesday closing =43.65−1.22

wednesday closing=42.43 .​

$42.43 is the price at which Revreg closed on Wednesday.

Page 56 Exercise 6 Answer

We are given : Annual dividend per share=$1.60

Friday closing=$44

Friday’s net change=$0.35

Dividend on Wednesday=$1.60.

We have to find the yield at Wednesday’s close .

Firstly, we will find the Thursday closing .

thursday closing = Friday closing −friday′s net change

thursday closing =$44−$0.35

thursday closing =$43.65 .

We will find the Wednesday closing .

wednesday closing = thursday closing − thursday′s net change

wednesday closing =$43.65−$1.22

wednesday closing=$42.43 .

​The yield at Wednesday’s close is given by :

yield on wednesday =dividend on wednesday price on wednesday yield on wednesday =1.6

42.3 yield on wednesday =0.0377

yield on wednesday =3.77%

yield on wednesday =3.8% .

3.8% was the yield at Wednesday’s close .

Page 56 Exercise 7 Answer

We are given : Annual dividend per share=$1.60

Friday closing=$44

Friday’s net change=$0.35.

By looking at the yields for Wednesday, Thursday, and Friday, we get that they are decreasing .

We have to explain why this decrease is not “bad news” to the investor who owns stock in Revreg.

The yield falls because of rising share prices, not because of falling dividends per share. Since

the dividend level has not changed, this is not “bad news” for the investor who owns shares in Revreg.

You will receive the same dividend amount regardless of the drop in performance.

And, another reason could be that the stock is a growth stock as its prices rise sharply every day.

Remember, growth stocks are bought for capital gains rather than dividends- buy low and sell high.

The decrease is not “bad news” to the investor who owns stock in Revreg because the amount of dividends did not actually decrease and there is possibly a growth stock .

Page 56 Exercise 8 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85.

We have to find the yield that was on Wednesday and then round answers to the nearest tenth of a percent.

As, the yield is the annual divided per share divided by the current price of one share .

Therefore, yield on Wednesday=dividend on Wednesday

price on Wednesday

yield on Wednesday=2/61

yield on Wednesday=0.0327

yield on Wednesday=3.27%

yield on Wednesday=3.3% .

​On Wednesday, the yield was :3.3%.

Page 56 Exercise 9 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85.

We have to find the price at which Zeescore closed on Tuesday .

We will be using the formula :

Tuesday closing = Wednesday closing − net change.

The Tuesday closing is given by :

Tuesday closing = Wednesday closing − net change

Tuesday closing=61−(−0.85)

Tuesday closing=61+0.85

Tuesday closing=$61.85 .

$61.85 is the price at which Zeescore closed on Tuesday .

Page 56 Exercise 10 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85 .

We have to find the yield that was on Tuesday and then round answers to the nearest tenth of a percent.

We will find the formula :Tuesday closing = Wednesday closing − net change.

Firstly, we will find the Tuesday closing .

Tuesday closing = Wednesday closing − net change

Tuesday closing =61−(−0.85)

Tuesday closing =61+0.85

Tuesday closing =$61.85 .

Now, we will find the yield at Tuesday’s close .

yield on Tuesday =dividend on Tuesday price on Tuesday

yield on Tuesday =2/61.85

yield on Tuesday =0.0323

yield on Tuesday=3.23%

yield on Tuesday=3.2% .

​3.2% is the yield at Tuesday’s close .

Page 56 Exercise 11 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85

Net change on Tuesday=−1.96.

We have to find the price at which Zeescore closed on Monday .

Firstly, we will find the Tuesday closing .

Tuesday closing = Wednesday closing − net change

Tuesday closing =61−(−0.85)

Tuesday closing =61+0.85

Tuesday closing =$61.85 .

Now, we will find the Monday closing .

Monday closing = Tuesday closing − net change

Monday closing =61.85−(−1.96)

Monday closing =61.85+1.96

Monday closing =$63.81 .

​$63.81 is the price at which Zeescore closed on Monday .

Page 56 Exercise 12 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85

Net change on Tuesday=−1.96.

We have to find the Monday’s yield .

We will be using the formula :Monday closing = Tuesday closing − net change.

Firstly, we will find the Tuesday closing .

Tuesday closing = Wednesday closing − net change

Tuesday closing =61−(−0.85)

Tuesday closing =61+0.85

Tuesday closing =$61.85.​

We will find the Monday closing .

Monday closing = Tuesday closing − net change

Monday closing =61.85−(−1.96)

Monday closing =61.85+1.96

Monday closing =$63.81.​

Now, we will find the Monday’s yield.

yield on monday = dividend on monday price on monday

yield on monday=2/63.81

yield on monday=0.0313

yield on monday=3.13%

yield on monday=3.1% .

​Monday’s yield is3.1% .

Page 56 Exercise 13 Answer

We are given : Annual dividend per share=$2

Wednesday closing=$61

Net change on Wednesday=−0.85 .

By looking at the yields for Monday, Tuesday, and Wednesday we get that they are increasing.

We have explain this increase is not “good news” to the investor who owns stock in

The yield is increasing and the reason is the price decrease of the stock, not because the dividends are increasing.

As the dividends are not increasing, the yield increase is not necessarily ‘good news’ to the investor who owns Zeescore stock because he or she will still receive the same amount of dividends despite the increase in yield.

The increase is not “good news” to the investor who owns stock in Zeescore because there is no increase in the amount of dividends.

Page 56 Exercise 14 Answer

We are given : Cost of shares=$1000

Annual Interest=6.34% or 6.34/100

=0.0634.

Time(in years)=1 year.

We have to find how much will Sascha receive in annual interest.

We will find the interest after one year.

Interest= Cost of shares×Annual interest×Time

Interest=1000×0.00634×1

Interest=63.4% .

Sascha will receive 63.4% in annual interest.

Page 56 Exercise 15 Answer

We are given: Cost of shares=$1000

Annual interest=6.34% or 6.34/100

=0.0634.

Time(in years)=14 years.

We have to find how much will Sascha receive in interest if she holds the bond for 14 years.

We will find the interest after 14 years.

Interest after14 years=14× interest after 1 year

Interest after 14 years=14×63.4

Interest after 14 years=88.76%.

Sascha will receive 88.76%  in interest if she holds the bond for 14 years.

Page 56 Exercise 16 Answer

We are given : Cost of shares=$1000

Annual interest=6.34% or 6.34/100

=0.0634

Price of stock=$46

Annual dividend=$2.

We have to find the yield.

We will find the yield.

yield =dividend

price of stock yield=2/46

yield=0.04347

yield=4.347%

yield=4.4% .​

The yield is 4.4%.

Page 56 Exercise 17 Answer

We are given: Cost of shares=$1000

Annual Interest​=6.34 %

=0.0634

Price of stock=$46

Annual dividend=$2

Time( in years)=14 years.

We have to determine which is higher, the yield on the stock or the interest rate on the bond.

Firstly, we will find the yield and then the interest.

We will find the yield.

yield =dividend price of stock

yield=2/46

yield=0.04347

yield=4.347% .

By rounding off ,we get :yield=4.4%.

We will find the interest after 14 years.

Interest after 14 years=14× interest after 1 year

Interest : =14×63.4

=88.76%.​

The interest rate on the bond is higher than the yield on the stock as the interest rate is 88.76%.

Page 56 Exercise 18 Answer

We are given: Cost of shares=$1000

Annual interest=6.34% or 6.34/100

=0.0634

Time(in years)=1,14 years.

Bond matures after one year when :

Amount received=initial amount+ interest of 1year

Amount received=1000+0.0634×1000×1

Amount received=1000+63.4

Amount received=$1063.4.

Bond matures after 14 years when :

Amount received=1000+14(0.0634)(1000)

Amount received=1000+14(0.0634)

Amount received=1000+887.6

Amount received=$1887.6.

Lewis Corp. pays to Sascha when the bond matures :

After1 year :$1063.4

After14 years:$1887.6 .

Page 56 Exercise 19 Answer

We are given: Annual dividend per share=$2.46

Selling price per share=$54.24.

Selling price per share underwent a 3-for-1 split.

We have to find the new price of one share after the split.

As, we are given3−for−1split, therefore, the selling price before split will be divided by 3.

Therefore,Selling price after split = Selling price before split ÷3 Selling price after split =$54.24÷3 Selling price after split =$18.08.​

The new price of one share after the split is$18.08.

Page 56 Exercise 20 Answer

We are given: Annual dividend per share=$2.46

Selling price per share=$54.24

Number of shares=100.

The number of shares underwent a 3-for-1 split.

We have to find the number of shares you own after the split.

As we are given3−for−1split, therefore, the number of shares is multiplied by 3.

Number of shares after split = Number of shares before split ×3

Number of shares after split=200×3

Number of shares after split =600.

​The number of shares you own after the split is600.

Page 56 Exercise 21 Answer

We are given : Annual dividend per share=$2.46

Selling price per share=$54.24.

The annual dividend per share underwent a 3-for-1 split.

We have to find the annual dividend per share after the As, the annual dividend per share underwent a3−for−1 split, therefore, the annual dividend is divided by3.

Annual dividend after split = Annual dividend before split ÷3

Annual dividend after split =$2.46÷3

Annual dividend after split =$0.82.

The annual dividend per share after the split is$0.82.

Chapter 1 Solving Linear Equations

• • Cengage Financial Algebra 1st Edition Chapter 1 Assessment The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.1 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.2 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.3 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.4 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.5 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.6 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.7 The Stock Market
  • Cengage Financial Algebra 1st Edition Chapter 1 Exercise 1.8 The Stock Market

Financial Algebra 1st Edition Chapter 3 Banking Services

Page 138 Problem 1 Answer

Given: Principal amount(P)dollars

Rate of interest(R) compounded annually. i.e.

Time(T) year

To find: Amount of interest

Solution: Interest

PRT dollars will earn dollars in one year at a rate of compounded annually

Page 139 Problem 2 Answer

Given: p=4000

r=5% compounded semi-annually

=0.05

To find: Balance after one year

Solution: Accounts that pay interest semiannually have the interest added on twice each year (every six months).

We will first find the balance after six monthsThen calculate interest on that amount

For the first six months,

t=6 months  =0.5 years

I=PRT

=4000×0.05×0.5

=100

Now we will add the interest to the principal,

New principal=4000+100

=$4,100

Now we will compute interest for the next six months

I=PRT

=4100×0.05×0.5

=102.5

Now, we will add the interest to find the balance after one year

Balance after one year=4100+102.5

=4,202.5

Alex deposits$4,000 in a savings account that pays 5% interest compounded semiannually, then the balance after one year will be$4,202.5

Page 139 Problem 3 Answer

Given: p=$3000

r=4% compounded quarterly i.e.0.04

To find: Earnings in six months

Solution: Interest will be calculated every three monthsWe will find interest for the period of first three months and then again for the next three months

For the first three months

I=prt

=3000×0.04×0.25     [Convert 3 months in years]

=30

New principal=3000+30

=$3030

For the next three months

I=prt

=3030×0.04×0.25

=30.3

Total Interest=30+30.3

=$60.3

For the next three months

I=prt

=3030×0.04×0.25

=30.3

Total Interest=30+30.3

=$60.3

$3,000 earns$60.3 in six months at an interest rate of 4% compounded quarterly

Page 139 Problem 4 Answer

Given;  x dollars earn in one day at an interest rate of 5 % compounded daily

To find; Express the answer algebraically.

Accounts that pay interest daily earn interest every day.There are 365 days in a year.

r=5

r=0.05

​t=1 day

t=1 day =1/365 years

simple interest I=prt

I=1,000×0.05×1/365

≈0.13 interest to the principal. 1,000+0.13=1,000.13

The first day’s interest is approximately 13 cents, so the new balance is 1,000.13.

This larger principal is used to compute the next day’s interest

Hence we conclude that The first day’s interest is approximately 13 cents, so the new balance is 1,000.13.

This larger principal is used to compute the next day’s interest

Page 140 Problem 5 Answer

Given that On January 7, Joelle opened savings account with 900. It earned 3 % interest compounded daily.

On January 8, she deposited her first paycheck of76.22.

To do: find what was her balance at the end of the day on January 8.

From the given data the starting principal amount is 900 $ and given that at an interest of 3 %

it is compounded daily then we get 900(0.03)(1/365) deposited a paycheck of 76.22 $ on adding up all of them we get

900+900(0.03)(1/365)(76.22)

=976.29 $ is the balance

Therefore, the total balance at the end of the day is 976.29 $

Page 141 Problem 6 Answer

We have to find that how might those words apply to what you learned in this lesson

In the old days, a man who would save money was called a miser, because they just saved their money and didn’t earn anything with this money.

Thus the man wouldn’t profit from saving that money and also wouldn’t use this money for some other purpose.

However, nowadays, it is very unique and a kind of miracle to find someone who saves money, because it is not profitable to start saving money at an older age (as time is crucial in making savings grow) and thus you won’t gain much on your money (we do earn some due to compound interest).

Thus if you want to profit the most of your savings, then we need to start saving at a young age, but most people don’t tend to start saving at a young age and thus it is a kind of miracle to find someone who do as such.

A miser just saved their money and didn’t earn anything with this money.

However, nowadays someone who saved money is a wonder, because they would have to start at a young age if they want to profit from their savings.

Page 141 Problem 7 Answer

Given: Jerome deposits $3,700 in a certificate of the deposit that pays 6×1/2 interest, compounded annually.

We have to find that how much interest does Jerome earn in one year

Convert rate of interest to a decimal.

r=6×1/2%=6.5%=0.065

Use the simple interest formula I=prt

Substitute I=3760×0.065×1=244.4

Jerome will earn $244.4. as interest at the end of 1 year.

Page 141 Problem 8 Answer

Sally deposits $4,000 in a certificate of deposit that pays 6×3/4% simple interest.

What is her balance after one year

Convert rate of interest to a decimal.

r=6×3/4%=6.75%=0.0675

Use the simple interest formula I=prt

Substitute I=4000×0.0675×1=270

Add interest to the principal. 4,000+270=4,270

Sally’s balance after one year will be$4,270.

Page 141 Problem 9 Answer

Given: Pierre deposits $9,000 in a certificate of deposit that pays 8% interest, compounded semiannually.

To find: How much interest does the account earn in the first six months?

What is the balance after six months

Convert rate of interest to a decimal. r=8%=0.08

Convert 6 months to years.t=6 months =0.5 years

Use the simple interest formula ​I=prt

By Substituting

I=9000×0.08×0.5=360

Pierre’s account will earn $360 as interest after 6 months.

The balance after 6 months will be $9,360​

Add interest to the principal. 9,000+360=9,360

The balance after 6 months will be $9,360

Page 141 Problem 10 Answer

Given: Kevin has x dollars in an account that pays 2.2% interest, compounded quarterly.

We have to Express his balance after one quarter algebraically.

Convert rate of interest to a decimal. r=2.2%=0.022

Step 2 Convert 3 month to years.

t=3 months =1/4 years =0.25 years Use the simple interest formula

By Substituting

I=x×0.022×0.25=0.0055x

Kevin’s balance after one quarter will be $1.0055x.​

Add interest to the principal. x+0.0055x=1.0055x

Kevin’s balance after one quarter will be $1.0055x.

Page 141 Exercise 1 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to Find the first quarter’s interest.

p=3500

r=7.5%=0.075

t=1 quarter =0.25yrs first quarter interest ∣=prt=35000.0750.25=65.625

We get, $65.625

Page 141 Exercise 2 Answer

Given: Liam deposits $3,500  in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the first quarter’s ending balance.

Given: Principal:$3,500

Result part (a): The first quarter interest is $65.625

First quarter interest =$65.625≈865.63

The first quarter’s ending balance if the principal increased by the first quarter interest:

First quarter’s ending balance = Principal + First quarter interest

​=$3,500+$65.63

=$3,565.63

​We get,

$3,565.63

Page 141 Exercise 3 Answer

Given: Liam deposits $3,500

in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the second quarter’s interest.

Given:

p= Principal =$3,565.63 (First quarter’s ending balance found in part (b))r= Interest rate =7×1/2 %=7.5%=0.075

t= Time in year=1

Quarter =1/4 years =0.25years

The interest is the product of the principal, interest rate and time expressed in years.

I=prt=3,565.63×0.075×0.25≈$66.86

We get,

$66.86

Page 141 Exercise 4 Answer

Given: Liam deposits $3,500

in a saving account that pays7x1/2% interest, compounded quarterly.

We have to find the second quarter’s ending balance.

Given: Principal=$3,565.63 (First quarter’s ending balance found in part (b))

Result part (c): The second quarter interest is $66.86

Second quarter interest =66.86

The second quarter’s ending balance is the principal increased by the second quarter interest:

Second quarter’s ending balance = Principal + Second quarter interest

=$3,565.63+$66.86

=$3,632.49

​We get,

$3,632.49

Page 141 Exercise 5 Answer

Given: Liam deposits $3,500  in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the third quarter’s interest.

Given: p= Principal =$3,632.49 (First quarter’s ending balance found in part (d)) r= Interest rate =7×1/2%=7.5%=0.075

t= Time in year=1 Quarter=1/4 years =0.25 years

The interest is the product of the principal, interest rate and time expressed in years.

I= prt =3,632.49×0.075×0.25≈$68.11

we get,

$68.11

Page 141 Exercise 6 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the third quarter’s ending balance.

Given: Principal =$3,632.49 (First quarter’s ending balance found in part (d))

Result part (e): The third quarter interest is $68.11

Third quarter interest =68.11

The third quarter’s ending balance is the principal increased by the third quarter interest:

Third quarter’s ending balance = Principal + Second quarter interest

​=$3,632.49+868.11

=$3,700.60

​The third quarter’s ending balance is : $3,700.60

Page 141 Exercise 7 Answer

Given: Liam deposits $3,500

in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find the fourth quarter’s interest.

Given:p= Principal r= (First quarter’s ending balance found in part (f))r= Interest rate =7×1/2%=7.5%=0.075

t= Time in year =1 Quarter=1/4 years =0.25 years

The interest is the product of the principal, interest rate and time expressed in years.

I=prt=3,700.60×0.075×0.25≈$69.39

We get,

$69.39

Page 141 Exercise 8 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

To find: What is the balance at the end of one year?

Given: Principal =$3,700.60 (First quarter’s ending balance found in part (f))

Result part (g): The fourth quarter interest is $69.39

Fourth quarter interest =69.39

The fourth quarter’s ending balance is the principal increased by the fourth quarter interest:

Fourth quarter’s ending balance = Principal + Second quarter interest

​=$3,700.60+$69.39

=$3,769.99

​The balance at the end of one year is : $5,769.99

Page 141 Exercise 9 Answer

Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.

We have to find that how much interest does the account earn in the first year?

Total interest in 1 st year = Balance after 1 year- original principal amount

=3769.98-3500=269.98

The amount earned in first year is : $269.98

Page 141 Exercise 10 Answer

Given : Janine opens a savings account with a deposit of $720.

The account pays3.4 percent interest, compounded daily.

We have to find first day’s interest.

We will use above formula .

Here, First convert rate of interest into decimal:r=3.4%=0.034​

Convert 1 day to years:t=1day =1/365 years​

Now, use simple interest formula: I=720×0.034×1/365≈0.07

Hence, the First day’s interest is approximately 7 cents.

Therefore, First day’s interest is approximately 7 cents.

Page 142 Exercise 11 Answer

We have given a table,

We have to find the missing amounts in the table.

We will find by using above information.

No principle amount initially thus

(1) Opening balance = 0

(1) Deposit on 10 August = $4,550

(3) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=0+4550−0

=4550

(4) The interest is the product of the principle, interest rate and time in years:

I=prt

=4550×0.04×1/365

≈0.50​

Thus interest rate is $0.5

(5)  Ending Balance = principal+interest

Ending Balance =4550+0.5=4550.50

ending balance is $4,550.50

(6)  Next day opening = previous day ending balance

Opening bal aug 11=4550.50

(7)  Deposit on August 11=300

(8)  principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=4550.50+300−0

=4850.50

​(9) The interest is the product of the principle, interest rate and time in years:

I=prt

=4850.50×0.04×1/365

≈0.53interest rate is $0.53

(10)  Ending Balance = principal+interest

Ending balance =4850.5+0.53=4851.03

(11)  Next day opening = previous day ending balance

Opening bal aug 12=4851.03

(12)  withdrawal=900

(13)  The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle= Opening balance + Deposit − Withdrawal

=4851.03+0−900

=3951.03

(14)  The interest is the product of the principle, interest rate and time in years:

I=prt

=3951.03×0.04×1/365

≈0.43

​(15)  Ending Balance = principal+interest

Ending Bal=3951.03+0.43=3951.46

The answers of the given blanks in the table are given above. All the answers are in USD.

Page 142 Exercise 12 Answer

Given: On December 18 of a leap year, Stacy opened a savings account by depositing $6,000.

The account pays 3.45% interest, compounded daily.

On December 19 she deposited $500, and on December 20 she withdrew $2,500.

By using this data we will fill all the blanks in the table.

(1)  no principal amount initially

Thus opening balance is = 0

(2) Deposit = $6000

(3)  Withdrawal =0

(4) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=0+6000−0

=6000

​(5)  The interest is the product of the principle, interest rate and time in years:

I=prt

=6000×0.0345×1/365

≈0.57

​(6)  Ending Balance = principal+interest

Ending Balance =6000+0.57=6000.57

(7)  Next day opening = previous day ending balance

Opening bal =6000.57

(8)  Deposit =500

(9)  Withdraw =0

(10)  The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=6,000.57+500−0

=6,500.57

(11)  The interest is the product of the principle, interest rate and time in years:

I=prt

=6500.57×0.0345×1/365

≈0.61

​(12) Ending balance =6500.57+0.61=6501.18

(13) Next day opening = previous day ending balance

Opening bal =6501.18

(14) deposit =0

(15)  withdrawal=2500

(16) ​ Principle = Opening balance + Deposit − Withdrawal

=6,501.18+0−2,500

=4,001.18

(17)  The interest is the product of the principle, interest rate and time in years:

I=prt

=4,001.18×0.0345×1/365

≈0.38

​(18)  Ending Balance = principal+interest

Ending Bal=4001.18+0.38=4001.56

The answers of the given blanks in the table are given above. All the answers are in USD.

Page 142 Exercise 13 Answer

Given : On May 29, Rocky had an opening balance of x dollars in an account that pays 3% interest, compounded daily. He deposits y dollars.

We have to express his ending balance on May30 algebraically.

We will use above formula .

Here, Consider that the deposit occurs on May 23

So, p=Principle

=Opening balance+Deposit

=$(x+y)

r=Interest rate

=3%

=0.03

t=Time in years

=1day

=1/365years (as there are 365 days in a year)

The interest is the product of the principle, interest rate and time in years:I=prt

=(x+y)×0.03×1/365

=x×3/100×1/365

=3(x+y)/36,500

The ending balance is then the principle increased by the interest: Ending balance

=p+I

=x+3(x+y)/36,500

=36,500x+3x+3y/36,500

=36,503x+3y/36,500

Now, p=Principle=ClosingbalanceMay29

t=$(36,503x+3y/36,500)

r=Interest rate

=3%

=0.03

The interest is the product of the principle, interest rate and time in years:I=prt

=(36,503x+3y/36,500)×0.03×1/365

=(36,503x+3y/36,500)×3/100×1/365

=(36,503x+3y/36,500)×3/36,500

=3(36,503x)+3(3y)/36,5002

=109,509x+9y/1,332,250,000​

The ending balance is then the principle increased by the interest:Ending balance =p+I

=36,503x+3y/36,500+109,509x+9y/1,332,250,000

=36,500(36,503x)+36,500(3y)/1,332,250,000+109,509x+9y/1,332,250,000

=1,332,359,500x+109,500y/1,332,250,000+109,509x+9y/1,332,250,000

=1,332,359,500x+109,500y+109,509x+9y/1,332,250,000

=1,332,469,009x+109,509y/1,332,250,000

Hence, The ending balance on May 30 is 1,332,469,009x+109,509y/1,332,250,000.

Therefore, The ending balance on May 30 is 1,332,469,009x+109,509y/1,332,250,000.

Page 142 Exercise 14 Answer

Given :Linda has d dollars in an account that pays 3.4% interest, compounded weekly. She withdraws w dollars.

We have to express her first week’s interest algebraically.

We will use above formula .

Here, p=Principle

=Opening balance−Withdrawal

=d−w​

r=Interest rate

=3.4%

=0.034

t=Timeinyears

=1week

=1 /52 years (as there are 52 weeks in a year)

Now, The interest is the product of the principle, interest rate and time in years: ​I

=prt

=(d−w)×0.034×1/52

=(d−w)×34/1000×1/52

=34(d−w)/52,000

=17(d−w)/26,000.

Hence, The first week’s interest is 17(d−w)/26,000.

Therefore, The first week’s interest is17(d−w)/26,000

Page 142 Exercise 15 Answer

Given: The table represents the compound interest calculations for an account that pays 2% interest compounded daily. Represent a–g algebraically.

Opening balance = P

Deposit   = D

Withdrawal = $0

By using this data we will fill the table .

1) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=P+D−0

=P+D

(2)  The interest is the product of the principle, interest rate and time in years:

I=prt

=(P+D)×0.02×1/365

=(P+D)×2/100×1/365

=2(P+D)/36,500

=P+D/18,250

(3)  The ending balance is then the principle increased by the interest:

Ending balance =p+I

=(P+D)+(P+D)/18,250

=(P+D)(1+1/18,250)

=(P+D)(18,250/18,250+1/18,250)

=(P+D)(18,251/18,250)

=18,251(P+D)/18,250

(4)  The opening balance on February 3 is the ending balance on the previous day and thus is the ending balance on February 2.

Opening balance on February 3= Opening balance on February 2

=18,251(P+D)/18,250

(5)  The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.

Principle = Opening balance + Deposit − Withdrawal

=18,251(P+D)/18,250+0−W

=18,251(P+D)/18,250−W

(6)  The interest is the product of the principle, interest rate and time in years:

I=prt

=(18,251(P+D)/18,250−W)×0.02×1/365

=18,251(P+D)−18,250W/18,250×2/100×1/365

=18,251(P+D)−18,250W/18,250×1/18,250

=18,251(P+D)−18,250W/18,2502

=18,251(P+D)−18,250W/333,062,500​

(7)  The ending balance is then the principle increased by the interest:

Ending balance =p+I

=18,251(P+D)/18,250−W+18,251(P+D)−18,250W/333,062,500

=18,251(P+D)−18,250W/18,250+18,251(P+D)−18,250W/333,062,500

=18,250(18,251)(P+D)−333,062,500W/333,062,500+18,251(P+D)−18,250W/333,062,500

=333,080,750(P+D)−333,062,500W/333,062,500/+18,251(P+D)−18,250W/333,062,500

=333,080,750(P+D)−333,062,500W+18,251(P+D)−18,250W/333,062,500

=333,099,001(P+D)−333,080,750W/333,062,500

Therefore,

1)The principle used to compute interest is the new balance is P+D

2)Interest = =P+D/18,250

3) Ending Balance =18,251(P+D)/18,250

4)Opening Balance =18,251(P+D)/18,250

5)The principle used to compute interest =18,251(P+D)/18,250−W

6)The interest is the product of the principle, interest rate and time in years:  =18,251(P+D)−18,250W/333,062,500

7)Ending Balance:- 333,099,001(P+D)−333,080,750W/333,062,500

Page 142 Exercise 16 Answer

Given: George had an opening balance of m dollars in an account that pays 2.25% interest compounded monthly.

On the last day of the month, he made a deposit equal to twice his opening balance.

We have tp express his ending balance on the last day of the month algebraically.

We will use above formula .

Here, Opening balance =m

Deposit =2m (twice the opening balance)

p  =Principle

= Opening balance + Deposit

=m+2m

=3m

​r= Interest rate

=2.25%

=0.0225

​t=Time in years

=1month

=1/12 years  (as there are 12 months in a year)

Now,The interest is the product of the principle, interest rate and time in years: ​I

=prt

=3m×0.0225×1/12

=3m×225/10000×1/12

=3m×9/400×1/12

=(3m)(9)/400(12)

=27m/4,800​

The ending balance is then the principle increased by the interest:Ending balance

=p+I

=3m+27m/4,800

=14,400m/4,800+27m/4,800

=14,400m+27m/4,800

=14,427m/4,800

=4,809m/1,600​

Hence,

The ending balance on the last day of the month is 4,809m /1,600.

Therefore,The ending balance on the last day of the month is 4,809m/1,600.