Technical and student resources

Introduction to coin problems and investment problems.

If you have 7 nickels, what is the value of your nickels in cents? If you have 4 dimes, what is the value of your dimes in cents? These are simple questions that you can readily answer, but they involve the general principle that is used in most "coin" problems: If you know the number of a certain kind of coin, and if you know the value of that coin in cents, then the number of coins multiplied by the value (in cents) of each coin is the total value (in cents) of the coins. Thus the value of 7 nickels is (7) (5) cents, or 35 cents; the value of 4 dimes is (4) (10) cents, or 40 cents; the value of x quarters is (x) (25), or 25x, cents.

If, instead of using cents as your unit of value, you used dollars, then 9 dimes would be worth (9)(.10) dollars; the value of 12 nickels would be worth (12)(.05) dollars; the value of y quarters would be .25y dollars.

ORAL EXERCISES

1. A man has three dollars and seventy-five cents. What number represents the value of the man's money in cents? What number represents the value of the man's money in dollars?

Answer :

The value of the man's money in cents is 375 cents, since there are 100 cents in each dollar.

The value of the man's money in dollars is $3.75, since 1 dollar is equal to 100 cents and 3 dollars and 75 cents is the same as 3 + 0.75 dollars.

2. A boy has $1.37. What number represents the value of his money in cents?

Answer :

Since there are 100 cents in each dollar, the value of the boy's money in cents is:

1 dollar = 100 cents 0.37 dollars = 0.37 x 100 cents = 37 cents

So, the total value of the boy's money in cents is:

$1.37 = 1 x 100 cents + 37 cents = 137 cents

Therefore, the value of the boy's money in cents is 137 cents.

3. A girl has 63 dimes. What number represents the value of the girl's money in cents? What number represents the value of the girl's money in dollars?

Answer :

Since there are 10 cents in each dime, the value of each dime is 10 cents.

To find the value of 63 dimes, we can multiply the number of dimes by the value of each dime:

63 dimes x 10 cents/dime = 630 cents

Therefore, the value of the girl's money in cents is 630 cents.

To find the value of the girl's money in dollars, we need to convert cents to dollars. Since there are 100 cents in each dollar, we can divide the number of cents by 100 to get the value in dollars:

630 cents / 100 cents/dollar = $6.30

Therefore, the value of the girl's money in dollars is $6.30.

4. A man has x quarters. How would you represent the value of these quarters in cents? How would you represent the value of these quarters in dollars?

Answer :

Each quarter is worth 25 cents. So, if a man has x quarters, the value of these quarters in cents would be:

25 cents/quarter x x quarters = 25x cents

To represent the value of these quarters in dollars, we need to divide the number of cents by 100, since there are 100 cents in each dollar:

25x cents / 100 cents/dollar = 0.25x dollars

Therefore, the value of the man's quarters in cents is 25x cents, and the value of the man's quarters in dollars is 0.25x dollars.

5. If x represents the number of nickels that a man has, what number represents the value of these nickels in cents?

Answer :

A nickel is worth 5 cents, so if a man has x nickels, the value of these nickels in cents would be:

5 cents/nickel x x nickels = 5x cents

Therefore, the value of the man's nickels in cents is 5x cents.

6. If 3x represents the number of half dollars that a man .has, how would you represent the value of his half dollars in cents?.

Answer :

A half dollar is worth 50 cents, so if a man has 3x half dollars, the value of these half dollars in cents would be:

50 cents/half dollar x 3x half dollars = 150x cents

Therefore, the value of the man's half dollars in cents is 150x cents.

7. If (3x -8) represents the number of dimes that a man has, how would you represent the value of these dimes in cents?

Answer :

A dime is worth 10 cents, so if a man has 3x - 8 dimes, the value of these dimes in cents would be:

10 cents/dime x (3x - 8) dimes = 30x - 80 cents

Therefore, the value of the man's dimes in cents is (30x - 80) cents.

8. If 5(2x +7) represents the number of quarters that a man has, how would you represent the value of these quarters in cents?

Answer :

If 5(2x + 7) represents the number of quarters that a man has, we can simplify this expression by distributing the 5:

5(2x + 7) = 10x + 35

Therefore, the man has 10x + 35 quarters.

Since each quarter is worth 25 cents, the value of these quarters in cents would be:

25 cents/quarter x (10x + 35) quarters = 250x + 875 cents

Therefore, the value of the man's quarters in cents is 250x + 875 cents.

9. If (a -b) represents the number of nickels that a man has, how would you represent the value of these nickels in cents?

Answer :

A nickel is worth 5 cents, so if a man has (a - b) nickels, the value of these nickels in cents would be:

5 cents/nickel x (a - b) nickels = 5a - 5b cents

Therefore, the value of the man's nickels in cents is (5a - 5b) cents.

10. A man has 18 coins consisting of dimes and quarters. If x represents the number of dimes, how would you represent the number of quarters that he has? How would you represent the value of these quarters in cents?

Answer :

If a man has 18 coins consisting of dimes and quarters, and x represents the number of dimes, then the number of quarters he has can be represented as:

Number of quarters = Total number of coins - Number of dimes Number of quarters = 18 - x

To represent the value of these quarters in cents, we need to know the value of a quarter. Since a quarter is worth 25 cents, the value of the man's quarters in cents would be:

25 cents/quarter x (18 - x) quarters = (450 - 25x) cents

Therefore, the value of the man's quarters in cents is (450 - 25x) cents.

EXERCISES: Arranging work in coin problems

Copy the following boxes onto a paper and fill in the blank spaces with the correct data.

ILLUSTRATIVE EXAMPLE: Coin problem

A man has 3 more quarters than nickels. If the value of these quarters and nickels is $3.45, how many coins of each kind has he?

Solution. Since we are given a relationship between the number of nickles and quarters that the man has, we can represent the number of nickels and the number of quarters in terms of a single unknown. Let the number of nickels be represented by x; then the number of quarters can be represented by x +3, and we can indicate this conveniently by using a box. Our equation will come from the fact that the man's coins are worth $3.45; hence we are interested in the value of the man's nickels and the value of the man's quarters. Since the "value" is to be represented by a number, we must decide on the unit of value which we are going to use; that is, we must express the value either in cents or in dollars. If we decide to use cents as our unit of value, then we will not have expressions involving decimals. Completing the box by filling in the "value" column:

But the value of the nickels and quarters is $3.45, or, in cents, 345.

Hence we can write the equation:

From the box we proceed to answer the question. The number of nickels that the man has, represented by x, is 9. The number of quarters that the man has, represented by x + 3, is 9 + 3, or 12.

EXERCISES: Coin problems

1. A man has twice as many dimes as he has nickels. If the value of his nickels and dimes together is $1.75, how many coins of each kind has he?

Let's use algebra to solve this problem. Let's say the number of nickels the man has is "x" and the number of dimes he has is "2x" (since he has twice as many dimes as nickels).

The value of a nickel is $0.05, and the value of a dime is $0.10.

The total value of the nickels can be expressed as 0.05x, and the total value of the dimes can be expressed as 0.10(2x) or 0.20x.

According to the problem, the total value of the nickels and dimes together is $1.75. So, we can set up an equation:

0.05x + 0.20x = 1.75

Now, let's solve for "x":

0.05x + 0.20x = 1.75

0.25x = 1.75

x = 1.75 / 0.25

x = 7

Now that we know the number of nickels (x = 7), we can find the number of dimes (2x = 2 * 7 = 14).

So, the man has 7 nickels and 14 dimes.

2. A man has 12 coins, some of them nickels and the rest quarters. If the value of his coins is $2.20, how many coins of each kind does he have?

Let's use algebra to solve this problem. Let's say the number of nickels the man has is "x" and the number of quarters he has is "12 - x" (since he has 12 coins in total).

The value of a nickel is $0.05, and the value of a quarter is $0.25.

The total value of the nickels can be expressed as 0.05x, and the total value of the quarters can be expressed as 0.25(12 - x) or 3 - 0.25x.

According to the problem, the total value of the nickels and quarters together is $2.20. So, we can set up an equation:

0.05x + 3 - 0.25x = 2.20

Now, let's solve for "x":

0.05x - 0.25x = 2.20 - 3 -0.20x = -0.80 x = -0.80 / -0.20 x = 4

Now that we know the number of nickels (x = 4), we can find the number of quarters (12 - x = 12 - 4 = 8).

So, the man has 4 nickels and 8 quarters.

3. A boy has 12 more dimes than he has quarters. If the value of his dimes and quarters is $6.80, how many coins of each kind has he?

Let's use algebra to solve this problem. Let's say the number of quarters the boy has is "x" and the number of dimes he has is "x + 12" (since he has 12 more dimes than quarters).

The value of a quarter is $0.25, and the value of a dime is $0.10.

The total value of the quarters can be expressed as 0.25x, and the total value of the dimes can be expressed as 0.10(x + 12) or 0.10x + 1.20.

According to the problem, the total value of the dimes and quarters together is $6.80. So, we can set up an equation:

0.25x + 0.10x + 1.20 = 6.80

Now, let's solve for "x":

0.25x + 0.10x + 1.20 = 6.80

0.35x + 1.20 = 6.80

0.35x = 6.80 - 1.20

0.35x = 5.60

x = 5.60 / 0.35

x = 16

Now that we know the number of quarters (x = 16), we can find the number of dimes (x + 12 = 16 + 12 = 28).

So, the boy has 16 quarters and 28 dimes.

4. A man has 7 more dimes than nickels and twice as many quarters as nickels. If the value of these coins is $5.90, how many coins of each kind has he?

Let's use algebra to solve this problem. Let's say the number of nickels the man has is "x".

According to the problem:

1. The number of dimes is 7 more than the number of nickels, so the number of dimes is "x + 7".
2. The number of quarters is twice the number of nickels, so the number of quarters is "2x".

The value of a nickel is $0.05, the value of a dime is $0.10, and the value of a quarter is $0.25.

The total value of the nickels can be expressed as 0.05x, the total value of the dimes can be expressed as 0.10(x + 7), and the total value of the quarters can be expressed as 0.25(2x) or 0.50x.

According to the problem, the total value of the nickels, dimes, and quarters together is $5.90. So, we can set up an equation:

0.05x + 0.10(x + 7) + 0.50x = 5.90

Now, let's solve for "x":

0.05x + 0.10x + 0.70 + 0.50 x = 5.90

0.65x + 0.70 = 5.90

0.65x = 5.90 - 0.70

0.65x = 5.20

x = 5.20 / 0.65

x = 8

Now that we know the number of nickels (x = 8), we can find the number of dimes (x + 7 = 8 + 7 = 15) and the number of quarters (2x = 2 * 8 = 16).

So, the man has 8 nickels, 15 dimes, and 16 quarters.

5. A man has three times as many quarters as half dollars and 2 more dimes than half dollars. If the value of these coins is $11, how many coins of each kind has he?

Let's say the number of half dollars the man has is "x".

According to the problem:

1. The number of quarters is three times the number of half dollars, so the number of quarters is "3x".
2. The number of dimes is 2 more than the number of half dollars, so the number of dimes is "x + 2".

The value of a half dollar is $0.50, the value of a quarter is $0.25, and the value of a dime is $0.10.

The total value of the half dollars can be expressed as 0.50x, the total value of the quarters can be expressed as 0.25(3x) or 0.75x, and the total value of the dimes can be expressed as 0.10(x + 2) or 0.10x + 0.20.

According to the problem, the total value of the half dollars, quarters, and dimes together is $11. So, we can set up an equation:

0.50x + 0.75x + 0.10x + 0.20 = 11

Now, let's solve for "x":

1.35x + 0.20 = 11

1.35x = 11 - 0.20

1.35x = 10.80

x = 10.80 / 1.35

x = 8

Now that we know the number of half dollars (x = 8), we can find the number of quarters (3x = 3 * 8 = 24) and the number of dimes (x + 2 = 8 + 2 = 10).

So, the man has 8 half dollars, 24 quarters, and 10 dimes.

6. A boy has $13.00 in nickels and dimes in a coin bank. He finds that the number of dimes is 2 less than five times the number of nickels. How many coins of each kind has he?

Let's say the number of nickels the boy has is "x".

According to the problem:

1. The value of a nickel is $0.05, so the total value of the nickels is 0.05x.
2. The value of a dime is $0.10, so the total value of the dimes is 0.10(5x - 2) or 0.50x - 0.20 (since the number of dimes is 2 less than five times the number of nickels).

The total value of the nickels and dimes together is $13. So, we can set up an equation:

0.05x + 0.50x - 0.20 = 13

Now, let's solve for "x":

0.55x - 0.20 = 13

0.55x = 13 + 0.20

0.55x = 13.20

x = 13.20 / 0.55

x = 24

Now that we know the number of nickels (x = 24), we can find the number of dimes (5x - 2 = 5 * 24 - 2 = 120 - 2 = 118).

So, the boy has 24 nickels and 118 dimes.

7. A man has 5 more half dollars than he has nickels, and the number of quarters is 3 less than twice the number of nickels. If the value of these coins is $17.50, how many coins of each kind has he?

Let's say the number of nickels the man has is "x".

According to the problem:

1. The number of half dollars is 5 more than the number of nickels, so the number of half dollars is "x + 5".
2. The number of quarters is 3 less than twice the number of nickels, so the number of quarters is "2x - 3".

The value of a nickel is $0.05, the value of a half dollar is $0.50, and the value of a quarter is $0.25.

The total value of the nickels can be expressed as 0.05x, the total value of the half dollars can be expressed as 0.50(x + 5) or 0.50x + 2.50, and the total value of the quarters can be expressed as 0.25(2x - 3) or 0.50x - 0.75.

According to the problem, the total value of the nickels, half dollars, and quarters together is $17.50. So, we can set up an equation:

0.05x + 0.50x + 2.50 + 0.50x - 0.75 = 17.50

Now, let's solve for "x":

1.05x + 1.75 = 17.50

1.05x = 17.50 - 1.75

1.05x = 15.75

x = 15.75 / 1.05

x = 15

Now that we know the number of nickels (x = 15), we can find the number of half dollars (x + 5 = 15 + 5 = 20) and the number of quarters (2x - 3 = 2 * 15 - 3 = 30 - 3 = 27).

So, the man has 15 nickels, 20 half dollars, and 27 quarters.

8. I have $7.20 in nickels, dimes, and quarters, with twice as many dimes as quarters, and the number of nickels is three times the number of dimes and quarters together. How many coins of each kind have I?

9. A man has 71 coins consisting of nickels, dimes, and quarters, with the number of nickels 8 more than twice the number of dimes. If the, coins have a value of $6.25, how many coins of each kind does he have?

Let's say the number of dimes the man has is "d".

According to the problem:

1. The number of nickels is 8 more than twice the number of dimes, so the number of nickels is "2d + 8".
2. The number of quarters can be calculated by subtracting the total number of nickels and dimes from the total number of coins, which is 71. So, the number of quarters is "71 - (2d + 8) - d" or "71 - 3d - 8" or "63 - 3d".

The value of a nickel is $0.05, the value of a dime is $0.10, and the value of a quarter is $0.25.

The total value of the nickels can be expressed as 0.05(2d + 8) or 0.10d + 0.40, the total value of the dimes can be expressed as 0.10d, and the total value of the quarters can be expressed as 0.25(63 - 3d) or 15.75 - 0.75d.

According to the problem, the total value of the nickels, dimes, and quarters together is $6.25. So, we can set up an equation:

0.10d + 0.10d + 0.40 + 15.75 - 0.75d = 6.25

Now, let's solve for "d":

0.20d + 16.15 - 0.75d = 6.25

-0.55d = 6.25 - 16.15

-0.55d = -9.90

d = -9.90 / -0.55

d = 18

Now that we know the number of dimes (d = 18), we can find the number of nickels (2d + 8 = 2 * 18 + 8 = 44) and the number of quarters (63 - 3d = 63 - 3 * 18 = 9).

So, the man has 44 nickels, 18 dimes, and 9 quarters.

10. A man has $12.00 consisting of nickels, dimes, and quarters. There are 8 more nickels than quarters, and the number of dimes is 10 less than twice the number of nickels. How many coins of each kind are there?

Answers :

• 1. 7 nickels, 14 dimes.
• 2. 4 nickels, 8 quarters.
• 3. 16 quarters, 28 dimes.
• 4. 8 nickels, 15 dimes, 16 quarters.
• 5. 8 half dollars, 10 dimes, 24 quarters.
• 6. 24 nickels, 118 dimes.
• 7. 15 nickels, 20 half dollars, 27 quarters.
• 8. 8 quarters, 16 dimes, 72 nickels.
• 9. 18 dimes, 44 nickels, 9 quarters.
• 10. 22 quarters, 30 nickels, 50 dimes.

Introduction to investment problems.

If you deposit $500 in a savings account at a bank that pays 2% interest annually, how much interest will you receive each year on this investment?

In arithmetic you learned to solve such problems as this by multiplying 500 (the principal) by .02 (the rate) to find the interest (or income). In algebra we state this fact by the formula

I= prt

where

• I represents the number of dollars of interest,
• p represents the number of dollars of principal,
• r represents the annual rate of interest, and 
• t represents the number of years the principal is invested.

Applying this formula to the above problem, we can write:

I = (500) (.02) (1)

Hence I=10

Since I is expressed in dollars, the interest is $10.

For the given problem: p = $500 (principal amount) r = 2% (annual interest rate expressed as a decimal, so r = 0.02) t = 1 year

Now, we can use the formula to calculate the interest:

I = (500) * (0.02) * (1) = 10

So, the interest on the $500 investment at an annual rate of 2% will be $10 per year. This means that after one year, the account balance will be $510 ($500 initial deposit + $10 interest).

EXERCISES: Investment problems

1. A man invests $200 at 5% interest. How would you represent the number of dollars of interest that he would receive annually? You are not required to perform the indicated multiplication.

I= (200)(.05) Answer

To represent the number of dollars of interest that the man would receive annually, we can use the formula:

I = prt

where: I = Number of dollars of interest p = Principal amount (initial investment) r = Annual rate of interest (in decimal form) t = Number of years the principal is invested

For this specific problem: p = $200 (principal amount) r = 5% (annual interest rate expressed as a decimal, so r = 0.05) t = 1 year (since it is annually)

Using the formula, we get:

I = (200) * (0.05) * (1)

I represents the number of dollars of interest that the man would receive annually. The value of "I" will be the amount of interest earned on the $200 investment at a 5% annual interest rate after one year. However, we don't perform the multiplication in this representation.

2. A man invests x dollars at 4% interest. How would you represent the number of dollars of interest that he would receive annually?

To represent the number of dollars of interest that the man would receive annually, we can use the formula:

I = prt

where: I = Number of dollars of interest p = Principal amount (initial investment) r = Annual rate of interest (in decimal form) t = Number of years the principal is invested

In this case, the principal amount (p) is "x" dollars, and the annual rate of interest (r) is 4% (which can be expressed as a decimal as r = 0.04). Since the problem does not specify the number of years (t) the principal is invested, we'll assume it's one year.

Using the formula, we get:

I = x * 0.04 * 1

I represents the number of dollars of interest that the man would receive annually. The value of "I" will be the amount of interest earned on the "x" dollar investment at a 4% annual interest rate after one year. However, we don't perform the multiplication in this representation. The value of "I" will depend on the actual value of "x" and the number of years the principal is invested.

3. A boy has a $1000 bond that pays 3½ % interest annually. How would you represent the number of dollars of interest that he would receive each year?

To represent the number of dollars of interest that the boy would receive each year from his $1000 bond, we can use the formula:

I = prt

where: I = Number of dollars of interest p = Principal amount (initial investment or bond value) r = Annual rate of interest (in decimal form) t = Number of years the principal is invested

In this case, the principal amount (p) is $1000, and the annual rate of interest (r) is 3.5% (which can be expressed as a decimal as r = 0.035). Since the problem doesn't specify the number of years (t) the bond is invested, we'll assume it's one year.

Using the formula, we get:

I = 1000 * 0.035 * 1

I represents the number of dollars of interest that the boy would receive each year. The value of "I" will be the amount of interest earned on the $1000 bond at a 3.5% annual interest rate after one year. However, we don't perform the multiplication in this representation. The value of "I" will be the actual amount of interest he would receive annually.

4. A man invests y dollars at 4½ % interest. How would you represent the number of dollars of interest that he would receive each year?

To represent the number of dollars of interest that the man would receive each year from his investment of "y" dollars at a 4½ % interest rate, we can use the formula:

I = prt

where:

• I = Number of dollars of interest
• p = Principal amount (initial investment or the value of "y" dollars)
• r = Annual rate of interest (in decimal form)
• t = Number of years the principal is invested

In this case, the principal amount (p) is "y" dollars, and the annual rate of interest (r) is 4.5% (which can be expressed as a decimal as r = 0.045). Since the problem doesn't specify the number of years (t) the investment is held, we'll assume it's one year.

Using the formula, we get:

I = y * 0.045 * 1

I represents the number of dollars of interest that the man would receive each year. The value of "I" will be the amount of interest earned on the "y" dollar investment at a 4½ % annual interest rate after one year. However, we don't perform the multiplication in this representation. The value of "I" will depend on the actual value of "y" and the number of years the principal is invested.

5. A girl invests (2x -100) dollars in stock that pays an interest of 6%. How would you represent the number of dollars of income that she receives each year from this investment?

6. How would you represent the annual interest received from an investment of (10,000 -x) dollars at 3% interest?

7. A man invests x dollars in stocks that pay an interest of 6½% and he invests 2x dollars in bonds that pay interest at 3%. How would you represent the number of dollars that he receives each year from the bonds? How would you represent the number of dollars that he receives each year from the stocks? How would you represent the number of dollars that he receives each year from the two investments?

8. A man invests x dollars in real estate and (15,000 -x) dollars in a savings account. If the real estate pays 8% interest and the savings account pays 3% interest, how would you represent the number of dollars in the man's annual income from these two investments?

Answers :

• 2. .04x.
• 3. (.035) (1000).
• 4. .045 y.
• 5. .06(2x -100).
• 6. .03(10,000 -x).
• 7. .06x, .065x, .125x.
• 8. .08x + .03(15,000 -x).

EXERCISES : Arranging work in investment problems

Copy the following boxes onto a paper and fill in the blank spaces with the correet data.

Answers :

• 1. 250.
• 2. .04x.
• 3. .06(7000 -x).
• 4. .06(1000 -x).
• 5. .05x, .06(10,000 -x).
• 6. .045x, .07(25,000 -x).
• 7. .025x, .06(2x + 500).
• 8. .09x, .05(x + 8000), .18x.

ILLUSTRATIVE EXAMPLE: Investment problem

A man invests $15,000, part in bonds which pay 4% interest and the rest in stocks which pay 5% interest. If his annual income from these two investments is $682.50, find the amount invested in bonds and the amount invested in stocks.

Solution. We can represent the number of dollars invested in bonds by x and the number of dollars invested in stocks by 15,000 -x. Filling these values in the box, and also filling in the given rates, we have:

We now answer the original question. The amount invested in bonds, x, is $6750; the amount invested in stocks, 15,000 -x, is 15,000 -6750, or $8250.

EXERCISES: Investment problems

1. A man invests some money at 4% and twice as much money at 5%. If is annual income from these two investments is $280, how much has he invested at each rate?

Let's use algebra to solve this problem. Let's say the amount of money the man has invested at 4% is "x" dollars.

According to the problem:

1. The amount of money invested at 5% is twice the amount invested at 4%, so the amount invested at 5% is "2x" dollars.

The annual income from the 4% investment can be calculated using the formula: Income from 4% investment = x * 0.04

The annual income from the 5% investment can be calculated using the formula: Income from 5% investment = 2x * 0.05 = 0.10x

The total annual income from both investments is given as $280. So, we can set up an equation:

Income from 4% investment + Income from 5% investment = 280 x * 0.04 + 0.10x = 280

Now, let's solve for "x":

0.04x + 0.10x = 280

0.14x = 280

x = 280 / 0.14

x = 2000

Now that we know the amount of money invested at 4% (x = 2000 dollars), we can find the amount of money invested at 5% (2x = 2 * 2000 = 4000 dollars).

So, the man has invested $2000 at 4% and $4000 at 5%.

2. A man invests a certain amount of money at 6%, and he invests $1000 more than this amount at 4%. If his annual income from these two investments is $315, how much has he invested at each rate?

Let's set up the equations to solve this problem.

Let's assume the amount the man initially invests at 6% is x dollars. Then, the amount he invests at 4% would be (x + $1000) dollars.

Now, we can calculate the income from each investment.

Income from the investment at 6%: Income_6percent = x * 0.06

Income from the investment at 4%: Income_4percent = (x + $1000) * 0.04

The total income from both investments is given as $315, so we can write the equation:

Income_6percent + Income_4percent = $315

Now, substitute the expressions for Income_6percent and Income_4percent:

x * 0.06 + (x + $1000) * 0.04 = $315

Now, solve for x:

0.06x + 0.04x + $40 = $315

Combine the x terms:

0.10x + $40 = $315

Subtract $40 from both sides:

0.10x = $275

Divide both sides by 0.10:

x = $2750

Now that we have the value of x, we can find the amount invested at each rate:

Amount invested at 6% = x = $2750

Amount invested at 4% = x + $1000 = $2750 + $1000 = $3750

So, the man invested $2750 at 6% and $3750 at 4%.

3. A man invests $5000, part at 4%, and the rest at 5%. If his annual income from these two investments is $215, how much has he invested at each rate?

Let's set up the equations to solve this problem.

Let x be the amount invested at 4% and y be the amount invested at 5%.

According to the problem, the total amount invested is $5000, so we have the first equation:

x + y = $5000 ---(1)

The annual income from the investment at 4% is given by:

Income_4percent = x * 0.04

The annual income from the investment at 5% is given by:

Income_5percent = y * 0.05

The total annual income from both investments is $215, so we can write the second equation:

Income_4percent + Income_5percent = $215

Substitute the expressions for Income_4percent and Income_5percent:

x * 0.04 + y * 0.05 = $215 ---(2)

Now, we can solve the system of equations (1) and (2) to find the values of x and y.

To eliminate decimals, let's multiply equation (2) by 100:

4x + 5y = $21500 ---(3)

Now we have the following system of equations:

1. x + y = $5000
2. 4x + 5y = $21500

To solve this system, we can use the method of substitution or elimination. Let's use the elimination method.

Multiply equation (1) by 4:

4x + 4y = $20000 ---(4)

Now, subtract equation (4) from equation (3):

(4x + 5y) - (4x + 4y) = $21500 - $20000

This simplifies to:

y = $1500

Now, we can find the value of x by substituting the value of y into equation (1):

x + $1500 = $5000

Subtract $1500 from both sides:

x = $5000 - $1500

x = $3500

So, the man invested $3500 at 4% and $1500 at 5%.

4. A man invests $12,000, part at 3½% and the rest at 5%. If the annual income on the 5% investment exceeds the annual income on the 3½% investment by $260, how much has he invested at each rate?

Let x be the amount invested at 3½% and y be the amount invested at 5%.

According to the problem, the total amount invested is $12,000, so we have the first equation:

x + y = $12,000 ---(1)

The annual income from the investment at 3½% is given by:

Income_3_5percent = x * 0.035

The annual income from the investment at 5% is given by:

Income_5percent = y * 0.05

According to the problem, the annual income on the 5% investment exceeds the annual income on the 3½% investment by $260, so we have the second equation:

Income_5percent - Income_3_5percent = $260

Substitute the expressions for Income_5percent and Income_3_5percent:

y * 0.05 - x * 0.035 = $260 ---(2)

Now, we can solve the system of equations (1) and (2) to find the values of x and y.

Let's use the given solution to verify:

x = $4000, y = $8000

Income_3_5percent = $4000 * 0.035 = $140

Income_5percent = $8000 * 0.05 = $400

Now, let's check if the annual income on the 5% investment exceeds the annual income on the 3½% investment by $260:

$400 - $140 = $260

As we can see, the difference is indeed $260, which confirms that the given solution is correct.

So, the man invested $4000 at 3½% and $8000 at 5%.

5. A man invests a certain amount of money at 6% and $1800 more than this amount at 4%. If his annual income from these two investments is $492, how much is invested at each rate?

Let x be the amount invested at 6%. Then, the amount invested at 4% would be (x + $1800).

Now, we can calculate the income from each investment.

Income from the investment at 6%: Income_6percent = x * 0.06

Income from the investment at 4%: Income_4percent = (x + $1800) * 0.04

The total income from both investments is given as $492, so we can write the equation:

Income_6percent + Income_4percent = $492

Now, substitute the expressions for Income_6percent and Income_4percent:

x * 0.06 + (x + $1800) * 0.04 = $492

Now, solve for x:

0.06x + 0.04x + $72 = $492

Combine the x terms:

0.10x + $72 = $492

Subtract $72 from both sides:

0.10x = $420

Divide both sides by 0.10:

x = $4200

Now that we have the value of x, we can find the amount invested at each rate:

Amount invested at 6% = x = $4200 Amount invested at 4% = x + $1800 = $4200 + $1800 = $6000

So, the man invested $4200 at 6% and $6000 at 4%.

6. A man invests a certain amount of money at 4½% and twice as much at 5%. If his annual income from the two investments is $522, how much has he invested at each rate?

7. A man invests a certain amount of money at 4% and a sum $2000 less than this amount at 6%. If his annual income from these two investments is $380, how much has he invested at each rate?

8. A man invests $15,000, part at 5½% and the rest at 4%. If the interest for one year on the 4% investment is $75 less than the interest for two years on the 5½%  investment, how much has he invested at each rate?

9. A man invests a certain sum of money at 6%, twice that sum at 3½% , and twice the second sum at 4%. If the total annual income from the three investments is $1189, how much does he invest at each rate?

10. A man invests a certain amount of money at 4%, a second amount $1500 greater than the first at 4½%, and a third amount $1250 less than the first at 3½%. If the yearly income from the three investments is $923.75, how much has he invested at each rate?

Answers :

• 1. $2000 at 4%, $4000 at 5%.
• 2. $2750 at 6%, $3750 at 4%.
• 3. $3500 at 4%, $1500 at 5%.
• 4. $4000 at 3½%, $8000 at 5%.
• 5. $4200 at 6%, $6000 at 4%.
• 6. $3600 at 4½%, $7200 at 5%.
• 7. $5000 at 4%, $3000 at 6%.
• 8. $10,500 at 4%, $4500 at 5½%.
• 9. $4100 at 6%, $8200 at 3½%, $16,400 at 4%.
• 10. $7500 at 4%, $9000 at 4½%, $6250 at 3½%.