# Coin Word Problems

Coin Word Problems

1. Phil T. Rich has some coins in his pocket consisting of dimes, nickels, and pennies. He has two more nickels than dimes, and three times as many pennies as nickels. How many of each kind of coin does he have if the total value is 52 cents?

Solution

First try to determine which type of coin he has fewest of. This is often a good way to find what to let x stand for. Here he has fewer dimes than nickels or pennies.

The question asks how many of each kind of coin (not how much they are worth). That is, what number of each kind of coin does he have? So, let

x = number of dimes

Go back to one fact at a time. He has two more nickels than dimes.

x + 2 = number of nickels

Another fact is that he has three times as many pennies as nickels.

3(x + 2) = number of pennies

Next the information left for the equation is that the total value is 52 cents. You can’t say that the total number of coins equals 52 cents. Number of must be changed to value. If you had two dimes, you would have 20 cents. You multiplied how many coins by how much each is worth. So let’s change our how many to how much. If you have the number of dimes, multiply by 10 to change to cents. Multiply the number of nickels by 5 to change to cents. The number of pennies is the same as the number of cents.

 Number of Coins Value in Cents x = number of dimes 10x = number of cents in dimes x + 2 = number of nickels 5(x + 2) = number of cents in nickels 3(x + 2) = number of pennies 3(x+ 2) = number of cents in pennies

Now we can add the amounts of money. If you make it all pennies, there are no decimals.

10x + 5(x + 2) + 3(x + 2) = 52

10x +5x + 10 + 3x + 6 = 52

18x + 16 = 52

18x = 36

x = 2

So,

x = 2

x + 2 = 4

3(x + 2) = 12

So,

2 dimes

4 nickels

12 pennies

1. Ernest Worker had a collection of silver coins worth \$205. There were five times as many quarters as half-dollars (50-cent pieces) and 200 fewer dimes than quarters. How many of each kind of coin did the Ernest have?

Solution

First try to determine which type of coin he has fewest of. This is often a good way to find what to let x stand for. Here he has fewer dimes than nickels or pennies.

The question asks how many of each kind of coin (not how much they are worth). That is, what number of each kind of coin does he have? So, let

x = number of dimes

 Number of Coins Value in Cents x = number of half-dollars 50x = number of cents in half-dollars 5x = number of quarters 25(5x) = number of cents in quarters 5x – 200 = number of dimes 10(5x – 200) = number of cents in dimes

Remember, everything has been changed to cents. So \$205 has to be changed to cents by multiplying by 100. \$205 has to be changed to cents by multiplying by 100. \$205 = 20,500 cents.

50x + 25(5x) + 10(5x – 200) = 20,500

50x + 125x + 50x – 2000 = 20,500

225x = 22,500

x = 100

5x = 500

5x – 200 = 300

So, Ernest has 100 half-dollars, 500 quarters and 300 dimes.

1. Kay Oss bought \$32.56 worth of stamps. She bought 20 more 19-cent stamps than 50-cent stamps. She bought twice as many 32-cent stamps as 19-cent stamps. How many of each kind did she buy?

Solution

 Number of Stamps Value in Cents x = number of 50-cent stamps 50x x + 20 = number of 19-cents 19(x + 20) 2(x + 20) = number of 32-cent stamps 32[2(x+ 20)]

Equation

Remember, \$32.56 equals 3256 cents.

50x + 19(x + 20) + 32[2(x + 20)] = 3256

50x + 19x + 380 + 64x + 1160 = 3256

133x = 1596

x = 12

x + 20 = 32

2(x + 20) = 64

So,

Kay purchased 12 50-cent stamps, 32 19-cent stamps and 64 32-cent stamps

Check

12(50) + 19(32) + 64(32) = 3256

600 + 608 + 2048 = 3256

3256 = 3256

1. A collection of coins has a value of 64 cents. There are two more nickels than dimes and three times as many pennies as dimes. How many of each kind of coin are there?

Solution

 Number of Coins Value in Cents x = number of dimes 10x = value of dimes x + 2 = number of nickels 5(x + 2) = value of nickels 3x = number of pennies 3x = number of pennies

The total value of the dimes, nickels, and quarters equals 64 cents.

Equation

10x + 5(x + 2) + 3x = 64

10x + 5x + 10 + 3x = 64

18x = 54

x = 3

x + 2 = 5

3x = 9

So,

There are 3 dimes, 5 nickels and 9 pennies.

Check

3(10 cents) = 30

5(5 cents) = 25

9(1 cent) = 9

______

Total = 64 cents

1. Lotta Spences has ten bills in her wallet. She has a total of \$40. If she has one more \$5 bill than \$10 bills, and two more \$ bills than \$5 bills, how many of each does she have? (There are two ways of working this problem. See if you can do it both ways.)

Solution

 Number of Bills Values in dollars x = number of \$10 bills 10x = value of \$10 bills x + 1 = numbers of \$5 bills 5(x +1) = value of \$5 bills (x + 1) + 2 = number of \$1 bills 1(x + 3) = value of \$1 bills

Equation

10x + 5(x + 1) + (x + 3) = 40

10x + 5x + 5 + x + 3 = 40

16x = 32

x = 2

x + 1 = 3

x + 3 = 5

So,

Lotta has 2 ten dollar bills, 3 five dollar bills and 5 one dollar bills.

Check

2(\$10) = \$20

3(\$5) = 15

5(\$1) = 5

____

Total = \$40

1. Dan D. Lyons bought \$21.44 worth of stamps at the post office. He bought 10 more 4-cent stamps than 19-cent stamps. The number of 32-cent stamps was three times the number of 19-cent stamps. He also bought two \$1 stamps. How many of each kind of stamp did he purchase?

Solution

 Number of Stamps Value in Cents
 x = number of 19-cent stamps 19x x + 10 = number of 4-cent stamps 4(x + 10) 3x = number of 32-cent stamps 32(3x) 2 = number of \$1 stamps 2(100)

Equation

19x + 4(x + 10) + 32(3x) + 2(100) = 2144

19x + 4x + 40 +96x + 200 = 2144

119x + 240 = 2144

119x = 1904

199x = 1904

x = 16

x + 10 = 26

3x = 48

So,

Dan purchased 16 19-cent stamps, 26 4-cent stamps and 48 32-cent stamps.

Check

19(16) + 4(26) + 32(48) + 200 =2144

304 + 104 + 1536 + 200 = 2144

2144 = 2144

1. Nick O’ Time purchases a selection of wrenches for his shop. His bill is \$78. He buys the same number of \$1.50 and \$2.50 wrench­es, and half that many of \$4 wrenches. The number of \$3 wrenches is one more than the number of \$4 wrenches. How many of each did he purchase? (Hint: if you have not worked with fractions, use decimals for all fractional parts.)

Solution

Number of Wrenches

x = number of \$4.00 wrenches

2x = number of \$1.50 wrenches

2x = number of \$2.50 wrenches

x + 1 = number of \$3.00 wrenches

Value in Dollars

(x)(4.00) = value of \$4.00 wrenches

(2x) (1.50) = value of \$1.50 wrenches

(2x)(2.50) = value of \$2.50 wrenches

(x + 1)(3.00) = value of \$3.00 wrenches

Equation

x(4.00) + 2x(1.50) + 2x(2.50) + (x + 1)(3.00) = 78

Note in this problem that the decimal is removed when you clear parentheses.

4x + 3x + 5x + 3x + 3 = 78

15x = 75

x = 5

2x = 10

x + 1 = 6

So,

Nick purchased 5 \$4.00 wrenches,10 \$1.50 wrenches, 10 \$2.50 wrenches and 6 \$3.00 wrenches.

Check

5(\$4.00) = \$20

10(\$1.50) = 15

10(\$2.50) = 25

6(\$3.00) = 18

____

Total = \$78

1. Phoebe Small at the XYZ Department Store receives \$15 in change for her cash drawer at the start of each day. She receives twice as many dimes as fifty-cent pieces, and the same number of quarters as dimes. She has twice as many nickels as dimes and a dollar’s worth of pennies. How many of each kind of coin does she receive?

Solution

 Number of Coins Value in Cents x = number of 50-cent pieces 50x = value of 50-cent pieces
 2x = number of 10-cent pieces 10(2x) = value of 10-cent pieces 2x = number of 25-cent pieces 25(2x) = value of 25-cent pieces 4x = number of 5-cent pieces 5(4x) = value of 5-cent pieces 100 = number of 1-cent pieces 1(100) = value of 1-cent pieces

Equation

Fifteen dollars equals 1500 cents.

50x + 2x(10) + 2x(25) + 4x(5) + 100 =1500

50x + 20x + 50x 20x + 100 = 1500

140x = 1400

x = 10

2x = 20

2x = 20

4x = 40

So,

Phoebe has 10 50 cents pieces, 20 10 cents pieces, 20 25 cents pieces, 40 5 cents pieces and 100 1 cent pieces.

Check

10(50 cents) = \$5

20(10 cents) = 2

20(25 cents) = 5

40(5 cents) = 2

100(1 cent) = 1

____

Total = \$15

1. A collection of 36 coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes than quarters. How many of each kind of coin are there?

Let  x = number of quarters (there are fewer quarters)

x + 3 = number of nickels

x + 6 = number of dimes

Here we do not change to cents because the number of coins is given.

Equation

x + (x + 3) + (x + 6) = 36

3x + 9 = 36

3x = 27

x = 9 quarters

x + 3 = 12 nickels

x + 6 = 15 dimes

Check

9 +12 + 15 = 36

36 = 36

1. The cash drawer of the Greasy Spoon Cafe contains \$227 in bills. There are six more \$5 bills than \$10 bills. The number of \$1 bills is two more than 24 times the number of \$10 bills. How many bills of each kind are there?

Let: x= number of dollars in \$10 bills.

5(x + 6) = number of dollars in \$5 bills.

24x + 2 = number of dollars in \$1 bills.

Equation

10x + 5(x + 6) + 24x + 2 = 227

10x + 5x + 30 + 24x + 2 = 227

39x + 32 = 227

39x = 195

x = 5

x + 6 = 11

24x + 2 = 122

Check

10(5) + 5(11) + 24(5) + 2 = 227

50 + 55 + 120 + 2 = 227

227 = 227

1. Lois Lane went to the drugstore. She bought a bottle of aspirin and a bottle of Tylenol. The aspirin cost \$1.25 more than the Tylenol. She also bought cologne which cost twice as much as the total of the other two combined. How much did each cost if her total (without tax) was \$24.75?

Let x = cost in cents of Tylenol

x + 125 = cost in cents of aspirin

2(x + x + 125) = cost in cents of cologne

Equation

x + x + 125 + 2(2x + 125) = 2475

2x + 125 +4x + 250 + 2475

6x + 375 + 2475

6x = 2100

x = 350

x + 125 = 475

2(2x + 125) = 1650

Check

3.50 + 4.75 + 16.50 = 24.75

24.75 = 24.75

1. Superman bought some gum and some candy. The number of packages of gum was one more than the number of mints. The number of mints was three times the number of candy bars. If gum was 24 cents a package, mints were 10 cents each, and candy bars were 35 cents each, how many of each did he get for \$5.72?

Let x = number of 35 cent candy bars

3x = number of 10 cent mints

3x + 1 = the number of 24 cent packages of gum

Value in Cents

35x = cents for candy bar

10(3x) = cents for mints

24(3x+1)= cents for gum

Equation

35x + 10(3x) + 24(3x+1) = 572

35x + 30x + 72x + 24= 572

137x + 24 = 572

137x = 548

x = 4

3x = 12

3x + 1 = 13

Check

4(35) + 12(10) + 13(24) = 572

140 + 120 + 312 + = 572

572 = 572

1. Clay Potts had \$50 to buy his groceries. He needed milk at \$1.95 a carton, bread at \$2.39 a loaf, breakfast cereal at \$3.00 a box, and meat at \$5.39 a pound. He bought twice as many cartons of milk as loaves of bread and one more package of cereal than loaves of bread. He also bought the same number of pounds of meat as packages of cereal. How many of each item did he pur­chase if he received \$12.25 in change?

Let x = number of loaves of bread at \$2.39 each

2x = number of cartons of milk at \$1.95 each

x + 1 =number of packages of cereal at \$3.00 each

x + 1 = number of pounds of meat at \$5.39 each

Cost in cents

195(2x) = cost of milk

300(x + 1) = cost of cereal

539(x + 1) = cost of meat

Equation

239x + 195(2x) + 300(x + 1) + 539(x + 1) = 5000 – 1225

239x + 390x + 300x + 300 + 539x + 539 = 3775

1468x + 839 = 3775

1468x = 2936

x = 2

2x = 4

x + 1 = 3

x + 1 = 3

Check

169(2) + 130(4) + 200(3) + 359(3) = 3500 – 1225

338 + 520 + 600 + 1077 = 2535

2535 = 2535

1. Mr. Merrill has 3 times as many nickels as dimes. The coins have a total value of \$1.50. How many of each coin does he have?

Let n = number of nickels.

Let d = number of dimes.

n = 3d

.05n + .10d = 1.50

Multiply through by 100

5n + 10d = 150

Use (n = 3d) and sub in 3d for n

5n + 10d = 150

5(3d) + 10d = 150

15d + 10d = 150

25d = 150

d = 6

So,

There are 6 dimes and 18 nickels

6 dimes equals 60 cents and 18 nickels equals 90 cents

60 cents and 90 cents = \$1.50

1. Ms. Lynch has 21 coins in nickels and dimes. Their total value is \$1.65. How many of each coin does she have?

Let n = number of nickels.

Let d = number of dimes.

n + d = 21

.05n + .10d = \$1.65

Multiply through by 100

5n + 10d = 165

Multiply first equation by -5

-5n + -5d = -105

5n + 10d = 165

+

-5n -5d = -105

=

5d = 60

d = 12

So,

Ms. Lynch has 12 dimes and 9 nickels.

12 dimes equals \$1.20 and 9 nickels equals 45 cents

\$1.20 and 45 cents = \$1.65.

1. A vending machine that takes only dimes and quarters contains 30 coins, with a total value of \$4.20. How many of each coin are there?

Let d = number of dimes.

Let q = number of quarters.

d + q = 30

.10d + .25q = \$4.20

Multiply through by 100

10d + 25q = 420

Multiply first equation by -10

-10d + -10q = -300

10d + 25q = 420

+

-10d + -10q = -300

=

15q = 120

q = 8

So,

The vending machine has 8 quarters and 22 dimes.

8 quarters equals \$2.00 and 22 dimes equals \$2.20

\$2.00 and  \$2.20 equals \$4.20

1. The total value of the \$1 bills and \$5 bills in a cash box is \$124. There are 8 more \$5 bills than \$1 bills. How many of each are there?

Let x = number of one dollar bills

Let f = number of five dollar bills.

1x + 5f = 124

x + 8 = f

Substitute (x+ 8) for f in the first equation.

1x + 5(x + 8) = 124

x + 5x + 40 = 124

6x + 40 = 124

6x = 84

x = 14

So,

There are 14 one dollar bills and 22 five dollar bills.

\$14 + \$110 = \$124

1. A collection of nickels and quarters amounts to \$2.60. There are 16 coins in all. How many of each coin are there?

Let n = number of nickels

Let q = number of quarters.

n + q = 16

.05n + .25q = \$2.60

Multiply through by 100

5n + 25q = 260

Multiply first equation by -5

-5n + – 5q = -80

5n + 25q = 260

+

-5n + – 5q = -80

=

20q = 180

q = 9 and n = 7

So,

The collection contains 9 quarters (\$2.25) and 7 nickels (.35)

\$2.25

+

.35

=

\$2.60

1. Joe Lick bought some 20-cent and 25-cent stamps. He bought 32 stamps in all, and paid \$7.40 for them. How many stamps of each kind did he buy?

Let x = number of 20 cent stamps

Let y = number of 25 cent stamps

x + y = 32

.20x + .25y = \$7.40

Multiply through by 100

20x + 25y = 740

Multiply first equation by -20

-20x + -20y = -640

20x + 25y = 740

+

-20x + -20y = -640

=

5y = 100

y = 20 and x = 12

So,

Joe lick has 20 25 cent stamps (5.00) and 12 20 cent stamps (\$2.40)

\$5.00

+

\$2.40

=

\$7.40

1. For a school play, 340 tickets valued at \$810 were sold. Some cost \$2 and some cost \$3. How many tickets of each kind were sold?

Let x = number of two dollar tickets.

Let y = number of three dollar tickets.

x + y = 340

\$2x + \$3y = \$810

Multiply first equation by -2

-2x + -2y = -680

2x + 3y = 810

+

-2x + -2y = -680

=

y = 130 and x = 210

So,

The school play sold 210 two dollar tickets (\$420) and 130 three dollar tickets (\$390)

\$420

+

\$390

=

\$810

1. Romeo bought a mixture of 20-cent, 35-cent, and 50-cent valentines. The number of 20-cent valentines was 1 more than twice the number of 35-cent valentines, and the number of 50-cent valentines was 2 less than the number of 35-cent ones. If he spent \$4.20 all together, how many valentines of each kind did he buy?

Let x = number of 20 cent stamps.

Let y = number of 35 cent stamps

Let z = number of 50 cent stamps

x = 2y + 1

z = y – 2

.20x + .35y + .50z = 4.20

Multiply third equation through by 100

20x + 35y + 50z = 420

Substitute x = 2y + 1 into third equation

Substitute z = y – 2 into third equation

20x + 35y + 50z = 420

20(2y + 1) + 35y + 50(y – 2) = 420

40y + 20 + 35y + 50y – 100 = 420

125y – 80 = 420

125y = 500

y = 4 and x = 9  and z = 2

So,

Romeo purchased nine 20 cent stamps (1.80), four 35 cent stamps (1.40), and two 50 cent stamps (1.00).

1.80

+

1.40

+

1.00

=

4.20 or \$4.20