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Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation
6.8 Word Problems - Coins

Example:
Jill has 26 coins in her purse. They are all dollar and quarter coins. If they add up to $18.50. How many of each coin does she have.
Solution:
Let 'd' stand for the number of dollar coins and 'q' stand for the number of quarters.
The rest of the coins are quarters.
Jill has 26 coins in her purse. They are all dollar and quarter coins. Then d + q = 26
1 dollar = 100 cents
1 quarter = 25 cents
If they add up to $18.50. Therefore 100d + 25q = 1850
Solving two equations we have
d + q = 26 -------------- equation 1
100d + 25q = 1850 ----- equation 2
d + q = 26
q = 26 - d
substituting q in equation 2
100d + 25(26 - d) = 1850
100d + 650 - 25d = 1850
75d = 1850 - 650
75d = 1200
d = 1200/75 = 16
q = 26 - d = 26 - 16 = 10
Jill has 16 dollar coins and 10 quarter coins in her purse.

Example:
There are 33 coins. They are nickels, dimes and quarters. They total to a value of $3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels. How many coins of each kind are there?
Solution:
Let 'n' stand for number of nickels, 'd' stand for number of dimes and 'q' for the number of quarters.
Then n + d + q = 33
n = 3q
d = 0.5n = 0.5(3q) = 1.5q
substituting n and d in the equation n + d + q = 33
3q + 1.5q + q = 33
5.5q = 33
q = 6
n = 3q = 3 x 6 = 18
d = 1.5q = 1.5 x 6 = 9
There are 6 quarters 18 nickles and 9 dimes.

Example:
A box contain the same number of pennies, nickels and dimes. The coins total $1.44. How many of each type of coin does the box contain?
Solution:
Let 'p' stand for number of pennies, 'n' stand for number of nickles and 'd' stand for number of dimes.
There are equal number of each of these implies that
number of pennies is p, number of nickles is p and number of dimes is p
The value of the coin is the number of cents for each coin times the number of that type of coin, therefore
value of pennies = 1p
value of nickles = 5p
value of dimes = 10p
The total value is $1.44
Therefore 1p + 5p + 10p = 144
16p = 144
p = 9
q = 9
d = 9
There are 9 of each type of coin in the box.

Directions: Solve the following word problems.
Q 1: Tom has 7 more quarters than dimes and has a total of $6.80. The number of quarters and dimes is 38. How many quarters and dimes does Terry have?
18 dimes and 20 quarters
8 dimes and 12 quarters
14 dimes and 16 quarters

Q 2: Eric's wallet has one-dollar bills, five-dollar bills, and ten-dollar bills. The total amount is $43. He has four times as many one-dollar bills as ten-dollar bills. All together, there are 13 bills in his wallet. How many of each bill does he have?
5 ten-dollar bills, 7 one-dollar bills, and 3 five-dollar bills.
3 ten-dollar bills, 8 one-dollar bills, and 4 five-dollar bills.
2 ten-dollar bills, 8 one-dollar bills, and 3 five-dollar bills

Q 3: Kelly has 25 coins in nickels and dimes only. They add up to $1.65. How many each coins does she have? (Hint: the equations can be written as: n+d=25 and 5n+10d=165 solve for n and d)
12 nickels and 18 dimes
10 nickels and 2 dimes
17 nickels and 8 dimes

Q 4: In a basketball game, 1000 tickets were sold. Adult tickets costs $8.50, children's costs $4.50, and a total of $7700 was collected. How many tickets of each kind were sold? (Hint: Let x be number of adult tickets and y be the number of children tickets sold. The two equations can be written as x + y = 1000, 8.5x + 4.5y = 7700 solve for x and y)
adult tickets sold are 600 and children tickets sold is 300
adult tickets sold are 800 and children tickets sold is 200
adult tickets sold are 600 and children tickets sold is 400

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