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Online Quiz (Worksheet A B C D)

Questions Per Quiz = 2 4 6 8 10

Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation
6.4 Coins Word Problems - 2

Example:
Tim has 11 coins in his pocket that have a total value of $1. If these coins consists of nickels and dimes only, how many of each kind are there?
Solution:
Let 'n' be the number of nickels and 'd' be the number of dimes. The word problem can be translated into equation
n+d=11
There are 5 cents in a nickel and 'n' nickels equals an amount of 5n.
There are 10 cents in a dime and 'd' dimes equals to an amount of 10d.
The total value is $1 or 100 cents. Another equation that can be written as 5n+10d=100
Solve these two equations for n and d to find the number of nickels n and dimes d
n+d=11 implies n = 11 - d
5n+10d=100
substituting n = 11 - d in the equation
5n+10d=100
5(11 - d)+10d=100
55 - 5d + 10d = 100
55 + 5d = 100
5d = 100 - 55
5d = 45
d = 9
n = 11 - d = 11 - 9 = 2
Tim has 9 dimes and 2 nickels
Verification:
substitute value of n and d in the equation n + d = 11 which is 9 + 2 = 11
or substitute value of n and d in the equation 5n+10d=100 which is 5n+10d= 5x2 + 10x9 = 10 + 90 = 100

Example:
Kim has 10 coins that total $2.10. The coins are nickels and quarters only. How many of each kind are there?
Solution:
Let 'n' be the number of nickels and 'q' be the number of quarters. The word problem can be translated into this equation as
n+q=10
Another equation that can be written is
5n+25q=210
Solve these two equations for n and q
n+q=10 implies that n = 10 - q
5n+25q=210
5(10 - q)+25q=210
50 - 5q + 25q = 210
50 + 20q = 210
20q = 160
q = 8
n = 10 - q = 10 - 8 = 2
There are 8 quarters and 2 nickles

Example:
Lilly has 21 coins that total $4.50. The coins are dimes and quarters only. How many of each kind are there?
Solution:
Let 'd' be the number of dimes and 'q' be the number of quarters. The word problem can be translated into this equation
d+q=21
Another equation that can be written is
10d+25q=450
Solve these two equations for d and q
d+q=21
10d+25q=450
d+q=21 implies d = 21 - q substituting in the equation 2 we have
10d+25q=450
10(21 - q)+25q=450
210 - 10q + 25q = 450
210 + 15q = 450
15q = 240
q = 16
d = 21 - q = 21 - 16 = 5

Directions: Solve the following word problems.
Q 1: Eric has 29 coins that total $6.50. The coins are dimes and quarters only. How many of each kind are there?
q = 22 and d = 7
q = 24 and d = 8
q = 24 and d = 5

Q 2: Peter has 40 coins in his pocket that have a total value of $3.50. If these coins consists of nickels and dimes only, how many of each kind are there?
d = 30 and n = 20
d = 30 and n = 10
d = 20 and n = 20

Q 3: Dan has 22 coins that total $2.50. The coins are dimes and quarters only. How many of each kind are there?
d = 18 and q = 4
d = 20 and q = 2
d = 2 and q = 20

Q 4: Mary has 20 coins in his pocket that have a total value of $1.75. If these coins consists of nickels and dimes only, how many of each kind are there?
d = 16 and n = 6
d = 12 and n = 8
d = 15 and n = 5

Q 5: Peter has 31 coins in his pocket that have a total value of $2.80. If these coins consists of nickels and dimes only, how many of each kind are there?
d = 25 and n = 6
d = 20 and n = 6
d = 21 and n = 8

Q 6: Harry has 12 coins that total $2.40. The coins are dimes and quarters only. How many of each kind are there?
q = 8 and d = 4
q = 6 and d = 6
q = 4 and d = 8

Q 7: Ron has 38 coins that total $6.70. The coins are nickels and quarters only. How many of each kind are there?
q = 24 and n = 10
q = 24 and n = 14
q = 21 and n = 14

Q 8: Kim has 15 coins that total $2.35. The coins are nickels and quarters only. How many of each kind are there?
q = 8 and n = 7
q = 5 and n = 10
q = 10 and n = 5

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Question 10: This question is available to subscribers only!


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