Algebra 1
Coin Word Problems
1) Stan’s pocket is filled with nickels and
dimes. He has 6 more dimes than nickels.
All together Stan has $1.20 in coins. How
many nickels and dimes does he have?
-Let’s start with the sentence “He has 6 more
dimes than nickels.”
-Ask yourself “Which coin does he have more
of?”
-The answer is dimes.
d=n+6
Quantity Equation
All together Stan has $1.20 in
coins.
• Each coin has a different value.
• Each nickel is worth 5 cents. Thus 5n is
the amount all his nickels are worth.
• Each dime is worth 10 cents. Thus 10d is
the amount all his dimes are worth.
• $1.20 is worth 120 cents…
• 5n + 10d = 120
The value equation.
Substitute and solve
• 10d + 5n = 120 and
• 10(n + 6) + 5n = 120
• 10n + 60 + 5n = 120
• 15n + 60 = 120
• 15n = 60
•n=4
d = (n + 6)
Complete the Problem
• He has 4 nickels.
•d=n+6
•d=4+6
• d = 10
• Stan has 4 nickels and 6 dimes.
Check
• 10 dimes is worth 10(.10) = $1.00
• 4 nickels is worth 4(.05) = $0.20
Total = $1.20
4) She gave the sitter 6 more
nickels than dimes.
•
•
•
•
•
•
•
She has more…
Nickels.
n=d+6
Quantity Equation.
She has 3 times as many quarters as dimes.
She has more…
Quarters.
q =3d
Quantity Equation.
All together Mrs. Ryan paid the
babysitter $7.50.
• Total nickel value in cents…
• 5n
• Total dime value in cents…
• 10d
• Total quarter value in cents…
• 25q
• 5n + 10d + 25q = 750
Substitute and Solve
•
•
•
•
•
•
•
•
n=d+6
q = 3d
5n + 10d + 25q = 750
5(d + 6) + 10d + 25(3d) = 750
5d + 30 + 10d + 75d = 750
90d + 30 = 750
90d = 720
d=8
She paid 8 dimes.
Solve for Nickels and Quarters
•d=8
•n=d+6
•n=8+6
• n = 14
• q = 3d
• q = 3(8)
• q =24
Check
• 8 dimes is worth
8($0.10) = $0.80
• 14 nickels is worth 14($0.05) = $0.70
• 24 quarters is worth 24($0.25) = $6.00
Total = $7.50