Lesson Coin problems

Coin problems

This lesson and the lesson next to this, More Coin problems, show you some typical Coin problems and methods to solve them.
The givens in this type of problems usually are
    a) an amount of money composed by the coins collection, and
    b) some information about the coins collection, for example, the number of coins.
The collection of coins in these problems consists typically of two, rarely of three types of coins.
The method to solve coin problems is to reduce them to the linear equation (if possible), or to the system of linear equations.
This lesson and the lesson next to this, More Coin problems, consider different sample problems to cover a variety of conditions that may be imposed to the coin collections.

Problem 1

Sue has $1.15 in nickels and dimes, totally 16 coins.
How many nickels and how many dimes does Sue have?

Solution 1
Lets us denote as the number of nickels Sue has.
Then the number of dimes is equal to .
So, Sue has cents in nickels and cents in dimes.
Since the total amount Sue has is equal to $1.15, or 115 cents, this leads to the equation
.

Simplify this equation:
(after brackets opening)
      (after collecting like terms at the left side and moving to the right side with the opposite sign)
            (after collecting like terms at the right side)
                    (after dividing both sides by ).

So, the number of nickels is equal to 9.
The number of dimes is equal to 16 minus the number of nickels, that is
.

Check.
The total number of coins is equal to .
The total amount is equal to .
The solution is correct.

Answer. Sue has 9 nickels and 7 dimes.

Solution 2
This solution is reducing to the system of two linear equations in two unknowns.

Denote the number of nickels Sue has as and the number of dimes as .
Since the total number of coins is equal to 16, you can write the first equation
.

Sue has cents in nickels and cents in dimes.
Since the total amount Sue has is equal to $1.15 = 115 cents, you can write the second equation
.

So, you have the system of two linear equations in two variables


Solve it using the substitution method.
Isolate the variable from the first equation
.

Substitute it to the second equation. You get
.

Simplify this equation (similar to that was done is the Solution 1):
(after brackets opening)
      (after collecting like terms at the left side and moving to the right side with the opposite sign)
            (after collecting like terms at the right side)
                    (after dividing both sides by ).

So, the number of nickels is equal to 9.
The number of dimes is equal to 16 minus the number of nickels, that is
.

Check.
The total number of coins is equal to .
The total amount is equal to .
The solution is correct.

Answer. Sue has 9 nickels and 7 dimes.

Problem 2

Michael has $1.95 totally in his collection, consisting of quarters and nickels.
The number of nickels is in three more than the number of quarters.
How many nickels and how many quarters does Michael have?

Solution 1
Denote as the number of quarters Michael has.
Then the number of nickels is equal to .
So, Michael has cents in quarters and cents in nickels.
Since the total amount Michael has is equal to $1.95, or 195 cents, this leads to the equation
.

Simplify this equation:
(after brackets opening)
      (after collecting like terms at the left side and moving to the right side with the opposite sign)
            (after collecting like terms at the right side)
                  (after dividing both sides by ).

So, the number of quarters is equal to 6.
The number of nickels is in three more, that is
.

Check.
The difference between the number of nickels and the number of quarters is equal to .
The total amount is equal to .
The solution is correct.

Answer. Michael has 6 quarters and 9 nickels.

Solution 2
This solution is reducing to the system of two linear equations in two unknowns.

Denote the number of quarters Michael has as and the number of nickels as .
Since the number of nickels is in three more than the number of quarters, you can write the first equation
.

Michael has cents in nickels and cents in quarters.
Since the total amount Michael has is equal to $1.95, or 195 cents, you can write the second equation
.

So, you have the system of two linear equations in two variables


Solve it using the substitution method.
Isolate the variable from the first equation
.

Substitute it to the second equation. You get
.

Simplify this equation:
(after brackets opening)
      (after collecting like terms at the left side and moving to the right side with the opposite sign)
            (after collecting like terms at the right side)
                  (after dividing both sides by ).

So, the number of quarters is equal to 6.
The number of nickels is in three more, that is
.

Check.
The difference between the number of nickels and the number of quarters is equal to .
The total amount is equal to .
The solution is correct.

Answer. Michael has 6 quarters and 9 nickels.