Algebra: Coin Problems

This tutorial explains the solution to two coin problems.

The first problem gives the total amount of cents and asks for the number of each type of coin.

The second problem gives the total amount of coins and asks for the total number of each type of coin.

PROBLEM 1

Fred has $1.80. He has twice as many quarters as he has dimes. How many dimes does he have? How many quarters does he have?

SOLUTION

let x = the number of dimes

let y = the number of quarters

We know that the total is $1.80.

(25 cents per quarter * number of quarters) + (10 cents per dime * number of dimes) = 180 cents

25y + 10x = 180

Fred has twice as many quarters as dimes.

Hence

number of quarters = 2 * number of dimes

y = 2x

(25 cents per quarter) * (number of dimes * 2) + (10 * number of dimes) = 1.80

50x + 10x = 180

60x = 180

x = (180)/60 = 3

number of dimes = 3

y = 2x = 2*3 = 6

number of quarters = 6

PROBLEM 2

Sam has 20 coins. He has two times as many quarters as he has half dollars

He also has three times as many half dollars as he has nickels.

How man half dollars does he have?

How many quarters does he have?

How many nickels does he have?

SOLUTION

let x = number of quarters

let y = number of half dollars

let z = number of nickels

x + y + z = 20

He has two times as many quarters as he has half dollars

Number of quarters = 2 * number of half dollars

x = 2y

He also has three times as many half dollars as he has nickels.

Number of half dollars = 3 * number of nickels

y = 3z

z = y/3

x + y + z = total

Substituting 2y for x and

y/3 for z

we have

2y + y + y/3 = 20

3y + y/3 = 20

9y + y = 60

10 y = 60

y = 6 half dollars

x = 2y

x = 12 = 12 quarters

z = y/3

z = 6/3 = 2 nickels

References:
I have a Bachelor of Science in Electrical Engineering.

Algebra with Trigonometry for College Students
Charles P McKeague
ISBN 0-03-096561-6