Problem 174 - Elementary Algebra

Question

Julio has \(\$ 2.75\) in his pocket in nickels and dimes. The number of dimes is 10 less than twice the number of nickels. Find the number of each type of coin.

Step-by-Step Solution

Answer

15 nickels and 20 dimes.

Step 1 - Define Variables

Let the number of nickels be denoted by . Then, the number of dimes is twice the number of nickels minus 10, so let the number of dimes be denoted by 2n - 10.

Step 2 - Express Values in Cents

Since nickels are worth 5 cents, the total value of the nickels is 5n cents. Since dimes are worth 10 cents, the total value of the dimes is 10(2n - 10) cents.

Step 3 - Set Up the Equation

The total value of all the coins combined is 275 cents (since \$2.75 = 275\ cents). Thus, the equation to find n is: \[5n + 10(2n - 10) = 275.\]

Step 4 - Solve the Equation

Simplify and solve the equation from Step 3: \[5n + 20n - 100 = 275,\] \[25n - 100 = 275,\] \[25n = 375,\] \[n = 15.\]

Step 5 - Find the Number of Dimes

Since the number of dimes is 2n - 10, substitute the value of n to find the number of dimes: \[2(15) - 10 = 20.\]

Step 6 - Conclusion

The number of nickels is 15, and the number of dimes is 20.

Key Concepts

Variables

In algebra, variables are symbols that represent unknown quantities. They allow us to write equations and solve real-world problems systematically.
Consider the problem where Julio has some nickels and dimes. To find out how many nickels and dimes he has, we use variables to represent these unknowns:
Let’s denote the number of nickels as . Therefore, we can express the number of dimes in terms of . According to the problem, the number of dimes is 10 less than twice the number of nickels: it can be represented as 2n - 10.
Using variables helps you set up relationships between quantities and solve the problem easily.

Equations

An equation is a mathematical statement that asserts the equality of two expressions. By setting up equations, we can solve for the unknown variables.
In Julio's coin problem, the total value of his coins in cents is given. We convert the dollar amount (2.75) to cents (275 cents) for easier calculation:
We know the values of nickels (5 cents) and dimes (10 cents) and can express their total value in terms of the variable n:
5n + 10(2n - 10) = 275.
This equation combines all the given information into a single, solvable expression.

Arithmetic Operations

Arithmetic operations such as addition, subtraction, multiplication, and division are basic mathematical tools that help in manipulating equations to find the value of variables.
To solve Julio's coin problem, we perform several arithmetic operations on the set equation:
First, we distribute the multiplication over addition inside the parentheses:
5n + 20n - 100 = 275.
Next, we combine like terms and isolate the variable by performing addition and subtraction:
25n - 100 = 275,
25n = 375.
Finally, we solve for the variable n by performing division:
n = 15.
These operations help to transform the equation step by step into a solution.

Coin Problems

Coin problems involve determining the number of each type of coin given certain conditions like total value and relationships between the quantities.
For example, Julio's problem provides the following information:

• Total value of coins (2.75 or 275 cents)
• Relationship between dimes and nickels (dimes are 10 less than twice the number of nickels)

By defining variables and setting up an equation based on the given conditions, we solve for the number of each type of coin:
We find that Julio has 15 nickels and 20 dimes.
Coin problems teach us how to apply algebraic techniques to everyday scenarios involving money and quantity relationships.