Question

Fred’s wallet contains coins with a total worth of $1.45. Suppose he has three times as many dimes as quarters and has exactly 7 nickels and no pennies. How many dimes does he have? (A) 2 (B) 3 (C) 6 (D) 9

Short answer

Fred has 6 dimes. Answer: (C) 6.

Step-by-step solution

Identify the value of each coin

A dime is worth \(0.10, a quarter is worth \)0.25, and a nickel is worth $0.05.

Set up an equation for the total worth of coins

Let d represent the number of dimes, and q represents the number of quarters that Fred has. The total worth of the coins is given by: \(0.10d + 0.25q + 0.05(7) = 1.45\) We also know that Fred has three times as many dimes as quarters, so: \(d = 3q\) Now we have two equations with two unknowns: \(0.10(3q) + 0.25q + 0.05(7) = 1.45\) \(d = 3q\)

Solve the system of equations

First, simplify the first equation and solve for q: \(0.30q + 0.25q + 0.35 = 1.45\) \(0.55q = 1.1\) \(q = 2\) Now, substitute q in the d equation to solve for d: \(d = 3 \cdot 2\) \(d = 6\)

Check the answer

We found d = 6 and q = 2. Let's check if this satisfies the condition that the total worth of coins is $1.45: \(0.10(6) + 0.25(2) + 0.05(7) = 0.60 + 0.50 + 0.35 = 1.45\) The result is the correct total worth of $1.45, therefore Fred has 6 dimes. So, the correct answer is (C) 6.

Key concepts

System of Equations

When tackling a problem involving a system of equations, we are dealing with two or more equations that we must solve together because they share common variables. In the context of a GED math practice problem like the coin value question involving Fred's wallet, we set up a system to represent the relationships between the number of coins and their total value.

To solve a system of equations, we can use a variety of methods such as substitution, elimination, or graphing. For this particular problem, substitution is efficient because one equation gives us a direct relationship between the number of dimes and quarters. Once we determine the value of one variable, we substitute it into the other equation to find the second variable. It's akin to solving a puzzle; each piece must fit perfectly to complete the picture, just as each coin's value and quantity must align to reach the total value in Fred's wallet.

Coin Value Problem

GED math often includes a coin value problem, where students must use algebra to find the number of coins of different denominations based on the total value and other given constraints. These problems are practical applications of algebra in real-life situations—in this case, dealing with money.

This type of problem typically involves creating algebraic expressions that represent the value of each coin type and setting up an equation or a system of equations to solve for the unknown quantities. The challenge here is to keep track of the denominations and ensure that the final solution makes cents, quite literally! For example, knowing a dime is worth \(0.10, a quarter \)0.25, and a nickel $0.05, we can convert this into algebraic equations to solve for the quantity of each type of coin in Fred's wallet.

Algebraic Expressions

An algebraic expression is a mathematical phrase that can include numbers, variables, and arithmetic operations. In GED math practice problems, mastering the use of algebraic expressions is crucial to translating words into mathematical language. For instance, stating that Fred has 'three times as many dimes as quarters' can be represented algebraically as d = 3q, where d is the number of dimes and q the number of quarters.

To solve problems, it's important to first define the variables and then create expressions that relate these variables to each other and to the known quantities. Always remember, algebraic expressions are the building blocks for equations; they express relationships without using equality or inequality signs. In our coin problem, combining algebraic expressions based on the value and number of each coin type leads us directly to a solvable system of equations.