Problem 49 - Algebra for College Students

Question

Solve each equation. A collection of 70 coins consisting of dimes, quarters, and half-dollars has a value of \(\$ 17.75\). There are three times as many quarters as dimes. Find the number of each kind of coin.

Step-by-Step Solution

Answer

15 dimes, 45 quarters, and 10 half-dollars.

Step 1: Define Variables

Let \( d \) represent the number of dimes, \( q \) represent the number of quarters, and \( h \) represent the number of half-dollars. According to the problem, we know there are three times as many quarters as dimes, so \( q = 3d \).

Step 2: Set Up Equations for Number of Coins

We are given that the total number of coins is 70. Therefore, we can write the equation:\[ d + q + h = 70 \]

Step 3: Set Up Equations for Value of Coins

The total value of the coins is \( \$ 17.75 \), which we need to express in cents for convenience, so 1775 cents. A dime is worth 10 cents, a quarter 25 cents, and a half-dollar 50 cents. The value equation can be written as:\[ 10d + 25q + 50h = 1775 \]

Step 4: Substitute and Simplify

Substitute \( q = 3d \) into both equations:- For the number of coins equation: \[ d + 3d + h = 70 \] Which simplifies to: \[ 4d + h = 70 \]- For the value equation: \[ 10d + 25(3d) + 50h = 1775 \] Simplifying this gives: \[ 10d + 75d + 50h = 1775 \] \[ 85d + 50h = 1775 \]

Step 5: Solve Simultaneous Equations

We have two simplified equations:1. \( 4d + h = 70 \)2. \( 85d + 50h = 1775 \)Solve equation 1 for \( h \):\[ h = 70 - 4d \]Substitute \( h \) in equation 2:\[ 85d + 50(70 - 4d) = 1775 \]\[ 85d + 3500 - 200d = 1775 \]\[ -115d + 3500 = 1775 \]\[ -115d = 1775 - 3500 \]\[ -115d = -1725 \]\[ d = 15 \]

Step 6: Find Number of Quarters and Half-Dollars

Substitute \( d = 15 \) back into the equations to find \( q \) and \( h \):- \( q = 3d = 3 \times 15 = 45 \)- From \( h = 70 - 4d = 70 - 4 \times 15 = 10 \)

Step 7: Verify Solution

Check the results to ensure they satisfy both conditions:- Total coins: \( 15 + 45 + 10 = 70 \)- Total value in cents: \( 10 \times 15 + 25 \times 45 + 50 \times 10 = 1775 \) cents, which is \( \$ 17.75 \).Both conditions are satisfied, confirming the solution.

Key Concepts

coin problems

Coin problems are a common type of word problem in algebra. They involve determining the number of each type of coin given certain conditions. These problems help practice setting up and solving equations, which are essential skills in mathematics.
Coin problems often involve a collection of coins with a total number or total value specified. In this exercise, we were given 70 coins consisting of dimes, quarters, and half-dollars, with a combined value of $17.75.

• To solve coin problems, start by defining variables to represent the number of each type of coin.
• Use the information in the problem to create two main equations: one for the total number of coins and another for their combined value.
• Additional conditions, such as a relationship between the number of coins, help simplify the equations. For instance, knowing the number of quarters is three times the number of dimes was crucial.

Coin problems not only practice algebraic skills but also improve logical thinking through strategic planning in setting up equations.

algebra

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It is fundamental in solving problems such as coin problems. Algebra involves creating equations based on relationships expressed in natural language, and then manipulating those equations to find solutions.
In the given exercise, algebraic methods were applied by first defining variables for dimes (\("d"\)), quarters (\("q"\)), and half-dollars (\("h"\)). Using these variables, equations were formed to reflect the total number of coins and their cumulative monetary value.

• The process starts with understanding the problem and identifying the variables.
• Next, translate the problem into algebraic equations, using given relationships to substitute variables and simplify.
• Expression manipulation, such as combining like terms, is crucial to get simpler forms of the equations which are easier to solve.

Algebra teaches how to model and solve real-world problems, as seen in converting a word problem into solvable mathematical equations.

equation solving

Equation solving is the process of finding the values of variables that satisfy given equations. It plays a central role in problems that require determining unknown quantities. In the exercise above, two equations were developed:

• A total coin equation: \[ d + q + h = 70 \]
• A value of coins equation: \[ 10d + 25q + 50h = 1775 \]

Equation solving involves steps like substitution and simplification. By expressing the number of quarters (\("q"\)) in terms of dimes (\("d"\)), the problem becomes easier to handle due to fewer variables.

• Simplifying these equations reduces complexity and helps visualize the solution.
• Solving involves isolating a variable using substitution or elimination methods.
• Check your solutions by substituting back into original equations to verify if they are correct.

Equation solving is a systematic approach that promotes critical thinking and accuracy in dealing with numerical relationships described in problems.