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Chapter 3: Problem 17

An application of three equations in three unknowns: A bag of coins contains nickels, dimes, and quarters. There are a total of 21 coins in the bag, and the total amount of money in the bag is \(\$ 3.35\). There is one more dime than there are nickels. How many dimes, nickels, and quarters are in the bag?

Short Answer

Expert verified
There are 5 nickels, 6 dimes, and 10 quarters in the bag.

Step by step solution

01

Assign Variables

Let \( n \) be the number of nickels, \( d \) be the number of dimes, and \( q \) be the number of quarters.
02

Set Up Equations from Problem Statements

From the problem, we have the following equations: 1. The total number of coins: \( n + d + q = 21 \). 2. The relation between nickels and dimes: \( d = n + 1 \). 3. The total value of the coins: \( 0.05n + 0.10d + 0.25q = 3.35 \).
03

Substitute

Substitute the equation from Step 2 that relates \( d \) and \( n \) into the other equations. So replace \( d \) with \( n + 1 \) in the total number of coins and the total value equations.New equations: - \( n + (n + 1) + q = 21 \) simplifies to \( 2n + q = 20 \).- \( 0.05n + 0.10(n + 1) + 0.25q = 3.35 \).
04

Simplify the Value Equation

Distribute and simplify the value equation from the previous step: \( 0.05n + 0.10n + 0.10 + 0.25q = 3.35 \)Combine like terms: \( 0.15n + 0.25q = 3.25 \) by subtracting 0.1 from both sides.
05

Solve the System of Equations

Use the two simplified equations: 1. \( 2n + q = 20 \)2. \( 0.15n + 0.25q = 3.25 \)From the first equation, solve for \( q \): \( q = 20 - 2n \). Substitute \( q = 20 - 2n \) into the second equation: \( 0.15n + 0.25(20 - 2n) = 3.25 \).
06

Simplify and Solve for One Variable

Simplify the equation from the substitution: \( 0.15n + 5 - 0.5n = 3.25 \).Combine terms to get \( -0.35n + 5 = 3.25 \).Subtract 5 from both sides: \( -0.35n = -1.75 \).Divide by -0.35 to solve for \( n \): \( n = 5 \).
07

Find Values of Other Variables

Now substitute \( n = 5 \) back into the equations:- For \( d \): \( d = n + 1 = 5 + 1 = 6 \).- For \( q \): \( q = 20 - 2n = 20 - 2(5) = 10 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Modeling
Algebraic modeling is a powerful tool used to describe real-world problems through mathematical expressions. In this coin problem, algebraic modeling allows us to convert the given conditions into mathematical equations, making the problem structured and solvable. First, we assign variables to the unknown quantities:
  • Let \( n \) represent the number of nickels.
  • Let \( d \) represent the number of dimes.
  • Let \( q \) represent the number of quarters.
Next, we translate the conditions of the problem into equations:
1. The total number of coins equals 21, modeled as: \[ n + d + q = 21 \].
2. There's one more dime than nickels, represented by: \[ d = n + 1 \].
3. The total value of all coins sums up to $3.35, given by: \[ 0.05n + 0.10d + 0.25q = 3.35 \].
Algebraic modeling breaks down complex situations into manageable mathematical equations, laying the groundwork for finding precise solutions.
Three Variables
In this problem, we deal with three variables, each representing a different type of coin: nickels, dimes, and quarters. Managing multiple variables can seem daunting, but breaking it down can simplify the process.
The key to handling three variables is utilizing substitution and elimination techniques. Here, we applied substitution by expressing the number of dimes \( d \) as \[ n + 1 \]. By substituting this expression in other equations, we eliminated one variable, reducing complexity.
Step by step, combining \( n + (n+1) + q = 21 \) simplifies to \[ 2n + q = 20 \]. Meanwhile, substituting \( d \) into the value equation gives \[ 0.05n + 0.10(n + 1) + 0.25q = 3.35 \], leading us to \[ 0.15n + 0.25q = 3.25 \].
By reducing two variables to one in the later stages, it becomes feasible to solve the equations systematically. Handling multiple variables with caution and planning is crucial in unraveling solutions efficiently.
Coin Problem
This coin problem is an example of a practical application of systems of equations in algebra. It reflects a real-life situation where you determine the quantity of each type of coin given certain constraints.
The steps involve understanding the problem constraints thoroughly, setting up the equations, and then solving them using algebraic techniques. Here's the summary of our approach:
  • Established the total number of coins, the relation between dimes and nickels, and the total monetary value.
  • Used substitution to simplify the system to two equations involving two variables.
  • Solved for one variable by further simplification, then found the others through substitution.
In our solution:
  • By solving \[ -0.35n = -1.75 \], we found \( n = 5 \) (nickels).
  • Substituting back, we calculated \( d = 6 \) (dimes) and \( q = 10 \) (quarters).
Through this systematic approach, a seemingly complex problem becomes easy to handle, demonstrating the power of algebraic methods in problem solving.

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Most popular questions from this chapter

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