个
Question 13 of 15
This test: 15 points) possible
This question: 1 points) possible
Submit test
A vending machine accepts nickels, dimes, and quariers. Exact change is nooded to make a purchase. How many ways can a person with five nickels, three dimes, and two quarters make a 80 cent purchase from the machina? For this problom, treat all coins within a particular value as if they are the same. That is, do not count the number of ways that a single coln (a nickel, a dime, or a quarter) could be chosen.
There are
ways to make the purchase,

We are asked to determine how many ways a person can make an 80-cent purchase using exactly five nickels, three dimes, and two quarters, considering all coins within a particular value to be identical.

### Step 1: Write down the values of the coins.
- Each **nickel** is worth 5 cents.
- Each **dime** is worth 10 cents.
- Each **quarter** is worth 25 cents.

### Step 2: Define variables.
Let:
- \( x \) be the number of nickels used.
- \( y \) be the number of dimes used.
- \( z \) be the number of quarters used.

We want the total value of the coins used to be 80 cents:
\[
5x + 10y + 25z = 80
\]
Additionally, the constraints are:
- \( 0 \leq x \leq 5 \) (you have at most 5 nickels),
- \( 0 \leq y \leq 3 \) (you have at most 3 dimes),
- \( 0 \leq z \leq 2 \) (you have at most 2 quarters).

### Step 3: Solve for possible combinations of \( x \), \( y \), and \( z \).
We now systematically check for different values of \( z \) (quarters) and solve for the corresponding \( x \) (nickels) and \( y \) (dimes).

#### Case 1: \( z = 2 \)
If you use 2 quarters, the value from the quarters is:
\[
25 \times 2 = 50 \text{ cents}.
\]
This leaves:
\[
80 - 50 = 30 \text{ cents}.
\]
Now solve \( 5x + 10y = 30 \):
- If \( y = 0 \), \( 5x = 30 \Rightarrow x = 6 \) (not possible because \( x \leq 5 \)).
- If \( y = 1 \), \( 5x + 10 = 30 \Rightarrow 5x = 20 \Rightarrow x = 4 \) (valid).
- If \( y = 2 \), \( 5x + 20 = 30 \Rightarrow 5x = 10 \Rightarrow x = 2 \) (valid).
- If \( y = 3 \), \( 5x + 30 = 30 \Rightarrow 5x = 0 \Rightarrow x = 0 \) (valid).

Thus, for \( z = 2 \), the valid combinations are:
\[
(4, 1, 2), (2, 2, 2), (0, 3, 2).
\]

#### Case 2: \( z = 1 \)
If you use 1 quarter, the value from the quarter is:
\[
25 \times 1 = 25 \text{ cents}.
\]
This leaves:
\[
80 - 25 = 55 \text{ cents}.
\]
Now solve \( 5x +

A vending machine accepts nickels, dimes, and quarters. Exact change is needed to make a purchase. How many ways can a person with five nickels, three dimes, and two quarters make an 80 cent purchase from the machine? For this problem, treat all coins within a particular value as if they are the same. That is, do not count the number of ways that a single coin (a nickel, a dime, or a quarter) could be chosen.

Explore how to solve the problem of making an 80-cent purchase using a combination of nickels, dimes, and quarters. This step-by-step solution covers combinatorics, algebra, and the use of linear equations to find all valid coin combinations. Ideal for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation