A newspaper carrier has $2.80 in change. He has two more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he have? O 00 How many dimes does he have? How many quarters does he have? O * How many nickels does he have? ØTES s
A newspaper carrier has $2.80 in change. He has two more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he have? O 00 How many dimes does he have? How many quarters does he have? O * How many nickels does he have? ØTES s
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Coin Problem
A newspaper carrier has $2.80 in change. He has two more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he have?
---
**Questions:**
- How many dimes does he have?
- [ ]
- How many quarters does he have?
- [ ]
- How many nickels does he have?
- [ ]
---
To solve this problem, set up equations based on the relationships given:
1. Let \( d \) represent the number of dimes.
2. Then, the number of quarters \( q \) is \( d + 2 \).
3. The number of nickels \( n \) is \( 3q \).
The total value equation is:
\[ 0.10d + 0.25q + 0.05n = 2.80 \]
Substitute the values:
\[ 0.10d + 0.25(d + 2) + 0.05(3(d + 2)) = 2.80 \]
Solve this system of equations to find the values of \( d \), \( q \), and \( n \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F543c40ca-509d-4a54-93c4-2ee368b8fc52%2Fa9ead63d-3adc-43c0-8c73-6b1b2eb1df02%2Fxrl8is_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Coin Problem
A newspaper carrier has $2.80 in change. He has two more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he have?
---
**Questions:**
- How many dimes does he have?
- [ ]
- How many quarters does he have?
- [ ]
- How many nickels does he have?
- [ ]
---
To solve this problem, set up equations based on the relationships given:
1. Let \( d \) represent the number of dimes.
2. Then, the number of quarters \( q \) is \( d + 2 \).
3. The number of nickels \( n \) is \( 3q \).
The total value equation is:
\[ 0.10d + 0.25q + 0.05n = 2.80 \]
Substitute the values:
\[ 0.10d + 0.25(d + 2) + 0.05(3(d + 2)) = 2.80 \]
Solve this system of equations to find the values of \( d \), \( q \), and \( n \).
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