0. A coin jar contains 425 coins all are pennies, nickels and dimes (1c, 5cand 10c) with total value $21.12. If there are 100 more nickels than dimes, find the nnumber of each types of coins in the jar.
0. A coin jar contains 425 coins all are pennies, nickels and dimes (1c, 5cand 10c) with total value $21.12. If there are 100 more nickels than dimes, find the nnumber of each types of coins in the jar.
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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![**Problem 10:**
A coin jar contains 425 coins, all of which are pennies, nickels, and dimes (1¢, 5¢, and 10¢ respectively) with a total value of $21.12. If there are 100 more nickels than dimes, find the number of each type of coin in the jar.
---
**Explanation:**
To solve this problem, we can set up equations based on the information given:
1. Let \( p \) be the number of pennies, \( n \) be the number of nickels, and \( d \) be the number of dimes.
2. The total number of coins equation:
\[
p + n + d = 425
\]
3. The total value equation in cents:
\[
1p + 5n + 10d = 2112 \quad (\text{since $21.12 = 2112 cents})
\]
4. The relationship between nickels and dimes:
\[
n = d + 100
\]
Using these equations, the problem can be solved using algebraic methods to find the values of \( p \), \( n \), and \( d \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4661dd34-4628-4619-9963-a3eb70f9d6a4%2Ffc6358d1-6e4b-4136-95a6-5b575b8aac63%2F0ky5i7o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 10:**
A coin jar contains 425 coins, all of which are pennies, nickels, and dimes (1¢, 5¢, and 10¢ respectively) with a total value of $21.12. If there are 100 more nickels than dimes, find the number of each type of coin in the jar.
---
**Explanation:**
To solve this problem, we can set up equations based on the information given:
1. Let \( p \) be the number of pennies, \( n \) be the number of nickels, and \( d \) be the number of dimes.
2. The total number of coins equation:
\[
p + n + d = 425
\]
3. The total value equation in cents:
\[
1p + 5n + 10d = 2112 \quad (\text{since $21.12 = 2112 cents})
\]
4. The relationship between nickels and dimes:
\[
n = d + 100
\]
Using these equations, the problem can be solved using algebraic methods to find the values of \( p \), \( n \), and \( d \).
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