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A box contains 78 coins, only dimes and nickels. The amount of money in the box is $4.90. How many dimes and how many nickels are in the box? The number of dimes is The number of nickels is

A box contains 78 coins, only dimes and nickels. The amount of money in the box is $4.90. How many dimes and how many nickels are in the box? The number of dimes is The number of nickels is

Algebra and Trigonometry (6th Edition)
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 Statement A box contains 78 coins, only dimes and nickels. The amount of money in the box is $4.90. How many dimes and how many nickels are in the box? --- ### Solution 1. Formulating Equations: - Let \( d \) be the number of dimes. - Let \( n \) be the number of nickels. 2. Given Information: - Total number of coins: \( d + n = 78 \) - Total value of coins: \( 0.10d + 0.05n = 4.90 \) 3. Solving for \( d \) and \( n \): - From equation 1: \( d + n = 78 \) \( \Rightarrow n = 78 - d \) - Substitute \( n \) in equation 2: \( 0.10d + 0.05(78 - d) = 4.90 \) - Simplify: \( 0.10d + 3.90 - 0.05d = 4.90 \) - \( 0.05d + 3.90 = 4.90 \) - \( 0.05d = 1.00 \) - \( d = 20 \) 4. Finding \( n \): - Substitute \( d \) back in \( n = 78 - d \) - \( n = 78 - 20 \) - \( n = 58 \) ### Conclusion - **The number of dimes is:** \( \boxed{20} \) - **The number of nickels is:** \( \boxed{58} \)
Transcribed Image Text:### Problem Statement A box contains 78 coins, only dimes and nickels. The amount of money in the box is $4.90. How many dimes and how many nickels are in the box? --- ### Solution 1. Formulating Equations: - Let \( d \) be the number of dimes. - Let \( n \) be the number of nickels. 2. Given Information: - Total number of coins: \( d + n = 78 \) - Total value of coins: \( 0.10d + 0.05n = 4.90 \) 3. Solving for \( d \) and \( n \): - From equation 1: \( d + n = 78 \) \( \Rightarrow n = 78 - d \) - Substitute \( n \) in equation 2: \( 0.10d + 0.05(78 - d) = 4.90 \) - Simplify: \( 0.10d + 3.90 - 0.05d = 4.90 \) - \( 0.05d + 3.90 = 4.90 \) - \( 0.05d = 1.00 \) - \( d = 20 \) 4. Finding \( n \): - Substitute \( d \) back in \( n = 78 - d \) - \( n = 78 - 20 \) - \( n = 58 \) ### Conclusion - **The number of dimes is:** \( \boxed{20} \) - **The number of nickels is:** \( \boxed{58} \)
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