A collection of dimes and quarters has a total value of five dollars and contains 29 coins.

A collection of dimes and quarters has a total value of five dollars and contains 29 coins. how many dimes are there in the collection?

The Correct answer and Explanation is:

This problem involves solving a system of equations to determine how many dimes and quarters make up the collection. Let’s break it down step by step:

Step 1: Define Variables

Let:

  • ddd represent the number of dimes in the collection.
  • qqq represent the number of quarters in the collection.

Step 2: Set Up Equations

  1. The total number of coins is 29: d+q=29d + q = 29d+q=29
  2. The total value of the coins is 5 dollars (or 500 cents, since dealing with cents makes calculations simpler). Since each dime is worth 10 cents and each quarter is worth 25 cents, the total value equation can be expressed as: 10d+25q=50010d + 25q = 50010d+25q=500

Step 3: Solve the System of Equations

We have two equations:

  1. d+q=29d + q = 29d+q=29
  2. 10d+25q=50010d + 25q = 50010d+25q=500

Solve the first equation for ddd:

d=29−qd = 29 – qd=29−q

Substitute this into the second equation:

10(29−q)+25q=50010(29 – q) + 25q = 50010(29−q)+25q=500

Simplify the equation:290−10q+25q=500290 – 10q + 25q = 500290−10q+25q=500 290+15q=500290 + 15q = 500290+15q=500 15q=500−29015q = 500 – 29015q=500−290 15q=21015q = 21015q=210 q=21015=14q = \frac{210}{15} = 14q=15210​=14

Find ddd:

Substitute q=14q = 14q=14 back into the equation d+q=29d + q = 29d+q=29:d+14=29d + 14 = 29d+14=29 d=29−14=15d = 29 – 14 = 15d=29−14=15

Step 4: Conclusion

There are 15 dimes in the collection.

Explanation

We used algebra to solve a system of equations based on two key conditions: the total number of coins and their total value. By substituting one equation into the other, we simplified the problem to a single variable, which allowed us to determine the number of quarters and then the number of dimes. These types of problems are common in algebra and teach important concepts like equation substitution and solving linear systems. The key to solving this was setting up the correct relationships based on the problem’s information and then using substitution to find the answer.

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