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Coin Problems

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As with most algebra 'word problems', let's start by looking at what we know.
First, we are told that Tony has 11 more nickels than quarters. We can call the number of quarters Tony has We're also told that Tony has $2.65. To solve the problem, we need to know that a dollar is 100 cents so $2.65 = 265 cents, a quarter is worth 25 cents (a quarter of a dollar), and a nickel is worth 5 cents. We can now use the following equation: 25 cents x no. of quarters + 5 cents x no. of nickels = 265 cents Translating this into symbols: 25q + 5(q +11) = 265 25q + 5q + 55 = 265 (distributing) 30q + 55 = 265 (adding like terms) 30q = 210 (subtracting 55 from each side) q = 7Since q = the number of quarters Tony has, this means that Tony has 7 quarters. Since we also know that Tony has 11 more nickels than quarters, he must have
Thus Tony has a total of
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