Answer:
Step-by-step explanation:
This sort of problem can be solved by writing a system of equations, one for each of the given relations. You are given a total number of coins, and a total of their value.
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Let q and n represent the numbers of quarters and nickels, respectively. The given relations can be written ...
q + n = 13 . . . . . . number of coins
0.25q + 0.05n = 2.45 . . . . . . value of coins
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It is usually convenient to eliminate the variable representing the smaller value coin. We can do this by using the first equation to write an expression for n that substitutes into the second equation:
n = 13 -q . . . . . . . . . . . . . expression for n
0.25q +0.05(13 -q) = 2.45 . . . . substitute for n
0.20q = 1.80 . . . . . . . . . subtract 0.65 and collect terms
q = 9 . . . . . . . . . . . . divide by 0.20; number of quarters
n = 13 -9 = 4 . . . . number of nickels
Calvin has 9 quarters and 4 nickels.
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Additional comment
Once you see how this substitution works, you can skip the need to write the equation for n, and jump right to the equation after substitution:
0.25q +0.05(13 -q) = 2.45 . . . . . where 13-q is the number of nickels
And, you can actually go further toward solving this mentally by recognizing the next step subtracts the value of 13 nickels from the total value. That difference in values is divided by the difference in coin value to give the number of higher-value coins.
When we describe this in words, we say ...
If all the coins were nickels, the value would be $0.65. The actual value is $2.45-0.65 = $1.80 more than that. Replacing a nickel by a quarter adds $0.20 to the value of the coins, so there must have been $1.80/0.20 = 9 such replacements. There are 9 quarters and 4 nickels.
