Nicole has 16 nickels and dimes. if the value of her coins is 1.05, how many of each coin does she have?

Step-by-step explanation:

If all were dimes, she would have $1.60, which is $0.55 more than she actually has. Replacing a dime with a nickel reduces the total by .05, so there must be .55/.05 = 11 such replacements. This is the number of nickels. The number of dimes makes up the rest of the 16 coins.

Nicole has 11 nickels and 5 dimes.

You can write equations for this several ways. My personal favorite is to use a single variable to represent the number of highest-value coin: d = number of dimes. Then the number of nickels is 16-d and the total value is ...

  0.10d + 0.05(16 -d) = 1.05

  0.05d + 0.80 = 1.05 . . . eliminate parentheses, collect terms

  0.05d = 0.25 . . . . . . . . . subtract 0.80

  d = 0.25/0.05 = 5 . . . . . number of dimes

  16-d = 16-5 = 11 . . . . . . . .number of nickels

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If you let the variable match the number of nickels, the solution looks a lot like the one described in words, above.

  0.05n + 0.10(16 -n) = 1.05

  -0.05n +1.60 = 1.05

  -0.05n = -0.55 . . . . . . . subtract 1.60

  n = -0.55/-0.05 = 11 . . . number of nickels

  16-n = 16-11 = 5 . . . . . . . number of dimes

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Comparing these two solutions, you see that we work with negative numbers in the second solution. The reason for choosing the variable to represent the higher-value coin is to keep the numbers positive, as they are in the first solution. (It's no big deal, if you're not bothered by negative number arithmetic.)

it would take 7/10 of an hour too fill the pool.

step-by-step explanation: