Joe has 37 coins consisting of nickels, dimes, and quarters. There are four more nickels than dimes and two more quarters than nickels. What is the total number of quarters that joe has?

  15

Step-by-step explanation:

Let n, d, q represent the numbers of nickels, dimes, and quarters. The problem statement tells us ...

  n +d +q = 37

  n = d +4

  q = n +2

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Rearranging the second equation gives ...

  d = n -4

Substituting that into the first, we get ...

  n + (n -4) +q = 37

  2n +q = 41 . . . . . . . add 4 and simplify

Rearranging the third original equation gives ...

  n = q -2

Substituting into the equation we just made, we get ...

  2(q -2) +q = 41

  3q = 45 . . . . . . . . add 4 and simplify

  q = 15 . . . . . . . . . divide by 3

Joe has 15 quarters.

Check

The number of nickels is 2 fewer, so is 13. The number of dimes is 4 fewer than that, so is 9. The total number of coins is 15 + 13 + 9 = 37, as required.