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Alex Ar [27]
3 years ago
9

Joe has 37 coins consisting of nickels, dimes, and quarters. There are four more nickels than dimes and two more quarters than n

ickels. What is the total number of quarters that joe has?
Mathematics
1 answer:
astraxan [27]3 years ago
5 0

Answer:

  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

___

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.

_____

<em>Check</em>

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.

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