**Quote:**

Ken has a group of coins - nickels, dimes and quarters only - worth $6.70. He has four times as many nickels as dimes, and twenty fewer quarters than nickels. How many quarters does he have?

A)19

B)18

C)16

D)15

E)14

Let N be the number of nickels, D the number of dimes, Q the number of quarters and let us consider CENTS as the "problem´s unit".

Doing so, we have:

\(N = 4D\,\,\,\,\, \Rightarrow \,\,\,\,5N = 20D\)

\(Q = N - 20\,\,\,\)

\(? = Q = N - 20\)

\(5N + 10D + 25Q = 670\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,2} \,\,\,10N + 5N + 50\left( {N - 20} \right) = 2 \cdot 670\)

\(65N = 2 \cdot 670 + 1000\, = 20\left( {67 + 50} \right)\,\,\,\,\mathop \Rightarrow \limits^{:\,\,5\,\,} \,\,\,\,13N = 4\left( {9 \cdot 13} \right)\,\,\,\,\,\, \Rightarrow \,\,\,N = 36\)

\(? = 16\,\,\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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