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 ::
GMATH method creator (Math for the GMAT)
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