1985 AJHSME Problems/Problem 14

Revision as of 18:37, 13 January 2009 by Waffle (talk | contribs) (New page: ==Solution== The most straightforward method would be to calculate both prices, and subtract. But there's a better method... Before we start, it's always good to convert the word problem...)
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Solution

The most straightforward method would be to calculate both prices, and subtract. But there's a better method...

Before we start, it's always good to convert the word problems into formulas, or at least expressions, we can solve.

So we know that the price of the object after a 6.5% increase will be $20 \times 6.5%$ (Error compiling LaTeX. Unknown error_msg), and the price of it after a 6% increase will be $20 \times 6$. And what we're trying to find is $6.5% \times 20 - 6 \times 20$ (Error compiling LaTeX. Unknown error_msg), and if you have experience in the field of algebra, you'll notice that both of the items have a common factor, 20, and we can factor the expression into...

$(6.5% - 6%) \times 20$ (Error compiling LaTeX. Unknown error_msg), which simplifies to $.5%$ (Error compiling LaTeX. Unknown error_msg). We know that $.5 = \frac{1}{20}$, so $.5% = \frac{1}{2000}$ (Error compiling LaTeX. Unknown error_msg). $20 \times \frac{1}{2000} = \frac{1}{100} = .01$, so the answer is 1 cent.

1 cent = $$.01$ = (D)