2003 CEMC Pascal Problems/Problem 6

Revision as of 13:54, 20 June 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== Which of the following is ''not'' equal to <math>\frac{18}{5}</math>? <math> \text{ (A) }\ \frac{6^2}{10} \qquad\text{ (B) }\ \frac{1}{5}[6(3)]) \qquad\text{ (C)...")
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Problem

Which of the following is not equal to $\frac{18}{5}$?

$\text{ (A) }\ \frac{6^2}{10} \qquad\text{ (B) }\ \frac{1}{5}[6(3)]) \qquad\text{ (C) }\ \frac{18 + 1}{15 + 1} \qquad\text{ (D) }\ 3.6 \qquad\text{ (E) }\ \sqrt{\frac{324}{5}}$

Solution

We can simply try answer choices until we get one that isn't equal to $\frac{18}{5}$:

$\frac{6^2}{10} = \frac{36}{10} = \frac{36 \div 2}{10 \div 2} = \frac{18}{5}$

$\frac{1}{5}[6(3)]) = \frac{1}{5} \times 18 = \frac{18}{5}$

$\frac{18 + 1}{15 + 1} = \frac{19}{16} \neq \frac{18}{5}$

This shows us that the answer is $\boxed {\textbf {(C) } \frac{18 + 1}{15 + 1}}$.

We can also verify that the others are equal to $\frac{18}{5}$:

$3.6 = \frac{36}{10} = \frac{18}{5}$

From $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we have:

$\sqrt{\frac{324}{25}} = \frac{\sqrt{324}}{\sqrt{25}} = \frac{18}{5}$

~anabel.disher