2005 CEMC Pascal Problems/Problem 8

Problem

Which of the numbers $-5, \frac{3}{2}, 2, \frac{3}{5}, 8$ is larger than its square?

$\text{ (A) }\ -5 \qquad\text{ (B) }\ \frac{3}{2} \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ \frac{3}{5} \qquad\text{ (E) }\ 8$

Solution 1

We can calculate the squares of the values of the numbers and compare them to the number itself:

$(-5)^2 = 25$ and $(-5) < 25$ , so it is not the answer

$\frac{3}{2} = 1.5$ , and $(\frac{3}{2})^{2} = \frac{3^2}{2^2} = \frac{9}{4} = 2.25$ . We can see that $1.5 < 2.25$ , so it is not the answer

$2^2 = 4$ . We can see that $2 < 4$ , so this is not the answer.

$\frac{3}{5} = 0.6$ and $(\frac{3}{5})^2 = \frac{3^2}{5^2} = \frac{9}{25} = 0.36$ . We can see that $0.6 > 0.36$ , so $\boxed {\textbf {(D) } \frac{3}{5}}$

~anabel.disher

Solution 2 (answer choices)

Without having to calculate the squares of the values of the numbers, we can notice that the square of a number is only less than the number itself when the number that is being squared is between $0$ and $1$ (exclusive).

We can see that $\frac{3}{5}$ is the only number that matches this, as the numerator is less than the denominator. Thus, the answer is $\boxed {\textbf {(D) } \frac{3}{5}}$

~anabel.disher

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2005 CEMC Pascal Problems/Problem 8

Problem

Solution 1

Solution 2 (answer choices)