1999 CEMC Gauss (Grade 7) Problems/Problem 15

Problem

A box contains $36$ pink, $18$ blue, $9$ green, $6$ red, and $3$ purple cubes that are identical in size. If a cube is selected at random, what is the probability that it is green?

$\text{(A)}\ \frac{1}{9} \qquad \text{(B)}\ \frac{1}{8} \qquad \text{(C)}\ \frac{1}{5} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{9}{70}$

Solution 1

To find the probability that the cube is green, we can divide the number of cubes that are green by the total number of cubes.

The total number of cubes is $36 + 18 + 9 + 6 + 3$ = $72$ .

This means that the probability that the cube selected is green is:

$\frac{9}{72} = \boxed{\textbf {(B)} \frac{1}{8}}$

Solution 2 (answer choices)

As found in solution 1, the total number of cubes is $72$ . This means that the denominator of the results must be a factor of $72$ , eliminating C and E.

We can then multiply the other answer choices by $72$ to check if we get the number of green cubes stated in the original problem. This gives $\boxed{\textbf {(B)} \frac{1}{8}}$ as the answer.

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1999 CEMC Gauss (Grade 7) Problems/Problem 15

Problem

Solution 1

Solution 2 (answer choices)