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Problem 298

Combinatorics Difficulty 1.6 Multiple choice CEMC Gauss (Grade 7) · Canada · 2024

A circular spinner is divided into 1212 identical unshaded sections and 33 identical shaded sections, as shown.Figure 0Each unshaded section is 33 times
the size of each shaded section. An arrow is attached to the centre of
the spinner. The arrow is spun once. What is the probability that the
arrow stops in a shaded section?

Pick one

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Official solution

Since each unshaded section is 33 times the size of each shaded section, then together the size of 33 shaded sections is equal to the size of 11 unshaded section. Thus, the combined size of all sections is equal to 12+1=1312+1=13 unshaded sections. We can now imagine the spinner as having 1313 equal sized sections of which 11 is shaded. The probability that the arrow stops in a shaded section is equal to the fraction of the spinner’s area comprised of shaded sections, which is 113\frac{1}{13}.

Figure for this problem

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.