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CEMC Gauss (Grade 8), 2026

Algebra Difficulty 2.6 Junior Find the answer Canada

The Dollop sells ice cream cones. Each cone sold is exactly one
of three different types: chocolate or vanilla or twist (chocolate and
vanilla combined). Of all cones sold, $\frac{11}{17}$ contained some chocolate
and $\frac{10}{17}$ contained some
vanilla. If $85$ ice cream cones were
sold in total, how many were twist cones?

Pick one

Solutions — 2

Solution 1

Of all $85$ cones sold, $\dfrac{11}{17}\times85=55$ cones contained
some chocolate. Of all $85$ cones
sold, $\dfrac{10}{17}\times85=50$
cones contained some vanilla. The total $55+50=105$ is more than the number of
cones $(85)$ because it counts each
twist cone exactly twice. Thus, $105-85=20$ must equal the number of twist
cones. That is, $20$ cones were
twist, $55-20=35$ cones were
chocolate, and $50-20=30$ cones were
vanilla, as shown in the Venn diagram.

Solution 2

Of all $85$ cones sold, $\dfrac{11}{17}\times85=55$ cones contained
some chocolate.

Of all $85$ cones sold, $\dfrac{10}{17}\times85=50$ cones contained
some vanilla.

The total $55+50=105$ is more than
the number of cones $(85)$ because it
counts each twist cone exactly twice. Thus, $105-85=20$ must equal the number of twist
cones. That is, $20$ cones were
twist, $55-20=35$ cones were
chocolate, and $50-20=30$ cones were
vanilla, as shown in the Venn diagram.

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