CEMC Fermat, 2014

Algebra Difficulty 2.7 Junior Find the answer Canada

A bin contains 10 kg of peanuts. 2 kg of peanuts are removed and 2 kg of raisins are added and thoroughly mixed in. Then 2 kg of this mixture are removed and 2 kg of raisins are added and thoroughly mixed in again. What is the ratio of the mass of peanuts to the mass of raisins in the final mixture?

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Solution

When 2 kg of the 10 kg of peanuts are removed, there are 8 kg of peanuts remaining.

Since 2 kg of raisins are added, then there are 2 kg of raisins in the bin.

The peanuts and raisins are thoroughly mixed.

Since 2 kg of this mixture is removed and this is one-fifth of the total mass, then one-fifth of the mass of peanuts (or $\frac{8}{5}$ kg) is removed and one-fifth of the mass of raisins (or $\frac{2}{5}$ kg) is removed.

This leaves $8 - \frac{8}{5} = \frac{32}{5}$ kg of peanuts, and $2-\frac{2}{5} = \frac{8}{5}$ kg of raisins.

When 2 kg of raisins are added, the mass of raisins becomes $\frac{8}{5}+2=\frac{18}{5}$ kg.

There are $\frac{32}{5}$ kg of peanuts and $\frac{18}{5}$ kg of raisins in the bin.

Therefore, the ratio of the masses is $\frac{32}{5}:\frac{18}{5} = 32:18=16:9$ .

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