Olympiad Maths Prep

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Problem 713

AIME late
Algebra Difficulty 5.3 Find the answer

4. Let A=A= the set of red balls, B=B= the set of yellow balls, C=C= the set of green balls, where A,B,CA, B, C are pairwise disjoint.

We have: card⁡A+cardB⁡=17,cardB⁡+card⁡C=29\operatorname{card} A+\operatorname{cardB}=17, \operatorname{cardB}+\operatorname{card} C=29 (2p)

Subtract the first relation from the second and obtain cardC - cardA =12=12. .(1p)

But cardC⁡=2⋅card⁡A⇒card⁡A=12,cardB⁡=5,card⁡C=24\operatorname{cardC}=2 \cdot \operatorname{card} A \Rightarrow \operatorname{card} A=12, \operatorname{cardB}=5, \operatorname{card} C=24 (3p)

## MATHEMATICS OLYMPIAD LOCAL PHASE, 09.02.2013 5th Grade

Official solution

4. (7p) In a box, there are red, yellow, and green balls. Only 17 of them are not green and only 29 are not red. The red ones are half as many as the green ones. Determine the number of balls of each color.

Liviu Cocariu-Ardelean

Note: All questions are mandatory.

Actual working time: 2 hours.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.