Maths Olympiad Prep

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, 2012

Algebra Difficulty 1.0 Junior Prove it Canada

John buys 10 bags of apples, each of which contains 20 apples. If he eats 8 apples a day, how many days will it take him to eat the 10 bags of apples?


Determine the value of
sin⁡(0∘)+sin⁡(60∘)+sin⁡(120∘)+sin⁡(180∘)+sin⁡(240∘)+sin⁡(300∘)+sin⁡(360∘)\sin(0^\circ)+\sin(60^\circ)+\sin(120^\circ)+\sin(180^\circ)+\sin(240^\circ)+\sin(300^\circ)+\sin(360^\circ)


A set of integers has a sum of 420, and an average of 60. If one of the integers in the set is 120, what is average of the remaining integers in the set?

Solution

Since John buys 10 bags of apples, each of which contains 20 apples, then he buys a total of 10×20=20010 \times 20 = 200 apples.

Since he eats 8 apples a day, then it takes him 200÷8=25200 \div 8 = 25 days to eat these apples.
Evaluating, sin⁡(0∘)+sin⁡(60∘)+sin⁡(120∘)+sin⁡(180∘)+sin⁡(240∘)+sin⁡(300∘)+sin⁡(360∘)=0+32+32+0+(−32)+(−32)+0=0\begin{aligned} \sin(0^\circ)+\sin(60^\circ)+\sin(120^\circ)+\sin(180^\circ)+\sin(240^\circ)+\sin(300^\circ)+\sin(360^\circ) & = 0 + \tfrac{\sqrt{3}}{2}+\tfrac{\sqrt{3}}{2}+ 0 + (-\tfrac{\sqrt{3}}{2})+(-\tfrac{\sqrt{3}}{2})+0 \\ & = 0\end{aligned} Alternatively, we could notice that sin⁡(60∘)=−sin⁡(300∘)\sin(60^\circ) = -\sin(300^\circ) and $sin⁡(120∘)=−sin⁡(240∘)\$\sin(120^\circ)=-\sin (240^\circ)and and sin⁡(0∘)=sin⁡(180∘)=sin⁡(360∘)=0\sin(0^\circ)=\sin(180^\circ)=\sin(360^\circ)=0,sothesumis, so the sum is 0$.
Since the set of integers has a sum of 420 and an average of 60, then there are 420÷60=7420\div 60 = 7 integers in the set.

Since one integer is 120, then the remaining 6 integers have a sum of 420−120=300420-120=300 and so have an average of 300÷6=50300 \div 6 = 50.

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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.