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Algebra Difficulty 2.1 Junior Prove it Canada

By finding a common denominator, we see that $\dfrac{1}{3}$ is greater than $\dfrac{1}{7}$ because $\dfrac{7}{21}>\dfrac{3}{21}$ . Similarly, we see that $\dfrac{1}{3}$ is less than $\dfrac{1}{2}$ because $\dfrac{2}{6}<\dfrac{3}{6}$ . Determine the integer $n$ so that $\dfrac{n}{40}$ is greater than $\dfrac{1}{5}$ and less than $\dfrac{1}{4}$ . Determine all possible integers $m$ so that $\dfrac{m}{8}$ is greater than $\dfrac{1}{3}$ and $\dfrac{m+1}{8}$ is less than $\dfrac{2}{3}$ . Fiona calculates her win ratio by dividing the number of games that she has won by the total number of games that she has played. At the start of a weekend, Fiona has played 30 games, has $w$ wins, and her win ratio is greater than 0.5. During the weekend, she plays five games and wins three of these games. At the end of the weekend, Fiona’s win ratio is less than 0.7. Determine all possible values of $w$ .

Solution

Chords $DE$ and $FG$ intersect at $X$ , and so $(DX)(EX)=(FX)(GX)$ or $(DX)(8)=(6)(4)$ . Solving this equation for $DX$ , we get $DX=\dfrac{(6)(4)}{8}=\dfrac{24}{8}=3$ . The length of $DX$ is 3. Chords $JK$ and $LM$ intersect at $X$ , and so $(JX)(KX)=(LX)(MX)$ or $(8y)(10)=(16)(y+9)$ . Solving this equation, we get $80y=16y+144$ or $64y=144$ , and so $y=\dfrac{144}{64}=\dfrac{9}{4}$ . Chords $PQ$ and $ST$ intersect at $U$ , and so $(PU)(QU)=(SU)(TU)$ . Since $TU=TV+UV$ , then $TU=6+n$ . Substituting values into $(PU)(QU)=(SU)(TU)$ , we get $(m)(5)=(3)(6+n)$ , and so $5m=18+3n$ . Chords $PR$ and $ST$ intersect at $V$ , and so $(PV)(RV)=(TV)(SV)$ . Since $SV=SU+UV$ , then $SV=3+n$ . Substituting values into $(PV)(RV)=(TV)(SV)$ , we get $(n)(8)=(6)(3+n)$ , and so $8n=18+6n$ or $2n=18$ or $n=9$ . Substituting $n=9$ into $5m=18+3n$ , we get $5m=18+3(9)$ or $5m=45$ , and so $m=9$ . Therefore, $m=9$ and $n=9$ .

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