Steve answered this for you last night.
http://www.jiskha.com/display.cgi?id=1432686891
At the end of Week 5 of my basketball team's season, our record is 3 wins and 17 losses. Each week after that, we win 3 games and lose 1 game. At the end of which week will my team have won at least 65% of its games total for the first time in the season?
3 answers
steve is wrong... the answer is 30
Let $x$ be the number of weeks we play after week 5, at which point we have won $3+ 3x$ games (3 in the first 5 weeks plus 3 per week afterwards) out of $20+ 4x$ (20 in the first 5 weeks plus 4 per week afterwards). So, we must have
\[\frac{3+3x}{20+4x} \ge \frac{65}{100}.\]Simplifying the right-hand side gives
\[\frac{3+3x}{20+4x} \ge \frac{13}{20}.\]Multiply both sides by 4 gives
\[\frac{3+3x}{5+x} \ge \frac{13}{5}.\]Multiplying both sides by $5(5+x)$ -- which is positive, so we do not have to worry about the direction of the inequality -- gives
\[5(3+3x) \ge 13(5+x),\]so \[15+15x \ge 65+13x,\]or $2x \ge 50$. Therefore, $x \ge 25$.
Checking our work, we see that after 25 additional weeks, we have $3 + 25\cdot 3 = 78$ wins out of $20 + 4\cdot 25=120$ games, and $\frac{78}{120} = \frac{13}{20} = 65\%$. Since we play for 25 weeks after Week 5, we hit the $65\%$ mark at the end of Week $\boxed{30}$.
Let $x$ be the number of weeks we play after week 5, at which point we have won $3+ 3x$ games (3 in the first 5 weeks plus 3 per week afterwards) out of $20+ 4x$ (20 in the first 5 weeks plus 4 per week afterwards). So, we must have
\[\frac{3+3x}{20+4x} \ge \frac{65}{100}.\]Simplifying the right-hand side gives
\[\frac{3+3x}{20+4x} \ge \frac{13}{20}.\]Multiply both sides by 4 gives
\[\frac{3+3x}{5+x} \ge \frac{13}{5}.\]Multiplying both sides by $5(5+x)$ -- which is positive, so we do not have to worry about the direction of the inequality -- gives
\[5(3+3x) \ge 13(5+x),\]so \[15+15x \ge 65+13x,\]or $2x \ge 50$. Therefore, $x \ge 25$.
Checking our work, we see that after 25 additional weeks, we have $3 + 25\cdot 3 = 78$ wins out of $20 + 4\cdot 25=120$ games, and $\frac{78}{120} = \frac{13}{20} = 65\%$. Since we play for 25 weeks after Week 5, we hit the $65\%$ mark at the end of Week $\boxed{30}$.
Let $x$ be the number of weeks we play after week 5, at which point we have won $3+ 3x$ games (3 in the first 5 weeks plus 3 per week afterwards) out of $20+ 4x$ (20 in the first 5 weeks plus 4 per week afterwards). So, we must have
\[\frac{3+3x}{20+4x} \ge \frac{65}{100}.\]Simplifying the right-hand side gives
\[\frac{3+3x}{20+4x} \ge \frac{13}{20}.\]Multiply both sides by 4 gives
\[\frac{3+3x}{5+x} \ge \frac{13}{5}.\]Multiplying both sides by $5(5+x)$ -- which is positive, so we do not have to worry about the direction of the inequality -- gives
\[5(3+3x) \ge 13(5+x),\]so\[15+15x \ge 65+13x,\]or $2x \ge 50$. Therefore, $x \ge 25$.
Checking our work, we see that after 25 additional weeks, we have $3 + 25\cdot 3 = 78$ wins out of $20 + 4\cdot 25=120$ games, and $\frac{78}{120} = \frac{13}{20} = 65\%$. Since we play for 25 weeks after Week 5, we hit the $65\%$ mark at the end of Week $\boxed{30}$.
\[\frac{3+3x}{20+4x} \ge \frac{65}{100}.\]Simplifying the right-hand side gives
\[\frac{3+3x}{20+4x} \ge \frac{13}{20}.\]Multiply both sides by 4 gives
\[\frac{3+3x}{5+x} \ge \frac{13}{5}.\]Multiplying both sides by $5(5+x)$ -- which is positive, so we do not have to worry about the direction of the inequality -- gives
\[5(3+3x) \ge 13(5+x),\]so\[15+15x \ge 65+13x,\]or $2x \ge 50$. Therefore, $x \ge 25$.
Checking our work, we see that after 25 additional weeks, we have $3 + 25\cdot 3 = 78$ wins out of $20 + 4\cdot 25=120$ games, and $\frac{78}{120} = \frac{13}{20} = 65\%$. Since we play for 25 weeks after Week 5, we hit the $65\%$ mark at the end of Week $\boxed{30}$.
4 answers