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On this year's Westchester basketball team, the players are all either
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05 Sep 2018, 09:43
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70% (02:27) correct 30% (01:51) wrong based on 56 sessions
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On this year's Westchester basketball team, the players are all either 5,7,or 11 years of age. If the product of ages of the players on the team is 18,865 , then what is the probability that a randomly selected team member will not be 7 A \(\frac{3}{7}\) B \(\frac{2}{5}\) C \(\frac{16}{37}\) D \(\frac{3}{5}\) E \(\frac{49}{55}\)
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On this year's Westchester basketball team, the players are all either
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Updated on: 06 Sep 2018, 08:42
This is a time consuming but not impossible question. Here is how I would solve this question. Understand the Question  Behind this word problem lurks a pretty standard Number Properties / Primes and divisibility question. The core skill being tested is the ability to prime factor a number, in this case 18,865. The good news is that they tell you the three factors  5, 7 and 11. One of these numbers, 5, is easy to factor out. One of these numbers, 11, has a little remembered rule. Lastly, one of these numbers, 7, has a super complicated algorithm that is better off ignored. So, a simple restatement of this question is "How many factors of 5, 7, and 11 are in 18,865?" Understand the Answer Choices  The denominator of the whichever answer choice is correct will at maximum be equal to the total number of 5s, 7s, and 11s in 18,865  the numerator will be the number of 5s and 11s. This is from the probability equation = (total number of desired outcomes)/(total number of outcomes). This makes answer choices C and E pretty unlikely as the denominators are way too high. Eliminate. Plan  what's the fastest way to attack this question? Well, you know by the question that there is at least one 5, 7, and 11. So, dividing by these numbers is a reasonable way to start. I would divide by 5 or 11 first as these are faster. If you happen to be slow at division, guessing among A, B, and D is also reasonable. Solve  18865 / 5 = 3,773. No more factors of 5 (Do you know why?)(Does not end in 0 or 5) 3,773 / 11 = 343 No more factors of 11 (Do you know why?) (the outside numbers do not add to the inside number) 343 / 7 = 49 49 = 7 * 7 So the prime factorization of 18,865 is 5*7*7*7*11 Eliminate A as 7 will not be the denominator Desired outcomes = 2 (one 5 and one 11) Total Outcomes = 5 B is correct Jayson Beatty Indigo Prep
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Re: On this year's Westchester basketball team, the players are all either
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06 Sep 2018, 00:09
jaysonbeatty12 wrote: This is a time consuming but not impossible question. Here is how I would solve this question.
Understand the Question  Behind this word problem lurks a pretty standard Number Properties / Primes and divisibility question. The core skill being tested is the ability to prime factor a number, in this case 18,865. The good news is that they tell you the three factors  5, 7 and 11. One of these numbers, 5, is easy to factor out. One of these numbers, 11, has a little remembered rule. Lastly, one of these numbers, 7, has a super complicated algorithm that is better off ignored. So, a simple restatement of this question is "How many factors of 5, 7, and 11 are in 18,865?"
Understand the Answer Choices  The denominator of the whichever answer choice is correct will at maximum be equal to the total number of 5s, 7s, and 11s in 18,865  the numerator will be the number of 5s and 11s. This is from the probability equation = (total number of desired outcomes)/(total number of outcomes). This makes answer choices C and E pretty unlikely as the denominators are way too high. Eliminate.
Plan  what's the fastest way to attack this question? Well, you know by the question that there is at least one 5, 7, and 11. So, dividing by these numbers is a reasonable way to start. I would divide by 5 or 7 first as these are faster. If you happen to be slow at division, guessing among A, B, and D is also reasonable.
Solve  18865 / 5 = 3,773. No more factors of 5 (Do you know why?)(Does not end in 0 or 5) 3,773 / 11 = 343 No more factors of 11 (Do you know why?) (the outside numbers do not add to the inside number) 343 / 7 = 49 49 = 7 * 7
So the prime factorization of 18,865 is 5*7*7*7*11 Eliminate A as 7 will not be the denominator Desired outcomes = 2 (one 5 and one 7) Total Outcomes = 5 B is correct
Jayson Beatty Indigo Prep The solution is correct , there is however a small typo , it should be 11 instead of 7



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Re: On this year's Westchester basketball team, the players are all either
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06 Sep 2018, 09:57
CounterSniper wrote: On this year's Westchester basketball team, the players are all either 5,7,or 11 years of age. If the product of ages of the players on the team is 18,865 , then what is the probability that a randomly selected team member will not be 7?
A \(\frac{3}{7}\) B \(\frac{2}{5}\) C \(\frac{16}{37}\) D \(\frac{3}{5}\) E \(\frac{49}{55}\)
\(? = P\left( {{\text{choose}}\,{\text{at}}\,{\text{random}}\,\,{\text{not}}\,{\text{a}}\,\,7{\text{y}}\,{\text{player}}} \right)\) There are only 2 real "problems" here: 1. To factorize 18,865 quickly to get: \(18865 = 5 \cdot {7^3} \cdot 11\) 2. To know how many players are in a basketball team: 5 (*) so that we may conclude that: (*) https://www.google.com.br/search?q=numb ... e&ie=UTF8There are one 5y , three 7y and one 11y players. Answer: 2/5. To factorize 18,865 I suggest: (a) Obviously divisible by 5, to get (in less than a minute): \(\frac{{18865}}{5} = \frac{{18 \cdot 1000 + 500 + 300 + 50 + 15}}{5} = 18 \cdot 200 + 100 + 60 + 10 + 3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,18865 = 5 \cdot 3773\) (b) 3773 is OBVIOUSLY divisible by 11 (because 37+73 = 0) :: check this: https://www.math.hmc.edu/funfacts/ffiles/10013.5.shtmlHence: \(\frac{{3773}}{{11}} = \frac{{3300 + 33 + 44 \cdot 10}}{{11}} = 300 + 3 + 40 = 343\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,3773 = 11 \cdot 343\,\,\,\) (c) Now the hard part (how would we imagine 343 is divisible by 7?)... but the examiner gave us the hint(only 5y, 11y... and 7y players!): Hence: \(\frac{{343}}{7} = \frac{{350  7}}{7} = 49\,\,\,\,\,\, \Rightarrow \,\,\,\,\,3773 = 7 \cdot 49 = {7^3}\,\,\) Finally, let´s group all together: \(18865 = 5 \cdot 11 \cdot {7^3}\) This powerful breaking numbers technique is part of our method, by the way! This solution follows the notations and rationale taught in the GMATH method. Regards, fskilnik.
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Re: On this year's Westchester basketball team, the players are all either
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11 Sep 2018, 07:56
This is a time consuming but not impossible question. Here is how I would solve this question.
Understand the Question  Behind this word problem lurks a pretty standard Number Properties / Primes and divisibility question. The core skill being tested is the ability to prime factor a number, in this case 18,865. The good news is that they tell you the three factors  5, 7 and 11. One of these numbers, 5, is easy to factor out. One of these numbers, 11, has a little remembered rule. Lastly, one of these numbers, 7, has a super complicated algorithm that is better off ignored. So, a simple restatement of this question is "How many factors of 5, 7, and 11 are in 18,865?"
Understand the Answer Choices  The denominator of the whichever answer choice is correct will at maximum be equal to the total number of 5s, 7s, and 11s in 18,865  the numerator will be the number of 5s and 11s. This is from the probability equation = (total number of desired outcomes)/(total number of outcomes). This makes answer choices C and E pretty unlikely as the denominators are way too high. Eliminate.
Plan  what's the fastest way to attack this question? Well, you know by the question that there is at least one 5, 7, and 11. So, dividing by these numbers is a reasonable way to start. I would divide by 5 or 11 first as these are faster. If you happen to be slow at division, guessing among A, B, and D is also reasonable.
Solve  18865 / 5 = 3,773. No more factors of 5 (Do you know why?)(Does not end in 0 or 5) 3,773 / 11 = 343 No more factors of 11 (Do you know why?) (the outside numbers do not add to the inside number) 343 / 7 = 49 49 = 7 * 7
So the prime factorization of 18,865 is 5*7*7*7*11 Eliminate A as 7 will not be the denominator Desired outcomes = 2 (one 5 and one 11) Total Outcomes = 5 B is correct
Jayson Beatty Indigo Prep
I understood the factorization part. I did not understand the part about why a certain number can/cannot be in the numerator.



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On this year's Westchester basketball team, the players are all either
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11 Sep 2018, 10:06
shardul171 wrote: This is a time consuming but not impossible question. Here is how I would solve this question.
Understand the Question  Behind this word problem lurks a pretty standard Number Properties / Primes and divisibility question. The core skill being tested is the ability to prime factor a number, in this case 18,865. The good news is that they tell you the three factors  5, 7 and 11. One of these numbers, 5, is easy to factor out. One of these numbers, 11, has a little remembered rule. Lastly, one of these numbers, 7, has a super complicated algorithm that is better off ignored. So, a simple restatement of this question is "How many factors of 5, 7, and 11 are in 18,865?"
Understand the Answer Choices  The denominator of the whichever answer choice is correct will at maximum be equal to the total number of 5s, 7s, and 11s in 18,865  the numerator will be the number of 5s and 11s. This is from the probability equation = (total number of desired outcomes)/(total number of outcomes). This makes answer choices C and E pretty unlikely as the denominators are way too high. Eliminate.
Plan  what's the fastest way to attack this question? Well, you know by the question that there is at least one 5, 7, and 11. So, dividing by these numbers is a reasonable way to start. I would divide by 5 or 11 first as these are faster. If you happen to be slow at division, guessing among A, B, and D is also reasonable.
Solve  18865 / 5 = 3,773. No more factors of 5 (Do you know why?)(Does not end in 0 or 5) 3,773 / 11 = 343 No more factors of 11 (Do you know why?) (the outside numbers do not add to the inside number) 343 / 7 = 49 49 = 7 * 7
So the prime factorization of 18,865 is 5*7*7*7*11 Eliminate A as 7 will not be the denominator Desired outcomes = 2 (one 5 and one 11) Total Outcomes = 5 B is correct
Jayson Beatty Indigo Prep
I understood the factorization part. I did not understand the part about why a certain number can/cannot be in the numerator. Once we are done with factorization we are left with finding the probability what do we need ? probability that a randomly selected team member will not be 7 total possible outcomes = number of (prime factors of 18865=5*7*7*7*11) = 5 required outcome = total possible outcomes  number of 7's =2 p(not being 7) = 2/5




On this year's Westchester basketball team, the players are all either &nbs
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