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27 Aug 2021, 08:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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(hard)
Question Stats:
58% (02:37) correct
42%
(02:27)
wrong
based on 356
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{x, y, 1, 2, 3}
If x and y are positive integers less than 10, what is the median of the list above?
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number
If x and y are positive integers less than 10, what is the median of the list above?
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number
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27 Aug 2021, 08:35
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{x, y, 1, 2, 3}
If x and y are positive integers less than 10, what is the median of the list above?
x and y can be anything from 1 to 9.
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number
1,1,1,2,3…..Median is 1
1,2,3,3,3…..Median is 3
Insufficient
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number
We have 2 prime numbers, so there should be more than 2 multiples of 3.
1,2,3,3,3….Median is 3
1,2,3,6,6…..Median is 3
As the lowest 3 numbers are 1,2,3, the median will always be 3.
Sufficient
B
If x and y are positive integers less than 10, what is the median of the list above?
x and y can be anything from 1 to 9.
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number
1,1,1,2,3…..Median is 1
1,2,3,3,3…..Median is 3
Insufficient
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number
We have 2 prime numbers, so there should be more than 2 multiples of 3.
1,2,3,3,3….Median is 3
1,2,3,6,6…..Median is 3
As the lowest 3 numbers are 1,2,3, the median will always be 3.
Sufficient
B
21 Sep 2021, 09:21
Kudos
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Bunuel
{x, y, 1, 2, 3}
If x and y are positive integers less than 10, what is the median of the list above?
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number
If x and y are positive integers less than 10, what is the median of the list above?
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number
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Official Solution:
\(\{x, \ y, \ 1, \ 2, \ 3\}\)
If \(x\) and \(y\) are positive integers less than 10, what is the median of the list above?
Recall that the median of a list is the middle term, when arranged in ascending or descending order.
(1) When one number is chosen at random from the list, the probability of selecting a multiple of 2 is less than the probability of selecting a non-prime number.
The above means that there are more non-prime numbers in the list than multiples of 2. Since without \(x\) and \(y\) there are equal numbers of non-prime numbers and multiples of 2 in the list (non-prime is 1 and multiple of two is 2), then in \(\{x, \ y \}\) must be more non-primes than multiples of 2.
We could have several different cases. For example, if \(\{x, \ y\} =\{1, \ 1\}\), the the list would be \(\{1, \ 1, \ 1, \ 2, \ 3 \}\) and the median would be 1 but if \(\{x, \ y\}=\{9, \ 9 \}\), the the list would be \(\{1, \ 2, \ 3, \ 9, \ 9 \}\) and the median would be 9. Two different answers. Not sufficient.
(2) When one number is chosen at random from the list, the probability of selecting a multiple of 3 is greater than the probability of selecting a prime number.
The above means that there are more multiples of 3 in the list than primes. Since without \(x\) and \(y\) there are less multiples of 3 than primes (multiple of 3 is 3 and primes are 2, and 3), then both \(x\) and \(y\) must be multiples of 3 but not primes. So, each of \(x\) and \(y\) could be 6 or 9. Now, if both \(x\) and \(y\) are more that 3 (6 or 9), then the list in ascending order would be \(\{1, \ 2, \ 3, x, \ y\}\), which means that the median is 3. Sufficient.
Answer: B
