Re: GMAT CLUB OLYMPICS: If x and y are positive integers less than 10, wha
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28 Aug 2021, 06:35
Answer: B
{x, y, 1, 2, 3}
If x and y are positive integers less than 10, what is the median of the list above?
If both x and y equals 1, e.g. {1, 1, 1, 2, 3} then median = 1
If One of x or y equals 2, e.g. {1, 2, 1, 2, 3} then median = 2
If both x and y greater than or equal to 3, e.g. {3, 3, 1, 2, 3} then median = 3
With this in mind, lets look at the statements
Statement 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
Multiple of 2 = 2, 4, 6, 8
P(multiple of 2) = 1/5, 2/5 or 3/5
In the set, 2 and 3 are already prime, if x and y both are non-prime then the set can have maximum 3 non-prime numbers.
therefore, P(select non-prime) can be equal to 3/5, 2/5 or 1/5
Lets take few examples and see if we can get different medians for the given condition, P(multiple of 2) < P(select non-prime):
Ex. 1 --> {1, 1, 1, 2, 3}
P(multiple of 2) = 1/5 < P(select non-prime) = 3/5
Median = 1
Ex. 2 --> {9, 9, 1, 2, 3}
P(multiple of 2) = 1/5 < P(select non-prime) = 3/5
Median = 3
Hence, we can have different medians.
So, statement 1 by itself is not sufficient.
Statement 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
P(multiple of 3) > P(select prime)
P(multiple of 3) = 1/5, 2/5 or 3/5
Since we already have two prime numbers in the set, P(select prime) = 2/5, 3/5 or 4/5
Therefore for condition to be true, P(multiple of 3) cannot be 1/5 or 2/5. It can only be 3/5.
So, both x and y in the set needs to be >= 3. so P(multiple of 3) = 3/5
As we already know, If both x and y greater than or equal to 3 in the set, then median = 3
Therefore, statement 2 is sufficient by itself.