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
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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

{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
Multiple of 2 = {2}
Non-prime numbers = {1,x/y}
At least one of x or y is odd non-prime number = {1,9}
If the list = {1,1,2,3,6} : median = 2
If the list = {1,2,3,6,9}: median = 3
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
Prime numbers = {2,3}
Multiple of 3 = {3, x, y} where x = 3k1 and y =3k2 ; k1 & k2 are positive integers other than 1
The list in ascending order = {1,2,3,3k1,3k2}
Median = 3
SUFFICIENT

IMO B
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Statement 1:
P(multiple of 2)<P(non prime)
x & y both have to be non prime so that 1,x, y can be non prime. So x & y both can be 1 or between 4 & 9.
In the first case the median is 1 and in the second case it is 3. Hence insufficient.

Statement 2:
P(multiple of 3)>P(prime)
For that to happen x & y have to be both multiples of 3.
In any case the median will be 3. Hence, sufficient.
Answer is B

Posted from my mobile device

My answer is B).

x and y could be any integer # from 1 to 9. There are also no restrictions on whether the numbers in the set can repeat.

Statement 1: x and y cannot be both even numbers, since 2 and 3 are already prime numbers, so the maximum # of non-prime numbers is 3. If both x and y are even numbers, then statement 1 cannot be true.

Consider if x is even and y is odd, so that there are 2 even numbers. We would need both x and y to be non-prime to have, together with 1, 3 non-prime numbers. But notice that y, being an odd number and non-prime, could be either 1 or 9. So if y = 1, x = 4, the set is {1,1,2,3,4}, and median is 2.

But if y = 9, x = 4, the set is {1,2,3,4,9}, and median is 3. Insufficient.

Statement 2: 2 and 3 are already prime numbers, so x and y must both be multiples of 3 (and not 3), so that, together with 3, we have 3 multiples of 3 vs. 2 prime numbers. So x and y can be either 6 or 9. Under either case, the median of the set would be 3. So sufficient.

B).

its not specified whether x,y are distinct or not
{x, y, 1, 2, 3}
x,y can be (1,2,3,4,5,6,7,8,9)

#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

case 1:
(1,1,2,3,4)
multiple of 2 ; 2/5
non prime ; 3/5
median is 2

case 2
(1,2,3,6,9)
multiple of 2 ; 2/5
non prime ; 3/5
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

possible at ( 1,2,3,6,9)
multiple of 3 = 3/5
prime = 2/5
median = 3
sufficient

OPTION B is correct

(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 2: 2,4,6,8
Non-Prime, 1,4,6,8,9

x,y,1,2,3

1,1,2,3,9

It is the only possible way.

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

Multiple 3 = 3,6,9
Prime = 2,3,5,7

We can arrange:
1,2,3,3,9
1,2,3,6,9

Median still 3

Sufficient

IMO, D

From Statement 1-

There exists multiple possbilities

1- x,y both multiples of 2 other than 2 - this will violate the condition
2- x,y one is multiple of 2 other is odd except 9 - this will violate the condition
3 - x,y one is 2 other is 9 - this will violate the condition - this will violate the condition
4 - x,y one is multiple of 2 except for 2 other is 9 - this will satisfy and median is 3
5 - x,y one is 1 other is 9 - this will satisfy the condition but median is 2



hence statement 1 is insufficient to answer.

From Statement 2-

it will only be possible if both x,y are multiple of 3 other than 3

x,y belongs to (6,9)

Hence median will be 3

Hence statement 2 is sufficient to answer

Hence Ans B

{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

Multiples of 2 - 2,4,6,8

11123 - Median 1 ( 1/5 < 3/5)

11234 - Median 2 (2/5 < 3/5)

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

Multiples of 3 - 3,6,9

But if x,y = 3 then probability of selecting prime also goes up

Median = 3

Sufficient

B

(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

so, the set is {x,y, 1,2,3}
From the given statement -> to have a probability of selecting an even number less than non-prime -> we need to introduce odd non-primes less than 10. to do that, we have only one scenario -> select x and y as 1. so, the median would be 1. 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

From this statement -> we can decipher that we should have more multiples of 3 than prime numbers in the set. We already have two prime numbers -> 2,3.
so, if we want a greater probability of multiples of 3 -> we can have 6 or 9 as x, y. we cannot have another 3 because then primes would increase to 3 and the probability can maximum be equated not made greater for multiples of 3. so, x, y are 6,9 which both are greater than 1,2,3 -> hence median is 3. Sufficient.

Ans. D.
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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

1 is not prime but 2 and 3 are prime. So we have 2 prime numbers, and 1 multiple of 2, hence the above is already satisfied which is not very helpful. Both of these numbers could be neither, and be 1 or 9, making the median 1 or 3 in each case. Insuff.

(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 primes and 1 multiple of 3. Great! that means we need 2 more multiples of three that aren't prime to satisfy the conditions. Whether we pick 3, 6, or 9 for any of x or y, the median will be 3. Hence this is sufficient to tell us what we need to know.

B is my answer.

S1:
we have one 2 and one non-prime number - so x,y can be 1,1 or 1,9 or 9,1 or 9,9 - different median - not suff

S2:
we have one 3 and two prime numbers - thus other two numbers has to be a multiple of 3 which is greater than 3 since 3 is prime
x,y can be 6,6 or 6,9 or 9,6 suff

Option B

Hello everyone

The correct answer is C

Statement 1

Numbers x,y,1,2,3

P(multiple of 2) < P (non-prime)

=> Let's say x and y = 1
then P (non-prime) = 3/5
P(multiple of 2) = 1/5

Median {1,1,1,2,3} = 1

=> Let's say x = 1 and y = 9
the p(non-prime) = 3/5
P(multiple of 2) = 1/5

Median {1,1,2,3,9} = 2

Statement 1 is not sufficient


Statement 2

P (multiple of 3) > P (prime numbers)

Numbers are 1,2,3,x,y

=> Min P (prime numbers) = 2/5 when x & y both are not prime

so x & y can not be 3. x & y has to be 6 & 9 or 9 & 6

Median (1,2,3,6,9) = 3

Statement 2 is sufficient



Hence, OA should be B

{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
x or y should not be either 2 as per the stmt. But x, y can be 1,1, or 4,4 (non-primes). So we cannot decide on the median

Not suff.

(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
AS per stmt, 3 should be atleast repeating once. So other numb can be 1 or 4. Which will then give diff median

Not suff

1 + 2

one number of x,y is 3 and other should be a non prime. But it could be 4 or 1 as an example. still we cannot arrive on one median solution.

IMO : E

The answer is B.

Statement 1 : 1, 9 satisfy the above condition. If x=1, y=9, we get median = 2. If x=y=9, then median = 3. Nowhere does it say x and y are distinct integers. So we can use the same values for both.
Insufficient.

Statement 2 : There are are 2 prime nos. and 1 multiple of 3. So x and y should be non-prime multiples of 3 to satisfy the above condition. Only 6,9 satisfy the condition. The median is always 3 even if you repeat the numbers.
Sufficient.

(B) is the answer IMO

x,y can from set {1,2,3,4,5,6,7,8,9}


Given set : {x, y, 1, 2, 3}
If x and y are positive integers less than 10, what is the median of the list above?


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

Given set : {x, y, 1, 2, 3}

Count Multiple of 2 in the set : 1 (number 2)
Count of Non prime : 1( number 1 )

We need to choose x and y i.e two numbers ; Probability of choosing non prime will be more than multiple of 2 in following two cases

Case 1: X and y both are non primes
Set can be {1,1,1,2,3} Median =1
set can be {1,2,3,9,9} Median =3

Case 2: one of (x,Y) is non prime and the other odd prime
Set can be {1,1,2,3,3} Median =2
set can be {1,2,3,7,8} Median =3

different answers : 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

Count Multiple of 3 in the set : 1 (number 3)
Count of prime : 2( number 2and 3 )

To have probability choosing multiple of 3 more than that of prime both x and y must be multiple of 3

Minimum value of such number is 3

set becomes (1, 2, 3, 3, 3)

x, y can be among 3, 6 or 9

in any case median will be 3

Sufficient

(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 other 2 numbers could be 5,5 or 5,8 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
since there are already 2 prime numbers for the multiple of 3 greater than prime numbers we need 2 more multiples it coulb be either 6,6 or ,9,9 6,9

either way the median can be uniquely defined

Therefore IMO B

(1) implies x,y can both take non prime values and it would still hold;
(2) implies that x,y must both be 9 (the only positive multiple of 3 which is non-prime and < 10);
hence B.