GMAT Club World Cup 2022 (DAY 8): Set A consists of nine consecutive

given that Set A consists of nine consecutive even integers and set B consists of six consecutive even integers
sum of elements in A = sum of elements in B

(1) All elements of set B are also in set A.

2x-2+2x+2x+2..+2x+12 =2x-2+2x+2x+2..
we get 6x+36=0
x=-6
Set A = -14,-12,-10,-8,-6,-4,-2,0,2
Set B = -14,-12,-10,-8,-6,-4
median of set A = -6
sufficient
#2
The median of set B is a prime number.
Set B = -12,-6,0,6,12,18 median is 3
or set B = -8,-4,0,4,8,12 median is 2
but since sum is same for set A and set B
Set B = -14,-12,-10,-8,-6,-4 with median as -11
set A = Set A = -14,-12,-10,-8,-6,-4,-2,0,2
this is not true as -ve 11 is not prime
OPTION A is correct

Clueless. Guess B and move on

Using Statement 1, we can conclude that:
The sum of those 3 extra even integers in Set A will be 0. So, the sets A and B can be as follows:
Alt 1:
A = {-2, 0, 2, 4, 6, 8, 10, 12, 14}
B = {4, 6, 8, 10, 12, 14}
Alt2:
A = {-14, -12, -10, -8, -6, -4, -2, 0, 2}
B = {-14, -12, -10, -8, -6, -4}

Hence, insufficient.

Using Statement 2,
We know that the median of Set B is a prime number. But as we know, for a set with even numbers, the median shall be the average of the 3rd and 4th terms when arranged in ascending order. There are so many values possible.

Combining both statements, we get that neither of the scenarios possible (Alt 1 and Alt 2) as per Statement 1 are feasible, since the median of set b is 9 which is not a prime number.

Hence, option E is the correct answer.

Set A consists of nine consecutive even integers and set B consists of six consecutive even integers.
If the sum of the elements of set A is equal to the sum of the elements of set B,
what is the value of the median of set A ?

(1) All elements of set B are also in set A.
Set A =(-2,0,2,4,6,8,10,12,14) and Set B = (4,6,8,10,12,14).. Median of Set A is 6
but if
Set A =(2,0,-2,-4,-6,-8,-10,-12,-14) and Set B = (-4,-6,-8,-10,-12,-14).. Median of Set A is -
st 1 is not sufficient
(2) The median of set B is a prime number.
Set A = (-6,-4,-2,0,2,4,6,8,10) set B = (-2,0,2,4,6,8) median of set B = 3..prime number
sum of set A = sum of setB
median of setA = 2..
for any other combination we cant meet the twin condition of sum of set A n B to be equal and median of setB to be prime number
therefore, st2 is sufficient
Answer B

Correct answer : Choice B

From question stem and statement 1 :

A = {-6,-4,-2,0,2,4,6,8,10} = sum = 18
B = {-2,0,2,4,6,8} = sum = 18
is true
and hence median = 2
and

A= {-10,-8,-6,-4,-2,0,2,4,6} = -18
B = {-8,-6,-4,-2,0,2} = - 18
is true

and hence median = -2
Since there are two values, Statement 1 is not sufficient

The question stem and from statement 2 :
There is only one possible solution for which median = 2
Hence answer is choice B

Let A = {2a, 2a+2, ... 2a+16} and B = {2b, 2b+2, ... , 2b+10}

Sum of elements of A = Sum of elements of B
9x2a + 72 = 6 x 2b + 30
3a + 12 = 2b + 5
3a = 2b - 7

Median of A = 2a + 8
We need to have a unique value of \(a\) to know median of A

(1) All elements of set B are also in set A.
This means b needs to be negative. This means first three terms of A will be -2, 0 and 2,
Now there can be two options
A = { -2,0,2,4,... 14} or {-14,-12,...,2,0,-2}
B= {4,6,...,14} or {-14,-12,...,-4} ,
Two value of a is possible.
Insufficient
(2) The median of set B is a prime number.
2b+5 = prime, b has many possible value
Insufficient

Both
median of B = 9 or -9 = Not prime
No value of b is possible. So no value of a is possible.
No median is possible

Ans E

Set A consists of nine consecutive even integers and set B consists of six consecutive even integers. If the sum of the elements of set A is equal to the sum of the elements of set B, what is the value of the median of set A ?

(1) All elements of set B are also in set A.
(2) The median of set B is a prime number.


Statement 1 - A should be (-2, 0,2,4,6,8,10,12,14)
B should be (4,6,8,10,12,14)
so median of A - 6

Statement 2 - value of set B will be vary with A
so median of B should be vary with A

not sufficient

Answer A

sum of elements of set a ( which are 9 consecutive even integers) is equal to set b consisting of six consecutive even integers

st.1

let 9 consecutive integers be (a-8)(a-6)(a-4)(a-2)a(a+2)(a+4)(a+6)(a+8), as per statement all elements of b are also in A,

now assume six consecutive integers are (a-8)(a-6)(a-4)(a-2)a(a+2), thus 9a= 6a-18
3a= -18
a= -6

thus series A will be -14, -12,-10,-8,-6,-4,-2,0,2, and series B will be -14,-12,-10,-8,-6,-4

Also if we assume six consecutive terms in b to be (a-2)a(a+2)(a+4)(a+6)(a+8), then 9a = 6a+18
then a= 6

A (-2,0,2,4,6,8,10,12,14) , and B( 4,6,8,10,12,14)

since different values of median are possible hence a is not sufficient

St. 2

let 9 consecutive integers be (a-8)(a-6)(a-4)(a-2)a(a+2)(a+4)(a+6)(a+8)
now the only series for which median of b is prime number is (4,6,8,10,12,14)

hence correct answer should be B

(1) All elements of set B are also in set A.
Since B is subset of A, and per stem the sum of both Set A and Set B are equal sum of remaining 3 consecutive integers should be 0
The numbers that satisfy the conditions are on -2,0,2, implies set A = [-2,0,2,4,6,8,10,12,14]
This is sufficient for finding Median of A

(2) The median of set B is a prime number.
Does not provide any information about A

correct answer choice is A

[quote="Bunuel"]Set A consists of nine consecutive even integers and set B consists of six consecutive even integers. If the sum of the elements of set A is equal to the sum of the elements of set B, what is the value of the median of set A ?

(1) All elements of set B are also in set A.
(2) The median of set B is a prime number.

Statement 1)Suppose the elements in set A=n-10,n-8,n-6,n-4,n-2,n,n+2,n+4,n+6
Sum of all terms of set A=9n -18
Then there are few possibilities for the elements in set B={n-4,n-2,n,n+2,n+4,n+6},{n-6,n-4,n-2,n,n+2,n+4}{n-8,n-6,n-4,n-2,n,n+2}
{n-10,n-8,n-6,n-4,n-2,n}
Now as per the problem
Considering the 1st option of set B =6n+6=9n -18 =>3n=24
=>n=8
Thus set A ={-2,0,2,4,6,8,10,12,14}, Thus median of set A=6
Now the 2nd option 6n-6=9n-18 =>12=3n=>n=4
Thus the set A={-6,-4,-2,0,2,4,6,8,10} Thus median of set =2
Thus with different combination the medians are different. Thus 1 is not sufficient.
2)Suppose we consider Set A={-6,-4,-2,0,2,4,6,8,10}
Set B={-2,0,2,4,6,8} - Only for this case median of set B is (2+4)/2=3 which is a prime number
Thus the median of set A is 2.If we try other combinations as mentioned above the median of set B won't be a prime number.
Thus 2 is sufficient. Hence B is the answer.

Please refer to the attachment for the solution.

(1) All elements of set B are also in set A.

Case 1
Set A = -2 0 2 4 6 8 10 12 14

Set B = -2 0 2 4 6 8 10 12 14

Median of A = 6

Case 2
Set A = -2 0 2 4 6 8 10 12 14

Set B = -14 -12 -10 -8 -6 -4 -2 0 2

Median of A = -6

A is not sufficient

Median of A = 6

(2) The median of set B is a prime number

Set A = -6 -4 -2 0 2 4 6 8 10

Median of B = 3
Median of A = 2

IMO B

IMO C

Set A consists of nine consecutive even integers and set B consists of six consecutive even integers. If the sum of the elements of set A is equal to the sum of the elements of set B, what is the value of the median of set A ?

(1) All elements of set B are also in set A.
(2) The median of set B is a prime number

Imo D

Let's say A has { a, a+2, a+4, a+6...and so on till 9 consecutive even integers}
B has { x, x+2, x+4... and so on till 6 consecutive even integers)
Given Sum of both the sets A and B is equal.
Implies 3a + 14 = 2x -------(1)

Statement 1: All elements of set B are also in set A.
Implies a = x, and thus from eq. 1 we have a =-14
For one unique value of a we have one unique set A and a unique median.
Sufficient

Statement 2 The median of set B is a prime number.
From equation 1
For a = 0,4 x will be 7, 13 [ Not possible as x should be even integer]
For a = 2, 6 x will be 10, 16 respectively [not possible as Median of Set B is not prime]
For a = -6 x will be -2
So, Set A will be { -2, 0, 2, 4, 6, 8, 10, 12, 14}
Set B will be {-2, 0, 2, 4, 6, 8} [Median is 3]
For set -10 and onwards, x will be -8 and so on
Set B = -8,-6,-4,-2,0,2 [Median will be negative and not a prime]
Hence there is only one unique value of Median for Set A.
Sufficient

Set A consists of nine consecutive even integers and set B consists of six consecutive even integers. If the sum of the elements of set A is equal to the sum of the elements of set B, what is the value of the median of set A ?

(1) All elements of set B are also in set A.

Let's say the elements are
Set A = 2k - 8, 2k - 6, 2k - 4, 2k - 2, 2k, 2k + 2, 2k + 4, 2k + 6, 2k + 8
Sum of Set A = 18k

Set B can be { 2k + 4, 2k + 6, 2k + 8, 2k + 10, 2k + 12, 2k + 14 }
Sum = 12k + 54

Now Sum(A) = Sum(B)
=> 18k = 12k + 54
=> k = 9

Now Set B can also be {2k - 14, 2k - 12, 2k - 10, 2k - 8, 2k - 6, 2k - 4}
Now 18K = 12k - 54
=> 6K = -54
=> k = -9

Hence there are two such Sets possible. Insufficient

(2) The median of set B is a prime number.
Now let's say we take the same set A as in statement 1 we have the sum as 18k
Let's say the median value of Set B is P
Then the sum of Set B = 6P
Now we have 18K = 6P
=> 3K = P
For P to hold up as a prime number we k can only have one value, i.e. 1. Any other value would make P composite
hence k = 1
and P = 3

Set A becomes {-6, -4, -2, 0, 2, 4, 6, 8, 10}
Set B becomes {-2, 0, 2, 4, 6, 8}
The sum of both of these is 18 hence the original condition satisfies too
Sufficient

IMHO Option B

Statement 1:
Suppose Set B={-2, 0, 2, 4, 6, 8}; sum=18
So, Set A={-6, -4, -2, 0, 2, 4, 6, 8, 10}; sum=18; Median=2

Suppose Set B={-8, -6, -4, -2, 0, 2}; sum=-18
So, Set A={-10, -8, -6, -4, -2, 0, 2, 4, 6}; sum=-18; Median=-2

As we have 2 different values of median above, this statement alone is not sufficient.

Statement 2:
Median of set B is a prime number. It has to be an odd number because median of 6 consecutive even integers will always be odd. Median of set A will always be even because the median of 9 consecutive even integers will always be even.
Because sum of set A=sum of set B, median of set B should be as close as possible to 0, so that adding 3 numbers to set A for arriving at set B should nullify each other.

If median of set B = 3 (odd prime number), set B={-2, 0, 2, 4, 6, 8}; sum=18
For sum of set A also to be 18, all the numbers of set B will definitely comprise into set A and we need to add more 3 numbers whose sum should equal 0. So, we can add 10, -4 and -6. Median of set B will be 2.

If median of set B=5, set B={0, 2, 4, 6, 8, 10}; sum=30
There is no 9 consecutive even numbers whose sum = 30
Similar is the observation if median of set B is 7 or 11 etc...

The only possible value for median of set B is 3, given the constraint that sum of set A=sum of set B.
Hence, this statement is sufficient.

Answer B.

Answer: B

Set A consists of nine consecutive even integers and set B consists of six consecutive even integers. If the sum of the elements of set A is equal to the sum of the elements of set B, what is the value of the median of set A ?

Let Set A = {a-8, a-6, a-4, a-2, a, a+2, a+4, a+6, a+8}
Set B = {b-5, b-3, b-1, b+1, b+3, b+5}

Median of A = a ... needs to be calculated.
Median of B = b

Sum of elements of A = Sum of elements of B
=> 9a = 6b
=> 3a = 2b
=> a is a multiple of 2, a is even,
and b is a multiple of 3

(1) All elements of set B are also in set A.
When a = 0, b = 0
A = {-8, -6, -4, -2, 0, 2, 4, 6, 8}
B = {-5, -3 ... etc}. ...doesn't satisfy the condition
When a = 2, b= 3
A = {-6, -4, -2, 0, 2, 4, 6, 8, 10}
B = {-2, 0, 2, 4, 6, 8}
Median of a = 0
When a = 6, b = 9
A = {-2, 0, 2, 4, 6, 8, 10, 12, 14}
B = {4, 6, 8, 10, 12, 14}
Median of a = 6
Since we get different values of a,
Not sufficient.

(2) The median of set B is a prime number.
Since b is a multiple of 3,
Median b (prime) = 3 is the only possibility.
When b = 3, a = 2
=> Median of set A = 2
Sufficient.