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

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.

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

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.

Set A: 9 consecutive even integers, Set B: 6 consecutive even integers.
We need to find median (5th term of ascending order) of Set A

Sum(Set A) = Sum(Set B). Now if one set contains 6 numbers and another one 9 and both contain consecutive even integers, their sum can only be equal if the sum of the additional 3 elements in Set A equal to 0, which implies that there are negative even integers involved as well

Consecutive even integers, some negative terms and sum of additional 3 terms of Set A equals 0. Taking all these constraints into consideration, we can only come up with 2 cases

Case 1: B = {-2,0,2,4,6,8}, A = {-6,-4,-2,0,2,4,6,8,10}, Sum = 18 (for both) [We added -6,-4 and 10 as the three additional elements, thereby maintaining all constraints]
Case 2: B = {-8,-6,-4,-2,0,2}, A = {-10,-8,-6,-4,-2,0,2,4,6}, Sum = -18 (for both) [We added 6,4 and -10 as the three additional elements, thereby maintaining all constraints]
Now let us look at the statements:

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

Yes, both possible cases above attest to this. In one, median = 2, and in the other, median = -2. NOT SUFFICIENT

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

Now, here its mentioned that median of Set B is a prime number. Let us check both possible cases:
Case 1: Median B = 3, which is a prime number
Case 2: Median B = -3, which is not a prime number
So only Case 1 suffices in this. Which means that median A = 2. SUFFICIENT

Answer - B

Lets say that A (2k, 2K+2,2k+4, ... 2k+16)
B (2d, 2d+2, ...2d+10)
if the two sums are equal then :18k+72=12d+30=>3k+7=2d(*)

(1) All elements of set B are also in set A.
if k=-1then d=2 and A=(-2, 0, 2, 4, 6, 8, 10, 12, 14) and B=(4, 6, 8, 10, 12, 14) here median of B=9
if k=-3 then d=-1 and A=(-6, -4, -2, 0, 2, 4, 6, 8, 10) and B=(-2, 0, 2, 4, 6, 8) here the median of B =3
in both cases set B is included in Set A and the median is different, So statement 1 is not sufficient
(2) The median of set B is a prime number.
the median of set b is 4d+10)/2 = P (prime)
the prime number could be 2, 3, 5, 7, 11, ...

d could be -1, 0, 1, 3, 4, 6, 7, 9, 12, 13, 16, 18...
only the number -1 is possible, because according to (*), k=(2d-7)/3, only d=-1 leads to K integer, otherwise, k wouldn't be integer, then
if d=-1
B=(-2, 0, 2, 4, 6, 8), the median of B=3
sufficient,

Answer is B

IMO answer choice is (B)