GMAT Club World Cup 2022 (DAY 7): If a, b, c, and d are distinct

in A if range is less than 4 means its consecutive integer ex. 1,2,3,4 or 2,3,4,5 in this scenario it always not divisible by 3 so its not multiply by 12
in B if integer is x-4,x-2,x+2,x+4> its sum has 4x means it may or may not divisible by 12
MY ANSWER IS A

Statement 1:
If the range is less than 4, then we can conclude that all 4 numbers are consecutive. Let's assume 2, 3, 4, and 5. Their sum is 14. Let's take the numbers 10, 11, 12, 13, or the numbers, 11, 12, 13, 14. As you can see, their sum will never be a multiple of 12.

Hence this statement is sufficient

Statement 2:
Consider two sets of numbers: 9, 11, 13, 15 and 11, 12, 13, 14. For the first one, the sum is in the form of 12k, whereas for the latter one, the sum is not in the form of 12k

Hence, this statement is insufficient.

Option A is the correct answer.

S1: It means a,b,c,d are consecutive integers. Hence, its summation can't be a multiple of 12. Hence Sufficient
S2: It means a,b,c, and d are in arithmetic progression, which is the same case as above. Its summation can't be a multiple of 12. Hence Sufficient.

Hence Both statements are sufficient alone, Option D is the right answer

Asked: If a, b, c, and d are distinct integers, is a + b + c + d a multiple of 12 ?

(1) The range of {a, b, c, d} is less than 4.
Since a, b, c & d are distinct integers and their range is less than 4, they are 4 consecutive integers and their range is 3.
a + b + c + d = a + (a+1) + (a+2) + (a+3) = 4a + 6
4a + 6 = {6,10,14,18,22,26,....}
4a + 6 is NOT a multiple of 12.
SUFFICIENT

(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).
Arithmetic mean = (a+b+c+d)/4
Median = (b+c)/2 ; if a, b , c & d are arrange in increasing/decreasing order.
Case 1: a=1;b=2;c=3;d=4; Arithmetic mean = 2.5; Median = (2+3)/2 = 2.5; a+b+c+d=10 is NOT a multiple of 12.
Case 2: a=1; b=2; c=4; d=5; Arithemic mean = 3; Median = 3; a+b+c+d=12 is a multiple of 12.
NOT SUFFICIENT

IMO A

Correct answer : Choice A

Statement 1 says : The range of {a, b, c, d} is less than 4.

This basically means , they are consecutive integers.
so if we take the first integer as x, then we get the set as {x, x+1, x+2, x+3}

we are checking if ( x + x+ 1 + x + 2 + x +3)/12 is an integer
=> (4x + 6)/12
=> (2x + 3 )/ 6
The above equation does not yeild an integer. Hence choice A is sufficient to answer this question.

Statement 2 says: The median of {a, b, c, d} is equal to its average (arithmetic mean).

according to the above satatement : { 10,11,13, 14} , this set satisfies the condition and the sum of it is divisible by 12
{1,2,3,4} this set satisfies the condition but the sum of it is not divisible by 12.
hence this statement is not sufficient .

Hence Choice A is correct

IMO answer A

1.
If range <4 and a, b, c and are distinct it means that a, b, c and d are consecutive numbers.
So a +b+c+d = a+a+1+a+2+a+3= 4k+6 (or 4k+2)

4k+2 is never a multiple of 12 (as it is not a multiple of 4)
Therefore 1 is sufficient to conclude.

2. Let's take 2 sets where median=mean

{1,2,4,5}
Median=(2+4)/2=3
Mean= 1+2+4+5=12/4=3
1+2+4+5=12 is a multiple of 12.

{2,3,7,8}
Median= (3+7)/2= 5
Mean= 2+3+7+8=20/4=5
2+3+7+8=20 not a multiple of 12.
Therefore 2 is not sufficient.

ANSWER A

Posted from my mobile device

without loss of generality we assume a<b<c<d

(1) The range of {a, b, c, d} is less than 4.
This means that a,b,c,d are consecutive integers. So sum will be 4a+6. This cannot be a multiple of 4 and hence 12 -- Sufficient
(2) The median of {a, b, c, d} is equal to its average (arithmetic mean). --
Let average = mean = b+c/2 = x = a+d/2
a+b+c+d = 4x, x can or cannot be multiple of 3. Insufficient

Ans A

(1) The range of {a, b, c, d} is less than 4.

Insufficient

(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).

Insufficient

Ans- E
Solution attached

Posted from my mobile device

Now we know that a, b, c, and d are distinct integers.

Question

a + b + c + d = 12 * some integer ; say 12x

Statement 1

The range of {a, b, c, d} is less than 4.

Let's assume that a < b < c < d.

As per the statement, d - a can be either 0 or 1 or 2 or 3. However as all the digits are distinct, we cannot have the range as 0 or 1 or 2. So, the range in this case has to be 3.

If that's the case, then the four numbers a, b , c and d are consecutive numbers.

We can represent a = n - 3; b = n-2; c = n-1 and d = n

Question thus becomes is (n-3 + n-2 + n-1 + n) = 12x

4n - 6 = 12x

n = \(\frac{12x - 6 }{ 4}\)

we see that n is not an integer in this case, hence we can answer the question with a definite No.

Hence A is sufficient.

Statement 2

The median of {a, b, c, d} is equal to its average (arithmetic mean).

Let's assume that a < b < c < d.

From the statement we know that -

\( \frac{b + c }{ 2}\) = \( \frac{a + b + c + d}{4}\)

Simplifying this we get -

a + d = b + c

Case 1

a = 1
d = 6
b = 3
c = 4

a + b + c + d = 14 / 4

is a + b + c + d a multiple of 12 -- No

Case 2

a = 2
d = 10
b = 5
c = 7

a + b + c + d = 24 / 4

is a + b + c + d a multiple of 12 -- Yes

Hence, B is not sufficient

IMO A

If a, b, c, and d are distinct integers, is a + b + c + d a multiple of 12 ?

(1) The range of {a, b, c, d} is less than 4.
(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).

statement 1 - many variable satisfies this, some are multiple of 12, some are not

not sufficient

statement 2 - same as statement 1

statement 1 + statement 2 - there is no value that is divisible by 12

sufficient

Answer C

IMO Option A is the answer.

Statement 1 only shows that the distinct integers must also be consecutive integers.
Try 1+2+3=4= 10 Not divisible by 12
Try 2+3+4+5 = 14 Not divisible by 12.
Try 3+4+5+6= 18 Not divisible by 12.
I can see a pattern that the sum will be increasing by 4, and the sum will not be divisible by 12.

I can conclude Statement I alone is sufficient because I can see a pattern.

Statement 2

Test 4 different distinct integers

1+2+3+4 = 10. Mean=Median = 2.5 . Sum is not divisible by 12
9+11+13+15 = 48. Mean = Median =12 . Sum is divisible by 12.
I can conclude Statement II alone is NOT sufficient because I can see different results.

Therefore option A is the answer IMO.

S1
"The range of {a, b, c, d} is less than 4."
This is only possible if distinct integers a, b, c, and d are consecutive integers.
So let a=n-1, b=n, c=n+1 and d=n+2
Thus a+b+c+d=4n+2
Obviously this is cannot be a multiple of 12 [As to be a multiple of 12 a number must be divisible by 4. 4n+2 is not divisible by 4]

Hence statement is SUFFICIENT

S2
"The median of {a, b, c, d} is equal to its average (arithmetic mean)."
This can be possible if distinct integers a, b, c and d are equidistant from each other.
Consider following cases,

Case-1
a=1,b=3,c=5,d=7
Mean=4
Median=4
a+b+c+d=16 NOT a multiple of 12

Case-2
a=3,b=5,c=7,d=9
Mean=6
Median=6
a+b+c+d=24 is a multiple of 12

Hence statement is INSUFFICIENT

Ans A

[quote="Bunuel"]If a, b, c, and d are distinct integers, is a + b + c + d a multiple of 12 ?

(1) The range of {a, b, c, d} is less than 4.
(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).


Statement 1) {a,b,c,d} range is less than 4 and they are distinct integers .Thus a,b,c,d are consecutive integers.
Suppose a=n, b=n+1,c=n+2,d=n+3
a+b+c+d=n+n+1+n+2+n+3=4n+6
4n+6 is not divisible by 12.
if we consider a=n-1,b=n,c=n+1,d=n+2 ,Thus a+b+c+d=4n+6.
Thus the sum is never divisible by 4.Statement 1 is sufficient.
Statement 2) median = (b+c)/2 average =(a+b+c+d)/4
(b+c)/2=(a+b+c+d)/4
a+d=b+c
Thus the a,b,c,d are evenly spaced integers.
Now if a= 3n
b=3n+2,c=3n+4,d=3n+6
a+b+c+d=12n+12 which is divisible by 12
but if we consider a=n ,b=n+1,c=n+2,d=n+3
then a+b+c+d =4n+6 which is not divisible by 12.
Thus 2 is insufficient.
Hence A is the answer

If a, b, c, and d are distinct integers, is a + b + c + d a multiple of 12 ?

(1) The range of {a, b, c, d} is less than 4.
Implies a, b, c and d are consecutive integers, and hence their sum will be of the form 4a + 6, this number will never be a multiple of 12.
Sufficient

(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).
Implies the numbers are consecutive integers, and hence their sum will be of the form 4a + 6, this number will never be a multiple of 12.
sufficient

Choice D is the correct answer since statements are independently sufficient to answer

Imo A

Statement 1: The range of {a, b, c, d} is less than 4.
Implies d-a =4 if arranged in an ascending order.
Since a b c d are distinct integers, there is only one possibility that
b=a+1; c=a+2; d=a+3
a + b + c + d = a + a+1 + a+2 + a+3 = 4a + 6
Lets say 4a + 6 is a multiple of 12, then 4a + 6 = 12k
On solving further a = 3(2k -1)/2 => (3 * odd number) /2
Hence for none of the values of a, that a + b + c + d is a multiple of 12. A definite 'No'
Sufficient

Statement 2: The median of {a, b, c, d} is equal to its average (arithmetic mean).
Median = (b + c) /2
Mean = (a + b + c + d) / 4
Median = Mean Implies b + c = a + d
a + b + c + d = 2 (b + c)
Let's take b = 2, c = 4 then the sum is divisible by 12.
The series will be a=1, b=2,c =4, d=5. Mean and median both are 3.
Let's take b=2, c=3 then the sum is not divisible by 12.
The series will be a=1, b=2, c=3, d=4. Mean and median both are 5/2
Insufficient

If a, b, c, and d are distinct integers, is a + b + c + d a multiple of 12 ?
It implies the sum must have factors 3 and 4

(1) The range of {a, b, c, d} is less than 4.
it implies a,b,c,d are consecutive numbers
hence, a+b+c+d can be written as n+n+1+n+2+n+3 = 4n + 6 = 2*(2n+3). but 2n+3 will always be an odd number, hence we dont have a factor of 4 in the sum. Hence sufficient to say that the sum is not a multiple of 12
(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).
This also implies that a,b,c,d are consecutive numbers
Hence sufficient to say that the sum is not a multiple of 12
Answer:D