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

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

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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

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

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

In order to be a multiple of 12, a number needs to be a multiple of both 4 and 3

Statement 1: a, b, c and d are consecutive integers.
Eg. -1 + 0 + 1 + 2 = 2: Not a multiple of 12
Eg. 0 + 1 + 2 + 3 = 6: Not a multiple of 12
Eg. 1 + 2 + 3 + 4 = 10: Not a multiple of 12
If we see the trend above, the RHS is increasing by 4 and none of the same is/ will be a multiple of 4. Hence, the answer to the required ques is a definite NO. Hence, this statement alone is sufficient.

Statement 2: Median=Mean
Eg. 1+2+3+4=10. Median=Mean=2.5. Answer will be NO.
Eg. 1+2+4+5=12. Median=Mean=3. Answer will be YES.
Hence, this statement is not sufficient.

Answer is A.

Answer: 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.
=> a, b, c, d will be four consecutive integers. and range will be 3
let the numbers be x, x+1, x+2, and x+3
=> 4x + 6 should be divisible by 12 or not.
For the expression to be a multiple of 12,
4x + 6 = 12k where k will be an integer
=> 2x +3 = 6k
For this to hold true 2x needs to be odd which is not possible.
Hence, a+b+c+d will not be a multiple of 12.
Sufficient.

(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).
=> distribution of numbers across median will be same.
For four consecutive integers, we already saw that it will not be a multiple of 12.
Lets consider x, x+2, x+4, and x+6
mean = median = x+3
a+b+c+d = 4x+ 12
for a multiple of 12, 4x +12 = 12k where k will be an integer.
when x = 3, this holds true.
i.e. for 3, 5, 7, 9
3+5+7+9 = 24 is a multiple of 12
Insufficient.

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.
This essentially suggests that the set consists of consecutive integers
For e.g. we have {-1, 0, 1, 2}, or {-6, -5, -4, -3} or {4,5,6,7}
The numbers are essentially (a-1), a, (a+1), (a+2)
The adding these numbers we get 4a + 2. Any number of the form 4a+12 would always have 2 as remainder when divided by 12.
Sufficient. We get NO as an answer

(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).
We can have the following examples
Set A is {9, 11, 13, 15} - 9+11+13+15 is a multiple of 12
Set A is {3, 4, 5, 6} - then 3+4+5+6 is not a multiple of 12
We get both YES and NO as answers. Insufficient

IMHO Option A

(1) The range of {a, b, c, d} is less than 4.
let's assume that a> b> c>d , if the range is less than 4 then, d-a <3, and since the a, b, cand d are integers, then the difference between a and b and between b and c and between c and d could not be more than 1, so the numbers are consecutive,
Then a+b+c+d=a+a+1+a+2+a+3=4a+6
first number possible is when a=0 then :4a+6=6 (left a remainder of 6 when divided by 12)
2nd number possible is when a=1 then :4a+6=10 (left a remainder of 10 when divided by 12)
3rd number possible is when a=2 then :4a+6=14 (left a remainder of 2 when divided by 12)
4th number possible is when a=3 then :4a+6=18 (left a remainder of 6 when divided by 12)

and by going like that, the sum would never be divisible by 12,
Answer to "is a + b + c + d a multiple of 12" is NO
Sufficant


(2) The median of {a, b, c, d} is equal to its average (arithmetic mean).
this statement gives, (b+c)/2 = (a+b+c+d)/4
=>b+c=a+d
a+b+c+d = 2(b+c)
cannot say whether it's divisible by 12 or not,
not sufficiant,
Our answer is 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).

Using (1)

If range is less than 4 , and distinct integers --> consecutive integers

But, sum of 4 consecutive integers in not divisible by 4

e.g. -2+ (-1) + 0 + 1 = -2
1 + 2 + 3 + 4 = 10

Hence, NOT divisible by 12..Hence, NO is the answer

C and E are out

Using (2)

If median = average

eg, (-4, -2, 2, 4) here, Median = Mean = 0 ; but a+b+c+d = 0 ( multiple of 12)..YES is the answer
(1, 2, 3, 4) ; here, Median = Mean = 2.5; but a+b+c+d = 10 ( NOT a multiple of 12)..NO is the answer

Unique YES/NO not found

B and D are out

(A) is the correct answer

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

If range of 4 distinct integers is less than 4, it is safe to imply that they are consecutive integers because otherwise there is no way the range will be less than 4
Now 4 consecutive integers, their sum will NEVER equal to a multiple of 12. Sum will always be 2 less or 2 more than the sum of 12
Example: 1,2,3,4 = 10: 2,3,4,5 = 14: 3,4,5,6 = 18 and so on

SUFFICIENT

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

Although there is a property of evenly spaced sets which states that for any evenly spaced set, the mean = median BUT that property is not always true in the opposite direction meaning that if given that the median is equal to mean of a set, it does not necessarily mean that the set is an evenly spaced set
Let us look at cases
Case 1: {2,4,6,8}: Mean=5, Median=5, Sum=20, NOT A MULTIPLE OF 12
Case 2: {1,2,4,5}: Mean=3, Median=3, Sum=12, YES, A MULTIPLE OF 12

NOT SUFFICIENT

Answer - A

From st.1 we have the range of {a, b, c, d} is less than 4, where a, b, c, and d are distinct integers.
For this to be true we can say that a, b, c, d are consecutive integers. Otherwise their range might be greater than 4.
If they are consecutive then, for example: {1, 2, 3, 4} then range is 4 – 1 = 3 < 4
If they are not consecutive then e.g {1, 5, 7, 12} then range is 12 – 1 = 11 > 4
Now if they are consecutive then say they are: n, n+1, n+2, n+3
So their sum = 4n + 6. Here the term 6 is never divisible by 12.
Hence a + b + c + d is never a multiple of 12. So this statement is sufficient

From st. 2 we have the median of {a, b, c, d} is equal to its average (arithmetic mean). This is possible when we have a set of consecutive numbers (integers, evens, odds, multiples)
So if e.g. it is {12, 24, 36, 48} then the median = mean and the sum is divisible by 12.
But if it is {3, 6, 9, 12} then the median = mean but the sum is not divisible by 12.
Hence this statement is not sufficient.

Hence answer choice (A)

Hi,

In statement 1, how on the basis of range of 4 numbers is less than 4, are we inferring that these 4 numbers are consecutive? Given the numbers are distinct integers and not positive integers, will not have any impact of our conclusion that 4 numbers are consecutive integers?

Can somebody explain?

Thanks in advance

I tried to reason this out in my solution. So instead of typing that all over, sharing the link https://gmatclub.com/forum/gmat-club-world-cup-2022-day-7-if-a-b-c-and-d-are-distinct-395338.html#p3046161

If you have questions let me know, would be happy to clarify.

I'm confused about S1)
What if the sequence is:
-1,0,1,2 i.e. 4 distinct integers
Range => 2-(-)1=3 which is less than 4.

but
-1 + 0 +1 +2 = 2 which isn't a multiple of 12.
Why do we rule that out?

That's precisely it !

Using statement 1 you will get a definite answer

Is a + b + c + d a multiple of 12 - Ans : No

I don't suppose you can formulate a scenario in which (a + b + c + d) is a multiple of 12.

Hence statement 1 is sufficient.