Is the mean of Set A equal to mean of Set B?

1) Let's say that A=B U {x}. There's no constraint for x other than being different from all numbers in the set, so x can either be negative or positive or zero. Let's suppose that Sb is the sum of all terms in set B. Then if we wanted the means to be equal, we'd need to find an x that:
(Sb+x)/7=Sb/6
here x= Sb/6 >0
As we can seen the term to be added for the means to be equal is equal to the mean of set B, which is clearly greater than the lowest member in B. Therefore, there is no x than satisfy both the question and the first statement.
Clearly this statement is sufficient.
2) sum of all numbers in set A <sum of all numbers in set B. If we divide each term by 7, we have
(sum of all numbers in set A)/7 <(sum of all numbers in set B)/7
But (sum of all numbers in set B)/7 <(sum of all numbers in set B)/6 [because the sum of all numbers in set B is positive]. Then,
(sum of all numbers in set A)/7 <(sum of all numbers in set B)/6
mean(A)<mean(B)
Clearly this statement is sufficient.
Answer: D

Qs : Is the mean of Set A equal to mean of Set B?
which can be written as Mean A= Mean B?
Given : A is a set containing 7 different numbers. B is a set containing 6 different positive numbers, all of which are members of set A.

let set A = [n, 2,3,4,5,6,7] and set B= [2,3,4,5,6,7]
however it is not mentioned that all numbers in set A are positive. we can derive from set B that only 6 no. are positive. so the remaining number n can be positive or negative in set A. this number n will decide the total sum.

1) Range of set A is greater than the range of set B.

as we know that range = largest number - smallest number
Range of A, RA= 7-1 = 6 if n=1
RA= 7-(-1)= 8 if n = -1
therefore mean A may or may not be greater than mean B.

Mean A= [n+ 2+3+4+5+6+7]/7= [n+ 27]/7 so mean of A depends on number n.
as range of A is greater than range of set B then this "n" must be less than the smallest value of set B i.e. 2 in above case considering the largest number is same. therefore the mean of A and mean of B will not be same. hence it is possible to get unique answer. this condition is sufficient.

2) Sum of all numbers in set A < Sum of all the numbers in set B
mean = total sum/number of terms
as total number of terms is more in set A, the mean of A will be less.

[Sum of set A]/7 < [sum of set B]/6
mean A< mean B
hence sufficient to answer the question.

The answer is D.

Is the mean of Set A equal to mean of Set B?
A is a set containing 7 different numbers. B is a set containing 6 different positive numbers, all of which are members of set A.

Immediately, I can tell that the only way this would be true is if the 7th number of Set A is equal to the mean, and every other number are numbers in Set B. Ex: A{1,2,3,4,5,6,7} | B{1,2,3,5,6,7} . This is true because Set A contains Set B, and each number is different.

1) Range of set A is greater than the range of set B. This means that the 7th "extra" number of Set A is either the first or last number, and so not equal to the mean of the rest of the set. Ex: A{1,2,3,5,6,7,8} | B{1,2,3,5,6,7} Sufficient (to disprove)

2) Sum of all numbers in set A < Sum of all the numbers in set B. Because Set A contains 6 positive numbers, all of which are also in set B, this means that the 7th extra number is less than 0, making sum of Set A < Set B. Ex: A{-1,1,2,3,5,6,7} B{1,2,3,5,6,7}. Sufficient (to disprove)

Answer: D

Is the mean of Set A equal to mean of Set B?
A is a set containing 7 different numbers. B is a set containing 6 different positive numbers, all of which are members of set A.

1) Range of set A is greater than the range of set B.
2) Sum of all numbers in set A < Sum of all the numbers in set B

The question says,
A is having 7 different numbers (positive or negative integer or 0 or any non integer)
B is having 6 different +ve numbers, all of which is are members of A. In this question there will be 3 scenarios. Lets say,
A = [1,2,3,4,5,6,X]
B = [1,2,3,4,5,6]
Scenario-1,if X > the largest no of the set (say 7) , then [A]Mean > [B]Mean
Scenario-2,if X =0 then [A]Mean < [B]Mean
Scenario-3,if X< the smallest no of the set (say -2), then [A]Mean < [B]Mean

Now, option (1) says, [A]range > [B]range. here Scenario-1 & 3 shall apply. Not Sufficient
Option (2) says, [A]sum < [B]sum, here Scenario 2&3 shall apply. Sufficient.

Correct Answer is (B)

#1
Range of set A is greater than the range of set B
A has larger value than B ; so the mean value of A is not same as B ; sufficient
#2
Sum of all numbers in set A < Sum of all the numbers in set B
since sum of integers is less than set B ; so mean is not same ; sufficient
IMO D
Is the mean of Set A equal to mean of Set B?
A is a set containing 7 different numbers. B is a set containing 6 different positive numbers, all of which are members of set A.

1) Range of set A is greater than the range of set B.
2) Sum of all numbers in set A < Sum of all the numbers in set B

Answer is B.
Given attachment for explanation

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