The range of set A is R. A number having a value equal to R is added

I think we all agree that neither statement is sufficient alone.

Now, let's consider the statements together.

Since {Range}={Largest}-{Smallest}, then the range will increase if R is either the smallest or the largest element of the new set.

Now, can R be the smallest element of the new set? If it is, then according to the second statement, we would have that the mean of the new set is smaller than the smallest element of the set, which is not possible: the mean of a set cannot be smaller than its smallest element.

Next, can R be the largest element of the new set? We are given that for set A, the range, R={Largest}-{Smallest}. Since from the first statement we know that all numbers in Set A are positive, then R must be less than the largest element of set A, hence R cannot be the largest element of the new set.

So, we have that R is neither the smallest nor the largest element of the set. Therefore, the range of the new set will not be greater than the range of set A. Sufficient.

Answer: C.

Hope it's clear.
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Yes it was pretty difficult. I had to guess on this question.

Official Explanation:
Disclosure: The following explanation has been typed word to word from an 800score practise test.


The range of a set is the difference between the largest and smallest elements of the set. Let's denote e and E as the smallest and largest element of the set, respectively. (if there is only one element, or if all elements are same, then e=E)

R, the range of the set = E-e

The questions asks whether the range of the Set A will increase when R becomes a member of the set. If R is less than the smallest member,e, or greater than the largest number,E, then the range will increase.

However, if E>R>e, then adding R to the set won't increase the range of the set.

Statement (1) Tells us that all the members of the set are positive, Since R= E-e and both E and e are positve then R must be less than E.

This makes it possible for R to be either between e and E, or less than e. In former case, the range of the new set won't increase but in the latter case it will. (see my example) Thus Statement 1 is insufficient.

Statement 2 tells us that the mean of the new set is smaller than R.

The mean of the original set A = Sum of elements /number of elements

Since the new set is just A plus the addition of R, there will be one addition element in the new set.
Thus mean of new set = (sum of elements of A + R) / Number of elements +1

Since R> mean of new set.
R> (sum of A + R) / (number of elements +1)

Multiplying both sides of the equation by (number of elements + 1)

(Number of elements +1) (R) > Sum of A + R, which is equal to

(Number of elements)(R) + (R) > Sum of A + R, subtracting R from both sides.

(Number of elements)(R)>Sum of A, dividing both sides by (number of elements)

R> Sum of A/number of elements, thus R> mean of A



Thus if know that the mean of the new set is smaller than R, we also know that the mean of the old set is smaller than R.

Furthermore, we know that the mean of any group of numbers is at least as largest as the smallest number so the mean of the old set is greater than of equal to e.

So, statment 2 tells us the R>mean of A>e. This is insufficient because R might be greater than E and, and increase the range, or it might be between e and E.

Combined two statements are sufficient. Statement (1) tells us that R is less than E and Statement (2) tells us that R is greater than e. This implies that adding R to the set will not increase the range of the set.

Final answer : (C)

The range of set A is R. A number having a value equal to R is added to set A. Will the range of set A increase?

Good question. +1.

Let's use the notations proposed by mainhoon:
\(a\) - smallest number in the set;
\(b\) - largest number in the set;
\(r\) - the range, so \(b-a=r\);
\(n\) - # of elements in set A;
\(m\) - the mean of set A.

The range of new set will NOT increase if \(a\leq{r}\leq{b}\), because new range will still be \(b-a\).

(1) All the numbers in set A are positive --> as all numbers are positive \(r\) can note be more than \(b\), the largest number in the set, so \(r<b\).

But \(r\) can still be less than \(a\) (example \(A=\{5,6\}\), \(r=1\) --> \(A_2=\{1,5,6\}\), \(r_2=5>r=1\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (example \(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO. Not sufficient.

(2) The mean of the new set is smaller than R --> \(m_2=\frac{m*n+r}{n+1}<r\) --> \(m<r\) (so R is also more than mean of set A) --> as \(r\) is more than mean of A, then \(r\) can note be less than \(a\), the smallest number in the set, so \(a<r\), (the mean is between the largest and smallest element of the set: \(a\leq{m}\leq{b}\) as \(r>m\), then \(a<r\)).

But \(r\) can still be more than \(b\) (example \(A=\{-5,0\}\), \(r=5\) --> \(A_2=\{-5,0,5\}\), \(r_2=10>r=5\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (\(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO. Not sufficient.

(1)+(2) From (1) \(r<b\) and from (2) \(a<r\) --> \(a<r<b\) --> new range will still be \(b-a\), so the answer to the question is NO. Sufficient.

Answer: C.
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Excellent! The trick is to realize that if R > mean then R > a. Great!

Posted from my mobile device

Bunuel awesome explanation ! Thanks

a A (1,1) gives (0,1,1) range increases.
(1,2) gives (1,1,2) range remains same.
Not sufficient.

b A(-2,-1) gives (-2,-1,1) range increases always for negative values.

A(1,5) gives (1,4,5) range remains same.
Not sufficient.

a+b (1,5) giving (1,4,5) range remaining same.
Hence C.

Because we are given a set of numbers, we can assume that all numbers are distinct, and that we have at least two numbers.

Let's denote by \(M\) the largest value of the set, and by \(m\) its smallest value. Then the range is \(R = M - m\).
The given question can be reformulated as "Is \(R < m\) or \(R>M\)?" If \(m\leq{R}\leq{M},\) the range of the set A will remain the same after adding \(R.\)

(1) If \(R=M-m<m,\) then \(\,\,M<2m.\)
If \(R=M-m>{m},\) then \(\,\,M>{2m}.\)
All the numbers in the set A being positive, both inequalities are possible. In addition, \(R=M-m<M.\)
So, \(R\) can be smaller or greater than the smallest element in the set, but for sure, it is smaller than the largest element.
Not sufficient.

(2) Let's denote by \(A_n\) the average of the set A before we add \(R\) to it.
Then \(\frac{A_n\cdot{n}+R}{n+1}<R,\) from which \(A_n\cdot{n}+R<nR+R\), or \(A_n<R.\)
Since the average \(A_n\geq{m}\) we can deduce that \(R>m.\)
But if \(m\) is negative, \(R=M-m>M\). If \(m\) is positive, then necessarily \(R=M-m<M.\)
Not sufficient.

(1) and (2) together: from the above analysis, we can see that if \(m>0,\) then \(R\) is between \(m\) and \(M,\) and the range will remain the same.
Sufficient.

Answer C.
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is zero a positive number ? can we consider zero in set A having all positive number ?

Zero is nether positive not negative. So, 0 cannot be in set A.
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