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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
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Bunuel awesome explanation ! Thanks
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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.
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The Range of Set A is R. A number having equal value to R, is added to set A. Will the range of Set A increase ?

(1) All numbers in Set A are positive.
(2) The mean of the new set is smaller than R.

Source: 800Score Tests.

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 ?
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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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