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(1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(a-b\) and \(b-a\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient.
(2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
In option 2, The range (largest - Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, b-a, a-b ?
It's explained in the solution above:
If a = b = 0, then S = {0, 0, 0} --> the range = 0 - 0 = 0 = 2b. If a = 0 and b = 1, then S = {-1, 1, 1} --> the range = 1 - (-1) = 2 = 2b.
_________________
a+b - (a-b) = 2b, which is true for a=b=0 as well. Hence that makes b-a as the median. From first option I got a+b as the median. What am I missing here?
Range= Highest Number - Lowest Number R=2b In order for R to be 2b, the only possible combination is (b+a)-(a-b). Thus, (a-b)≤(b-a)<(b+a) or (a-b)<(b-a)≤(b+a) So b-a is the median.
Range= Highest Number - Lowest Number R=2b In order for R to be 2b, the only possible combination is (b+a)-(a-b). Thus, (a-b)≤(b-a)<(b+a) or (a-b)<(b-a)≤(b+a) So b-a is the median.
Where am I wrong?
(2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
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Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.
Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.
A set, by definition, is a collection of elements without any order. (While, a sequence, by definition, is an ordered list of terms.)
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.
as explained even if a=b=0 or a=1, b=0, mean is b-a.
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.
as explained even if a=b=0 or a=1, b=0, mean is b-a.
The point is that in data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.
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Statement 1 : Given that mean is a+b. (a+b)/3=a+b ; i.e a+b=0. This means that the set is -2b,0,2b. The median in this case becomes zero. We get a definite answer.
Statement 2 : Range is 2b. The middle value is b-a. This is not a definite value. Hence not sufficient.
(1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(a-b\) and \(b-a\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient.
(2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
(1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(a-b\) and \(b-a\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient.
(2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
close
50% (02:01) correct 
