Is the range of a combined set (S,T) bigger than the sum of

The range of a set is the difference between the largest and smallest elements of a set.

\(range_t=t_{max}-t_{min}\);
\(range_s=s_{max}-s_{min}\);

Question: \(range_{t&s}>t_{max}-t_{min}+s_{max}-s_{min}\)?

(1) The largest element of T is bigger than the largest element of S --> \(t_{max}>s_{max}\), so the largest element of combined set is \(t_{max}\) but we still don't know which is the smallest element of combined set:

If it's \(t_{min}\) then the question becomes is \(t_{max}-t_{min}>t_{max}-t_{min}+s_{max}-s_{min}\) --> is \(0>s_{max}-s_{min}\) and the answer would be NO;
If it's \(s_{min}\) then the question becomes is \(t_{max}-s_{min}>t_{max}-t_{min}+s_{max}-s_{min}\) --> is \(t_{min}>s_{max}\) and the answer would be sometimes NO and sometimes YES.

Not sufficient.

(2) The smallest element of T is bigger than the largest element of S --> \(t_{min}>s_{max}\), so the largest of combined set is \(t_{max}\) and the smallest element of combined set is \(s_{min}\).

So the question becomes is \(t_{max}-s_{min}>t_{max}-t_{min}+s_{max}-s_{min}\) --> is \(t_{min}>s_{max}\) and the answer is YES.

Sufficient.

Answer: B.
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Thanks Bunuel great explaination

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the range of a combined set (S,T) bigger than the sum of ranges of sets S and T ?

(1) The largest element of T is bigger than the largest element of S.
(2) The smallest element of T is bigger than the largest element of S.


Modify the original condition and the question. In order to satisfy the question ‘Is the range of a combined set (S,T) bigger than the sum of ranges of sets S and T?,’ the 2 sets of S and T shouldn’t be overlapped. For instance, if S={1,2} T={100,101}, it is yes. Thus, if you look at 2), there is no overlap, which is sufficient.
Therefore, the answer is B.


 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Let S= {x,...,y} where x is the smallest element and y is the largest element in set S
Let T= {z,...,a} where z is the smallest element and a is the largest element in set T
Let Range of Set S and T combined = \(R_{st}\)
Let Range of Set S = \(R_s\)
Let Range of Set T = \(R_t\)

Question : \(R_{st}\)> \(R_s\) + \(R_t\)

1. a>y

\begin{array}{c|c|c|c|c|c|c|c|c}
x & y & z & a & \(R_s\) & \(R_t\) & \(R_s\)+\(R_t\) & \(R_{st}\) & \(R_{st}\) > \(R_s\)+\(R_t\) \\
\hline
1 & 2 & 1 & 3 & 1 & 2 & 3 & 2 &2>3? No \\
1 & 1 & 2 & 2 & 0 & 0 & 0 & 1 &1>0? Yes \\
\hline
\end{array}

Yes and No. Hence Not Sufficient
2. z>y
\begin{array}{c|c|c|c|c|c|c|c|c}
x & y & z & a & \(R_s\) & \(R_t\) & \(R_s\)+\(R_t\) & \(R_{st}\) & \(R_{st}\) > \(R_s\)+\(R_t\) \\
\hline
1 & 2 & 3 & 3 & 1 & 0 & 1 & 2 & 2>1 ? Yes \\
1 & 1 & 2 & 2 & 0 & 0 & 0 & 1 & 1>0 ? Yes \\
1 & 2 & 3 & 4 & 1 & 1 & 2 & 3 & 3>2 ? Yes \\
\hline
\end{array}

Always Yes. Hence Sufficient

Hence answer is B.

Refer to the picture

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