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15 Sep 2014, 23:50
Official Solution: 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: \(range_{t \text{ and } s} \gt (t_{max}t_{min})+(s_{max}s_{min})\)? (1) The largest element of \(T\) is bigger than the largest element of \(S\). Given: \(t_{max} \gt s_{max}\), 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 \(t_{max}t_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(0 \gt s_{max}s_{min}\) and the answer would be NO; If it's \(s_{min}\) then the question becomes is \(t_{max}s_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(t_{min} \gt s_{max}\) 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 \(S\). Given: \(t_{min} \gt s_{max}\), so the largest element of the combined set is \(t_{max}\) and the smallest element of the combined set is \(s_{min}\). So the question becomes is \(t_{max}s_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(t_{min} \gt s_{max}\)? And that is given to be true, so the answer is YES. Sufficient. Answer: B
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Re: M1326
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09 Oct 2014, 14:07
Is there any way, we can present the above statements with some examples?
Some other way in which we don't have to resort to variables such as tmax and tmin.



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22 Oct 2014, 18:06
imagine the two sets as two line segments in a straight line. the lines can overlap each other, one inside another or be in different place altogether. to answer the question, we need to know if the lines overlap or there is a finite distance between them? The 1st point says the right end of line t is on the right side of right end of line s  this doesn't answer our question. The 2nd point says left end of line t is on the right side of right end of line s > there is a finite distance between the lines. We just got our answer



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31 Aug 2016, 03:46
I think this is a highquality question and I agree with explanation.



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Re: M1326
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23 Oct 2016, 05:19
Bunuel wrote: Official Solution:
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: \(range_{t \text{ and } s} \gt (t_{max}t_{min})+(s_{max}s_{min})\)? (1) The largest element of \(T\) is bigger than the largest element of \(S\). Given: \(t_{max} \gt s_{max}\), 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 \(t_{max}t_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(0 \gt s_{max}s_{min}\) and the answer would be NO; If it's \(s_{min}\) then the question becomes is \(t_{max}s_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(t_{min} \gt s_{max}\) 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 \(S\). Given: \(t_{min} \gt s_{max}\), so the largest element of the combined set is \(t_{max}\) and the smallest element of the combined set is \(s_{min}\). So the question becomes is \(t_{max}s_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(t_{min} \gt s_{max}\)? And that is given to be true, so the answer is YES. Sufficient.
Answer: B Hi Bunuel, Wont the hypothesis break down in the following screnario T=[1,2,3,4,5] S=[0,0,0,0] so the answer to the question would be "E" cannot be determined. Kindly let me know if i have misunderstood.



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23 Oct 2016, 05:40
rt1601 wrote: Bunuel wrote: Official Solution:
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: \(range_{t \text{ and } s} \gt (t_{max}t_{min})+(s_{max}s_{min})\)? (1) The largest element of \(T\) is bigger than the largest element of \(S\). Given: \(t_{max} \gt s_{max}\), 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 \(t_{max}t_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(0 \gt s_{max}s_{min}\) and the answer would be NO; If it's \(s_{min}\) then the question becomes is \(t_{max}s_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(t_{min} \gt s_{max}\) 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 \(S\). Given: \(t_{min} \gt s_{max}\), so the largest element of the combined set is \(t_{max}\) and the smallest element of the combined set is \(s_{min}\). So the question becomes is \(t_{max}s_{min} \gt t_{max}t_{min}+s_{max}s_{min}\). Or: is \(t_{min} \gt s_{max}\)? And that is given to be true, so the answer is YES. Sufficient.
Answer: B Hi Bunuel, Wont the hypothesis break down in the following screnario T=[1,2,3,4,5] S=[0,0,0,0] so the answer to the question would be "E" cannot be determined. Kindly let me know if i have misunderstood. For this case you still have an YES answer to the question for (2). The range of combined set is 5 and the sum of the ranges of T and S is 4 + 0 = 4.
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Re: M1326
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16 Jun 2017, 20:52
what if s =( 2,3,40) & t= (41,41,41)



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25 Mar 2018, 14:57
I think this is a highquality question and I agree with explanation.



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Re: M1326
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17 Apr 2018, 07:18
What about both S and T are single element set? Let S = {0} and T = {1} . T's smallest element is larger than S's biggest element . S and T can merge into {0,1} with range to be 1. The sum of the range of S and T is 1. So (2) is insufficient. What's wrong with this logic?



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17 Apr 2018, 12:52



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10 Jun 2018, 23:24
Let us take 2 sets: 1) S = {1,2,3} 2) T = {4,6}
This satisfies statement 2. Range of S = 2 Range of T = 2 Total = 4 Range of S,T = 5
It satisfies, so far so good.
Now take t ={4,5}, with S remaining same. Now Range of S,T is 4 and not greater than the sum. Answer should be (E) I believe



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11 Jun 2018, 04:39
AkarshS wrote: Let us take 2 sets: 1) S = {1,2,3} 2) T = {4,6}
This satisfies statement 2. Range of S = 2 Range of T = 2 Total = 4 Range of S,T = 5
It satisfies, so far so good.
Now take t ={4,5}, with S remaining same. Now Range of S,T is 4 and not greater than the sum. Answer should be (E) I believe In case S = {1, 2, 3} and T = {4, 5}. The sum of ranges of sets S and T = 2 + 1 = 3, while the range of combined set is 4. So, you'd still have an YES answer. The correct answer to the question is B, as explained above.
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Re: M1326
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01 Aug 2018, 06:41
What if it is a single set elements . It didnt talk about number of elements .



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21 Aug 2018, 02:56
Hi Bunuel, Let's say: S: 2 T: 4,6 Range of (S,T) will not be greater than individual ranges. Let;s say: S:1 T:7,9 Range of (S,T) will be greater than individual ranges. Please guide



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21 Aug 2018, 03:47
Milind27 wrote: Hi Bunuel, Let's say: S: 2 T: 4,6 Range of (S,T) will not be greater than individual ranges. Let;s say: S:1 T:7,9 Range of (S,T) will be greater than individual ranges. Please guide S: 2 > range = 0 T: 4,6 > range = 2 Combined {2, 4, 6} > range = 4.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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