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Is the range of a six-number set greater than 3?

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Is the range of a six-number set greater than 3?

by vinni.k » Mon Jan 22, 2018 10:20 am
Is the range of a six-number set greater than 3?

(1) The sum of the five largest numbers is greater than 16.
(2) The average of the five smallest numbers is less than 2.6.

OA is C

The following is my working:-
(1). suppose there are 6 numbers - a,b,c,d,e,f. The sum of 5 largest numbers > 16
So, b + c + d + e + f > 16
2 + 3 + 4 + 5 + 6 > 16 and suppose a = 1 or -10 or 0 etc. Now, this clearly indicates that range is > 3.
Even if i take, -5 - 4 - 3 - 2 + 31 = -14 + 31 =17 > 16 . a = -6. Now here also range will be 31 - (-6) > 3.
Not sure how to get different answers for s(1). According to me s(1) is sufficient. (please clarify)

(2). a + b + c + d + e < 13
case 1: -1, 0, 1, 2, 3, 4. range : 5 > 3 (yes)
case 2: -3, -2, -1, -1, -1, 0 range : 3 = 3 (no)
clearly s(2) is insufficient.

Please let me know how to make s(1) insufficient. OA is C but from my analysis OA is A.
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by GMATGuruNY » Mon Jan 22, 2018 11:00 am
vinni.k wrote:Is the range of a six-number set greater than 3?

(1) The sum of the five largest numbers is greater than 16.
(2) The average of the five smallest numbers is less than 2.6.
Let the 6 numbers, in ascending order, be a, b, c, d, e, and f.
Range = biggest - smallest = f-a.
Question stem, rephrased:
Is f-a > 3?

Statement 1: b+c+d+e+f > 16
Let b=c=d=e=f=4, with the result that b+c+d+e+f = 5*4 = 20.
Case 1 : a=4
Here, f-a = 4-4 = 0, so the answer to the question stem is NO.
Case 2: a=0
Here, f-a = 4-0 = 4, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2: The average of a, b, c, d and e is less than 2.6
Let a=b=c=d=e=2, with the result that the average of a, b, c, d and e is 2.
Case 1 : f=2
Here, f-a = 2-2 = 0, so the answer to the question stem is NO.
Case 2: f=6
Here, f-a = 6-2 = 4, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statements combined:
If the average of a, b, c, d, and e were equal to 2.6, then their sum = (5)(2.6) = 13.
Since the average of a, b, c, d and e is LESS THAN 2.6, a+b+c+d+e < 13.
Adding together 13 > a+b+c+d+e and b+c+d+e+f > 16, we get:
13 + b+c+d+e+f > a+b+c+d+e + 16
13 + f > a + 16
f-a > 16-13
f-a > 3.
Thus, the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.
Last edited by GMATGuruNY on Mon Jan 22, 2018 12:52 pm, edited 1 time in total.
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by ceilidh.erickson » Mon Jan 22, 2018 12:26 pm
vinni.k wrote:Is the range of a six-number set greater than 3?

(1) The sum of the five largest numbers is greater than 16.
(2) The average of the five smallest numbers is less than 2.6.

OA is C

The following is my working:-
(1). suppose there are 6 numbers - a,b,c,d,e,f. The sum of 5 largest numbers > 16
So, b + c + d + e + f > 16
2 + 3 + 4 + 5 + 6 > 16 and suppose a = 1 or -10 or 0 etc. Now, this clearly indicates that range is > 3.
Even if i take, -5 - 4 - 3 - 2 + 31 = -14 + 31 =17 > 16 . a = -6. Now here also range will be 31 - (-6) > 3.
Not sure how to get different answers for s(1). According to me s(1) is sufficient. (please clarify)

(2). a + b + c + d + e < 13
case 1: -1, 0, 1, 2, 3, 4. range : 5 > 3 (yes)
case 2: -3, -2, -1, -1, -1, 0 range : 3 = 3 (no)
clearly s(2) is insufficient.

Please let me know how to make s(1) insufficient. OA is C but from my analysis OA is A.
You're making a very common assumption here: you're assuming that the terms must be DISTINCT, but the question never said that.

As Mitch pointed out, the easiest way to prove insufficiency on each statement is to ask "what if all the terms are the same?" In fact, this is always the FIRST case you should test on statistics DS questions (if applicable). If will often be the fastest way to prove insufficiency.
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Is the range of a six-number set greater than 3?

by vinni.k » Mon Jan 22, 2018 7:41 pm
GMATGuruNY wrote: Statement 1: b+c+d+e+f > 16
Let b=c=d=e=f=4, with the result that b+c+d+e+f = 5*4 = 20.
Case 1 : a=4
Here, f-a = 4-4 = 0, so the answer to the question stem is NO.
Case 2: a=0
Here, f-a = 4-0 = 4, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
Thanks Mitch and Ceilidh,

But here is my question for s(1). If i take a = 4, then i am considering "a" as the largest number, that means all 6 numbers are the largest numbers.

b+c+d+e+f
4+4+4+4+4
Now, i take "a" as 4, 4+4+4+4+4+4, then it is largest too. then what is the difference between 5 largest number and a?
What i thought was b,c,d,e,f are the largest so i understand that they can all be 4, but if i make a = 4, then we don't have five largest numbers; however, now we have 6 largest numbers. Please clarify.

Same questions for s(2), though i was able to get s(2) when i tried.
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by ceilidh.erickson » Mon Jan 22, 2018 8:42 pm
vinni.k wrote:
GMATGuruNY wrote: Statement 1: b+c+d+e+f > 16
Let b=c=d=e=f=4, with the result that b+c+d+e+f = 5*4 = 20.
Case 1 : a=4
Here, f-a = 4-4 = 0, so the answer to the question stem is NO.
Case 2: a=0
Here, f-a = 4-0 = 4, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
Thanks Mitch and Ceilidh,

But here is my question for s(1). If i take a = 4, then i am considering "a" as the largest number, that means all 6 numbers are the largest numbers.

b+c+d+e+f
4+4+4+4+4
Now, i take "a" as 4, 4+4+4+4+4+4, then it is largest too. then what is the difference between 5 largest number and a?
What i thought was b,c,d,e,f are the largest so i understand that they can all be 4, but if i make a = 4, then we don't have five largest numbers; however, now we have 6 largest numbers. Please clarify.

Same questions for s(2), though i was able to get s(2) when i tried.
I see your point. To my mind, though, "the 5 largest" does not have to imply that 5 are larger, 1 is smaller. Consider: if I have the set [3, 3, 3, 3, 3], the median is 3. But which one of those 3's is the median? Irrelevant! Likewise with 6 identical terms, we can still have 5 largest - any 5 of the 6. I could see someone else reasonably interpreting it the other way, though.

Here's the good news on the GMAT - I've never seen a question actually hinge on this kind of semantic distinction. I doubt one ever would (because then some high-level scorers like me might get it wrong, and the question would give them bad data and they'd throw it out).

Either way on this problem, the set [3, 4, 4, 4, 4, 4] would prove insufficiency. So you should be thinking of counterexamples where some/all of the terms are the same.
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by GMATGuruNY » Tue Jan 23, 2018 5:06 am
vinni.k wrote: If i take a = 4, then i am considering "a" as the largest number, that means all 6 numbers are the largest numbers.
The OG defines range as the largest value minus the smallest value.
{4, 4, 4, 4, 4, 4}.
Here, the largest value and the smallest value are THE SAME -- 4 -- with the result that the range = 4-4 = 0.
Even if all of the data points are equal, the set is still considered to have a largest value and a smallest value.

I would abide by the following definitions:
a, b and c may be considered the smallest numbers in a set if no number in the set is smaller than a, b or c.
d, e and f may be considered the largest numbers in a set if no number in the set is larger than d, e or f.

Statement 1: The sum of the five largest numbers is greater than 16.
Here:
b, c, d, e, and f may be considered the five largest numbers as long as no number in the set is greater than b, c, d, e, or f.
Case 1: a=b=c=d=e=f=4
This case is valid because no number in the set is greater than b, c, d, e, or f and because the sum of b, c, d, e and f is greater than 16.
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by vinni.k » Tue Jan 23, 2018 11:01 am
Mitch and Ceilidh,

Thanks for enlightening me. Now, i have understood this problem clearly. Yes, i had some reservations on this question, making me difficult to understand, but as i am fine now, so this learning will help me in other similar kind of questions.

I don't see any upvotes button. "Thanks" button was better or earlier version was far better.

Regards
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