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Bunuel
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A is a set of 11 consecutive integers. Is the sum of the terms of set 22 Nov 2018, 04:11
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45% (medium)

Question Stats:

65% (01:51) correct 35% (01:44) wrong based on 274 sessions

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A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A

M36-124

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A is a set of 11 consecutive integers. Is the sum of the terms of set 16 Nov 2021, 04:37
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Bunuel
A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A

M36-124
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Official Solution:


A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive

The set at hand is an evenly spaced, so, its median equals to its mean. The sum of the terms is \(Mean*Number\ of \ terms = positive*positive=positive\). Sufficient.

(2) The range of set A is less than the largest term in set A

The range of any set of 11 consecutive integers is 10. So, the minimum value of the largest term is 11, which makes all the numbers in the set positive. The sum is obviously positive. Sufficient.


Answer: D
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A is a set of 11 consecutive integers. Is the sum of the terms of set 22 Nov 2018, 05:08
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Bunuel
A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A
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Let the Nos be X, (X+1), ............... , (X+4) , .......................... , (X+10) {11 Consequtive no.}
Sum= 11X+55

Statement(1)- Median of A Positive= (X+4)>0 => X>-4
Therefore, X= -3,-2,-1,0,1,2..................Infinity
For any value of X from above, Sum=11X+55 is positive {Try for X=-3, sum=22}
Statement(2)- Range= (X+10)-X= 10
X+10>10=> X>0
So all no. are postive, therefore sum positive.

Ans. D- Both statements are sufficient individually.
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A is a set of 11 consecutive integers. Is the sum of the terms of set 23 Nov 2018, 03:29
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Bunuel
A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A
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From statement 1:
For stmtn 1 to be valid all the least range of integers can be from (-5 to 5) where median would be 0, so yes the sum of Set A would be positive always
sufficient

Frm 2:
For range to be < than largest term of Set A : all terms in set need to be +ve , sufficient else in case of -ve consective integers we would end up getting range < median...

sufficient

IMO D
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A is a set of 11 consecutive integers. Is the sum of the terms of set 24 Dec 2018, 01:45
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A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A
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Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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A is a set of 11 consecutive integers. Is the sum of the terms of set 08 Oct 2019, 05:34
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Bunuel
A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A
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Given: A is a set of 11 consecutive integers.
Asked: Is the sum of the terms of set A positive?

(1) The median of set A is positive
If median is positive; the sum of the terms of set A will be positive since they are 11 consecutive integers
SUFFICIENT

(2) The range of set A is less than the largest term in set A
10 < largest term in set A
the sum of the terms of set A will be positive
SUFFICIENT

IMO D
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A is a set of 11 consecutive integers. Is the sum of the terms of set 18 Oct 2019, 11:17
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Bunuel
A is a set of 11 consecutive integers. Is the sum of the terms of set A positive?

(1) The median of set A is positive
(2) The range of set A is less than the largest term in set A
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(1) The median of set A is positive: sufic.

consecutive integers, median=mean, median>0, then mean>0;

(2) The range of set A is less than the largest term in set A: sufic.

range=largest-smallest, r=tn-a;
r=tn-a<tn…tn-tn<a…a>0; set is all positive: mean>0;

Answer (D)
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