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# D01-19

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Math Expert
Joined: 02 Sep 2009
Posts: 52296

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15 Sep 2014, 23:12
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Difficulty:

45% (medium)

Question Stats:

66% (00:40) correct 34% (00:54) wrong based on 164 sessions

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Is the mean of set $$S$$ greater than its median?

(1) All members of $$S$$ are consecutive multiples of 3

(2) The sum of all members of $$S$$ equals 75

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Math Expert
Joined: 02 Sep 2009
Posts: 52296

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15 Sep 2014, 23:12
Official Solution:

First of all:

The median of a set with odd # of terms is just a middle term (when ordered in ascending/descending order).

The median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

Next, one of the most important properties of evenly spaced set (aka arithmetic progression):

In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\{7,11,15,19\}$$, $$mean=median=\frac{7+19}{2}=13$$.

(1) All members of $$S$$ are consecutive multiples of 3. This statement says that $$S$$ is an evenly spaced set, thus its mean equals to its median. Sufficient.

(2) The sum of all members of $$S$$ equals 75. Clearly insufficient, consider two sets $$\{25, 25, 25\}$$ and $$\{0, 0, 75\}$$.

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Manager
Joined: 22 Aug 2012
Posts: 73
Concentration: Technology
GMAT 1: 710 Q47 V40

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09 Oct 2015, 05:47
Bunuel wrote:
Official Solution:

First of all:

The median of a set with odd # of terms is just a middle term (when ordered in ascending/descending order).

The median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

Next, one of the most important properties of evenly spaced set (aka arithmetic progression):

In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\{7,11,15,19\}$$, $$mean=median=\frac{7+19}{2}=13$$.

(1) All members of $$S$$ are consecutive multiples of 3. This statement says that $$S$$ is an evenly spaced set, thus its mean equals to its median. Sufficient.

(2) The sum of all members of $$S$$ equals 75. Clearly insufficient, consider two sets $$\{25, 25, 25\}$$ and $$\{0, 0, 75\}$$.

Don't mean to sound picky but is this sentence correct? "evenly spaced set (aka arithmetic progression)" don't evenly spaced sets include geometric progressions and arithmetic progressions?
Math Expert
Joined: 02 Sep 2009
Posts: 52296

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09 Oct 2015, 07:09
Jonas84 wrote:
Bunuel wrote:
Official Solution:

First of all:

The median of a set with odd # of terms is just a middle term (when ordered in ascending/descending order).

The median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

Next, one of the most important properties of evenly spaced set (aka arithmetic progression):

In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\{7,11,15,19\}$$, $$mean=median=\frac{7+19}{2}=13$$.

(1) All members of $$S$$ are consecutive multiples of 3. This statement says that $$S$$ is an evenly spaced set, thus its mean equals to its median. Sufficient.

(2) The sum of all members of $$S$$ equals 75. Clearly insufficient, consider two sets $$\{25, 25, 25\}$$ and $$\{0, 0, 75\}$$.

Don't mean to sound picky but is this sentence correct? "evenly spaced set (aka arithmetic progression)" don't evenly spaced sets include geometric progressions and arithmetic progressions?

An evenly spaced set is one in which the gap between each successive element in the set is equal.

A geometric progression is NOT an evenly spaced set. For example, 2, 4, 8, 16, 32, ... is NOT evenly spaced because the gap between each successive element in the set is NOT equal.

Hope it's clear.
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Joined: 27 Feb 2015
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Concentration: General Management, Economics
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10 Oct 2016, 11:27
1
Bunuel
what if the set is {-6,-3,0,3} is this not consecutive multiples?? in this case mean=-2 and median is -1.5
I think Statement 1 should say positive consecutive multiples? OR is that in GMAT multiples of any no. are always taken as positive multiples??
thanks
Math Expert
Joined: 02 Sep 2009
Posts: 52296

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11 Oct 2016, 05:25
deepak268 wrote:
Bunuel
what if the set is {-6,-3,0,3} is this not consecutive multiples?? in this case mean=-2 and median is -1.5
I think Statement 1 should say positive consecutive multiples? OR is that in GMAT multiples of any no. are always taken as positive multiples??
thanks

We can consider negative integers as multiples too. For example, -4 is a multiple of 2.

The problem is that the mean of {-6,-3,0,3} is -1.5, not -2.
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Location: India
Concentration: Finance, Entrepreneurship
Schools: CBS '22, LBS MIF '20
GPA: 3.55
WE: Investment Banking (Investment Banking)

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10 Jun 2018, 06:11
Statement 1.. All the numbers could also be 0 right? Therefore, I'd go with C
Intern
Joined: 27 Apr 2015
Posts: 4
Location: Singapore
Concentration: General Management, Strategy
GMAT 1: 620 Q47 V28
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10 Jun 2018, 07:08
drithisunil wrote:
Statement 1.. All the numbers could also be 0 right? Therefore, I'd go with C

It's given in statement 1 that
All members of S are consecutive multiples of 3
And this doesn't seem to be S = {0,0,0,0,0,0....}
Intern
Joined: 28 Oct 2018
Posts: 3

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26 Dec 2018, 14:04
I do not understand how the answer is "A."
It depends on how many "members" we talk about.

(3+6+9)/3 = 6 (mean) and median 6
(3+6+9+12)/3 = 10 (mean) and 7.5 median

Therefore, I picked "C."
Math Expert
Joined: 02 Sep 2009
Posts: 52296

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26 Dec 2018, 22:26
viitalik wrote:
I do not understand how the answer is "A."
It depends on how many "members" we talk about.

(3+6+9)/3 = 6 (mean) and median 6
(3+6+9+12)/3 = 10 (mean) and 7.5 median

Therefore, I picked "C."

The mean of 3, 6, 9, and 12 is (3+6+9+12)/4 = 7.5. As it'd given in the solution: in ANY evenly spaced set the arithmetic mean (average) is equal to the median.
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Re: D01-19 &nbs [#permalink] 26 Dec 2018, 22:26
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# D01-19

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