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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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16 Sep 2014, 00:12
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 \(\{0, 0, 75\}\). Answer: A
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Re: D0119
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09 Oct 2015, 06:47
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 \(\{0, 0, 75\}\).
Answer: A 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?



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Re: D0119
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09 Oct 2015, 08:09
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 \(\{0, 0, 75\}\).
Answer: A 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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Re: D0119
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10 Oct 2016, 12:27
Bunuelwhat about negative multiples?? 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



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Re: D0119
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11 Oct 2016, 06:25
deepak268 wrote: Bunuelwhat about negative multiples?? 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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Statement 1.. All the numbers could also be 0 right? Therefore, I'd go with C



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Re: D0119
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10 Jun 2018, 08:08
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....}



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Re: D0119
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26 Dec 2018, 15:04
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."



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Re: D0119
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26 Dec 2018, 23:26
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: D0119
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22 Apr 2019, 08:29
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 \(\{0, 0, 75\}\).
Answer: A Hi Bunuel, Statement 1 says all members are consecutive multiples of 3. My question is, the numbers can repeat themselves and still be consecutive multiples of 3, isn't it? Like {0,0,3,6,9}. Here, the mean and median would be different, right? I first thought like you and was gonna go with A. Then, this thought struck me, and I ended up choosing E.
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Re: D0119
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22 Apr 2019, 09:00
OhsostudiousMJ 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 \(\{0, 0, 75\}\).
Answer: A Hi Bunuel, Statement 1 says all members are consecutive multiples of 3. My question is, the numbers can repeat themselves and still be consecutive multiples of 3, isn't it? Like {0,0,3,6,9}. Here, the mean and median would be different, right? I first thought like you and was gonna go with A. Then, this thought struck me, and I ended up choosing E. {0,0,3,6,9} is not a set of consecutive multiples of 3. 0 is a multiple of 3 but the next multiple is 3, it cannot be 0 again.
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Re: D0119
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22 Apr 2019, 10:45
Bunuel wrote: {0,0,3,6,9} is not a set of consecutive multiples of 3. 0 is a multiple of 3 but the next multiple is 3, it cannot be 0 again.
So, does a "set" mean distinct collection of numbers?
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