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A certain list consists of 3 different numbers. Does the med

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A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 02 Aug 2016, 05:58
Hi Bunuel,

The OG16 started the explanation for this question by saying this: "Let the numbers be x, y, and z so that x ≤ y ≤ z...." but the question stem said that: "A certain list consists of 3 different numbers..."

From my understanding, it should be the case that : x # y # z and x < y < z

Is the explanation for this question wrong or my reasoning is wrong?

P/S: This question is from OG16 DS Q135
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 21 Oct 2016, 21:59
EgmatQuantExpert wrote:
avgroh wrote:
Bunuel,

Could you please help with interpreting statement #2?

Thanks

Hi avgroh,

Let's assume three numbers {a, b, c} arranged in ascending order. In this case b will be the median.

Statement-II
St-II tells that the sum of these three number is equal to 3 times one of the numbers. Let's take all the possible cases:

1. a + b + c = 3a i.e. 2a = b + c. However we know that a > b and a > c, therefore 2a > b + c. Thus we can reject this case.

2. a + b + c = 3c i.e. 2c = a + b. However we know that c < a and c < b, therefore 2c < a + b. Thus we can reject this case too.

3. a + b + c = 3b i.e. 2b = a + c. We know that b < a but b > c, thus this is the only possible case.

Solving this would give us b = (a + c)/2 i.e. b is the mean of a & c. Since the only other number in the set is b, we can say that b is the mean of the set {a, b, c}.
As b is also the median of this set we can definitely say that mean of the set = median of the set.

Hence st-II is sufficient to answer the question.

Statement-I
Also adding the explanation for St-I here: St-I tells us that the range of 3 numbers is twice the difference between the greatest number and the median.

For the set {a, b, c} arranged in ascending order, range would be the difference between the greatest and the smallest number i.e. st-I tells us that a - c = 2(a -b) i.e. b = (a + c)/2 which again tells us that b is the mean of the numbers a & c. Since the only other number in the set is b, we can say that b is the mean of the set {a, b, c}.
As b is also the median of this set we can definitely say that mean of the set = median of the set.

Hence st-I is sufficient to answer the question.

Hope this helps :)

Regards
Harsh



i tried this way:

let a b c are three numbers. it is asked: (a+b+c)/3 =b?

stamnt. 1: range = 2(c-b)
i.e c-a = 2(c-b)
2b-a=c
now put the value of c in LHS of question stem
(a+b+2b-a)/3 =b
hence,, yes, average is equal to median.

stmnt two is possible only when the number are in AP. therefore Average=Median

So option D is ans.
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A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 13 Nov 2016, 02:33
Bunuel wrote:

A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?

Say the three numbers are x, y, and x, where x < y < z. The median would be y and the average would be (x + y + z)/3. So, the question asks whether y = (x + y + z)/3, or whether 2y = x + z.

(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median --> the range is the difference between the largest and the smallest numbers of the set, so in our case z - x. We are given that z - x = 2(z - y) --> 2y = x + z. Sufficient.

(2) The sum of the 3 numbers is equal to 3 times one of the numbers --> the sum cannot be 3 times smallest numbers or 3 times largest number, thus x + y + z = 3y --> x + z = 2y. Sufficient.

Answer: D.

Hope it's clear.


Bunuel

If Statement 1 ended up being say, 2y = 2x + z (instead of 2y=x+z), would the statement still be sufficient? In other words, when matching equations from the statements with the equation in the question, is it sufficient if you can get ONE single equation (even if it doesn't match), or is it only sufficient if the resulting equation from the statement matches the equation in the question exactly?
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 13 Nov 2016, 05:43
1
allthewayup16 wrote:
Bunuel wrote:

A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?

Say the three numbers are x, y, and x, where x < y < z. The median would be y and the average would be (x + y + z)/3. So, the question asks whether y = (x + y + z)/3, or whether 2y = x + z.

(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median --> the range is the difference between the largest and the smallest numbers of the set, so in our case z - x. We are given that z - x = 2(z - y) --> 2y = x + z. Sufficient.

(2) The sum of the 3 numbers is equal to 3 times one of the numbers --> the sum cannot be 3 times smallest numbers or 3 times largest number, thus x + y + z = 3y --> x + z = 2y. Sufficient.

Answer: D.

Hope it's clear.


Bunuel

If Statement 1 ended up being say, 2y = 2x + z (instead of 2y=x+z), would the statement still be sufficient? In other words, when matching equations from the statements with the equation in the question, is it sufficient if you can get ONE single equation (even if it doesn't match), or is it only sufficient if the resulting equation from the statement matches the equation in the question exactly?


In this case the statement would not be sufficient because 2y = 2x + z and 2y = x + z and x < y<z can simultaneously be correct.
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 10 Dec 2016, 09:31
Hi all,

I think that the approach that official guide offers seems to be too time consuming. While going through this question I simply used substitution approach.
First, the basic condition is that mean should be equal to median, so the number set must be evenly spaced.
(1) Let's take x=10,y=12,z=14 - evenly spaced set, then
14-10=2(14-12)
4=4
SUFFICIENT
(2)
10+12+14=3(12)
36=36
As long as it tells that "the sum of the 3 numbers is equal to 3 times one of the numbers" it doesn't say the the sum should be equal to 3 times of each of the number, therefore it is SUFFICIENT

Asking GMAT geeks, please advise if subject approach is eligible in terms of this particular question???

P.S. I cracked it namely above mentioned way.

Thank you in advance.
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A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 16 Apr 2017, 00:07
Bunuel wrote:
HBSdetermined wrote:
Bunuel plz help in the second statement how is that sufficient most of the explanations above are cryptic!


A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?

Say the three numbers are x, y, and x, where x < y < z. The median would be y and the average would be (x + y + z)/3. So, the question asks whether y = (x + y + z)/3, or whether 2y = x + z.

(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median --> the range is the difference between the largest and the smallest numbers of the set, so in our case z - x. We are given that z - x = 2(z - y) --> 2y = x + z. Sufficient.

(2) The sum of the 3 numbers is equal to 3 times one of the numbers --> the sum cannot be 3 times smallest numbers or 3 times largest number, thus x + y + z = 3y --> x + z = 2y. Sufficient.

Answer: D.

Hope it's clear.



men you're brilliant

I solved it by testing.

I just chose one number and then tried to fulfil our constraints with the other two numbers, then realised that the median equaled the average.

But I feel much better with your approach.

So thank you very much
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 17 Aug 2017, 03:20
1) let the 3 numbers be a,b,c for first case let us take in increasing order
c-a = 2( c-b)
c-a = 2c-2b
c= 2b-a

mean of abc is (a+b+c)/3
a+b+2b-a /3
3b/3
b

sufficient. A or D can be answer..

2) let 3 numbers be a,b,c not necessarily in increasing order

a+b+c = 3a. b+c =2a. now let us say a=2 b+c =2 in this case the 2 number can be 1,1 and the set will become 1,1,1 so the median will be 1. and avg of 1+1+1 / 3 = 1, in same case assume a=2 and b =0 and c=4 to satisfy equation then here we have numbers in increasing order as 0,2,4 here median is 2 and avg of 3 numbers is 0+2+4 / 3 = 6/3 = 2. same as median.

this process is valid for whether it is 3 times b or 3 times c .

therefore both are sufficent

D is the answer
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 01 Jun 2018, 03:27
Tried with plugin the values
Three numbers y z
1 says z-x=2(z-y)
So
by plugin in the random number to support the statement
10-6=2(10-8)
4=4

2
3*8=24

So D is the answer
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A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 04 Jun 2018, 21:19
Note to remember : If the number is in Arithmetic Progression series then Median=Average for odd number in series .

So, if we just prove that the number is in Arithmetic progression then our work is finish.

1. Option 1
The range of the 3 numbers is equal to twice the difference between the greatest number and the median.

Lets the three number is A,B,C, and is in increasing order

So, as per the option A

C-A = 2(C-B)
C-A = 2C - 2B
C-A = C+C - B -B
B-A = C-B


For a number to be in Arithmetic Progression , Difference between two consecutive number is constant.

So, from above , Difference between 2nd and 1st number ( B-A ) and difference between 3rd and 2nd number ( C-B) is same. Therefore A,B & C is in Arithmetic progression .

Thus, Median = Average. So Option 1 is Sufficient

Option 2 : The sum of the 3 numbers is equal to 3 times one of the numbers.

Lets the three number is A,B,C, and is in increasing order

Case 1:

A+B+C = 3A
B+C = A+A
B-A =A-C, So, number not in Arithmetic progression, (For Arithmetic progression, [b] B-A = C-B [/b] )
So, We can not prove B is median or Not.

Case 2:

A+B+C = 3B
A+C = B+B
C-B =B-A, So, number is in Arithmetic progression, (For Arithmetic progression, [b] B-A = C-B [/b] )
Hence, B is median.

Case 3:

A+B+C = 3C
A+B = C+C
C-A =B-C, So, number not in Arithmetic progression, (For Arithmetic progression, B-A = C-B )
So, We can not prove B is median or Not.

As, Case 2, gives OK result. So Option 2 is Sufficient

So, Option D, is correct.
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 19 Jun 2018, 01:52
Common ... is this plausible to be solved under 2 minutes?
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 19 Jun 2018, 04:59
karovd wrote:
A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?

(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median.
(2) The sum of the 3 numbers is equal to 3 times one of the numbers.


Let S = the smallest number, M = median, and L = the largest number
For the median of the 3 numbers to be equal to the average of the 3 numbers, the 3 numbers must be EVENLY SPACED.
Question stem, rephrased:
Are the 3 numbers evenly spaced?

Statement 1:
Case 1: M=2 and L=3
Range = twice the difference between the largest and the median = 2(3-2) = 2.
Since the smallest must be 2 less than the largest, S = 3-2 = 1.
The resulting set -- 1, 2, 3 -- is evenly spaced.
Thus, the answer to the rephrased question stem is YES.

Case 2: M=10 and L=15
Range = twice the difference between the largest and the median = 2(15-10) = 10.
Since the smallest must be 10 less than the largest, S = 15-10 = 5.
The resulting set -- 5, 10, 15 -- is evenly spaced.
Thus, the answer to the rephrased question stem is YES.

Cases 1 and 2 illustrate that -- given the condition in Statement 1 -- the resulting set must be evenly spaced.
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.

Statement 2:
In Case 1, the sum of the 3 numbers (1+2+3 = 6) is equal to 3 times the median (2).
In Case 2, the sum of the 3 numbers (5+10+15 = 30) is equal to 3 times median (10).
Implication:
Cases 1 and 2 also satisfy Statement 2.
In Cases 1 and 2, the answer to the rephrased question stem is YES.

Test one more case.
Case 3: Let one of the numbers = 20 and the sum of the 3 numbers = 3*20 = 60.
Since the sum of all 3 numbers = 60 and one of the numbers = 20, the sum of the other 2 numbers = 60-20 = 40.
The following sets are possible:
19, 20, 21
18, 20, 22
17, 20, 23
10, 20, 30
And so on.
In every case, the set is EVENLY SPACED.
Implication:
The answer to the rephrased question stem is YES.
SUFFICIENT.


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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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New post 17 Sep 2018, 08:30
Friends, a doubt.
If we consider three numbers (0,7,11), this does not work out. Please clarify.
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Re: A certain list consists of 3 different numbers. Does the med &nbs [#permalink] 17 Sep 2018, 08:30

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