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BSchool Moderator B
Joined: 27 Feb 2014
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A certain list consists of 3 different numbers. Does the med  [#permalink]

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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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EgmatQuantExpert wrote:
avgroh wrote:
Bunuel,

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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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.

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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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.

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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GMAT 1: 620 Q46 V29 Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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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

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

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

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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.

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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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

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

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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

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

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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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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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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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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  [#permalink]

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prav04 wrote:
Friends, a doubt.
If we consider three numbers (0,7,11), this does not work out. Please clarify.
Bunuel

prav04

Statement 1 : Lets try to see if this set you choose is correct

Range =11

11= 2( 11-7)
Is 11 = 2(4) No

So this is not true , this mean you choose wrong set.

You need to select numbers that will evaluate the Statement .

Statement 2 says :
0+7+11= 18
Its neither equal to 3(0) or 3(7) or 3(11)

So the set you choose is incorrect.

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

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The question is pretty straightforward if one is able to churn out what the gist is:

the median of the 3 numbers will equal to the average (arithmetic mean) of the 3 numbers only in 2 cases-

1) when the 3 numbers are in AP (Arithmetic Progression)
2) when the 3 numbers are all equal

case (2) is not possible since it is mentioned in the question stem that "a list consists of 3 different numbers"

Now, all we've got to do is check if the two statements provide us with an AP or not.

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

There are various approaches posted above; However, I will check the above condition for an AP and non-AP

i> considering, 2 4 6
Range---> 6-2=4
median = 4
difference b/w greatest number and median--> 6-4=2
condition apply : yes, range(4)=2*2

You can check the results for AP's and non-Ap's and the result will always be true only in case of an AP. Hence, statement 1 is sufficient.

Statement 2 can be similarly tested using an AP and non-AP, and the result will hold true only in case of an AP

statement 2) The sum of the 3 numbers is equal to 3 times one of the numbers.

ii> considering, 2 4 6
the sum of 3 numbers = 12
condition apply--> indeed in all AP's the sum is equal to thrice of the middle number

Statement 2 is also sufficient. Hence, the answer is D
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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prav04 wrote:
Friends, a doubt.
If we consider three numbers (0,7,11), this does not work out. Please clarify.
Bunuel

See the three numbers you've provided are not in AP and the statements will more or less always hold true only when the 3 numbers are in AP
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A certain list consists of 3 different numbers. Does the med  [#permalink]

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the most important part in solving this question is to know what question is exactly asking.
Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers === Does SET evenly spread?
now we need to check one simple thing the avg of greatest and smallest is equal to the middle number.
r:range, g: greatest, s:smallest, m:middle
st1: r=g-s=>2*(g-m)
g-s=>2g-2m
2m=>g+s
m=>(g+s)/2 which is the formula for an average of arithmetic mean which means it is evenly spread. hence sufficient.
st2: g+m+s=3*m (m is the only possible number logically)
g+s=2*m
m=>(g+s)/2 which is the formula for an average of arithmetic mean which means it is evenly spread. hence sufficient.
please give kudos. urgent need for kudos.

Originally posted by Abhinav kumar on 10 Aug 2019, 01:03.
Last edited by Abhinav kumar on 10 Aug 2019, 01:08, edited 1 time in total.
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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the most important part in solving this question is to know what question is exactly asking.
Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers === Does SET evenly spread?
now we need to check one simple thing the avg of greatest and smallest is equal to the middle number.
r:range, g: greatest, s:smallest, m:middle
st1: r=g-s=>2*(g-m)
g-s=>2g-2m
2m=>g+s
m=>(g+s)/2 which is the formula for an average of arithmetic mean which means it is evenly spread. hence sufficient.
st2: g+m+s=3*m (m is the only possible number logically)
g+s=2*m
m=>(g+s)/2 which is the formula for an average of arithmetic mean which means it is evenly spread. hence sufficient.
please give kudos. urgent need for kudos.
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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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.
Easiest thing is just to picture it visually on a number line:
s---me---L
it says the distance between L & s = twice L to me, so the distances between the 3 points are equal, thus the median will also equal the average.

Algebra:
|L - s| = 2|L - me|
L - s = 2L - 2me
2me = L + s
me = (L+s)/2
The RHS is the average, which is equal to the middle term (median).

(2) The sum of the 3 numbers is equal to 3 times one of the numbers.
Since n=3, we know the median is part of the set.
The only way this works is if they are consecutive in some way i.e. {1,2,3}, because as in 1) what it really states is that the the middle number will equal the entire set (since 1 number is lower and 1 is higher)
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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Statement 1 : Let us say the set is a,b and c and a<b<c. The statement says c-a=2(c-b). This gives us a+c=2b or b=a+c/2. Meaning b is equidistant from a and c . This means the set ofa ,b and c is an evenly spaced set.The mean of an evenly spaced set is always equal to the median of the set. Sufficient .

Statement 2 : Once again we will assume the set has a,b and c as its members.So a+b+c=3a/3b/3c. Either we get a in the middle of b and c or b in the middle of a and c or c in the middle of a and b . In all those 3 iterations ,we have an evenly spaced set .Sufficient
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Re: A certain list consists of 3 different numbers. Does the med  [#permalink]

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Thanks a ton!

Posted from my mobile device Re: A certain list consists of 3 different numbers. Does the med   [#permalink] 27 Nov 2019, 11:36

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