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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 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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Nov 2013, 01:16, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Got D here
1) ascertains that diff b/w largest-median and smallest-median number is same and answer would be yes
2) the only "one" number which would be equivalent to the sum of the other 3 would be the median. Hence, "yes", mean = median
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Paul wrote:
Got D here
1) ascertains that diff b/w largest-median and smallest-median number is same and answer would be yes
2) the only "one" number which would be equivalent to the sum of the other 3 would be the median. Hence, "yes", mean = median


Wow..I didn't understand anything while solving. Even after reading your explanation I can fully understand this. Can anyone shed some light :)

If we take the no. as x,y,z
1. z-x = 2(z-y)
z-x = 2z -2y
2y-x = z

Couldn't conclude anything

2. x+y+z = 3(x or y or z)

I was not sure what to do ahead. But still kind of realized that it could be y

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

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Hi b2bt! Here is my train of thoughts:
Median equals average when you have evenly spaced set of numbers.
(1) tells you that, so sufficient.
(2) if a is a multiplier and n1, n2, n3 = 1, 2, 3 than numbers in your (evenly spaced?) set can be represented as following: a*n1, a*n2, a*n3. Let's find their sum.
an1+an2+an3= a(n1+n2+n3)= a(n1+n1+1+n1+2)= a(3n1+3)= 3a(n1+1)= 3a*n2, so sufficient

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bogdanbond wrote:
Hi b2bt! Here is my train of thoughts:
Median equals average when you have evenly spaced set of numbers.
(1) tells you that, so sufficient.
(2) if a is a multiplier and n1, n2, n3 = 1, 2, 3 than numbers in your (evenly spaced?) set can be represented as following: a*n1, a*n2, a*n3. Let's find their sum.
an1+an2+an3= a(n1+n2+n3)= a(n1+n1+1+n1+2)= a(3n1+3)= 3a(n1+1)= 3a*n2, so sufficient


It took me a while but I understood! Thanks!
I think I lack conceptual clarity here..

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New post 18 May 2015, 14:15
Bunuel,

Could you please help with interpreting statement #2?

Thanks

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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
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Question: Is median = mean?

St1: tells us that the median is equidistant from the other two numbers on the number line. Thus, median = mean. Sufficient.
Attachment:
Untitled.png
Untitled.png [ 1.06 KiB | Viewed 17498 times ]

St2: sum of 3 no.s = 3* (one of the numbers)
we know that sum of 3 no.s = 3* (mean)
Thus, mean = one of the numbers. Now, as long as the 3 nos are distinct. the largest and smallest nos cannot be equal to mean. Thus, median = mean. Sufficient.

Answer D.

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New post 29 Jul 2015, 06:20
I think the solution provided in the OG is messed up for this question. Bunuel can you please check and tell

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Heres how to translate statement #2 in two steps:

The sum of the 3 numbers is equal to 3 times one of the numbers

=> The (sum of the 3 numbers) divided by 3 is equal to one of the numbers
=> The mean is equal to one of the numbers

=> Mean can only be equal to the median # in an oddly sized set, sufficient

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My approach for statement 2

Take any number as "one of the numbers"

Let 5 is that number. So the sum is 5*3 = 15.
The sum of other two numbers is 15-5 = 10
So, we get the following combinations of two other different numbers
(1,9),(2,8), (3,7), (4,6)------ In all the cases median and mean is 5. Sufficient

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Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.


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.

==> according to variable approach method, we have 3 variables listed a<M<b. Transforming the original condition and the question we have M=(a+M+b)/3? and thus 3M=a+M+b. Since this is same as 2), 2) itself is sufficient. Transforming again gives us 2M=a+b, and since a=2M-b -> range=b-a=b-(2M-b)=2b-2M=2(b-M) is the same as 1), this is also sufficient. Therefore the answer is D.
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Solved it by picking some numbers :
Median = Average in 2 cases: 1. Evenly spaced set of numbers 2. All numbers are equal

1. let's say our numbers are: 4,8,12
12-4=2(12-8) Ok, if it's not evenly spaced then use this case 2,5,7 --> 7-2=/2(7-5) Sufficient

2. This statement tells us directly that all the numbers are equal Sufficient (D)
Here we can pick some varriables in ascending order to check my solution:
x + y + z =3x,(y − x) + (z − x) = 0 -> y=x, z=x -> y=z=x we'll get the same solution for all 3 cases =3x, 3z or 3y
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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 descending 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 descending 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


Hi Harsh,

I believe in the above statements you mean three numbers {a, b, c} arranged in descending order? Looks like a typo.

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

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New post 06 Apr 2016, 12:36
Bunuel,please explain this question.Not able to understand how statement 2 is sufficient

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New post 07 Apr 2016, 07:29
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 the 3 numbers be x,y and z(in ascending order).

From statement 1, we have,

z-x = 2(z-y)
or ,

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New post 07 Apr 2016, 07:33
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 the 3 numbers be x,y and z(in ascending order).

From statement 1, we have,

z-x = 2(z-y)
or ,z-x = 2z - 2y
or, y = (z-x)/2.

Thus, y (the median) is also the average. Hence statement 1 is sufficient.

For statement 2, think of the given info conceptually.
The sum of the 3 numbers cannot be equal to 3 times the highest or the lowest number. Either the 3 numbers have to be equal or the only other possibility is that the sum of the 3 numbers is equal to 3 times the middle number.

Therefore, option (D) is correct.

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New post 30 Jul 2016, 05:11
Bunuel plz help in the second statement how is that sufficient most of the explanations above are cryptic!

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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.
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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?

(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.


Bunuel Thanks, more than getting it right I have always gained a new perspectives with most of your explanations!

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