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

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

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

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

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

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


Another approach(Logical) for 2

We know that sum of 3 numbers = 3* Average (one of the numbers is THE AVERAGE)

Now, since all the numbers are different, 2 other numbers will be smallest and largest number. So, that one number is the Median
Sufficient.

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.

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

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

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.

Let the three numbers be: S M L
Median = M
Average will be M too when deviation of S from M is the same as deviation of L from M i.e. S is as far away from M as L is away from M. In other words, if M = 10, and S = 8, then L = 12 and so on.

So we need to find whether S and L are equidistant from M or not.

(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median.
Range (diff between L and S) is twice the difference between L and M. So S is equidistant from M too.
Sufficient.

(2) The sum of the 3 numbers is equal to 3 times one of the numbers.
The sum = 3M (3S or 3L is not possible. Think why*.)
Average = 3M/3 = M
Sufficient

Answer (D)



*Can the sum be 3S?
3S = S + M + L
2S = M + L
We know that M and L are both greater than S. So M + L cannot add up to 2S.
Similarly sum cannot be 3L too.

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Solution:

Question Stem Analysis:

We need to determine whether the median of the 3 numbers of a certain list is equal to the average of the 3 numbers, given that the 3 numbers are different. We can let the 3 numbers be a, b, and c such that a < b < c. Therefore, we need to determine whether b = (a + b + c)/3.

Statement One Alone:

We see that c - a = 2(c - b). Simplifying, we have:

c - a = 2c - 2b

2b - a = c

Now, let’s substitute this for c in b = (a + b + c)/3:

b = (a + b + (2b - a))/3 ?

b = 3b/3 ?

b = b ?

We can see that the answer is yes. Statement one alone is sufficient.

Statement Two Alone:

If the sum of the three numbers is equal to 3 times one of the numbers, then the average of the three numbers is equal to one of the three numbers. The average cannot equal the smallest or the largest integer; thus the average must equal the median.

To see why the average cannot equal the smallest integer, notice that a < b and a < c. Adding these two inequalities, we obtain:

2a < b + c

Adding a to each side, we have:

3a < a + b + c

Dividing each side by 3, we see that a < (a + b + c)/3, which shows that the average is strictly greater than the smallest integer. It can be shown using similar arguments that the average is strictly less than the largest integer. Since we know the average is equal to one of the three numbers and since the average cannot equal the smallest or the largest number, it is equal to the median. Statement two alone is sufficient.

Answer: D

If a,b,c are the different numbers, then assuming a<b<c, we have b as the median

S1. Range = c-a = 2(c-b)
=> c - a = 2c - 2b
=> c = 2b-a

Mean = (a + b + c)/3 = (a + b + 2b-a)/3 = b = Median - Sufficient

S2. a+b+c = 3 * one number

If we take: a+b+c=3a => not possible since b and c are greater than a

If we take: a+b+c=3c => not possible since b and a are smaller than c

Thus, it must be: a+b+c=3b => Mean = (a+b+c)/3 = b = Median - Sufficient

Answer D

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Answer: Option D

Video solution by GMATinsight

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Hi Bunuel !

Thanks a lot for the explanation, it's concise and to the point - and these help a lot!

Could I ask one thing, though? In the second statement, you mention that the sum cannot be three times the smallest or largest number.
Why not? Is there a rule for series/sets that forbids this, or is this some logical deduction?

All the best,

Gil