D

for mean = median, the sequence must be evenly spaced.

II) lets take 1 2 3

1+2+3 = 6, which is also 3*2.

now lets take 1 4 5

can't replicate that. so II is sufficient

I) Hi - Low = 2(Hi - Median)

lets take 1 2 3 again.

3-1 = 2 (3 - 1), which is equal to 2. works.

lets take 1 4 5. can't replicate that. so I is also sufficient.

a, b, c are the numbers where a < b < c

median = b and average is (a+b+c)/3

Is b = (a+b+c)/3 ?
3b = a + c +b
2b = a + c
1:
(c-a) = 2(c-b)
c-a = 2c - 2b
2b=c +a

So, Statement 1 is true.

2:
a+b+c=3a or 3b or 3c

2 does not lead to statement above.

Answer: A

i go with A as well, as a+b+c cud be equal to 3a,3b or 3c !

It is D)

Stmt I
Let h be the highest number
Let the median be the middle number
Let l be the lowest number

h-l = 2(h-median)
h-l = 2h-2median

2median = h+l

median = (h+l)/2 (1)

mean = l+median+h/3

3 mean = l+h+median

3mean = l+ h + (l+h)/2

6 mean = 3(l+h)

mean = l+h/2 (2)

From (1) and (2) mean and median are the same

SUFF

Stmt II

Let 3 numbers be x,y,z

3x= x+y+z i.e 2x = y+z i.e x = y+z/2

(Or)

y = x+z/2 (similar to above)

(or)

z = x+y/2

Again this is telling us that there is a common difference between one number to the next number and is is the same SINCE ONE OF THE NUMBERS THAT FALLS EXACTLY BETWEEN 2 OTHERS NUMBERS IS THE AVERAGE OF THE 2 NUMBERS

Mean = Median (for consecutive numbers or for numbers where the common difference is the same from one number to thenext after arranging them in ascending order)

SUFF


Choose D)

You can think about statement 2 conceptually. The sum of three different numbers cannot equal 3 times the smallest one, because the sum will be more than 3 times the smallest since the other two are greater, by definition, than the smallest. The sum of the numbers cannot be 3x the biggest for the opposite reason. Therefore, if the sum equals 3x a number, it must be the middle number. But if you know that the sum is 3 times the middle, and you also know that is equal to the sum of all three, the larger and smaller must be equidistant from the middle number, and then the middle would therefore be the average as well as the median.

IMO D

Nice question!