satishchandra wrote:There are four numbers in sequence. The average of first three is 6; the average of the last three is 7 and
the last number is 3 more than the first. The average of second & third numbers is:
(A) 6 (B) 6.5 (C) 5.75 (D) 7 (E) Indeterminate
This question is tricky if we make assumptions about what a sequence is.
A sequence is an ordered list of numbers.
Note: A sequence need not have a "nice" pattern like {5, 10, 15, 20, etc}
For example, {4, -1, 11, 999999} is a sequence.
Let's say that our sequence looks like this: {w, x, y, z}
The average of first three is 6
So, (w+x+y)/3 = 6, which means w+x+y = 18
The average of the last three is 7
So, (x+y+z)/3 = 7, which means x+y+z = 21
When we take the top equation (w+x+y = 18) and subtract it from the bottom equation (x+y+z = 21), we get z-w=3. In other words, z is 3 more than w.
So, we're
already able to conclude that the last number is 3 more than the first without being told that this is true. Hmmm, this suggests that the answer may be
E
Consider the following sequences that meet the given conditions:
case a: the sequence is: 6, 6, 6, 9
Here, the average of the 2nd and 3rd numbers is
6
case b: the sequence is: 5, 6, 7, 8
Here, the average of the 2nd and 3rd numbers is
6.5
The correct answer is
E
Cheers,
Brent
