Lucky2783 wrote:
Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S .
This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?
a) 1/2
b) 1/6
c) 1/8
d) 1/9
c) 1/12
Kudos for correct solution . Thanks
Brilliant, Brandon!
I arrived at the answer after looking around a bit for a pattern.
Given is my train of thought for those who did not think of going the options route:
Set S: {1/96 , 2/96 , ... 95/96 , 96/96 }
Operation: 2ab - (a+b) + 1
My first thought was that the two numbers selected should be complementary such that they lead to a simple answer which would eventually lead to some kind of sequence.
It made sense to pick 1/96 and 95/96 so that a+b = 1. Similarly, 2/96 and 94/96 etc. We were left with 2ab from each operation. But the leftover values were too complicated to work with. Didn't work.
Next thought was using a number which simplified the operation such as: if one of the numbers picked was 1 (the last number 96/96), what would happen?
2*1*b - (1 + b) + 1 = b
Left with b - the other number picked up. Say b was 95/96. We put back 95/96 and now the set doesn't have 1 anymore. Back to square one.
The next logical choice was to try what happens when one of the numbers picked is 1/2 (the number 48/96) since that would take care of 2 of 2ab.
2*(1/2)b - 1/2 - b + 1 = 1/2
This was what worked! When one of the numbers picked was 1/2, it eventually led to 1/2. So every time, all you do is pick 48/96 with any other number and you will always put back 1/2. You keep doing this till you are left with just 1/2.
Answer (A)