Bunuel wrote:
If p, r, and s are consecutive integers in ascending order and x is the average (arithmetic mean) of the three integers, what is the value of x?
(1) Twice x is equal to the sum of p, r, and s.
(2) The sum of p, r, and s is zero.
I answered this question correctly but I took 3.03 minutes instead of less than 1 minute. I want to mention the mistakes besides my correct analogeis.
Given that p, r, and s consecutive numbers in ascending order. We know that for an arranged consecutive numbers the middle number is the average. So, \(r\) is the average, which is the \(x\) in this question.
So, x will be in the middle position, the left one will be \(x-1\) and the right one will be \(x+1\), now the arrangement is \(x-1, x, \ and / x+1\). Don't mistake considering \(p, r, \ and \ s \ as \ x, x+1,\) and \(x+2\) as the middle number is \(x+1\) which is not equal to \(x\). I did this mistake.
Thus, the bottom line is:
\(r=x\)
\(p=x-1\)
(as it's the previous number of x)\(s= x+1\)
(as it's the post number of x)The numbers are: \((x-1), \ x, \ (x+1)\)
(1) \(x-1+x+x+1=2x\)
\(3x=2x\)
\(x=0\); Sufficient.
Twice x is equal to the sum of p, r, and s.
(2) \(x-1+x+x+1=0\)
\(3x=0\)
\(x=0;\) Sufficient.
The answer is \(D\)
When I was stuck to solve the DS. I took a different approach. I looked at the conditions and saw that the second condition said that the total of the integers is 0. Look here, if the total is \(0\) then the average is \(0\), which is the middle number as per the given condition. We already got \(x=0\). For my further understanding, the previous number is \(0-1=-1\) and the post number is \(0+1=1,\) the numbers are \(-1,0,1.\)
Taking this example also satisfies the first condition. \(-1+0+1=0, \ then \ x=0. \)Sufficient.
The answer is \(D \)
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