When positive integer y is added to each of the first n non-negative i

Question

When positive integer y is added to each of the first n non-negative integers, which of the following statements is true?

I. If the median of the resulting numbers is  y+\frac{n−1}{2}  then n is odd
II. The arithmetic mean of the resulting numbers is equal to the median of the resulting numbers
III. The arithmetic mean of the resulting numbers is y units greater than the arithmetic mean of the first n positive integers.

A. I only
B. II only
C. III only
D. I, II and III
E. None of the above

CrackverbalGMAT · Answer

At the outset, is this a ‘true’ question or a ‘must be true’ question? That’s probably the most confusing aspect because the constraints given in the question statement make it look like it’s a ‘must be’ question but the question asks which of the statements are true.

Anyways, let’s take a simple case and try to prove the statements false. Whatever cannot be proven false has to be our answer.

The question statement mentions ‘the first n non-negative integers’ – this means we are dealing with consecutive numbers. Consecutive numbers are equally spaced numbers.  When there are equally spaced numbers ,

Mean =  \frac{First element + Last element }{ 2} 
Mean = Median.

Let n = 4 and y = 1. This means, we are assuming the first 4 non-negative integers i.e. 0,1,2 and 3. When we add 1 to each of them, we obtain 1,2,3 and 4. Let us now evaluate the statements.

Statement I: The median of the resulting numbers is 5/2. 5/2 can be written as 1+ fraction](4-1)/2   i.e. the median of the resulting numbers can be written as y +  \frac{(n-1)}{2} . But, n is 4 which is even. Therefore, statement II is not true. 
Answer options A and D can be eliminated.

Statement II: Since the resultant numbers are equally spaced values, the arithmetic mean of the numbers is equal to the median. Statement II is true. 
Answer options C and E can be eliminated. Only answer option B is left.

Statement III: The AM of the resulting integers is the same as the AM of the first 4 positive integers and NOT y more. Statement III is false. 
 The correct answer option is B .

As with a ‘must be true’ question, taking simple cases and disproving the statements is the best way in solving such questions. You also need to continuously eliminate the answer options as you disprove statements.

Hope that helps!

amoghhlgr · Answer

Bunuel  Can someone please explain how III is not true?

Thanks

JerryAtDreamScore · Answer

Breaking Down the Info:

Note that the first n non-negative integers start with 0, so that series would be 0, 1, 2, ...,   n - 1  . We end on  n - 1  since 0 is the first integer.

Also, recall that for arithmetic sequences, we have the property mean = median. The first n non-negative integers create an arithmetic sequence and adding y to each element still keeps it an arithmetic sequence.

I:  The median of the original sequence would be  \frac{0 + (n - 1)}{ 2} = \frac{n - 1}{2} . Adding y to every element would change that to  y + \frac{n - 1}{2} . Hence the median is always  y + \frac{n - 1}{2}  regardless of the value of n. FALSE.

II:  This is true for all arithmetic sequences. The resulting sequence is arithmetic, so this statement is TRUE.

III:  We mentioned earlier the median/mean of the resulting sequence is  y + \frac{n - 1}{2} .  The first n positive integers is 1, 2, ..., n.  So the arithmetic mean of the first n positive integers is  \frac{n+1}{2} . We can see the difference between  y + \frac{n - 1}{2}  and  \frac{n+1}{2}  is not y. FALSE.

Answer: B

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When positive integer y is added to each of the first n non-negative i

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When positive integer y is added to each of the first n non-negative i

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