If the arithmetic mean of n consecutive odd integers is 20, what is

VERITAS PREP OFFICIAL SOLUTION:

We have discussed mean in case of arithmetic progressions in the previous posts. If mean of consecutive odd integers is 20, what do you think the integers will look like?

19, 21 or
17, 19, 21, 23 or
15, 17, 19, 21, 23, 25 or
13, 15, 17, 19, 21, 23, 25, 27 or
11, 13, 15, 17, 19, 21, 23, 25, 27, 29
etc.

Does it make sense that the required numbers will represent one such sequence? The numbers in the sequence will be equally distributed around 20. Every time you add a number to the left, you need to add one to the right to keep the mean 20. The smallest sequence will have 2 numbers 19 and 21, the largest will have infinite numbers. Did you notice that each one of these sequences has a unique “range,” a unique “least number” and a unique “greatest number?” So if you are given any one statistic of the sequence, you will know the entire desired sequence.

Statement 1: Only one possible sequence: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 will have the range 18. The greatest number here is 29. This statement alone is sufficient.

Statement 2: Only one possible sequence: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 will have 11 as the least number. The greatest number here is 29. This statement alone is sufficient too.

Answer (D).

Ans: D

Solution: consecutive odd integers sequence with avg 20 means 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31,33 and so on. but to keep the average 20 we need to keep 19,21 at the centre of the series.

now Statement 1: range 18 means (difference of least and greatest element of sequence = 18) for a fix pattern sequence we can get the largest value. [Sufficient]
Statement 2: once again this statement is also sufficient [Sufficient]

IMO D,reading the term 'Arithmetic Mean' is really important, I somehow took some more time to read that

pretty simple

statement one says range is 18. so we can get actual value of n which turns out to be even and 20 is arithmetic mean so 20 is middle number between 19 & 21...

statement two says least is 11 and we have already got the value of n.so we can say with conviction about the final number in the series.

n consecutive odd integers

Case 1 - sequence contains even number of terms
Where n, the mean, is 20
smallest number = ( n - (2x+1) ), where x is a positive integer
largest number = ( n + (2x+1) ). where x is a positive integer
Range becomes 2(2x+1) or 2*some odd number,

Case 2 - sequence contains odd number of terms
This is not possible as the mean would also become part of the sequence. But the mean, which is 20, is even. 20 cannot be a part of the sequence that contains only odd numbers. Hence, we can rule out all cases where no. of terms is odd

Given statements
(1) The range of the n integers is 18.
- or, 2 * some odd number is 18, which implies 2x+1 = 9 or x = 4 (the only valid solution)
The largest number = 20 + 2(4) + 1 = 29
Statement 1 is sufficient

(2) The least of the n integers is 11.
Diff between the least and the mean = 9.
Therefore, the largest must be 9 more than mean = 29
Statement 2 is sufficient

Therefore both 1 and 2 are sufficient to answer

Statement 1: Range of n integers is 18-The nos are 11,13,15,17,19,21,23,25,27,29. Range=29-11=18. Sufficient
Statement 2: Least of n integers is 11. Again use same set of numbers. Sufficient
Answer D

Excellent Quality Question.
Here is my approach ->

Given data -->
N consecutive odd integers.
Consecutive odd integer set is an AP set with common difference =2
Hence Mean=Median=Average of the first and the last terms.
Thus,Median=20.
As median =Even => The number of terms must be even.So,the set will have two middle terms.

We are asked about the value of Greatest element.
Statment 1->
Range =18
Let p be the first element.
Hence the last element must be p+18 so that the range is 18.
Thus the average = p+p+18/2 => p+9
Hence p+9=20 => p=11
Hence the first term =11
And the median =20
The number of terms to the left of the median = number of terms to the right of the median.
Hone we can get the last term.
Hence sufficient

Statement 2-->
The first term =11
As we have the median and the first term -> we can get the last term by using the principle =>The number of terms to the left of the median = number of terms to the right of the median.
Hence Sufficient

Hence D

Although we can apply the formula avg of first term and last term = avg of evenly spaced set to conclude that statement 1 is sufficient, I liked the veritas approach to it which is formula independent. The only doubt here was on the sufficiency of statement 1.

Statement 2 is very clearly sufficient.

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