The average of six consecutive integers in increasing order of size is

Let the six consecutive integers be n, n+1,n+2,n+3,n+4,n+5.

Therefore its average would be (n+n+1+n+2+n+3+n+4+n+5)/6

(6n+15)/6 = 19/2 ----> n+5/2 = 19/2-----> n = 14/2 = 7. (1)

Now lets find out the average of the last three digits,

(n+3+n+4+n+5)/3 = (10+11+12)/3 = 11 after Substituting the value of n from (1).

Therefore the answer is Option D.

I like this question because it forced me to choose between a decent bit of algebra and logic/common sense.

List the numbers. We know 9.5 is the average, and because we have consecutive integers where mean = median, we know 9.5 is the "middle term."*

So thinking about a number line, 9.5 will be between the consecutive integers 9 and 10.

From there, you've used two of the six integers. They're consecutive. You need to use up the other four integers, and they need to "balance" so that 9.5 stays in the middle.

It's then easy to see that the numbers, which must be integers, are 7, 8, 9, 10, 11, 12

You can find the average of the last three numbers either by remembering that 10, 11, 12 is an arithmetic sequence, too, so median = mean (average), hence 11, OR

\(\frac{10 + 11 + 12}{3}\) = \(\frac{33}{3}\)= 11

*Also, with an even number of terms, the average will be the average of the two middle terms, so there will be three consecutive integer terms to the right of 9.5, and three to the left of 9.5. Even if this fact doesn't pop into your brain, once you start listing the first two numbers, that fact will become apparent.

Answer D. Hope it helps.

If a given set of numbers are in Arithmetic Progression (AP), then
their Arithmetic Mean (or Average) = their Median .
This is true for all AP series.

Now, we have 6 numbers in AP (with a common difference of 1).So Mean = 9.5 = Median
Median of a series having even number of terms (6 here) is found by taking average of middle 2 terms.
Since these are consecutive numbers and average of middle 2 terms is 9.5,
the middle 2 terms have to be 9 and 10

Thus, the 3rd and 4th terms of this consecutive number series are 9 and 10.
Since 4th term is 10, next two terms would be 11 and 12

For 10, 11, 12 (again an AP series), Average must be equal to median which is 11.

Hence answer is D.

If average is \(9 \frac{1}{2}\) or 19/2, since the average is
(sum of n numbers)/n .

Since we are given 6 consecutive integers, we need to get a base of 6. So 19/2 becomes 57/6.
If the sum of integers is 57,
5a+15 = 57 or a=7

Hence the last 3 integers are a+3,a+4 and a+5(10,11,12) and average is 11(Option D)

Theory

➡ In case of consecutive integers Mean = Median
➡ In case of even number of numbers in the set: Median is the mean of the two middle numbers (after the numbers are arranged in the increasing / decreasing order)
➡ In case of odd number of numbers in the set: Median is the middle number (after the numbers are arranged in increasing/ decreasing order )

The average of six consecutive integers in increasing order of size is \(9\frac{1}{2}\)

Since there are 6 numbers so Mean = Median = Mean of middle two consecutive integers = \(9\frac{1}{2}\) = 9.5
=> the two middle integers are 9 and 10

So, six numbers are 7, 8, 9, 10, 11, 12

The average of the last three integers

Last three integers are 10 ,11, 12
Since, there are 3 consecutive integers so Mean = Median = Middle term = \(2^{nd}\) term = 11

So, Answer will be D.
Hope it helps!

Watch the following video to Learn the Basics of Statistics

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