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The average of six consecutive integers in increasing order

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GeminiHeat
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The average of six consecutive integers in increasing order [#permalink] New post   07 Nov 2019, 22:38
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The average of six consecutive integers in increasing order of size is \(9\frac{1}{2}\). What is the average of the last three integers?

(A) 8

(B) \(9\frac{1}{2}\)

(C) 10

(D) 11

(E) 19
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D
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Re: The average of six consecutive integers in increasing order [#permalink] New post   08 Nov 2019, 04:07
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The avreage is 9.5

since the integers are consecutive they could be also all equal and at most 6

3*6=18 and the average is 9.5

D must be the answer without further calculation
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The average of six consecutive integers in increasing order [#permalink] New post   04 Aug 2022, 23:45
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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!

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The average of six consecutive integers in increasing order [#permalink]
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