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You'll actually have to calculate these, but it's an easy job: the average of any evenly-spaced multiple is just first + last over two. All three groups are evenly-spaced.

a is multiples of 3, so (63+99)/2 = 162/2 = 81
b is odds, so (61+99)/2 = 160/2 = 80
c is multiples of 7, so (63 + 98)/2 = 161/2 = 80.5

In this case, b < c < a, so the answer is (D).

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a is the average of the multiples of 3 between 60 and 100.
b is the average of the odd numbers between 60 and 100.
c is the average of the multiples of 7 between 60 and 100.

Which of the following is true?

A) a < b < c
B) a < c < b
C) b < a < c
D) b < c < a
E) c < b < a
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You could even just compare the numerators , because in each case the sum is divided by two
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Excellent point!

kevincan
You could even just compare the numerators , because in each case the sum is divided by two
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My doubt is if it isn't mentioned inclusive or not can we always assume that the values are not inclusive?
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The smallest and largest multiples of 3 between 60 and 100 are 63 and 99.

Consecutive multiples of three form an arithmetic sequence. A useful fact about arithmetic sequences is that the average of their terms is equal to the average of the first and last term. Thus the average of the multiples of three between 60 and 100 is 1/2 of (63 + 99 ).
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There are only two integers between 2 and 5. There are 4 integers between 2 and 5 inclusive
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