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BuggerinOn
30 May 2015, 17:47
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A
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Difficulty:
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72% (02:11) correct
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If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is
a) 517
b) 518
c) 519
d) 521
e) 525
a) 517
b) 518
c) 519
d) 521
e) 525
30 May 2015, 19:04
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A very helpful rule to know in arithmetic is the rule that in evenly spaced sets, average = median. By evenly spaced sets I mean any set where the numbers are evenly spaced from one another (they do not have to necessarily be even numbers, or spaced 2 apart, or adhere to any other rule). The sets (-3,-2,-1,0,1,2), (15,20,25), (12,14,16,18,20,22,24), and (-35,-28,-21,-14) are all examples of evenly spaced sets. Because the average will equal the median in these sets, then we quickly know that the median of this set of consecutive odd integer numbers is 534.
There are 18 numbers in the set, and in a set with an even number of terms the median is just the average of the two most median terms (here the 9th and 10th numbers in the set). This means that numbers 9 and 10 in this set are 533 and 535. Because we know that number 9 is 533, we know that the smallest number is 8 odd numbers below this, which means that it is 8*2 = 16 below this (every odd number is every other number). Therefore 533-16 = 517, answer choice A.
I hope this helps!
There are 18 numbers in the set, and in a set with an even number of terms the median is just the average of the two most median terms (here the 9th and 10th numbers in the set). This means that numbers 9 and 10 in this set are 533 and 535. Because we know that number 9 is 533, we know that the smallest number is 8 odd numbers below this, which means that it is 8*2 = 16 below this (every odd number is every other number). Therefore 533-16 = 517, answer choice A.
I hope this helps!
30 May 2015, 22:59
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Hello BuggerinOn
I would be most comfortable solving it in this fashion
\(S = \frac{n}{2} [ 2a + (n-1)d ]\)
(you would like to remember this formula which is the sum of an arthimetic progression. also notice number of terms multiplied by average of those numbers is the sum of the set)
\(534*18 = \frac{18}{2} [ 2a + (18-1)2 ]\)
\(1068 = (2a + 34)\)
\(a = 517\)
I would be most comfortable solving it in this fashion
\(S = \frac{n}{2} [ 2a + (n-1)d ]\)
(you would like to remember this formula which is the sum of an arthimetic progression. also notice number of terms multiplied by average of those numbers is the sum of the set)
\(534*18 = \frac{18}{2} [ 2a + (18-1)2 ]\)
\(1068 = (2a + 34)\)
\(a = 517\)
