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Interesting Question from Princeton Review [#permalink]
19 Jul 2007, 17:08
I hadn't seen a question quite like this one, so I thought I would share it with everyone. It tricked me.
The sum of all the integers between 101 and 147, exclusive, is how much greater than the sum of the odd integers between 115 and 133, inclusive?
A. 3224
B. 4216
C. 4340
D. 4588
E. 5580
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Director
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I got D.
101 to 147
101 + 147 = 248/2=124
147 - 101= 46 + 1=47
47*124=5828
115 to 133, odd
115 + 133=248/2=124
133 - 115=18/2=9+1=10
10*124=1240
5828-1240=4588, hence D.
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Manager
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You fell for one of the tricks...check the question again...
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Director
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I did! I was trying to wrap up a few things at work and missed exclusive. The answer should be C from a quick glance.
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Manager
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There you go. It got me too. I believe this is the first question I have ran across that uses "exclusive". I'm so used to seeing "inclusive" that I did the same thing you did.
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Sum of all integers = (47/2)(101+147) = 5828
Sum of all odd integers = 10/2(2(115) + (10-1)(2)) = 1240
Difference = 4588
Ans: D
For the first equation, it's the sum of an arithmetic progression with arithmetic difference of 1.
For the second equation, it's the sum of an arithmetic progression with arithmetic difference of 2.
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GMAT Club Legend
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ywilfred wrote: Sum of all integers = (47/2)(101+147) = 5828 Sum of all odd integers = 10/2(2(115) + (10-1)(2)) = 1240 Difference = 4588
Ans: D
For the first equation, it's the sum of an arithmetic progression with arithmetic difference of 1.
For the second equation, it's the sum of an arithmetic progression with arithmetic difference of 2.
okay, i just saw the word exclusive, so my answer D is wrong. I'll rework and post later.
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Senior Manager
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C is correct
101 - 147 exclusive = 102 - 146
# * avg = ((146-102)/2 +1) * (146+102)/2 = 45 * 124
odd integers from 115 through 133
# * avg = ((133-115)/2 +1) * (133 + 115)/2 = 10 * 124
ans = 45 * 124 - 10 * 124 = 4340
_________________
AimHigher
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Director
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The first sum (S1) = 102 + 103 + .. + 146
the second sum (S2) = 115 + 117 + .. + 133
Number of elements in S1 = 147 - 101 -1 = 45
Number of elements in S2 = (133 - 115)/2 + 1 = 10
For equally spaced terms in a series:
Average = sum / number of terms = The median
S1: 124 = S1 / 45 --> S1 = 5580
S2: 124 = S2 / 10 --> S2 = 1240
S1 - S2 = 5580 - 1240 = 4340
ANS: C
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Senior Manager
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Mishari wrote: The first sum (S1) = 102 + 103 + .. + 146 the second sum (S2) = 115 + 117 + .. + 133
Number of elements in S1 = 147 - 101 -1 = 45 Number of elements in S2 = (133 - 115)/2 + 1 = 10
For equally spaced terms in a series: Average = sum / number of terms = The median
S1: 124 = S1 / 45 --> S1 = 5580 S2: 124 = S2 / 10 --> S2 = 1240
S1 - S2 = 5580 - 1240 = 4340
ANS: C
mishari.
this is the second time i have seen someone use the forumla "(last term - first term)/2 +1 = number of odd or even terms in an arithemtic progression with an odd number of total terms. i think it is important to note it only works when the series is all positive numbers
0,1,2 = how many odd terms? just one. but with your formula (2-0)/2 +1 =2
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Director
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of course. It only works when all numbers in the series are either all positive or all negative.
If the ZERO is involved, then the whole thing changes because Zero doesn't affect the calculations, yet it counts for an addition element in the series.
Thanks for the tip pal
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Senior Manager
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Mishari wrote: of course. It only works when all numbers in the series are either all positive or all negative.
If the ZERO is involved, then the whole thing changes because Zero doesn't affect the calculations, yet it counts for an addition element in the series.
Thanks for the tip pal
no need to get snotty. there are a lot of people who read these boards for help and tips and take away formulas and tricks without understanding the concepts or exceptions to the rule or investigating; I think it is important to note them.
I have seen people using this formula and I didn't understand where it came from and no one really explained it so I started playing with it (it is what people should do with a new concept) and I quickly noted the exception and proper use.
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