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# Interesting Question from Princeton Review

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Manager
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Interesting Question from Princeton Review [#permalink]  19 Jul 2007, 16:08
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
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.
Manager
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You fell for one of the tricks...check the question again...
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.
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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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.
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
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AimHigher

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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
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
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
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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