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The sum of first N consecutive odd integers is N^2 . What is the sum of all odd integers between 13 and 39 inclusive? (A) 351 (B) 364 (C) 410 (D) 424 (E) 450 Source: GMAT Club Tests - hardest GMAT questions
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Hey,
First the sum of odd numbers between1 to 39, which represents 20 odd consecutive numbers
is 20*20 = 400, 400 - (1+ 3 + 5 + 7 +9 +11) = 364.
Hope this clear....
Jugolo1
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ventivish wrote: The sum of first N consecutive odd integers is N^2 . What is the sum of all odd integers between 13 and 39 inclusive?
(C) 2008 GMAT Club - m20#31
* 351
* 364
* 410
* 424
* 450 how about the way that is explained in the question: The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400 The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36 The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364
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GMAT TIGER wrote: ventivish wrote: The sum of first N consecutive odd integers is N^2 . What is the sum of all odd integers between 13 and 39 inclusive?
(C) 2008 GMAT Club - m20#31
* 351
* 364
* 410
* 424
* 450 how about the way that is explained in the question: The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400 The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36 The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364 Hi thank you, this is the OA However I used the same methodology, but I included 13 as in the sum of consecutive odd integers from 1 to 13 and subtracted that from 400, thereby getting 351 as my answer. Why do we only calculate the sum from 1 to 11? I know I am missing something, sorry if this is a stupid question!
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ventivish wrote: GMAT TIGER wrote: ventivish wrote: The sum of first N consecutive odd integers is N^2 . What is the sum of all odd integers between 13 and 39 inclusive?
(C) 2008 GMAT Club - m20#31
* 351
* 364
* 410
* 424
* 450 how about the way that is explained in the question: The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400 The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36 The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364 Hi thank you, this is the OA However I used the same methodology, but I included 13 as in the sum of consecutive odd integers from 1 to 13 and subtracted that from 400, thereby getting 351 as my answer. Why do we only calculate the sum from 1 to 11? I know I am missing something, sorry if this is a stupid question! "13 and 39 inclusive" so if you subtract 13 also then you are only finding the sum of all odd numbers from 15-39
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i dont understand the purpose of the first statement portion of the question. how is that relevant to what the question is asking?
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The way this question is written makes a person use "their" way (meaning the test maker's way) and in reality, just going through add adding them all up doesn't take more than 60 seconds. 13 15 17 19 21 23 25 27 29 31 33 35 37 39 I approach adding these up as in pairs: 11 & 39, 13 & 37, 15 & 35, 17 & 33, 19 & 31, 21 & 29, 23 & 27. Each of these 6 pairs = 50, so 50 * 6 = 300. Then you have 39 and 25 left, that's 64, so 364. Not the most "advanced" method, but it certainly won't take over 60 seconds to complete this way.
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Intern
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20² - 6² = 400 - 36
= 364
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A faster way to do this is
Number of odd numbers 39 inclusive = (39 - 13)/2 + 1 = 13 + 1 = 14
Average = (13+39)/2 = 26
26 * 14 = 364
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Whats the relevance of first part of the question to the second..any idea?
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Manager
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th sum of first N odd integers = N^2
so the sum of 1 to 39 inclusive
N=(39-1)/2+1=20
so N^2=400
so the sum of 1 to 11 inclusive
N=(11-1)/2+1 =6
so N^2=36
so the sum of 13 to 39 inclusive = 400-36 =364
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jallenmorris wrote: The way this question is written makes a person use "their" way (meaning the test maker's way) and in reality, just going through add adding them all up doesn't take more than 60 seconds.
13
15
17
19
21
23
25
27
29
31
33
35
37
39
I approach adding these up as in pairs: 11 & 39, 13 & 37, 15 & 35, 17 & 33, 19 & 31, 21 & 29, 23 & 27. Each of these 6 pairs = 50, so 50 * 6 = 300. Then you have 39 and 25 left, that's 64, so 364.
Not the most "advanced" method, but it certainly won't take over 60 seconds to complete this way. I did this as well since I just froze up but the numbers were manageable. If the set was any larger I would have been done for.
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(13+39) = 52
52/2 = 26
26x14 = 364
Answer is B.
I would estimate this question has to be on the far easier side.
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nitya34 wrote: S=13+15+.......+37+39
39=13+(n-1)2
=>n=14
S=0.5(14)(13+39)=364 can someone please explain me this line - 39=13+(n-1)2 where can i c a summary of ways to know the amount of N in series? thanks guys
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144144 wrote: nitya34 wrote: S=13+15+.......+37+39
39=13+(n-1)2
=>n=14
S=0.5(14)(13+39)=364 can someone please explain me this line - 39=13+(n-1)2 where can i c a summary of ways to know the amount of N in series? thanks guys So the problem states that it is every odd number in between 13 and 39 inclusive. To find N integers you just count off how many are included in this series (14). The line in question just sets up an equation to easily solve for N. Subtract 13, then divide by two then add one and you get N=14. To check just count off. By and large it is better to memorize that equation since there are some instances were counting off via a check is too time consuming. Hope this helps.
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jallenmorris wrote: The way this question is written makes a person use "their" way (meaning the test maker's way) and in reality, just going through add adding them all up doesn't take more than 60 seconds.
13
15
17
19
21
23
25
27
29
31
33
35
37
39
I approach adding these up as in pairs: 11 & 39, 13 & 37, 15 & 35, 17 & 33, 19 & 31, 21 & 29, 23 & 27. Each of these 6 pairs = 50, so 50 * 6 = 300. Then you have 39 and 25 left, that's 64, so 364.
Not the most "advanced" method, but it certainly won't take over 60 seconds to complete this way. I agree with you on this method. It is fast and easy to do rather than to think of other ways to solve as mentioned by others and then solve which will take more time.
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Total no of terms = n=((39-13)/2)+1 = 14
so sum = 14/2(39+13) = 364 so OA is B
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