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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.
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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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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
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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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dakhan wrote: i dont understand the purpose of the first statement portion of the question. how is that relevant to what the question is asking? The purpose of the first sentence is two fold; 1) save you time adding consecutive odd integers 2) test quantitative reasoning. That being said, it is not necessary to solve the problem, but it will save you a minute. Here is how it breaks down:The sum of first N consecutive odd integers is N^2.There are 20 consecutive odd integers from 0-39. Therefore, The sum of the first 20 consecutive odd integers is 20^2Expressable as, The sum of 1+2...39 = 400 or 20^2 = 400 What is the sum of all odd integers between 13 and 39 inclusive?What is the sum of 13+14+15...39? Since 400 = [1+2...11] + [13+15...39] We can reason that 364 = [6^2] - [20^2]
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topmbaseeker wrote: 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 That's the exact way I did it. Not sure if it was the fastest method but it worked.
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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? (A) 351 (B) 364 (C) 410 (D) 424 (E) 450 Source: GMAT Club Tests - hardest GMAT questions list the odd numbers from 1 to 39 which counts to 20 and from 1 to 13 counts 6 ( excluding 13) so from the given definition we can write.. 20sq - 6sq 400 - 36 364
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Re: Sum of Consecutive Integers [#permalink]
 06 Jun 2012, 14:17
Hi
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 13 and 39 inclusive?
In this case OA = 351
because there 20 odd terms between 1 and 39 .
and between 1 and 13 . There are 7 odd terms
Since given sum of n consecutive odd terms terms is n^2 so sum of first 20 odd terms = 20^2
and sum of first 7 odd terms = 7^2
so sum of odd terms between 13 and 39 = 20^2 - 7^2= 400 - 49 = 351 = Answer A
on the other hand if
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 11 and 39 inclusive.
Then in this case odd # of integers between 1 and 11 is 6 .
so 20^2 - 6^2 = 400 - 36= 364 = ans = B
Hope that helps , please correct me if I am wrong .
Last edited by Joy111 on 07 Jun 2012, 00:00, edited 1 time in total.
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Re: Sum of Consecutive Integers [#permalink]
06 Jun 2012, 19:30
Joy111 wrote: Hi
Please change the question or change the OA.
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 13 and 39 inclusive?
In this case OA = 351
because there 20 odd terms between 1 and 39 .
and between 1 and 13 . There are 7 odd terms
Since given sum of n consecutive odd terms terms is n^2 so sum of first 20 odd terms = 20^2
and sum of first 7 odd terms = 7^2
so sum of odd terms between 13 and 39 = 20^2 - 7^2= 400 - 49 = 351 = Answer A
on the other hand if
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 11 and 39 inclusive.
Then in this case odd # of integers between 1 and 11 is 6 .
so 20^2 - 6^2 = 400 - 36= 364 = ans = B
Hope that helps , please correct me if I am wrong . I don't agree - you have removed the 13 which should be counted in your first example, and the 11 which should be counted in your second example. The question says inclusive of 13, i.e. 13 + 15 + ... + 39, which is most definitely 364 ( see)
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Re: Sum of Consecutive Integers [#permalink]
06 Jun 2012, 19:49
(39-13)/2+1 * (13+39)/2 = 364
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Re: Sum of Consecutive Integers [#permalink]
06 Jun 2012, 23:56
pike wrote: Joy111 wrote: Hi
Please change the question or change the OA.
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 13 and 39 inclusive?
In this case OA = 351
because there 20 odd terms between 1 and 39 .
and between 1 and 13 . There are 7 odd terms
Since given sum of n consecutive odd terms terms is n^2 so sum of first 20 odd terms = 20^2
and sum of first 7 odd terms = 7^2
so sum of odd terms between 13 and 39 = 20^2 - 7^2= 400 - 49 = 351 = Answer A
on the other hand if
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 11 and 39 inclusive.
Then in this case odd # of integers between 1 and 11 is 6 .
so 20^2 - 6^2 = 400 - 36= 364 = ans = B
Hope that helps , please correct me if I am wrong . I don't agree - you have removed the 13 which should be counted in your first example, and the 11 which should be counted in your second example. The question says inclusive of 13, i.e. 13 + 15 + ... + 39, which is most definitely 364 ( see) @pike Thank you for the clarification : because it is sum of odd from 13 to 39 so we have to subtract the sum of odd from 1 to 11 and not 1 to 13 between 1 to 11 there are 6 odd ones so 400 - 36= 364 = answer B similarly in the second example 11 must be included while adding, because it is sum of odd from 11 to 39 so we have consider odd from 1 to 10 and not from 1 to 11 while subtracting from the total. so between 1 - 10 there are 5 odd ones so 400 - 25 = 375 Thank you for the correction , +1 to all of you , for the realization that I don't make the same silly mistake during exams .
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jallenmorris wrote: Then you have 39 and 25 left, that's 64, so 364. I think you meant: Then you have 11 and 25 left (NOT 39 as mentioned coz you already counted it)!
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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 The sum of all odd integers between 13 and 39, inclusive equals to the sum of all integers from 1 to 39, inclusive minus the sum of all integers from 1 to 11, inclusive. Since there are 20 odd integers from 1 to 39, inclusive then the sum of all integers from 1 to 39, inclusive is 20^2; Since there are 6 odd integers from 1 to 11, inclusive then the sum of all integers from 1 to 11, inclusive is 6^2; So, the required sum is 20^2-6^2=364. Answer: B.
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LGOdream wrote: Bunuel 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?
A. 351
B. 364
C. 410
D. 424
E. 450
The sum of all odd integers between 13 and 39, inclusive equals to the sum of all integers from 1 to 39, inclusive minus the sum of all integers from 1 to 11, inclusive.
Since there are 20 odd integers from 1 to 39, inclusive then the sum of all integers from 1 to 39, inclusive is 20^2;
Since there are 6 odd integers from 1 to 11, inclusive then the sum of all integers from 1 to 11, inclusive is 6^2;
So, the required sum is 20^2-6^2=364.
Answer: B. " The sum of all odd integers between 13 and 39, inclusive equals to the sum of all integers from 1 to 39, inclusive minus the sum of all integers from 1 to 11, inclusive." Should be: " The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive." The explanation in the problem M20-31 from GMAT Club Tests is most likely copied from Bunuel's explanation and presents the same problem. Thank you. Typo edited in the tests.
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