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M20-31

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M20-31  [#permalink]

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New post 16 Sep 2014, 01:09
1
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A
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C
D
E

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  45% (medium)

Question Stats:

72% (02:04) correct 28% (02:18) wrong based on 149 sessions

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Re M20-31  [#permalink]

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New post 16 Sep 2014, 01:09
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Official Solution:

The sum of first \(N\) consecutive positive 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 odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.

Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd 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 odd 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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Re: M20-31  [#permalink]

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New post 03 Jan 2016, 04:31
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HI.
we can restate the problem as : what is the summation of evenly spaced series (arithmatic progression) 13,15,17...,39
no. of terms of the series= (39-13)/2 +1=14 and avg of the series =(1st term+last term)/2=52/2=26. so summation =26*14=364
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Re: M20-31  [#permalink]

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New post 29 Jun 2017, 12:19
I think this is a high-quality question and I don't agree with the explanation. it says between 13 and 39 inclusive, not between 11 and 39.
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Re: M20-31  [#permalink]

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New post 29 Jun 2017, 14:46
rer05 wrote:
I think this is a high-quality question and I don't agree with the explanation. it says between 13 and 39 inclusive, not between 11 and 39.


Please re-read the solution. It says that 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.
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Re: M20-31  [#permalink]

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New post 07 Aug 2017, 03:06
Why are we taking 'the sum of all odd integers from 1 to 39, inclusive ' even though the question asks for odd numbers 'BETWEEN' 13 and 39?
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Re: M20-31  [#permalink]

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New post 07 Aug 2017, 16:38
chaitalip wrote:
Why are we taking 'the sum of all odd integers from 1 to 39, inclusive ' even though the question asks for odd numbers 'BETWEEN' 13 and 39?


Please re-read the solution. It says that 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.
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Re: M20-31  [#permalink]

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New post 07 Aug 2017, 21:53
right my bad. thank you!
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Re: M20-31  [#permalink]

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New post 17 Sep 2017, 05:10
Hi Bunuel,
Please help me understand this question. As per my reasoning goes, there are total (39-13)/2+ 1 odd terms. i.e. N= 14. thus according to the question sum of n is n2 which is 14*14.
What is the link am missing here ?
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Re: M20-31  [#permalink]

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New post 17 Sep 2017, 05:18
jyotipes21@gmail.com wrote:
Hi Bunuel,
Please help me understand this question. As per my reasoning goes, there are total (39-13)/2+ 1 odd terms. i.e. N= 14. thus according to the question sum of n is n2 which is 14*14.
What is the link am missing here ?


The question says that: The sum of first N consecutive consecutive odd integers is N^2, not that the sum of any N consecutive integers is N^2.
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Re: M20-31  [#permalink]

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New post 17 Sep 2017, 09:05
How do we know the sum of all the integers 1 to 39 is 400? Is that derived a formula, trick or just something that a person should know. What about other numbers or evens?

Thanks.
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Re: M20-31  [#permalink]

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New post 17 Sep 2017, 09:27
Plunkster82 wrote:
How do we know the sum of all the integers 1 to 39 is 400? Is that derived a formula, trick or just something that a person should know. What about other numbers or evens?

Thanks.


The sum of 20 odd integers from 1 to 39 (20 integers) is 20^2 = 400. You can derive it from arithmetic progression formula: https://gmatclub.com/forum/math-sequenc ... 01891.html
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Re: M20-31  [#permalink]

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New post 26 Oct 2017, 07:52
Great explanations above. This is how I solved it .. Sum of all odd numbers from 13 to 39 is ((14/2) * (13+39)) This comes to 364. In other words, we are concerned about the sum of all numbers in an AP with a CD of 2 and the first term as 13. The answer is option B
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Re: M20-31  [#permalink]

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New post 13 Dec 2017, 01:22
I used the formulae :n/2 * 2a+(n-1)d
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M20-31  [#permalink]

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New post 30 Oct 2018, 19:54
\(\frac{39-13}{2}=13+1=14\)

\(\frac{14}{2} = 7\) numbers so then: 13 / 15 / 17 / 19 / 21 / 23 / 25 / Average / ...other 7 odd consecutive numbers

Average= 26

Sum: 14*26 = 364 --> Ans B
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Re: M20-31  [#permalink]

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New post 09 Feb 2019, 11:42
Hi Bunuel

I don't get how you arrive to the conclusion that "It says that 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" from the question... would you pls further explain?
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New post 10 Feb 2019, 01:46
patto wrote:
Hi Bunuel

I don't get how you arrive to the conclusion that "It says that 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" from the question... would you pls further explain?


Take a simple example: the sum of all odd integers between 5 and 9, inclusive (5 + 7 + 9) equals to the sum of all odd integers from 1 to 9, inclusive (1 + 3 + 5 + 7 + 9) minus the sum of all odd integers from 1 to 3, inclusive (1 + 3).
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M20-31  [#permalink]

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New post 24 Sep 2019, 07:04
Bunuel wrote:
Official Solution:

The sum of first \(N\) consecutive positive 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 odd integers from 1 to 39, inclusive[/b] minus the sum of all odd integers from 1 to 11, inclusive.

Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd 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 odd integers from 1 to 11, inclusive is \(6^2\);

So, the required sum is \(20^2-6^2=364\).


Answer: B


Hi Bunuel,
I am not able to understand this 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 [b]the sum of all odd integers from 1 to 11, inclusive
Where does this come from.Is there any basis, 11 is specific for this case or for every case
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Re: M20-31  [#permalink]

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New post 24 Sep 2019, 07:26
Aadi01 wrote:
Bunuel wrote:
Official Solution:

The sum of first \(N\) consecutive positive 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 odd integers from 1 to 39, inclusive[/b] minus the sum of all odd integers from 1 to 11, inclusive.

Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd 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 odd integers from 1 to 11, inclusive is \(6^2\);

So, the required sum is \(20^2-6^2=364\).


Answer: B


Hi Bunuel,
I am not able to understand this 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 [b]the sum of all odd integers from 1 to 11, inclusive
Where does this come from.Is there any basis, 11 is specific for this case or for every case


It's simple logic:

1 + 3 + 5 +7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39

RED (the sum of all odd integers between 13 and 39, inclusive) equals to TOTAL (the sum of all odd integers from 1 to 39, inclusive) minus BLUE (the sum of all odd integers from 1 to 11, inclusive).

Does this make sense?
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Re: M20-31   [#permalink] 24 Sep 2019, 07:26
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