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# M20Q31

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M20Q31 [#permalink]

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18 Nov 2008, 04:46
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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

SOLUTION: m20q31-73009-20.html#p1106480

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Re: M20Q31 [#permalink]

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18 Nov 2008, 13:30
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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
(give kudos)

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Re: M20Q31 [#permalink]

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20 Nov 2008, 08:44
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S=13+15+.......+37+39

39=13+(n-1)2
=>n=14

S=0.5(14)(13+39)=364
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Re: M20Q31 [#permalink]

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20 Nov 2008, 21:55
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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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Senior Manager
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Re: M20Q31 [#permalink]

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23 Nov 2008, 12:34
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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Re: M20Q31 [#permalink]

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04 Dec 2008, 18:30
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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Re: M20Q31 [#permalink]

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01 Jun 2009, 15:22
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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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Re: M20Q31 [#permalink]

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02 Feb 2010, 10:29
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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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Re: M20Q31 [#permalink]

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02 Feb 2010, 22:19
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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Re: M20Q31 [#permalink]

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07 Feb 2011, 07:51

20^2-6^2
400-36
364
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Re: M20Q31 [#permalink]

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07 Feb 2011, 13:23

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Re: M20Q31 [#permalink]

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16 Feb 2011, 20:20
(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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Re: M20Q31 [#permalink]

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16 Feb 2011, 23:39
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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Re: M20Q31 [#permalink]

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17 Feb 2011, 07:15
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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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Re: M20Q31 [#permalink]

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09 Feb 2012, 06:22
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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Re: M20Q31 [#permalink]

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09 Feb 2012, 08:05
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 20^2
Expressable 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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Re: M20Q31 [#permalink]

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29 Apr 2012, 00:20
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

[Reveal] Spoiler: OA
B

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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Sum of Consecutive Integers [#permalink]

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06 Jun 2012, 11:45
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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?

A 351
B 364
C 410
D 424
E 450

[Reveal] Spoiler:
OA : B

I know its a straight forward question. But somehow I was unable to solve it within 2 minutes.
I took some time to figure out that
[Reveal] Spoiler:
39 = 20th term and 11 = 6 th term

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Re: Sum of Consecutive Integers [#permalink]

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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]

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06 Jun 2012, 19:24
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manulath:

Fool proof
2-minute method, list the integers from 13 to 39 and sum them.

Better method
Let's use the equation given, we are looking for:

13 + 15 + ... + 39 = (1+3+ ... + 39) - (1+3+ ... + 11) = sum of odd numbers from 1 to 39 less sum of odd numbers from 1 to 11

Sum of odd numbers from 1 to 39 = 20^2
- there are twenty odd numbers: (39-1)/2+1 = 20
- use the equation given, 20^2

Sum of odd numbers from 1 to 11 = 6^2
- there are six odd numbers (11-1)/2+1 = 6

20^2 - 6^2 = 364
B)

Robust method
Question is for an arithmetic series, starting number = 13, constant difference = 2, number of items = 14
sum = 14*(13+(14-1)/2*2) = 14*13 = 364

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Re: Sum of Consecutive Integers   [#permalink] 06 Jun 2012, 19:24

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