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# M20Q31

Author Message
Intern
Status: not enough sleep
Joined: 09 Feb 2012
Posts: 3
Location: United States
Concentration: Technology, Finance
GMAT Date: 04-24-2012
GPA: 2.35
WE: Operations (Investment Banking)
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Kudos [?]: 5 [0], given: 2

Re: M20Q31 [#permalink]  09 Feb 2012, 07:05
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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Status: I will not stop until i realise my goal which is my dream too
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Kudos [?]: 38 [0], given: 16

Re: M20Q31 [#permalink]  28 Apr 2012, 23:20
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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Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

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Re: Sum of Consecutive Integers [#permalink]  06 Jun 2012, 13:17
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 06 Jun 2012, 23:00, edited 1 time in total.
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Re: Sum of Consecutive Integers [#permalink]  06 Jun 2012, 18:30
1
This post was
BOOKMARKED
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, 18:49
(39-13)/2+1 * (13+39)/2 = 364
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Re: Sum of Consecutive Integers [#permalink]  06 Jun 2012, 22:56
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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Re: M20Q31 [#permalink]  21 Jul 2012, 05:21
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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Kudos [?]: 57260 [0], given: 8811

Re: M20Q31 [#permalink]  22 Jul 2012, 02:34
Expert's post
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 $$20^2-6^2=364$$.

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Kudos [?]: 57260 [0], given: 8811

Re: M20Q31 [#permalink]  13 Mar 2013, 02:29
Expert's post
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 $$20^2-6^2=364$$.

"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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Re: M20Q31 [#permalink]  21 Apr 2014, 04:10
A=(13+15+...+39) = (1+3+...+39) - (1+3+...+11)
Among 2 sums on the right side, the 1st sum has (39-1)/2+1=20 elements -> the 1st sum = 20^2=400
The 2nd has (11-1)/2+1=6 elements -> 6^2=36
=> A= 400-36=364

Choose B
Re: M20Q31   [#permalink] 21 Apr 2014, 04:10

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# M20Q31

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