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M20Q31

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Re: Sum of Consecutive Integers [#permalink] New post 06 Jun 2012, 18:30
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] New post 06 Jun 2012, 18:33
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manulath wrote:
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


Method of operation:

1. What topics are being tested here? (A) Evenly spaced sets and (B) Statistics

2. What do I know about evenly spaced sets?

(A) average = ( first term + last term ) / 2 = (39+13)/2 = 26
(B) # of terms in the set = [last term - first term)/multiple (which is 2) + 1 = (39-13)/2 + 1 = 14 terms

As such

Sum = 14 * 26 = 364

Answer (B)
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Re: Sum of Consecutive Integers [#permalink] New post 06 Jun 2012, 18:49
(39-13)/2+1 * (13+39)/2 = 364
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Re: Sum of Consecutive Integers [#permalink] New post 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] New post 07 Jun 2012, 22:03
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it's the best that you write the whole matrix out, like -

1 3 5 7 9
11 13 15 17 19
21 .. .. .. ..
31 .. .. .. ..
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Re: M20Q31 [#permalink] New post 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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Re: M20Q31 [#permalink] New post 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.

Answer: B.
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Re: M20Q31 [#permalink] New post 12 Mar 2013, 07:56
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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.
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Re: M20Q31 [#permalink] New post 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.

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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M20Q31 [#permalink] New post 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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