GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 18:39

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

M20-31

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

16 Sep 2014, 01:09
1
18
00:00

Difficulty:

45% (medium)

Question Stats:

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

HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

16 Sep 2014, 01:09
1
2
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 $$20^2-6^2=364$$.

_________________
Intern
Joined: 13 Oct 2014
Posts: 2

Show Tags

03 Jan 2016, 04:31
4
1
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
Intern
Joined: 27 Jun 2017
Posts: 1

Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

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.
_________________
Intern
Joined: 01 Jul 2017
Posts: 6

Show Tags

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?
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

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.
_________________
Intern
Joined: 01 Jul 2017
Posts: 6

Show Tags

07 Aug 2017, 21:53
Intern
Joined: 20 Sep 2016
Posts: 21

Show Tags

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

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.
_________________
Manager
Joined: 09 Aug 2017
Posts: 59
Location: United States
Concentration: Technology
GMAT 1: 640 Q44 V33
GMAT 2: 630 Q47 V29
WE: Research (Investment Banking)

Show Tags

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.
_________________
I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Also, I appreciate any kudos.
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

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
_________________
Senior Manager
Joined: 08 Jun 2015
Posts: 420
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

Show Tags

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
_________________
" The few , the fearless "
Intern
Joined: 18 Mar 2015
Posts: 12

Show Tags

13 Dec 2017, 01:22
I used the formulae :n/2 * 2a+(n-1)d
Manager
Joined: 08 Sep 2017
Posts: 77
Location: Colombia
GMAT 1: 710 Q49 V39

Show Tags

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
_________________
Kudos please if you liked my post

Thanks!
Manager
Joined: 22 Jun 2017
Posts: 168
Location: Argentina
Schools: HBS, Stanford, Wharton
GMAT 1: 630 Q43 V34

Show Tags

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?
_________________
The HARDER you work, the LUCKIER you get.
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

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).
_________________
Manager
Joined: 11 Sep 2019
Posts: 53
Schools: Great Lakes '21

Show Tags

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 $$20^2-6^2=364$$.

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
_________________
Please be generous. Hit +1 Kudos if you find this post helpful.

Check out the Special RC Collection- https://gmatclub.com/forum/trouble-in-rc-listen-to-your-rc-friend-15-days-rc-challenge-day-305898.html
Math Expert
Joined: 02 Sep 2009
Posts: 58320

Show Tags

24 Sep 2019, 07:26
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 $$20^2-6^2=364$$.

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?
_________________
Re: M20-31   [#permalink] 24 Sep 2019, 07:26
Display posts from previous: Sort by

M20-31

Moderators: chetan2u, Bunuel