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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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]
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, inclusiveminusthe 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 \(6^2\);
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, inclusiveminusthe 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. _________________
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