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:

S is a set of positive integers. The average of the terms in S is equal to the range of the terms in S. What is the sum of all the integers in S? (1) The range of S is a prime number that is less than 11 and is not a factor of 10. (2) S is composed of 5 different integers.

My answer is E. Based on option A - prime numbers less than 11 are 2, 3, 5, 7. 2 & 5 is rejected because they are factors of 10. we get two numbers - 3 & 7.- Insuff. Based on option B - multiple combination possible - Insuff.

Together - 3 & 7 both doesn't fall under creteria. Answer E.

Cheers! _________________

----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------

S is a set of positive integers. The average of the terms in S is equal to the range of the terms in S. What is the sum of all the integers in S? (1) The range of S is a prime number that is less than 11 and is not a factor of 10. (2) S is composed of 5 different integers.

A - average (1) Primes less than 11 are 2, 3, 5, and 7. Not factor of 10, we are left with 3 and 7. If A = 3, we can have {2, 2, 3, 5} or {1, 2, 3, 4, 4, 4}. Not sufficient.

(2) For 5 evenly spaced numbers, k - 2d, k - d, k, k + d, k + 2d the range is 4d and the average is k. We can simply take k = 4d, and we have infinitely many sets fulfilling the condition of the form {2d, 3d, 4d, 5d, 6d}. For example {2, 3, 4, 5, 6}, {20, 30, 40, 50, 60}... Not sufficient.

(1) and (2) together: We have seen above that the range can be either 3 or 7. If the range is 3, we cannot have 5 distinct integers in the set, so only 7 is left.

We know that the sum of the 5 integers is 5 * 7 = 35, the smallest number is k and the largest number is k + 7, where k is some positive integer. Necessarily 35 - 7 = 28 > 5k, so k must be not greater than 5. If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfill the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

Answer E.

CORRECTION:

What a miss! We have to find the sum of the numbers and not the set of numbers. Although there is more than one possibility, the total sum is clearly 7*5 = 35. Answer C.

And IanStewart is right about (1), in a set of numbers, each element is different. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 24 Sep 2012, 22:55, edited 2 times in total.

S is a set of positive integers. The average of the terms in S is equal to the range of the terms in S. What is the sum of all the integers in S? (1) The range of S is a prime number that is less than 11 and is not a factor of 10. (2) S is composed of 5 different integers.

A - average (1) Primes less than 11 are 2, 3, 5, and 7. Not factor of 10, we are left with 3 and 7. If A = 3, we can have {2, 2, 3, 5} or {1, 2, 3, 4, 4, 4}. Not sufficient.

(2) For 5 evenly spaced numbers, k - 2d, k - d, k, k + d, k + 2d the range is 4d and the average is k. We can simply take k = 4d, and we have infinitely many sets fulfilling the condition of the form {2d, 3d, 4d, 5d, 6d}. For example {2, 3, 4, 5, 6}, {20, 30, 40, 50, 60}... Not sufficient.

(1) and (2) together: We have seen above that the range can be either 3 or 7. If the range is 3, we cannot have 5 distinct integers in the set, so only 7 is left.

We know that the sum of the 5 integers is 5 * 7 = 35, the smallest number is k and the largest number is k + 7, where k is some positive integer. Necessarily 35 - 7 = 28 > 5k, so k must be not greater than 5. If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfills the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

Answer E.

@EvaJager - in (2) why are you considering "5 evenly spaced numbers"? This is not mentioned anywhere in the question. _________________

----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------

S is a set of positive integers. The average of the terms in S is equal to the range of the terms in S. What is the sum of all the integers in S? (1) The range of S is a prime number that is less than 11 and is not a factor of 10. (2) S is composed of 5 different integers.

A - average (1) Primes less than 11 are 2, 3, 5, and 7. Not factor of 10, we are left with 3 and 7. If A = 3, we can have {2, 2, 3, 5} or {1, 2, 3, 4, 4, 4}. Not sufficient.

(2) For 5 evenly spaced numbers, k - 2d, k - d, k, k + d, k + 2d the range is 4d and the average is k. We can simply take k = 4d, and we have infinitely many sets fulfilling the condition of the form {2d, 3d, 4d, 5d, 6d}. For example {2, 3, 4, 5, 6}, {20, 30, 40, 50, 60}... Not sufficient.

(1) and (2) together: We have seen above that the range can be either 3 or 7. If the range is 3, we cannot have 5 distinct integers in the set, so only 7 is left.

We know that the sum of the 5 integers is 5 * 7 = 35, the smallest number is k and the largest number is k + 7, where k is some positive integer. Necessarily 35 - 7 = 28 > 5k, so k must be not greater than 5. If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfill the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

Answer E.

@EvaJager - in (2) why are you considering "5 evenly spaced numbers"? This is not mentioned anywhere in the question.

It is the easiest to express the average when the numbers are evenly spaced. The goal is to find as easily as possible one/more examples which fulfill the condition. And it is nowhere stated that the numbers cannot be evenly spaced. Also, in (1) it was not mentioned that some of the numbers cannot be equal. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: S is a set of positive integers. The average of the terms in [#permalink]

Show Tags

24 Sep 2012, 22:37

2

This post received KUDOS

Expert's post

rohitgarg wrote:

S is a set of positive integers. The average of the terms in S is equal to the range of the terms in S. What is the sum of all the integers in S?

(1) The range of S is a prime number that is less than 11 and is not a factor of 10. (2) S is composed of 5 different integers.

Neither statement is sufficient alone.

From Statement 1, the range (and thus the average) is either 3 or 7. From Statement 2, we know that the range cannot be 3, since no set with a range of 3 could contain five different integers. So the mean is 7, and we have 5 things in our set, so the sum is 5*7=35 which is what the question asked for. The answer is C.

If the OA in the original source is 'E' here because the OA claims that S could contain repeated elements, the OA is wrong. The ambiguous wording of Statement 2 aside, the word 'set' has a precise definition in mathematics, and sets cannot contain repeated elements. You don't need to know that for the GMAT, but there's a reason GMAT questions always talk about 'lists of values' or 'data sets' or use some similar term in questions where repeated elements might be relevant. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfill the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

^^^ what's this that you have done , can you please explain thinking why did you calculate the above, I would like to mention that I understand the problem sufficiently clear.

If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfill the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

^^^ what's this that you have done , can you please explain thinking why did you calculate the above, I would like to mention that I understand the problem sufficiently clear.

Thanks..

I edited my first post, see the CORRECTION attached :

What a miss! We have to find the sum of the numbers and not the set of numbers. Although there is more than one possibility, the total sum is clearly 7*5 = 35. Answer C.

And IanStewart is right about (1), in a set of numbers, each element is different.

I was trying to find sets which fulfill the given conditions, which was in fact unnecessary. We were only asked what is the sum of the numbers. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfill the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

^^^ what's this that you have done , can you please explain thinking why did you calculate the above, I would like to mention that I understand the problem sufficiently clear.

Thanks..

I edited my first post, see the CORRECTION attached :

What a miss! We have to find the sum of the numbers and not the set of numbers. Although there is more than one possibility, the total sum is clearly 7*5 = 35. Answer C.

And IanStewart is right about (1), in a set of numbers, each element is different.

I was trying to find sets which fulfill the given conditions, which was in fact unnecessary. We were only asked what is the sum of the numbers.

EvaJager.... could you please explain the entire solution.....please do it independently and not as an extension of some other post...

S is a set of positive integers. The average of the terms in S is equal to the range of the terms in S. What is the sum of all the integers in S? (1) The range of S is a prime number that is less than 11 and is not a factor of 10. (2) S is composed of 5 different integers.

A - average (1) Primes less than 11 are 2, 3, 5, and 7. Not factor of 10, we are left with 3 and 7. If A = 3, we can have {2, 2, 3, 5} or {1, 2, 3, 4, 4, 4}. Not sufficient.

(2) For 5 evenly spaced numbers, k - 2d, k - d, k, k + d, k + 2d the range is 4d and the average is k. We can simply take k = 4d, and we have infinitely many sets fulfilling the condition of the form {2d, 3d, 4d, 5d, 6d}. For example {2, 3, 4, 5, 6}, {20, 30, 40, 50, 60}... Not sufficient.

(1) and (2) together: We have seen above that the range can be either 3 or 7. If the range is 3, we cannot have 5 distinct integers in the set, so only 7 is left.

We know that the sum of the 5 integers is 5 * 7 = 35, the smallest number is k and the largest number is k + 7, where k is some positive integer. Necessarily 35 - 7 = 28 > 5k, so k must be not greater than 5. If k = 5, the first and the last number are 5 and 12. The smallest 3 remaining integers are 6, 7, and 8 together with 5 and 12 will give a sum of 37. If k = 4, we find sets which fulfills the condition: {4, 5, 7, 8, 11} and also (4, 5, 6, 9, 11}. Obviously, not sufficient.

Answer E.

@EvaJager - in (2) why are you considering "5 evenly spaced numbers"? This is not mentioned anywhere in the question.

From statement 1 and statement 2, we have range = 7 ( cannot be 2,5 or 3), hence average is also 7, hence sum = 5*7 = 35

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...