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Re: Real Challenge [#permalink]
23 Sep 2012, 22:15
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.
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! -----------------------------------------------------------------------------------------
Re: Real Challenge [#permalink]
23 Sep 2012, 23:18
4
This post received KUDOS
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.
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, 21:55, edited 2 times in total.
Re: Real Challenge [#permalink]
24 Sep 2012, 00:17
EvaJager wrote:
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.
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! -----------------------------------------------------------------------------------------
Re: Real Challenge [#permalink]
24 Sep 2012, 02:45
1
This post received KUDOS
Capricorn369 wrote:
EvaJager wrote:
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.
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]
24 Sep 2012, 21:37
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
Re: Real Challenge [#permalink]
25 Oct 2012, 14:27
EvaJager wrote:
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.
Re: Real Challenge [#permalink]
26 Oct 2012, 06:26
himanshuhpr wrote:
EvaJager wrote:
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.
Re: Real Challenge [#permalink]
27 Oct 2012, 11:35
EvaJager wrote:
himanshuhpr wrote:
EvaJager wrote:
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...
Re: Real Challenge [#permalink]
20 Dec 2013, 23:27
Expert's post
Capricorn369 wrote:
EvaJager wrote:
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.
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 _________________
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