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A set of consecutive positive integers beginning with 1 is
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11 Nov 2008, 02:02
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40% (02:38) correct 60% (03:00) wrong based on 222 sessions
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A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased? A. 7 B. 8 C. 9 D. 15 E. 17
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Re: A set of consecutive positive integers beginning with 1 is
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02 Jan 2015, 13:31
Hi All, This question is layered with subtle "clues" as to how you can find the correct answer: 1) We're asked which number was removed from the list and the answers ARE numbers. We can use them against the prompt. 2) The average of the new group of numbers is 35 7/17. The denominator of THAT fraction tells us that the total number of terms in the new list MUST be a multiple of 17: 17, 34, 51, 68, 85, etc. 3) Since we're removing 1 number from a list of consecutive positive integers, we could quickly limit down the possible number of terms in the ORIGINAL list: If new = 17, old = 18, but the numbers from 1 to 18 would NOT have an average in the mid30s (it would be much smaller). ELIMINATE this option. If new = 34, old = 35, but we run into the same problem. The average won't match here either. ELIMINATE this option. If new = 51, old = 52, same problem here. ELIMINATE this option. If new = 68, old = 69....HERE we have a group of numbers that might just be what we're looking for. In the "old group", the average is the "middle term" = 35, which is REALLY close to the new average once we remove a number. Now, we just have to figure out which of the 5 answers was removed (and that changed the new average to 35 7/17). There are a couple of ways to do this math, but I'm going to use the answers to save myself some steps: Sum of 1 to 69, inclusive = 69(35) = 2415 Remove Answer B....2415  8 = 2407 New average = 2407/68 = 35 27/68 Now, we can compare 27/68 to 7/17 27/68 and 28/68 This is REALLY close, but is not the answer. Eliminate B. The answer is either A or C. Since 28/68 is a BIGGER fraction than 27/68, we need the sum (of the 68 integers) to be BIGGER...which means we need to remove a SMALLER number. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A set of consecutive positive integers beginning with 1 is
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08 Nov 2014, 11:48
A better and easier method is as follows: First of all, when ever a number is deleted from consecutive numbers, avg cant change more than 1/2. So in this case new avg is 35 7/17, so old average was 35. Now, 35 is avg of first 69 numbers, so previously there were 69 num, and now there are 68(after deletion). Now, 7/17 = 28/68 . So effectively, the number which was deleted gave 28/68 to each 68 num left, so the number should have been 28 less than 35(old avg) Ans= 3528=7. Advantage with this method is that you can do it verbally.




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Re: PS: Average
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11 Nov 2008, 03:38
This one sure looks tricky. I am not able to reach the solution, but this is what i think. I am sure you must have done all this by now, but no harm in putting it down. 35 7/17 , means 602/17 , earlier i thought there must be 18 numbers, which is 17 + 1 that is taken out. But if there were, then their sum is 171 which is far too low. The average of 35 shows that the numbers were little higher, ( their sum should be 602 ) . The answer choices are very low numbers. I don't know how this is possible at all.
Sorry mate, i really doubt if the qs is correct ! Can you pls provide the source of this qs and check whether the numbers you mentioned are all correct ?



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Re: PS: Average
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11 Nov 2008, 03:43
My method, I may not use this during the test, so i will still wait for another better solution.
set of consecutive positive integers beginning with 1, if n is the last number => SUM(1) = n(n+1)/2
If 1 number was taken out, so the SUM(2) then will be (n  1) * average = (n  1) * (35 + 7/17)
As SUM(2) is an integer so (n1) is divisible by 17, the original average is in this range (34+7/17, 36 + 7/17) => n is in (2*(34+7/17), 2*(36 + 7/17)) or n is in (68.8, 72.8)
(n1) is divisible by 17 and n is in (68.8, 72.8) => n = 69
The number was erased = SUM(1)  SUM(2) = 69*70/2  (691) * (35 + 7/17) = 7
Answer: A



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Re: PS: Average
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11 Nov 2008, 04:03
lylya4 wrote: My method, I may not use this during the test, so i will still wait for another better solution.
set of consecutive positive integers beginning with 1, if n is the last number => SUM(1) = n(n+1)/2
If 1 number was taken out, so the SUM(2) then will be (n  1) * average = (n  1) * (35 + 7/17)
As SUM(2) is an integer so (n1) is divisible by 17, the original average is in this range (34+7/17, 36 + 7/17) => n is in (2*(34+7/17), 2*(36 + 7/17)) or n is in (68.8, 72.8)
(n1) is divisible by 17 and n is in (68.8, 72.8) => n = 69
The number was erased = SUM(1)  SUM(2) = 69*70/2  (691) * (35 + 7/17) = 7
Answer: A Could you elaborate the highlighted step further?



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Re: PS: Average
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11 Nov 2008, 04:42
scthakur wrote: lylya4 wrote: My method, I may not use this during the test, so i will still wait for another better solution.
set of consecutive positive integers beginning with 1, if n is the last number => SUM(1) = n(n+1)/2
If 1 number was taken out, so the SUM(2) then will be (n  1) * average = (n  1) * (35 + 7/17)
As SUM(2) is an integer so (n1) is divisible by 17, the original average is in this range (34+7/17, 36 + 7/17) => n is in (2*(34+7/17), 2*(36 + 7/17)) or n is in (68.8, 72.8)
(n1) is divisible by 17 and n is in (68.8, 72.8) => n = 69
The number was erased = SUM(1)  SUM(2) = 69*70/2  (691) * (35 + 7/17) = 7
Answer: A Could you elaborate the highlighted step further? SUM (2) = (n  1) * (35 + 7/17) = 35(n1) + 7/17(n1) as SUM(2) is an integer => 7/17(n1) is integer or n1 is divisible by 17 The range for the original average is just for to limit the result of n, if the largest number n was taken out, the average will be decreased at max n / (n1) = 1 + 1/(n1) ~ 1, so to be safe I place the range for this average +1 to 35 7/17. This step can be ignored, instead we know that the average is ~35, so n should be around ~70 (average for consecutive integers = (n + 1)/2)



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Re: A set of consecutive positive integers beginning with 1 is
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02 Jan 2015, 11:24
BunuelCould you please help on this one ? this part is little confusing to me. . The range for the original average is just for to limit the result of n, if the largest number n was taken out, the average will be decreased at max n / (n1) = 1 + 1/(n1) ~ 1, so to be safe I place the range for this average +1 to 35 7/17. This step can be ignored, instead we know that the average is ~35, so n should be around ~70 (average for consecutive integers = (n + 1)/2) thanks
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Re: A set of consecutive positive integers beginning with 1 is
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02 Apr 2015, 23:33
EMPOWERgmatRichC wrote: Hi All, This question is layered with subtle "clues" as to how you can find the correct answer: 1) We're asked which number was removed from the list and the answers ARE numbers. We can use them against the prompt. 2) The average of the new group of numbers is 35 7/17. The denominator of THAT fraction tells us that the total number of terms in the new list MUST be a multiple of 17: 17, 34, 51, 68, 85, etc. 3) Since we're removing 1 number from a list of consecutive positive integers, we could quickly limit down the possible number of terms in the ORIGINAL list: If new = 17, old = 18, but the numbers from 1 to 18 would NOT have an average in the mid30s (it would be much smaller). ELIMINATE this option. If new = 34, old = 35, but we run into the same problem. The average won't match here either. ELIMINATE this option. If new = 51, old = 52, same problem here. ELIMINATE this option. If new = 68, old = 69....HERE we have a group of numbers that might just be what we're looking for. In the "old group", the average is the "middle term" = 35, which is REALLY close to the new average once we remove a number. Now, we just have to figure out which of the 5 answers was removed (and that changed the new average to 35 7/17). There are a couple of ways to do this math, but I'm going to use the answers to save myself some steps: Sum of 1 to 69, inclusive = 69(35) = 2415 Remove Answer B....2415  8 = 2407 New average = 2407/68 = 35 27/68 Now, we can compare 27/68 to 7/17 27/68 and 28/68 This is REALLY close, but is not the answer. Eliminate B. The answer is either A or C. Since 28/68 is a BIGGER fraction than 27/68, we need the sum (of the 68 integers) to be BIGGER...which means we need to remove a SMALLER number. Final Answer: GMAT assassins aren't born, they're made, Rich Thanks for providing this easy to understand solution. Just one step which can reduce effort at the end 69*35 = 2415 and our earlier average is 35 7/17 = 602/17, we know total numbers here in this case (after one got erased is 68) so 4*602/4*17 = 2408/68. now since total should have been 2415, the deleted number is 24152408=7



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Re: A set of consecutive positive integers beginning with 1 is
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25 Jul 2016, 22:36
tarek99 wrote: A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased?
A. 7 B. 8 C. 9 D. 15 E. 17 The average of consecutive positive integers will be either an integer (if we have odd number of integers) or integer.5 (if we have even number of integers). The maximum change that can happen to the average when one integer is deleted is 0.5. New average = 35 (7/17) Old average = Either 35 or 35.5 So there were either 69 consecutive integers starting from 1 (in which case avg was 35) or 70 consecutive integers starting from 1 (in which case avg was 35.5) . If the average was 35 (and number of integers was 69) and it increased to 35 (7/17), it means a small number was removed which was taking the average down by 7/17 for the rest of the 68 numbers. So it must be 68 *(7/17) = 28 less than the average. That is, the number removed must have been 35  28 = 7. It is one of the options so answer (A). For more details on this method, check: http://www.veritasprep.com/blog/2015/09 ... thegmat/
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Re: A set of consecutive positive integers beginning with 1 is
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20 Apr 2017, 14:47
VeritasPrepKarishma wrote: tarek99 wrote: A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased?
A. 7 B. 8 C. 9 D. 15 E. 17 The average of consecutive positive integers will be either an integer (if we have odd number of integers) or integer.5 (if we have even number of integers). The maximum change that can happen to the average when one integer is deleted is 0.5. New average = 35 (7/17) Old average = Either 35 or 35.5 So there were either 69 consecutive integers starting from 1 (in which case avg was 35) or 70 consecutive integers starting from 1 (in which case avg was 35.5) . If the average was 35 (and number of integers was 69) and it increased to 35 (7/17), it means a small number was removed which was taking the average down by 7/17 for the rest of the 68 numbers. So it must be 68 *(7/17) = 28 less than the average. That is, the number removed must have been 35  28 = 7. It is one of the options so answer (A). For more details on this method, check: http://www.veritasprep.com/blog/2015/09 ... thegmat/ VeritasPrepKarishmaSay the number removed from the consecutive integer set actually decreased the average by 7/17, then would we need to add 35 & 28?
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Re: A set of consecutive positive integers beginning with 1 is
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20 Apr 2017, 23:45
colorblind wrote: VeritasPrepKarishma wrote: tarek99 wrote: A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased?
A. 7 B. 8 C. 9 D. 15 E. 17 The average of consecutive positive integers will be either an integer (if we have odd number of integers) or integer.5 (if we have even number of integers). The maximum change that can happen to the average when one integer is deleted is 0.5. New average = 35 (7/17) Old average = Either 35 or 35.5 So there were either 69 consecutive integers starting from 1 (in which case avg was 35) or 70 consecutive integers starting from 1 (in which case avg was 35.5) . If the average was 35 (and number of integers was 69) and it increased to 35 (7/17), it means a small number was removed which was taking the average down by 7/17 for the rest of the 68 numbers. So it must be 68 *(7/17) = 28 less than the average. That is, the number removed must have been 35  28 = 7. It is one of the options so answer (A). For more details on this method, check: http://www.veritasprep.com/blog/2015/09 ... thegmat/ VeritasPrepKarishmaSay the number removed from the consecutive integer set actually decreased the average by 7/17, then would we need to add 35 & 28? I am not sure what you intend to keep and what you intend to change but if the previous avg was 35 (which means 69 integers) and by removing one integer, the average reduces by 7/17, it means the number was providing 7/17 to each of the other 68 numbers i.e. (7/17)*68 = 28. So the number removed must be 28 more than the average 35 and hence must be 35 + 28 = 63.
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A set of consecutive positive integers beginning with 1 is
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22 Jul 2017, 02:26
For the avg to be around 35 you need first 69 terms. N(N+1)/2*N=35 implies N to be 69.
The new avg is 35 + 7/17. It could also be written as 35 + 28/68.
If the avg had to be maintained (35) the number 35 would have been erased. But since the one erased gives 28 to the remaining lot (to remaining 68 numbers)
The answer should be 3528=7 Because 7 is the only number which would have taken an additional 28 from the entire lot to keep the avg 35.
When erased it gives that extra 28 back to group of 68.
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Re: A set of consecutive positive integers beginning with 1 is
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05 Aug 2018, 18:07
a much better and easier method is this one : let suppose consecutive integer 1 , 2 ,..... n , n+1 Now in new case n+1 is remove remaining are n consecutive integers whose AM is 357/17 so n ( n+1)/2 = 357/17 ( sum of consecutive integers = total integers (n) * ( first term + last term)/2 n(n+1) = 42 n^2 + n 42 = 0 (n6)(n+7) = 0 n cannot be negative so n =6 hence ( n+1) the number removed must be 6 + 1 = 7




Re: A set of consecutive positive integers beginning with 1 is &nbs
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