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Director  Joined: 21 Jul 2006
Posts: 953
A set of consecutive positive integers beginning with 1 is  [#permalink]

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7
32 00:00

Difficulty:   95% (hard)

Question Stats: 34% (02:43) correct 66% (02:42) wrong based on 175 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  [#permalink]

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7
1
5
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 mid-30s (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.

GMAT assassins aren't born, they're made,
Rich
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Intern  Joined: 08 Nov 2014
Posts: 1
Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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5
9
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= 35-28=7.
Advantage with this method is that you can do it verbally.
##### General Discussion
Manager  Joined: 14 Oct 2008
Posts: 101

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1
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 ?
Manager  Joined: 30 Sep 2008
Posts: 82

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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 (n-1) 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)

(n-1) 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 - (69-1) * (35 + 7/17) = 7

Director  Joined: 17 Jun 2008
Posts: 970

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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 (n-1) 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)

(n-1) 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 - (69-1) * (35 + 7/17) = 7

Could you elaborate the highlighted step further?
Manager  Joined: 30 Sep 2008
Posts: 82

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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 (n-1) 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)

(n-1) 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 - (69-1) * (35 + 7/17) = 7

Could you elaborate the highlighted step further?

SUM (2) = (n - 1) * (35 + 7/17) = 35(n-1) + 7/17(n-1) as SUM(2) is an integer => 7/17(n-1) is integer or n-1 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 / (n-1) = 1 + 1/(n-1) ~ 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)
Senior Manager  Joined: 07 Aug 2011
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Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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Bunuel

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 / (n-1) = 1 + 1/(n-1) ~ 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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Joined: 23 Aug 2014
Posts: 7
Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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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 mid-30s (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.

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 2415-2408=7
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Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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4
6
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 ... -the-gmat/
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Posts: 76
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Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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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 ... -the-gmat/

VeritasPrepKarishma

Say 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  [#permalink]

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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 ... -the-gmat/

VeritasPrepKarishma

Say 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  [#permalink]

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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)

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.

Posted from my mobile device
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Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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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
(n-6)(n+7) = 0
n cannot be negative so n =6 hence ( n+1) the number removed must be 6 + 1 = 7
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Re: A set of consecutive positive integers beginning with 1 is  [#permalink]

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_________________ Re: A set of consecutive positive integers beginning with 1 is   [#permalink] 08 Oct 2019, 07:46

# A set of consecutive positive integers beginning with 1 is  