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 350,000 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:

A set of consecutive positive integers beginning with 1 is [#permalink]
11 Nov 2008, 01:02

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

50% (01:12) correct
50% (02:58) wrong based on 32 sessions

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?

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 ?

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

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

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

Answer: A

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)

Re: A set of consecutive positive integers beginning with 1 is [#permalink]
08 Nov 2014, 10:48

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= 35-28=7. Advantage with this method is that you can do it verbally.

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

Re: A set of consecutive positive integers beginning with 1 is [#permalink]
02 Jan 2015, 12:31

2

This post was BOOKMARKED

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.

I´ve done an interview at Accepted.com quite a while ago and if any of you are interested, here is the link . I´m through my preparation of my second...

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...