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:

The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

09 Apr 2005, 01:13

1

This post received KUDOS

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

57% (01:30) correct
43% (01:22) wrong based on 810 sessions

HideShow timer Statistics

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

Even numbers between 1 and 100 are 2, 4, 6 ... 100 (50 numbers).

Numbers that are multiple of 3 are 3, 6, 9 ... 99. However, all even multiples of 3 have been counted already in even numbers. Therefore, this set includes 3, 9, 15, 21 ... 99. (17 numbers).

Thus total numbers that are to be excluded = 50 + 17 = 67.

Therefore probability that a randomly picked number is nither even nor a multiple of 3 = 33%
_________________

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than: "how many odd numbers are in the first 100 positive ones that are not multiple of 3?"

Careful theArch, that wall is rather fragile !!!

Btw what're you saying? Your answer matches mine?
_________________

My first answer was 67/100.
However, I was deceived by christoph, because I calculated 67/100 and then I looked at answers already given, and forgot to subtract. This wouldn't happen in the actual test

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than: "how many odd numbers are in the first 100 positive ones that are not multiple of 3?"

This happens with many of us....We get there to the answer and forget subtracting.......GMAT has both the answers as a trap and we should be careful enough not to fall in it........

Zem, we are trying to find the numbers that are nither divisible by 2 nor 3.

If you take the numbers that are divisible by 2 away, you've taken all even numbers between 1 and 100 away. They are (100-1)/2 = 50 numbers.

We could repeat the same process with 3. However, there are numbers that are common to 2 and 3, and we don't have to take them out (they were accounted for in the even numbers list anyway).

So we see what are these numbers we're looking at (odd multiples of 3).
They are 3, 9, 15, 21 and so on.

There're a plathora of ways to figuring out how many such numbers would exist between 1 and 100. I'd list out two here.

One is the way of sequences.

the last number in this list would be 99.
99 = 3 + (n - 1) 6 => n = 17.

Else do a 100/16 = 16.xxx. 16*6 = 96. but since it starts at 3, the 16th multiple would be 99. 16 multiples, and add one (for 3) and you've 17.

I think that in this problem it's easier to find directly the prob than to use the formula 1-(opposite prob)

from 1 to 100 : 100 numbers
even numbers : 50
multiples of 3 : 33 (3*1, 3*2, ...3*33) however in those 33 numbers there are already even numbers and you have to be careful to not repeat the even numbers you've already taken out so finally you will get 17 odd and 16 even (you can see that everytime 3 is multiply by an even number the final result is obviously even so 1/2 of the multiples of 3 will be even and 1/2 odd, howver the last one is odd so there is an additionnal odd one)

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

18 Mar 2012, 20:56

1

This post received KUDOS

Simple to solve using P(n) = 1-X concept

P = 1 - [ P(even 0r multiples of 2)+P(multiples of 3) - P(multiples of 2 and 3 or multiples of 6)] P = (1 - [ 50/100+33/100 - 16/100]) = 33%
_________________

Press +1 Kudos rather than saying thanks which is more helpful infact..

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83% B. 67% C. 50% D. 33% E. 17%

# of multiples of 2 in the range (100-2)/2+1=50 (check this: totally-basic-94862.html#p730075); # of multiples of 3 in the range (99-3)/3+1=33; # of multiples of both 2 and 3, so multiples of 6, in the range (96-6)/6+1=16 (to get the overlap of above two sets);

Hence there are total of 50+33-16=67 numbers from 1 to 100, inclusive, which are multiples of 2 or 3, which means that the probability of selecting neither multiple of 2 nor multiple of 3 is 1-67/100=33/100=33%.

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

30 Sep 2013, 11:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

10 Dec 2013, 06:19

Zem wrote:

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83% B. 67% C. 50% D. 33% E. 17%

Bunuel is this approach correct?

Ballparking from 1 to 10 there are only 3 numbers that satisfy these conditions: 1,5 and 7. Then 3/10 so approximately 30% only answer choice that is close is D

If you do it for other ranges you are basically going to have 4,3,4,3,4,3 so in theory it would be something like 3.5/10 or between 30% and 35%

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

10 Dec 2013, 06:22

Zem wrote:

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83% B. 67% C. 50% D. 33% E. 17%

Multiple of 3: 100/3 = 33 numbers Multiple of 2: 100/2 = 50 numbers

Multiple of 6: 100/6= 16 numbers

Total numbers divisible by 2 or 3: 50 + 33 - 16 = 67

Neither = 100 - 67 = 33

Probability = 33/100

Pushpinder Gill
_________________

Enroll for our GMAT Trial Course here - http://gmatonline.crackverbal.com/

Learn all PS and DS strategies here- http://gmatonline.crackverbal.com/p/mastering-quant-on-gmat

For more info on GMAT and MBA, follow us on @AskCrackVerbal

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

25 Dec 2014, 22:48

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

Show Tags

03 Jan 2016, 09:46

Even numbers = 100/2 = 50 Numbers divisible by 3 = 99/3 = 33 Even or divisible by 3 = 50 + 33 - duplicates(divisible by 3 and even) Duplicates = 32/2 = 16 Even or divisible by 3 = 50 + 33 - 16 = 67 Numbers that are not even or divisible by 3 = 100 - 67 = 33