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:

A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

Show Tags

01 Oct 2010, 11:19

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

56% (02:33) correct
44% (02:28) wrong based on 73 sessions

HideShow timer Statistics

A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg}. This set is good enough to weight any object having an integral weight betweem 1Kg and 63Kg (Eg. 19Kg = 16Kg + 2Kg + 1Kg). If the 4Kg weight is lost, how many weights between 1Kg & 63Kg can no longer be measured ?

A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg}. This set is good enough to weight any object having an integral weight betweem 1Kg and 63Kg (Eg. 19Kg = 16Kg + 2Kg + 1Kg). If the 4Kg weight is lost, how many weights between 1Kg & 63Kg can no longer be measured ?

A) 16 B) 24 C) 28 D) 32 E) 36

Consider the following example: how many different selections are possible from \(n\) people (including a subset with 0 members and a subset with all \(n\) members)?

\(C^0_n+C^1_n+C^2_n+...+C^n_n=2^n\) --> so, number of different subsets from a set with \(n\) different terms is \(2^n\) (this include one empty subset). Or another way: each person has 2 choices, either to be included or not to be included in the subset, so # of total subsets is \(2^n\).

Next, from a set {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg} obviously no term can be obtained by adding any number of other terms.

So, from a set with 6 different terms {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg} we can form \(2^6-1=63\) subsets each of which will have different sum (minus one empty subset) --> we can weight 63 different weights;

From a set with 5 different terms {1Kg, 2Kg, 8Kg, 16Kg, 32Kg} we can form \(2^5-1=31\) subsets each of which will have different sum (minus one empty subset) --> we can weight 31 different weights;

Which means that if 4Kg weight is lost 63-31=32 weights can no longer be measured.

Good solution, alluding to a binary notation of a number.

Here is another way to think of the problem :

We can form : 1 {1} 2 {2} 3 {1,2} But we cannot form any of 4 through to 7 without a 4. So 4 numbers we cannot form up till 7

Next look at the next set of numbers from 8 to 15 Notice that this is set is nothing but : 8+{0,1,2,3,..,7} But we know from this that we cannot form a 4,5,6,7 So if we count up till 15, there are 2x4=8 numbers in all that we cannot form

Next look at the numbers from 16 to 31 Same patter repeats This is 16+{0,1,2,...,15} And we know from 0,..,15 there are 8 numbers we cannot form. So from 16 to 31 there are 8 more ... hence 2x8=16 numbers between 1 and 31

There is one more way to do this, but that involves knowledge of a binary notation of a number (a concept not tested on the GMAT). Here is the solution :

In binary a number <=63 can be represented in 6 digits. Of this 4 represents the 3rd digit from the right. The number of 6 digit binary numbers possible forcing the 3rd digit to be 1 (all the numbers that need 4) is exactly 2^5 or 32
_________________

Good solution, alluding to a binary notation of a number.

Here is another way to think of the problem :

We can form : 1 {1} 2 {2} 3 {1,2} But we cannot form any of 4 through to 7 without a 4. So 4 numbers we cannot form up till 7

Next look at the next set of numbers from 8 to 15 Notice that this is set is nothing but : 8+{0,1,2,3,..,7} But we know from this that we cannot form a 4,5,6,7 So if we count up till 15, there are 2x4=8 numbers in all that we cannot form

Next look at the numbers from 16 to 31 Same patter repeats This is 16+{0,1,2,...,15} And we know from 0,..,15 there are 8 numbers we cannot form. So from 16 to 31 there are 8 more ... hence 2x8=16 numbers between 1 and 31

and this patter goes on ...

For numbers upto 63, it will be 2x16=32

Answer is (d)

I was doing it this way ... however, lost trach somewhere in the middle. Realize that rather than doing all the way from 1 - 63 in one go, it is better to take it in batches. Thanks.
_________________

A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg}. This set is good enough to weight any object having an integral weight betweem 1Kg and 63Kg (Eg. 19Kg = 16Kg + 2Kg + 1Kg). If the 4Kg weight is lost, how many weights between 1Kg & 63Kg can no longer be measured ?

A) 16 B) 24 C) 28 D) 32 E) 36

Originally there could be \(2*2*2*2*2*2=2^6=64\) combinations of weight. Now that we took 1 weight off, we get \(2*2*2*2*2=2^5=32\) combinations of weight.

What is lost? \(64-32=32\)

Answer: D

In case you are wondering why 2 was multiplied n times, it's because 2 represents two things: BEING SELECTED and NOT BEING SELECTED. _________________

Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

Show Tags

04 May 2016, 08:16

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: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

Show Tags

05 May 2016, 10:00

Here's how I thought about it: For each weight, there are two possibilities; either it's included or it's not. So, for 6 weights, there are (2*2*2*2*2*2) = 2^6 possibilities, minus 1 (the case where there are no weights) or 63 possibilities. Therefore, if we take away one of the weights, there are now 2^5 - 1 = 31 possibilities. 63-31 = 32.

Answer: D

Aside: Think about it similiarly to how you think about counting the total number of factors from a number's prime factorization: Either the factor is included or it's not.

Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

Show Tags

05 May 2016, 10:16

the first thought i had was 5c0 +5c1 + 5c2 + 5c3 +5c4 +5c5 = 32. This is because i need to choose a 4 and that can be selected in the following combinations. with each one. on solving it gives me 32. 32 sets will be lost. Hope this helps

gmatclubot

Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg,
[#permalink]
05 May 2016, 10:16

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...