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 scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]
01 Oct 2010, 10:19

1

This post received KUDOS

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

39% (02:11) correct
61% (02:24) wrong based on 46 sessions

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 ?

Re: Missing Weights [#permalink]
01 Oct 2010, 10:54

6

This post received KUDOS

Expert's post

shrouded1 wrote:

Here is an interesting question :

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.

Re: Missing Weights [#permalink]
01 Oct 2010, 11:02

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

Re: Missing Weights [#permalink]
01 Oct 2010, 16:37

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 _________________

Re: Missing Weights [#permalink]
01 Oct 2010, 19:19

shrouded1 wrote:

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. _________________

Re: Missing Weights [#permalink]
20 Dec 2012, 21:31

2

This post received KUDOS

shrouded1 wrote:

Here is an interesting question :

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. _________________

Impossible is nothing to God.

gmatclubot

Re: Missing Weights
[#permalink]
20 Dec 2012, 21:31

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...