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Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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11 May 2015, 07:16

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Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

This is a really tricky abstract question, and one that is probably best solved by backsolving. This kind of question is particularly dangerous because it is not only difficult to get correct, but it will bait test takers into spending far too long on it. Do not get "lost" in a problem, and know when to cut the cord on problems like this and make an educated guess. Veritas does a great job at teaching these high level strategies in their courses.

In terms of backsolving on this, I am using the term loosely because I mean plug in the answer choices and see what comes out of this formula...and if that means anything to you. For answer choice a (the correct answer), lets assign the value of 1/2 to the variable "a". The variable "b" will remain unknown. Plugging a = 1/2 into this equation, we have 2*1/2*b - 1/2 - b + 1...simplifying we get b - 1/2 - b + 1, or -1/2 + 1, or 1/2. This means that any time you use 1/2 for one of the variables in this equation, the equation will spit the result of 1/2 right back out. So when all is said and done, that is the only value that will remain.

Each of the other answer choices provide you a different output from input and keep a variable in the output (thus making the exact value impossible to know). Look at answer choice b for example. Doing the same thing, 2*1/6*b - 1/6 - b + 1 = -2/3b + 5/6. We of course cannot know what this is, as we do not know b, and it will change when b changes.

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

Brilliant, Brandon!

I arrived at the answer after looking around a bit for a pattern.

Given is my train of thought for those who did not think of going the options route:

Set S: {1/96 , 2/96 , ... 95/96 , 96/96 }

Operation: 2ab - (a+b) + 1

My first thought was that the two numbers selected should be complementary such that they lead to a simple answer which would eventually lead to some kind of sequence. It made sense to pick 1/96 and 95/96 so that a+b = 1. Similarly, 2/96 and 94/96 etc. We were left with 2ab from each operation. But the leftover values were too complicated to work with. Didn't work.

Next thought was using a number which simplified the operation such as: if one of the numbers picked was 1 (the last number 96/96), what would happen? 2*1*b - (1 + b) + 1 = b Left with b - the other number picked up. Say b was 95/96. We put back 95/96 and now the set doesn't have 1 anymore. Back to square one.

The next logical choice was to try what happens when one of the numbers picked is 1/2 (the number 48/96) since that would take care of 2 of 2ab. 2*(1/2)b - 1/2 - b + 1 = 1/2 This was what worked! When one of the numbers picked was 1/2, it eventually led to 1/2. So every time, all you do is pick 48/96 with any other number and you will always put back 1/2. You keep doing this till you are left with just 1/2.

Re: Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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11 May 2015, 21:52

VeritasPrepBrandon wrote:

This is a really tricky abstract question, and one that is probably best solved by backsolving. This kind of question is particularly dangerous because it is not only difficult to get correct, but it will bait test takers into spending far too long on it. Do not get "lost" in a problem, and know when to cut the cord on problems like this and make an educated guess. Veritas does a great job at teaching these high level strategies in their courses.

In terms of backsolving on this, I am using the term loosely because I mean plug in the answer choices and see what comes out of this formula...and if that means anything to you. For answer choice a (the correct answer), lets assign the value of 1/2 to the variable "a". The variable "b" will remain unknown. Plugging a = 1/2 into this equation, we have 2*1/2*b - 1/2 - b + 1...simplifying we get b - 1/2 - b + 1, or -1/2 + 1, or 1/2. This means that any time you use 1/2 for one of the variables in this equation, the equation will spit the result of 1/2 right back out. So when all is said and done, that is the only value that will remain.

Each of the other answer choices provide you a different output from input and keep a variable in the output (thus making the exact value impossible to know). Look at answer choice b for example. Doing the same thing, 2*1/6*b - 1/6 - b + 1 = -2/3b + 5/6. We of course cannot know what this is, as we do not know b, and it will change when b changes.

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

Brilliant, Brandon!

I arrived at the answer after looking around a bit for a pattern.

Given is my train of thought for those who did not think of going the options route:

Set S: {1/96 , 2/96 , ... 95/96 , 96/96 }

Operation: 2ab - (a+b) + 1

My first thought was that the two numbers selected should be complementary such that they lead to a simple answer which would eventually lead to some kind of sequence. It made sense to pick 1/96 and 95/96 so that a+b = 1. Similarly, 2/96 and 94/96 etc. We were left with 2ab from each operation. But the leftover values were too complicated to work with. Didn't work.

Next thought was using a number which simplified the operation such as: if one of the numbers picked was 1 (the last number 96/96), what would happen? 2*1*b - (1 + b) + 1 = b Left with b - the other number picked up. Say b was 95/96. We put back 95/96 and now the set doesn't have 1 anymore. Back to square one.

The next logical choice was to try what happens when one of the numbers picked is 1/2 (the number 48/96) since that would take care of 2 of 2ab. 2*(1/2)b - 1/2 - b + 1 = 1/2 This was what worked! When one of the numbers picked was 1/2, it eventually led to 1/2. So every time, all you do is pick 48/96 with any other number and you will always put back 1/2. You keep doing this till you are left with just 1/2.

Answer (A)

Great logic Karishma, thank you! As Karishma and I have just demonstrated, a lot of these problems (particularly the more complicated ones) can be approached from multiple angles...so do not be afraid to get creative!
_________________

Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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16 May 2015, 09:08

Hi karishma , I guess , By " So every time, all you do is pick 48/96 with any other number and you will always put back 1/2. You keep doing this till you are left with just 1/2." you mean to say So every time, all you do is pick 1/2, bcz 48 /96 is no more in the list since the first time its been picked, as the stem says selection without replacement...

Hi karishma , I guess , By " So every time, all you do is pick 48/96 with any other number and you will always put back 1/2. You keep doing this till you are left with just 1/2." you mean to say So every time, all you do is pick 1/2, bcz 48 /96 is no more in the list since the first time its been picked, as the stem says selection without replacement...

elegant solutions , keep that coming

Yes, you can call it what you like since 48/96 = 1/2. The only thing I wanted to point out was that the first 1/2 you get is because in the list you have 48/96 which is 1/2. You pick it up with another number and put only 1/2 ( = 48/96) back. Then you repeat this exercise till you are left with only 1/2.
_________________

Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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18 May 2015, 04:25

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Lucky2783 wrote:

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

This question is really easy if one understands the concept of "Equal Status" used in algebra questions. I will post some questions and solutions using the concept later. For this question consider the following:

Let us look at the numbers 1/96, 2/96...Which number is more special (higher status) than the other? 3/96? 7/96? in fact if we say any (prime number)/96 then there are many and so none should be the answer BY THIS CRITERIA. Similarly (a factor of 96)/96 ? Again we have many numbers and since there is only one correct choice (one special number) .. these numbers don't qualify BY THIS CRITERIA. Now how about the 1st number and last number? NOT in the option. Each of the option B,C,D and E have equal status --- if B can be the answer then why not C or D or E?. Option A, 1/2 =48/96 is the (around) middle number (note 47/96 will have same status as 49/96). I can safely GUESSS Option A as the answer. It splits the set into two parts.

Re: Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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18 May 2015, 04:50

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Lucky2783 wrote:

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

Algebraic Solution:

The answer is actually independent of the set!! It is just a function of the expression 2ab -a-b+1 As per the conditions let us assume that a is that number (we could take b as well) Thus we have 2ab-a-b+1 = a Or 2ab-b+1 = 2a Or 2ab-2a = b-1 Or 2a(b-1) = (b-1) ==> a = 1/2 and b can now be any number... not restricted to the given set!!

Re: Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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23 Sep 2015, 03:47

Can you please elaborate how you came to as assumption that "a is that number"

sagarsir wrote:

Lucky2783 wrote:

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

Algebraic Solution:

The answer is actually independent of the set!! It is just a function of the expression 2ab -a-b+1 As per the conditions let us assume that a is that number (we could take b as well) Thus we have 2ab-a-b+1 = a Or 2ab-b+1 = 2a Or 2ab-2a = b-1 Or 2a(b-1) = (b-1) ==> a = 1/2 and b can now be any number... not restricted to the given set!!

Re: Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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24 Sep 2015, 08:48

Lucky2783 wrote:

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

I was trying to find out a way to solve this rationally without simply plugging in the numbers. It took me close to 20 mins to figure it out and honestly I don't think this is a GMAT question. But it was fun. So here it is:

2ab-a+b+1=ab+(1-a)(1-b) The highest value this function can take when a an b are plugged in from the given set is when a and b or (1-a)and (1-b) are the highest. that's when one part of the addition sign becomes close to 1 and the other becomes zero. So the function becomes close to 1 (but a little less than that). On the other hand the lowest value this function can take happens when a and b are both close to 48/96 (or 1/2). The the function becomes 1/2.

So irrespective of the plugged values of a and b, the function adopts a value between 1 and 1/2. And as these limit values itself are within the range of the original set, the whole set will continue to converge somewhere between these 2 values ( 1 and 1/2) as we keep reducing the numbers.

The only option that falls in this range is a. So that's the answer.

Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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24 Sep 2015, 09:47

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Lucky2783 wrote:

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

If the numbers are exhausted in a set,a will be equal to b for that set. Putting a=b in given equation, \(2a^2 - 2a +1\) But since the set is exhausted, we would get same number on putting the number in above equation. So, \(2a^2 - 2a +1=a\) or \(2a^2 - 3a +1=0\) or \(a^2 - \frac{3a}{2} +\frac{1}{2}=0\) or \((a-\frac{1}{2})(a-1)=0\) or a=1,1/2 Since 1 is not in the options, our number is 1/2.

Re: Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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24 Sep 2015, 22:25

---I didnt assume... I just said to take that number as 'a'. And then we figure out the independency discussed in the solution. Its a sort of prethinking approach to such questions.

rohitmanglik wrote:

Can you please elaborate how you came to as assumption that "a is that number"

sagarsir wrote:

Lucky2783 wrote:

Set S is composed of following real numbers {1/96 , 2/96 , .. , 96/96 } . Two numbers 'a' and 'b' are selected at random without replacement and are used to calculate result according to equation 2*a*b-a-b+1 and this result is put back into the set S . This step is done repeatedly until all the numbers in the set are exhausted and set contains only 1 number. what is that last number ?

a) 1/2 b) 1/6 c) 1/8 d) 1/9 c) 1/12

Kudos for correct solution . Thanks

Algebraic Solution:

The answer is actually independent of the set!! It is just a function of the expression 2ab -a-b+1 As per the conditions let us assume that a is that number (we could take b as well) Thus we have 2ab-a-b+1 = a Or 2ab-b+1 = 2a Or 2ab-2a = b-1 Or 2a(b-1) = (b-1) ==> a = 1/2 and b can now be any number... not restricted to the given set!!

Re: Set S is composed of following real numbers {1/96 , 2/96 , .. , [#permalink]

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21 Dec 2017, 23:42

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