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For any numbers a and b, a#b=a + b - ab. If a#b=0, which of

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For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post Updated on: 11 Apr 2013, 05:05
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For any numbers a and b, a#b=a + b - ab. If a#b=0, which of the following CANNOT be a value of b?

A. 2
B. 1
C. 0
D. -1
E. -3/2

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Originally posted by Stiv on 05 Jul 2012, 02:33.
Last edited by Bunuel on 11 Apr 2013, 05:05, edited 2 times in total.
Edited the question.
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Re: For any numbers a and b, ab= a + b - ab. If ab=0, which of  [#permalink]

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New post 05 Jul 2012, 02:51
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Stiv wrote:
For any numbers a and b, ab= a + b - ab. If ab=0, which of the following CANNOT be a value of b?

A. 2
B. 1
C. 0
D. -1
E. -3/2


Some function (#) is defined for all numbers \(a\) and \(b\) as \(a#b= a + b - ab\).

Now, since given that \(a#b=0\), then \(a + b - ab=0\) --> \(a=\frac{b}{b-1}\) --> if \(b=1\) then the given expression is undefined so \(b\) cannot equal to 1.

Or: \(a + b - ab=0\) --> \((a-1)(1-b)+1=0\) --> \((a-1)(1-b)=-1\). If \(b=1\), then \((a-1)(1-b)=0\) not -1, so \(b\) cannot equal to 1.

Answer: B.

Hope it's clear.
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Re: For any numbers a and b, ab= a + b - ab. If ab=0, which of  [#permalink]

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New post 06 Jul 2012, 00:55
Can someone pls edit the question.? i was wondering where i was going wrong thinking. question was ab=a+b-ab => 2ab=a+b. :( and then saw bunuel's reply then understood what the question was! :|
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Re: For any numbers a and b, ab= a + b - ab. If ab=0, which of  [#permalink]

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New post 25 Jul 2012, 01:40
asax wrote:
Can someone pls edit the question.? i was wondering where i was going wrong thinking. question was ab=a+b-ab => 2ab=a+b. :( and then saw bunuel's reply then understood what the question was! :|

Even I got it wrong... :cry:
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Re: For any numbers a and b, ab= a + b - ab. If ab=0, which of  [#permalink]

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New post 11 Apr 2013, 04:59
asax wrote:
Can someone pls edit the question.? i was wondering where i was going wrong thinking. question was ab=a+b-ab => 2ab=a+b. :( and then saw bunuel's reply then understood what the question was! :|


Yes indeed, very annoying..
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Re: For any numbers a and b, ab= a + b - ab. If ab=0, which of  [#permalink]

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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 11 Apr 2013, 05:15
Stiv wrote:
For any numbers a and b, a#b=a + b - ab. If a#b=0, which of the following CANNOT be a value of b?

A. 2
B. 1
C. 0
D. -1
E. -3/2

------------------------------------------
for me the fastest way must be the use of answer choices to find the value of a if a will be an integer after putting the value of of b as given in the answer choices then the answer will be correct with no cumbersome calculations & its fast too ....!!
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 17 Jul 2014, 07:27
2
Can I do it this way?

I will just plug the number in to see if we can solve for a.

a#b=a+b-ab and a#b=0

=> a+b-ab=0

A. b=2 -> a=2
B. b=1 -> a+1-a=0 -> 1=0???
C. b=0 -> a=0
D. b=-1 -> a=1/2
E. b=-3/2 -> a= 3/5

So it's clear that b cannot be 1!
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 17 Jul 2014, 07:37
1
linhnd1492 wrote:
Can I do it this way?

I will just plug the number in to see if we can solve for a.

a#b=a+b-ab and a#b=0

=> a+b-ab=0

A. b=2 -> a=2
B. b=1 -> a+1-a=0 -> 1=0???
C. b=0 -> a=0
D. b=-1 -> a=1/2
E. b=-3/2 -> a= 3/5

So it's clear that b cannot be 1!


Yes, it's a valid approach.

Check other function questions in our Special Questions Directory:

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

Hope it helps.
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 01 Jan 2016, 19:24
Use answer choices to see whether a valid value for a can be determined.
B shows that 1=0 which is never true, so that is the answer.
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 18 Mar 2018, 17:45
Hi All,

This is an example of a Symbolism question (and you'll likely see 1 on Test Day). The idea is that you'll be given a "made up" math symbol, told what it "means" mathematically and then asked to solve some minor equation.

Here, we're told to substitute values in for A and B so that the equation….

A + B - AB = 0

We're asked which of the following answers CANNOT be the value of B? So 4 of the answers are POSSIBLE and one is IMPOSSIBLE. There are a couple of ways to approach this prompt. You could work mathematically or you can TEST THE ANSWERS. I'm going to use the answers to my advantage and find the 4 that are possible solutions and the one the creates an impossible situation:

If B = 2, then we'd have...
A + 2 - 2A = 0
2 = A
So B COULD be 2

If B = 1, then we'd have…
A + 1 - A = 0
1 = 0???????
B CANNOT equal 1

At this point, we could stop. I'll show you why the other answers are possible though:

If B = 0, then we'd have…
A + 0 - 0 = 0
0 = A
So B COULD be 0

If B = -1, then we'd have…
A -1 -(-1)(A) = 0
2A = 1
1/2 = A
So B COULD be -1

If B = -2/3, then we'd have…
A - 2/3 -(-2/3)(A) = 0
A + 2A/3 = 2/3
5A/3 = 2/3
A = 6/15
So B COULD be -2/3

Final Answer:

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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 08 Aug 2018, 09:00
Stiv wrote:
For any numbers a and b, a#b=a + b - ab. If a#b=0, which of the following CANNOT be a value of b?

A. 2
B. 1
C. 0
D. -1
E. -3/2


Please find here the simple solution in the picture.
Hope it helps.

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1533743971297718472570848051729.jpg
1533743971297718472570848051729.jpg [ 2.02 MiB | Viewed 5254 times ]


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For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 30 Oct 2018, 01:35
Stiv wrote:
For any numbers a and b, a#b=a + b - ab. If a#b=0, which of the following CANNOT be a value of b?

A. 2
B. 1
C. 0
D. -1
E. -3/2


Given:
a # b = a + b – ab.
and a # b = 0


i.e. a + b – ab = 0

(Adding 1 and subtracting 1 simultaneously in the left part of the equation)

i.e. a + b - ab -1 +1 = 0

i.e. a(1-b) -1 (1-b) = -1

i.e. (a-1)*(1-b) = -1

i.e. (a-1)*(b-1) = 1

i.e. either a = 0 and b = 0
or a = 2 and b = 2
or (a, b) = (3, 3/2) in any order etc.

Point to be Noted:
for \((a-1)*(b-1) = 1\)to be true none of the parts of equation can be zero i.e.
(a-1)≠0 as well as (b-1)≠0
i.e. a and b can NOT be 1


Answer: Option B
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 23 May 2019, 18:05
1
EMPOWERgmatRichC wrote:
Hi All,

This is an example of a Symbolism question (and you'll likely see 1 on Test Day). The idea is that you'll be given a "made up" math symbol, told what it "means" mathematically and then asked to solve some minor equation.

Here, we're told to substitute values in for A and B so that the equation….

A + B - AB = 0

We're asked which of the following answers CANNOT be the value of B? So 4 of the answers are POSSIBLE and one is IMPOSSIBLE. There are a couple of ways to approach this prompt. You could work mathematically or you can TEST THE ANSWERS. I'm going to use the answers to my advantage and find the 4 that are possible solutions and the one the creates an impossible situation:

If B = 2, then we'd have...
A + 2 - 2A = 0
2 = A
So B COULD be 2

If B = 1, then we'd have…
A + 1 - A = 0
1 = 0???????
B CANNOT equal 1

At this point, we could stop. I'll show you why the other answers are possible though:

If B = 0, then we'd have…
A + 0 - 0 = 0
0 = A
So B COULD be 0

If B = -1, then we'd have…
A -1 -(-1)(A) = 0
2A = 1
1/2 = A
So B COULD be -1

If B = -2/3, then we'd have…
A - 2/3 -(-2/3)(A) = 0
A + 2A/3 = 2/3
5A/3 = 2/3
A = 6/15
So B COULD be -2/3

Final Answer:

GMAT assassins aren't born, they're made,
Rich


Rich,

Thanks for the TEST the Answers method. I was trying to find answer through solving the function and took time whereas your method was very effective.
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 23 May 2019, 19:26
Hi akadiyan,

Since the GMAT is a big 'critical thinking test', you'll find that almost every question that you'll face can be approached using more than one approach. In that way, the Exam actually 'rewards' strong thinkers - if you know multiple ways to get to the correct answer, and choose the most efficient method for the question that's in front of you, then you can get to the solution with less work and in a shorter amount of time. By extension, it then becomes easier to score at a higher level overall.

Many Test Takers focus solely on questions that they get wrong - and that makes a certain amount of sense - but there's some significant potential benefits to reviewing questions that you answered correctly (since there might be faster methods that you could use or other 'efficiencies' that can help you to improve your Scores AND your pacing).

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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of  [#permalink]

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New post 11 Jun 2019, 20:23
1
Stiv wrote:
For any numbers a and b, a#b=a + b - ab. If a#b=0, which of the following CANNOT be a value of b?

A. 2
B. 1
C. 0
D. -1
E. -3/2


We are given that a#b=a + b - ab and that a#b=0. Thus we have

a + b - ab = 0

a + b = ab

Now let’s look at the choices:

A. If b = 2, then

a + 2 = a(2)

2 = a

So b can be 2.

B. If b = 1, then

a + 1 = a(1)

We see that a + 1 (1 more than a) can’t be equal to a itself. So b CANNOT be 1, and that is the number we are looking for.

Alternate Solution:

We are given that a#b=a + b - ab and that a#b=0. Thus we have

a + b - ab = 0

a + b = ab

b = ab - a

b = a(b - 1)

We observe here that b = 1 results in the inconsistent result of 1 = 0. Thus, the only value that b cannot assume is 1.

Answer: B
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Re: For any numbers a and b, a#b=a + b - ab. If a#b=0, which of   [#permalink] 11 Jun 2019, 20:23
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