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M06-24

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M06-24  [#permalink]

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New post 16 Sep 2014, 00:28
3
20
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

36% (01:14) correct 64% (01:15) wrong based on 228 sessions

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New post 16 Sep 2014, 00:28
2
4
Official Solution:


Notice that we cannot reduce \(ab=ac\) by \(a\) and write \(b=c\), since \(a\) can be zero and division by zero is not allowed.

Rearrange and factor out \(a\): \(a(b-c)=0\): either \(a=0\) or \(b=c\).

(1) \(c=1\). If \(a=0\) then \(b\) can take any value irrespective of the value of \(c\). Not sufficient.

(2) \(a\) is a prime number and \(c\) is NOT a prime number. Now, \(a=\text{prime}\) means that \(a \ne 0\), so it must be true that \(b=c\). Now, since also given that \(c \ne \text{prime}\) then \(b\) is also not equal to a prime number so it cannot equal a prime number 2. Sufficient.


Answer: B
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New post 04 Nov 2015, 19:26
Hi,

I'm struggling with understanding the answer explanation. For (2), if c is not a prime and thus b is not a prime either, why does that mean that b (and c) needs to be 2 necessarily? Can it not be any other non-prime numbers such as 4 or 6?
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Re: M06-24  [#permalink]

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New post 04 Nov 2015, 21:56
rammangst wrote:
Hi,

I'm struggling with understanding the answer explanation. For (2), if c is not a prime and thus b is not a prime either, why does that mean that b (and c) needs to be 2 necessarily? Can it not be any other non-prime numbers such as 4 or 6?



hi,
the stat 2 tells us


(2) a is a prime number and c is NOT a prime number

and it is given that " ab=ac "..
or ab-ac=0..
a(b-c)=0..
this means either a=0, b=c, or all three could be 0..
statement 2 gives us 'a' is a prime number, so' a' cannot be 0, therefore b=c...
also it is given that 'c' is not a prime number, so 'b' is also not a prime number, so 'b' cannot be 2....
therefore statement 2 is suff to ans the question with 'NO'..
hope it helped
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Re: M06-24  [#permalink]

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New post 17 Nov 2015, 17:26
1
B it is!
First step, Try solving or simplifying the question stem. When ab = ac, we cannot cancel out 'a' until we know the value of 'a'. So first thought should be what if if a=0? In this case we cannot divide both sides by a.

Statement 1: Value of 'c' only won't help to find out value of b. As explained above, if a=0, then LHS=RHS irrespective of values of b and c. Clearly insufficient.

Statement 2: a is a prime number and c is NOT a prime number.
This means a is not zero, rather a positive number. So let's cancel on both sides. Thus b=c. Now, c is given to be non-prime. So whatever be it's value, it is certainly not 2 or 3 or 5 or 7 or so on. Therefore, we can conclude that b (which is equal to c) is also not equal to 2. Sufficient.
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M06-24  [#permalink]

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New post 18 Nov 2015, 00:27
If ab=ac is b=2?


(1) c=1

(2) a is a prime number and c is NOT a prime number

Explanation-
ab=ac -->a(b-c)=0
1)c=1 ,from (1) we cant confirm whether b=2 or not.
2)a is a prime number and c is NOT a prime number
This implies a is not 0 and b=c .Further since c is NOT a prime number ,c is not equal to 2.Hence sufficient.

Ans-B
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Re: M06-24  [#permalink]

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New post 11 Jan 2016, 14:50
Statement 2 is tricky! At first I thought it was insufficient since C is infinite but on second glance it clearly answers the statement!
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M06-24  [#permalink]

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New post 16 Jun 2016, 19:14
Bunuel wrote:
Official Solution:


Notice that we cannot reduce \(ab=ac\) by \(a\) and write \(b=c\), since \(a\) can be zero and division by zero is not allowed.

Rearrange and factor out \(a\): \(a(b-c)=0\): either \(a=0\) or \(b=c\).

(1) \(c=1\). If \(a=0\) then \(b\) can take any value irrespective of the value of \(c\). Not sufficient.

(2) \(a\) is a prime number and \(c\) is NOT a prime number. Now, \(a=\text{prime}\) means that \(a \ne 0\), so it must be true that \(b=c\). Now, since also given that \(c \ne \text{prime}\) then \(b\) is also not equal to a prime number so it cannot equal a prime number 2. Sufficient.


Answer: B



From Stat 1, how can b take any value when it is given as c=1 ?
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New post 17 Jun 2016, 06:20
avdgmat4777 wrote:
Bunuel wrote:
Official Solution:


Notice that we cannot reduce \(ab=ac\) by \(a\) and write \(b=c\), since \(a\) can be zero and division by zero is not allowed.

Rearrange and factor out \(a\): \(a(b-c)=0\): either \(a=0\) or \(b=c\).

(1) \(c=1\). If \(a=0\) then \(b\) can take any value irrespective of the value of \(c\). Not sufficient.

(2) \(a\) is a prime number and \(c\) is NOT a prime number. Now, \(a=\text{prime}\) means that \(a \ne 0\), so it must be true that \(b=c\). Now, since also given that \(c \ne \text{prime}\) then \(b\) is also not equal to a prime number so it cannot equal a prime number 2. Sufficient.


Answer: B




From Stat 1, how can b take any value when it is given as c=1 ?


It says: for c=1, IF a=0, then b can take any value.
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New post Updated on: 16 Aug 2016, 00:26
from ab = ac; => a(b-c) = 0;
it means
either a=0 or b=c.

Now,
from stat 1 we do not know what would be the value of a ?
if a = 0 then b can be take any value and equation ab = ac will still satisfy.right !
so stat 1 not sufficient.


from stat 2 we are given that a is not zero.
now if a is not zero, we can say b = c; (from either a= 0 or b = c).
and c is not a prime number, then b also wont be a prime number.

is b = 2 ? NO! (2 is a prime)
stat 2 is sufficient.

Originally posted by minhaz3333 on 14 Aug 2016, 05:45.
Last edited by minhaz3333 on 16 Aug 2016, 00:26, edited 1 time in total.
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New post 15 Aug 2016, 05:16
I think this is a high-quality question and I agree with explanation.
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New post 30 Aug 2016, 05:11
I think this is a high-quality question and I agree with explanation.
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Re: M06-24  [#permalink]

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New post 31 Jan 2018, 12:15
Hi Bunuel,

From the question stem, we already got that a = 0 and b = c. This is what we derived and is valid.

But in statement 2, we have information that a is prime and obviously prime number is only positive and can't be zero. Ok.. now a is not equal 0 , but from question stem we have concluded that a is zero but this statement 2 is showing as a is +ve and this statement is contracting with the question stem.
Do we get such question really in GMAT.
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New post 31 Jan 2018, 12:25
msk0657 wrote:
Hi Bunuel,

From the question stem, we already got that a = 0 and b = c. This is what we derived and is valid.

But in statement 2, we have information that a is prime and obviously prime number is only positive and can't be zero. Ok.. now a is not equal 0 , but from question stem we have concluded that a is zero but this statement 2 is showing as a is +ve and this statement is contracting with the question stem.
Do we get such question really in GMAT.


From the stem we got that a = 0 OR b = c (OR not AND). So, if b = c, a is not necessarily 0. From (2) \(a=\text{prime}\) means that \(a \ne 0\), so it must be true that \(b=c\).
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New post 17 Feb 2018, 00:45
Good question. I missed to consider a=0 in option A and considered it as sufficient and marked C as answer incorrectly.

Thanks Bunuel for explanation
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New post 18 Mar 2019, 19:44
I think this is a high-quality question and I agree with explanation. Really good question !!!
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New post 29 May 2019, 03:40
Bunuel, if division by 0 is not allowed then why is it being assessed in this question?

I thought the only time we really need to worry about the sign or value of a variable is during inequalities, since we may need to flip the signs etc.
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New post 29 May 2019, 03:43
dcummins wrote:
Bunuel, if division by 0 is not allowed then why is it being assessed in this question?

I thought the only time we really need to worry about the sign or value of a variable is during inequalities, since we may need to flip the signs etc.


Can you please re-read the solution and then rephrase your question? Where are we dividing by 0? Where are we concerned about the sign?
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New post 29 May 2019, 04:00
Bunuel wrote:
dcummins wrote:
Bunuel, if division by 0 is not allowed then why is it being assessed in this question?

I thought the only time we really need to worry about the sign or value of a variable is during inequalities, since we may need to flip the signs etc.


Can you please re-read the solution and then rephrase your question? Where are we dividing by 0? Where are we concerned about the sign?



I re-read the solution and understand it, but I still think the phrase "since a can be zero and division by zero is not allowed" is confusing since it's more to do with the fact that a can be 0, making (0)b=(0)c, then it has to do with any "division by zero"
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New post 17 Jun 2019, 08:07
I think this is a high-quality question and I agree with explanation. Oh boi
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Re M06-24   [#permalink] 17 Jun 2019, 08:07

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