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M03-01

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New post 16 Sep 2014, 00:19
1
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A
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E

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New post 16 Sep 2014, 00:19
3
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Official Solution:


Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3.

Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10.

Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.


Answer: E
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New post 06 Oct 2014, 06:19
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Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3.

Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10.

Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.


Answer: E


I incorrectly selected A.
i thought that

X is an integer so X-1, X-2 and X-3 are 3 consecutive numbers and since product of 3 numbers is >0 so all 3 numbers should be either positive or any two can be -ve.

No any 2 can not be negative cause they are consecutive. if 2 are -ve and one is +ve then it would be -2, -1 and 0 but it is given that product can not be zero.
So we ultimately left with one solution of all 3 +ve no hence.
and if the 3 consecutive number less than X are positive then X will be greater than 3.

I got the way you solved it. I understood it before in one of your post but it did not striked in my mind when i was giving test. Can you plese tell me that what is the flaw in my thinking.
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New post 06 Oct 2014, 07:42
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him1985 wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3.

Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10.

Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.


Answer: E


I incorrectly selected A.
i thought that

X is an integer so X-1, X-2 and X-3 are 3 consecutive numbers and since product of 3 numbers is >0 so all 3 numbers should be either positive or any two can be -ve.

No any 2 can not be negative cause they are consecutive. if 2 are -ve and one is +ve then it would be -2, -1 and 0 but it is given that product can not be zero.
So we ultimately left with one solution of all 3 +ve no hence.
and if the 3 consecutive number less than X are positive then X will be greater than 3.

I got the way you solved it. I understood it before in one of your post but it did not striked in my mind when i was giving test. Can you plese tell me that what is the flaw in my thinking.


You assumed (incorrectly) that x is an integer.
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New post 07 Oct 2014, 17:20
Hello Bunuel. Can you please explain the solution in a more lay terms...i implemented similar logic as him1985. If x<3 then the product of the three terms will be negative - isn't it? and we are given that the product of these three terms is +ve...so doesn't that imply that x>3? Can you please point out the flaw in my logic? Thanks for your help.
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New post 08 Oct 2014, 01:52
p2bhokie wrote:
Hello Bunuel. Can you please explain the solution in a more lay terms...i implemented similar logic as him1985. If x<3 then the product of the three terms will be negative - isn't it? and we are given that the product of these three terms is +ve...so doesn't that imply that x>3? Can you please point out the flaw in my logic? Thanks for your help.


No. Try x = 1.5.

Check other solutions here: m03-70436.html#p771274

Below links might help:

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New post 06 Jan 2016, 07:33
I think range of X in here x>3 & 1<x<2. Am I right? I have little confusion about this. Would any like to ensure my question? thank you.
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New post 17 Mar 2016, 11:06
In my opinion.... :peek

The solution for statement one should be:


S1 therefore limits x to A) 1 > x < 2 B) x >=3

Instead of

S1 therefore limits x to (1,2)∪(3,∞) :rocket

Unlike the official solution, x cannot be 1 or 2 since the equation would then equal zero.
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New post 17 Mar 2016, 11:17
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eldin wrote:
In my opinion.... :peek

The solution for statement one should be:


S1 therefore limits x to A) 1 > x < 2 B) x >=3[/color]

Instead of

S1 therefore limits x to [color=#ed1c24](1,2)∪(3,∞) :rocket

Unlike the official solution, x cannot be 1 or 2 since the equation would then equal zero.


1. The highlighted part does not make any sense.
2. x cannot be 3 either.
3. (1,2) means that x is between 1 and 2, not inclusive.
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New post 18 Jul 2016, 08:19
I think this is a high-quality question and I agree with explanation. Hi, i used following logic to solve this question. Pls let me know whether my logic seems ok.

Statement 1 implies x-3>0, x-2>0 and x-1>0, and hence x>3, x>2 and x>1. So, x could be more than 3 and less than/equal to 3 (e.g 2,3). Not sufficient

Statement 2: x>1. Not sufficient

Combine: x could be 2, 3 or more than three. Not sufficient
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Re: M03-01  [#permalink]

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New post 26 Sep 2017, 17:10
hi Bunuel

if the question had said x is an integer, confirm A would have been suffient?
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New post 25 Oct 2017, 02:06
My solution. I always prefer to analyze from statement 2. So,

(2) Clearly insufficient. x>1 does not mean that x>3. X could be either 2 (the answer is no) or 4 (the answer is yes)
(1) Clearly insufficient. Roots of the polynomial are 1, 2 and 3. Could be less or greater than 3 as well.

(2)+(1). If x>1, it means that (x-3)(x-2) must be both positive or negative. If x=1.5, then (x-3)(x-2)>0. But x is less than 3, so the answer is NO. If x=4, then (x-3)(x-2) also greater than 0. And x greater than 3, so the answer is YES.

Answer: E
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New post 22 May 2018, 13:44
I think this is a high-quality question and I agree with explanation.
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New post 28 Dec 2018, 16:27
I think this is a high-quality question and I agree with explanation. tricky. I never even thought about fractions.
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Re: M03-01  [#permalink]

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New post 01 Apr 2019, 11:15
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3.

Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10.

Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.


Answer: E



Hi Bunuel

S1 says (x−3)(x−2)(x−1) > 0 
This implies that x > 3, x > 2 and x > 1

So, x can be any value (integer and non-integer), hence Answer is YES or NO.

As mentioned in the solution above for S1, can you please elaborate on the following?
"The polynomial (x−3)(x−2)(x−1)(x−3)(x−2)(x−1) has roots (1,2,3)(1,2,3). They are distinct, which means that the polynomial changes its sign around the roots. If xx is greater than 3, then it is positive. If xx is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits xx to (1,2)∪(3,∞)(1,2)∪(3,∞)"

i'm not exactly familiar with this ??

Would appreciate your support!
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New post 01 Apr 2019, 11:25
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JIAA wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3.

Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10.

Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.


Answer: E



Hi Bunuel

S1 says (x−3)(x−2)(x−1) > 0 
This implies that x > 3, x > 2 and x > 1

So, x can be any value (integer and non-integer), hence Answer is YES or NO.

As mentioned in the solution above for S1, can you please elaborate on the following?
"The polynomial (x−3)(x−2)(x−1)(x−3)(x−2)(x−1) has roots (1,2,3)(1,2,3). They are distinct, which means that the polynomial changes its sign around the roots. If xx is greater than 3, then it is positive. If xx is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits xx to (1,2)∪(3,∞)(1,2)∪(3,∞)"

i'm not exactly familiar with this ??

Would appreciate your support!
Thanks


I provided some links on the theory above. If that does not help, here is an alternative solution.

Is \(x>3\)?

(1) \((x-3)(x-2)(x-1)>0\)

The product of 3 numbers is positive if all three are positive (+++) OR two of them are negative and the third one is positive (+--).

Note that: out of 3 numbers \(x-3\) is the least one and \(x-1\) is the biggest one.

\((+)(+)(+)\) is when even the least one is positive so when \(x-3>0\) --> \(x>3\);
\((+)(-)(-)\) is when the biggest one is positive (\(x-1>0\) --> \(x>1\)) and the next one (hence the leas one too) negative (\(x-2<0\) --> x<2), so when \(1<x<2\);

So \((x-3)(x-2)(x-1)>0\) means that: \(x>3\) or \(1<x<2\) --> \(x\) may or may not be more than 3. Not sufficient.

(2) \(x>1\). Clearly insufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is the range we had in (1) \(x>3\) or \(1<x<2\), so \(x\) may or may not be more than 3. Not sufficient.

Answer: E.
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New post 02 Apr 2019, 00:32
Bunuel wrote:
JIAA wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. The polynomial \((x-3)(x-2)(x-1)\) has roots \((1, 2, 3)\). They are distinct, which means that the polynomial changes its sign around the roots. If \(x\) is greater than 3, then it is positive. If \(x\) is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits \(x\) to \((1, 2) \cup (3, \infty)\). \(x\) can be either greater or less than 3.

Statement (2) by itself is insufficient. S2 limits \(x\) to \((1, \infty)\), which means that \(x\) can equal either 2 or 10.

Statements (1) and (2) combined are insufficient. \(x\) can still be either less than 3 or greater than 3.


Answer: E



Hi Bunuel

S1 says (x−3)(x−2)(x−1) > 0 
This implies that x > 3, x > 2 and x > 1

So, x can be any value (integer and non-integer), hence Answer is YES or NO.

As mentioned in the solution above for S1, can you please elaborate on the following?
"The polynomial (x−3)(x−2)(x−1)(x−3)(x−2)(x−1) has roots (1,2,3)(1,2,3). They are distinct, which means that the polynomial changes its sign around the roots. If xx is greater than 3, then it is positive. If xx is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. S1 therefore limits xx to (1,2)∪(3,∞)(1,2)∪(3,∞)"

i'm not exactly familiar with this ??

Would appreciate your support!
Thanks


I provided some links on the theory above. If that does not help, here is an alternative solution.

Is \(x>3\)?

(1) \((x-3)(x-2)(x-1)>0\)

The product of 3 numbers is positive if all three are positive (+++) OR two of them are negative and the third one is positive (+--).

Note that: out of 3 numbers \(x-3\) is the least one and \(x-1\) is the biggest one.

\((+)(+)(+)\) is when even the least one is positive so when \(x-3>0\) --> \(x>3\);
\((+)(-)(-)\) is when the biggest one is positive (\(x-1>0\) --> \(x>1\)) and the next one (hence the leas one too) negative (\(x-2<0\) --> x<2), so when \(1<x<2\);

So \((x-3)(x-2)(x-1)>0\) means that: \(x>3\) or \(1<x<2\) --> \(x\) may or may not be more than 3. Not sufficient.

(2) \(x>1\). Clearly insufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is the range we had in (1) \(x>3\) or \(1<x<2\), so \(x\) may or may not be more than 3. Not sufficient.

Answer: E.



Thanks Bunuel for the alternate explanation! It really did help !
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Re: M03-01  [#permalink]

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New post 15 May 2019, 16:46
I couldn't see any 'theory' relating to polynomials in the links provided unless I overlooked it via the search function in my browser.

While the theory is noted via Bunuel's explanation, I think a more simplistic approach is to test non-integer values.

Statement 1

X= 1.5

produces (1.5-3)(1.5-2)(1.5-1) = +ve number... therefore x>0 --> is X>3 ? NO
X=4
produces (+)(+)(+)>0 is x>3? Yes

Therefore Insufficient

Statement (2) is also insufficient even without considering non-integer values
i.e. since "x>1" x can be 2 or 4 producing No and Yes answers.

(1+2) no new information --> Insufficient.
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New post 29 May 2019, 07:42
I don't know if my way of thinking is correct. However, I managed to answer it correctly.

I gave x a small number (3), a large number (10) and a negative number. X=3 gave me 0. x=10 gave me a number higher than 3. Therefore statement 1 could be either more or less than 3.

Statement two gave me a number that was any number (infinite) that 2 or bigger. It could be a 2 or it could be a number bigger than 3. That statement was also insufficient.

Together, both are insufficient.
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Re: M03-01   [#permalink] 29 May 2019, 07:42

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