First thing about DS questions, I love solving them by trying to show that the statement can give yes & no answers to the question and thereby eliminating them,
Lets take 2) first, since it is easier
Statement (2): x>1
We can easily eliminate since x can be 1.5 or 2 or 10, we can't say whether x is greater than 3 or not. So, there goes Statement 2) and answer choices (B) and (D)
Statement (1): (x-3)(x-2)(x-1) > 0
take x as 1.5 -> (1.5-3)(1.5-2)(1.5-1) -> (-ve)(-ve)(+ve) -> so the answer is positive, since two neagtives make a prositive and therefore we have proved that x can be less than 3
now, let us prove that x can be greater than 3
-> take x as 4 -> (4-3)(4-2)(4-1) -> which is obviously greater than zero.
So, statement 1 alone is also not sufficient and so there goes answer choice (A) as well
Combining both statements
X> 1 and (x-3)(x-2)(x-1) > 0
from our calculations for statement 2, we already showed that taking x=1.5 and x=4, we are able to satisfy the inequality and since these two values of x are greater than 1, we are thus unable to determine whether x>3.
Hope this helps.
is x > 3?
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ajith
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1. (x-3)(x-2)(x-1) > 0rahul.s wrote:is x > 3?
1) (x-3)(x-2)(x-1) > 0
2) x > 1
OA: E
the OE was complicated. could someone simplify it?
For numbers greater than 3 = +ve
Numbers between 2 and 3 = -ve
numbers between 1 and 2 = +ve
Numbers less than 1 = -ve
Not sufficient
2. x>1 is not sufficient to tell whether x>3 Not sufficient
Combining
Even if we combine 1 and 2 the x could be between 1 and 2 or x is greater than 3 Not sufficient again to conclude whether x>3
Hence E
Always borrow money from a pessimist, he doesn't expect to be paid back.
couldn't you just say that:
1) INS: the 3 possible solutions of (x-3)(x-2)(x-1)>0 are x > 3; x > 2; or x > 1. so, you cannot conclude x to be greater than 3. it could be 2 or 2.5
2) INS: x >1. x could be 2
Together: still insufficient b/c x could still be 2.
Am I looking at this logically?
1) INS: the 3 possible solutions of (x-3)(x-2)(x-1)>0 are x > 3; x > 2; or x > 1. so, you cannot conclude x to be greater than 3. it could be 2 or 2.5
2) INS: x >1. x could be 2
Together: still insufficient b/c x could still be 2.
Am I looking at this logically?
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Ian Stewart
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No, that isn't quite right; I think you're solving the inequality as though it were an equation. Here, we are multiplying three things (x-1, x-2 and x-3) and getting a positive result. That can happen in two ways:mstone wrote:couldn't you just say that:
1) INS: the 3 possible solutions of (x-3)(x-2)(x-1)>0 are x > 3; x > 2; or x > 1. so, you cannot conclude x to be greater than 3. it could be 2 or 2.5
* all three are positive
* one is positive, the other two negative
Notice that we know which of our three things is largest, and which smallest: x-3 < x-2 < x-1. So if all three are positive, the smallest surely must be positive, and x > 3. If two are negative, and one positive, then the largest must be positive, so x > 1, and the second largest must be negative, so x < 2. That is, if (x-3)(x-2)(x-1) > 0, then either x > 3 or 1 < x < 2. For any other value of x, then (x-3)(x-2)(x-1) will be negative or zero, as you can confirm by plugging in values for x.
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