saxenagarima
Bunuel
saxenagarima
If 4x-12 >= x + 9, which of the following must be true
a) X > 6
b) X < 7
C) X > 7
D) X > 8
E) X < 8
I doubt the OA in
MGMAT solutions
Given: \(4x-12\geq{x + 9}\) --> \(3x\geq{21}\) --> \(x\geq{7}\).
Only A is always true, as ANY \(x\) from the TRUE range \(x\geq{7}\) will be more than 6.
Answer: A.
But with A) x can be 6.1, which will not satisfy the given equation.....
shouldn't option C) be valid choice ....
It should be other way around.
We are given that \(x\geq{7}\). The question is: which of the following is true about \(x\)?
\(x>6\) is true about \(x\), because as \(x\) is more than (or equal to) 7 then it's definitely more than 6.
To elaborate more. Question uses the same logic as in the examples below:If \(x=5\), then which of the following must be true about \(x\):A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10
Answer is E (x>-10), because as x=5 then it's more than -10.
Or:
If \(-1<x<10\), then which of the following must be true about \(x\):A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120
Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.
Or:
If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\):A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1
As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.
\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.
On the other hand for example A is not always true as it says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.
Hope it's clear.