On the number line you can see that -5 <= x <= 3
you can use back solving here, from choice E.
|X+1|<=4
x + 1 <= 4
x <= 3
-|X+1|<=4
-x - 1 <= 4
-x <= 5
x >= -5
so it fits perfectly.
-5 <= x <= 3
Also, you can eliminate A and B b/c they are not closed intervals.
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Source: Beat The GMAT — Problem Solving |
I got both B and E fit into this. But why can't we use anything other than closed intervals? Even B gives the same output right?
B is
X>= -5 to X <= 5 (incorrect since it includes region from 3 to 5 as well)
E is
X+1 <= 4 ; X+1 >= -4
i.e. X <=3; x>=-5
this satisfies the whole region
Hope its clear now.
X>= -5 to X <= 5 (incorrect since it includes region from 3 to 5 as well)
E is
X+1 <= 4 ; X+1 >= -4
i.e. X <=3; x>=-5
this satisfies the whole region
Hope its clear now.
reasoning way:
|x-a| is the distance from any point x to point a.
just look at the number line and see the options:
A. |x| <= 3 distance from any point to 0 is less than or equal to 3-wrong
B. |X| <= 5 distance from any point to 0 is less than or equal to 5-wrong
C. |X-2| <= 3 distance from any point to 2 is less than or equal to 3-wrong
D. |X-1|<= 4 distance from any point to 1 is less than or equal to 4-wrong
E. |X+1|<=4 distance from any point to -1 is less than or equal to 4-right
|x-a| is the distance from any point x to point a.
just look at the number line and see the options:
A. |x| <= 3 distance from any point to 0 is less than or equal to 3-wrong
B. |X| <= 5 distance from any point to 0 is less than or equal to 5-wrong
C. |X-2| <= 3 distance from any point to 2 is less than or equal to 3-wrong
D. |X-1|<= 4 distance from any point to 1 is less than or equal to 4-wrong
E. |X+1|<=4 distance from any point to -1 is less than or equal to 4-right
The powers of two are bloody impolite!!
















