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the possible values of x?

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by sanju09 » Sat Oct 22, 2011 3:55 am
If |x - 2| < 5, what are the possible values of x?
(A) 0 < x < 5
(B) 0 < x < 2
(C) -3 < x ≤ 7
(D) -3 ≤ x < 7
(E) -3 < x < 7
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by Anurag@Gurome » Sat Oct 22, 2011 4:19 am
sanju09 wrote:If |x - 2| < 5, what are the possible values of x?
|x - 2| < 5 implies x is a number such that its distance from 2 is less than 5. The numbers which are at a distance of 5 units from 2 in both direction are (2 - 5) and (2 + 5), i.e. -3 and 7.

Hence, x must be a number between -3 and 7.

The correct answer is E.
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by krishnakumar.ks » Sat Oct 22, 2011 4:20 am
When we remove the modulus of a number, it will assume two possibilities. For example, |a| will assume two values for a : +a and -a.

Therefore when you remove modulus of the given equation we will have two cases.

Case1: x-2<5,
x<7.
Case2 : -(x-2)<5, multiply by - sign through out, the equation becomes, (x-2)>-5
=> x>-3

Combining both the cases, -3<x<7 is the answer. Hence D.
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by GmatKiss » Sat Oct 22, 2011 7:16 am
IMO:E
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by thestartupguy » Sun Oct 23, 2011 10:10 am
It's better to solve by putting the values in the options. We can get the answer in 15s :p

Ans: E
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by MBA.Aspirant » Sun Oct 23, 2011 10:41 am
If |x - 2| < 5, what are the possible values of x?


|x - 2| < 5 = -5< x - 2 <5

-3< x < 7, so E
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