If |x – 1/5|

Question

If |x – 1/5| < 3/5, then the number of integer values that x can take is

A) 0
B) 1
C) 2
D) 3
E) 4

IanStewart · Answer

|a - b| is the distance between a and b on the number line. So the inequality:

|x - 1/5| < 3/5

says in words that "the distance from x to 1/5 is less than 3/5". So x needs to be very close to 1/5, and the only integer less than 3/5 away from 1/5 is zero, so the answer is one.

ScottTargetTestPrep · Answer

Solution:

When x – 1/5 is positive, we have:

x - 1/5 < 3/5

x < 4/5

When x - 1/5 is negative, we have:

-x + 1/5 < 3/5

-x < 2/5

x > -2/5

Thus:

-2/5 < x < 3/5

So, x can only be one integer, namely 0.

Alternate Solution:

When an absolute value inequality is of the “less than” variety, we can drop the absolute value and create a “between” statement:

-3/5 < x - 1/5 < 3/5

-2/5 < x < 4/5

There is only 1 integer value that x can take on, which is 0.

Answer: B

BrushMyQuant · Answer

Given that |x – \frac{1}{5} | < \frac{3}{5} and we need to find the number of integer values that x can take To open |x – \frac{1}{5} | < \frac{3}{5} we need to take two cases (Watch this video to know about the Basics of Absolute Value ) Case 1 : Assume that whatever is inside the Absolute Value/Modulus is non-negative => x – \frac{1}{5} ≥ 0 => x ≥ \frac{1}{5} |x – \frac{1}{5} | = x – \frac{1}{5} (as if A ≥ 0 then |A| = A) => x – \frac{1}{5} < \frac{3}{5} => x < \frac{3}{5} + \frac{1}{5} => x < \frac{4}{5} And our condition was x ≥ \frac{1}{5} . So the answer will be part common in x ≥ \frac{1}{5} and x < \frac{4}{5} => \frac{1}{5} ≤ x < \frac{4}{5} is the solution 1by5 to 4by5.JPG But there are no integers in this range. Case 2 : Assume that whatever is inside the Absolute Value/Modulus is Negative => x – \frac{1}{5} < 0 => x < \frac{1}{5} |x – \frac{1}{5} | = -(x – \frac{1}{5} ) (as if A < 0 then |A| = -A) => -(x – \frac{1}{5} ) < \frac{3}{5} => -x + \frac{1}{5} < \frac{3}{5} => x > \frac{1}{5} - \frac{3}{5} => x > \frac{-2}{5} And our condition was x < \frac{1}{5} . So the answer will be part common in x < \frac{1}{5} and x > \frac{-2}{5} => \frac{-2}{5} < x < \frac{1}{5} is the solution -2by5 to 1by5.JPG Only integer in this range is 0 So, Answer will be B Hope it helps! Watch the following video to learn How to Solve Absolute Value Problems https://www.youtube.com/watch?v=rBWfl4AmQDI

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If |x – 1/5| < 3/5, then the number of integer values that x can take

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