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

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



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



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