Two properties involving absolute value inequalities:
Property #1: If |something| < k, then –k < something < k
Property #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive

Since the given equation is in the same form as property #1, we can write: -4/3 < p - 3 < 4/3
To provide a clear picture, let's express the fractions as decimal approximations: -1.33 < p - 3 < 1.33
Now add 3 to all parts of the inequality: 1.67 < p < 4.33

Since p is an integer, there are three possible values of p that satisfied the inequality 1.67 < p < 4.33.
The values are p = 2, p = 3, and p = 4

Answer: D
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Given that |p - 3| < \(\frac{4}{3}\) and we need to find how many integer values of p satisfy this equation

To open |p - 3| 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

=> p - 3 ≥ 0 => p ≥ 3

|p - 3| = p - 3 (as if A ≥ 0 then |A| = A)
=> p - 3 < \(\frac{4}{3}\)
=> p < \(\frac{4}{3}\) + 3
=> p < \(\frac{4+3*3}{3}\)
=> p < \(\frac{13}{3}\) ~ 4.33
=> But our condition was p ≥ 3. So solution will be the range common in both of them (refer below image)



=> 3 ≤ p < 4.33

So, possible integer values of p in this range are 3 and 4

Case 2: Assume that whatever is inside the Absolute Value/Modulus is Negative

p - 3 < 0 => p < 3

|p - 3| < \(\frac{4}{3}\) (as if A < 0 then |A| = -A)
=> -(p - 3) < \(\frac{4}{3}\)
=> -p + 3 < \(\frac{4}{3}\)
=> p > 3 - \(\frac{4}{3}\)
=> p > \(\frac{3*3 - 4}{3}\)
=> p > \(\frac{5}{3}\) ~ 1.67
=> But our condition was p < 3. So solution will be the range common in both of them (refer below image)



=> 1.67 < p < 3

So, possible integer values of p in this range is 2

=> Final values of p are 2, 3, 4 => 3 values

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems


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