How many integer values of p are there such that |p - 3|

Question

How many integer values of p are there such that |p - 3| < 4/3?

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

BrentGMATPrepNow · Accepted Answer

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

bv8562 · Answer

|p - 3| < 4/3 a) p<3 -(p-3) < \frac{4}{3} => -p + 3 < \frac{4}{3} => p > 3-\frac{4}{3} => p > \frac{5}{3} => p > 1.6\bar{6} b) p\geq{3} (p-3) < \frac{4}{3} => p < \frac{4}{3}+3 => p < \frac{13}{3} => p < 4.3\bar{3} Thus, 1.6\bar{6}

BrentGMATPrepNow · Answer

You'll find that p =  3  it's also a solution to the given inequality, since | 3  - 3| < 4/3 evaluates to become |0| < 4/3, which is true.

bv8562 · Answer

Yes, I realized it after looking at your solution. The second region in my solution should be  p\geq{3}  and not  p>3  because at p=3 value inside mod is non-negative. Is it correct?  BrentGMATPrepNow

BrentGMATPrepNow · Answer

I'm unfamiliar with your technique (it seems like it would be problematic if the algebraic expression for more complex than p - 3). 
That's said, your rationale sounds valid.

BrushMyQuant · Answer

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) temp3 to 4.33.JPG => 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 to 3.JPG => 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 https://www.youtube.com/watch?v=rBWfl4AmQDI

ashishpathak · Answer

Taking a number line approach 3 will be the centre point and 4/3 is the distance either side of the centre.4/3 equals 1.3 so range will be 3 -1.3 to 3 + 1.3 integers within this range will be 2 3 and 4 .
Is this approach correct please
Feel free to correct thanks

Posted from my mobile device

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How many integer values of p are there such that |p - 3| < 4/3?

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How many integer values of p are there such that |p - 3| < 4/3?

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