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How many integer values of p are there such that |p - 3| < 4/3?
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07 Apr 2022, 05:41
07 Apr 2022, 05:41
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How many integer values of p are there such that |p - 3| < 4/3?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
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Re: How many integer values of p are there such that |p - 3| < 4/3?
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07 Apr 2022, 05:51
07 Apr 2022, 05:51
Carcass wrote:
How many integer values of p are there such that |p - 3| < 4/3?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
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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How many integer values of p are there such that |p - 3| < 4/3?
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29 May 2022, 08:41
29 May 2022, 08:41
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 [ 22.07 KiB | Viewed 1800 times ]
=> 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 [ 21.76 KiB | Viewed 1819 times ]
=> 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
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)
Attachment:
temp3 to 4.33.JPG [ 22.07 KiB | Viewed 1800 times ]
=> 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)
Attachment:
1.67 to 3.JPG [ 21.76 KiB | Viewed 1819 times ]
=> 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
