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If \(|p – |p|| = 3\), how many values of \(p\) satisfy this equation?
A. 0
B. 1
C. 2
D. 3
E. 4
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A. 0
B. 1
C. 2
D. 3
E. 4
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Official Solution:
If \(|p – |p|| = 3\), how many values of \(p\) satisfy this equation?
A. 0
B. 1
C. 2
D. 3
E. 4
When \(p \geq 0\), then \(|p| = p\), so in this case we'd have \(|p - p| = 3\), which gives \(0 = 3\). This is obviously incorrect. Therefore, \(p\) cannot be 0 or positive.
Thus, \(p < 0\). In this case, \(|p| = -p\), so we'd have \(|p + p| = 3\), which gives \(|2p| = 3\). This simplifies to \(p = -\frac{3}{2}\) (since \(p = \frac{3}{2}\) is not possible because we know that \(p\) is negative). Therefore, one value of \(p\) satisfies the given equation.
Answer: B
_________________
If \(|p – |p|| = 3\), how many values of \(p\) satisfy this equation?
A. 0
B. 1
C. 2
D. 3
E. 4
When \(p \geq 0\), then \(|p| = p\), so in this case we'd have \(|p - p| = 3\), which gives \(0 = 3\). This is obviously incorrect. Therefore, \(p\) cannot be 0 or positive.
Thus, \(p < 0\). In this case, \(|p| = -p\), so we'd have \(|p + p| = 3\), which gives \(|2p| = 3\). This simplifies to \(p = -\frac{3}{2}\) (since \(p = \frac{3}{2}\) is not possible because we know that \(p\) is negative). Therefore, one value of \(p\) satisfies the given equation.
Answer: B
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- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
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General Discussion
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
Bunuel
Assume p = +ve
\(|p – |p|| = 3\)
\(|p – p| = 3\)
This is invalid. Hence, we can conclude that p is not positive
Assume p = -ve
\(|p – |p|| = 3\)
\(|p – (-p)| = 3\)
\(|2p| = 3\)
\(-2p = 3\)
\(p = \frac{-3}{2}\)
Hence, only for one value of p, the equation hold.
Option B
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Bunuel
LOGICAL APPROACH
\(|p – |p|| = 3\)
Observation 1: If p were a positive value then p and |p| would be equal and p – |p| would be zero
Since, \(|p – |p|| = 3\) i.e. p MUST be negative
Observation 2: since p is Negative therefore |p| must be positive and \(|p – |p||\) mist be equal to |2p|
Now since |2p| = 3
p must be -1.5 only hence ONLY ONE SOLUTION
Answer: Option B
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Given that |p – |p|| = 3 and We need to find how many values of p satisfy this equation?
So, there is one solution for p = -3/2
So, Answer will be B
Hope it helps!
Watch the following video to MASTER Absolute Values
_________________
| We have two cases as we have |p| | |
| -Case 1: p ≥ 0 => |p| = p => |p - p| = 3 => 0 = 3 => NO SOLUTION | -Case 2: p ≤ 0 => |p| = -p => |p - (-p)| = 3 => |2p| = 3 => 2*|p| = 3 => 2*-p = 3 => p = -3/2 The condition was p ≤ 0 and -3/2 ≤ 0 => p = -3/2 is a SOLUTION |
So, there is one solution for p = -3/2
So, Answer will be B
Hope it helps!
Watch the following video to MASTER Absolute Values
_________________
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GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
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