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Considering that both P & Q are integers, lets see how can 4 be written as (base)^exponent
4 = 2^2 Or (-2)^2 Or 4^1 Or (-4)^1

So we have cases:
P = 2, Q = 2
P = -2, Q = 2 (in both these cases, |P|=2)
P = 4, Q = 1
P = -4, Q = 1 (in both these cases, |P|=4)

as we can see, P can be both positive and negative, but Q can only be positive

Statement 1. |P| + |Q| = |P+Q|
If P is negative and Q is positive, then above statement cannot be true
But if P and Q are both positive, then above statement is definitely true

So we get that P is positive, thus it answers NO to the question asked. Sufficient

Statement 2. P^2 - 5|P| + 4 = 0
Note that P^2 can also be written as |P|^2

So we have |P|^2 - 5|P| + 4 = 0
Factorise this quadratic equation to get
(|P| - 1)(|P| - 4) = 0
So |P| = 1 OR |P| = 4

Now |P| = 1 is not possible if |P|^Q = 4
But for |P| = 4, we have |P| =4, Q = 1 so that 4^1 = 4

But still here we can have P = 4 or P = -4
Cant say whether P is positive or negative. Insufficient

Hence answer is A
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niteshwaghray
If P & Q are integers such that \(|P|^Q=4\), is P negative?

(1) \(|P|+|Q|=|P+Q|\)
(2) \(P^2−5|P|+4=0\)


In DS the question is the third condition - read it carefully.
Since P and Q are integer, Q, in the equation |P|^Q=4, must be always positive.

1 - True only if either P+ and Q+ or P- and Q-. As per question stem Q always positive. A is sufficient.
2 - (|P|-4)(|P|-1)=0 - clearly not suff
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