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pyzo
If \(p=\frac{x^{a+b}}{x^b}\), what is the value of positive integer p?


(1) \(x = 5\)
(2) \(a = 0\)

\(p=\frac{x^{a+b}}{x^b}\) = \(x^{a+b-b}\) = \(x^a\)

(1) NS - as we donot know the value of \(a\).
(2) Suff ... as the index value is \(0\), whatever is the base \(x\), the expression \(x^a\) yields a fixed result ......thus Ans B
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Given, P=x^(a+b) /x^b,
p= x^a

A) x= 5,
P=5^a, not sufficient.

B) a=0
P= x^0=1,sufficient

Hence, b.

Posted from my mobile device
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Bunuel
If \(p = \frac{x^{(a + b)}}{x^b}\), what is the value of positive integer p ?


(1) \(x = 5\)

(2) \(a = 0\)

Solution


Step 1: Analyse Question Stem


    • \(p = \frac{x^{(a+b)}}{x^b}\)
    • \(p = x^{(a+b-b)}\)
    • \(p= x^a\)
We need to find the value of p.
    • The value of p depends upon the value of x and a.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: \(x = 5.\)
    • \(p= 5^a\)
      o However, we don’t know the value of a.
      o Thus, we cannot find the value of p.
Hence, statement 1 is not sufficient, we can eliminate answer options A and B.
Statement 2: a = 0.
    • \(p = x^0\)
      o Now, irrespective of the value of x, \(x^0\) will be 1 only.
      o Therefore, \(p= x^0 = 1\).
Hence, statement 2 is sufficient, the correct answer is Option B.
Note that in this question many of the students will mark the answer as option C, this is the C-trap.
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pyzo
If \(p=\frac{x^{a+b}}{x^b}\), what is the value of positive integer p?


(1) \(x = 5\)
(2) \(a = 0\)

Hi Bunuel

Why is the answer B? Nothing is mentioned about the possibility of x not being 0. How can statement 2 alone be sufficient?

I read in the below article that
Quote:
Any nonzero number to the power of 0 is 1.

https://gmatclub.com/forum/exponents-an ... 74993.html
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pyzo
If \(p=\frac{x^{a+b}}{x^b}\), what is the value of positive integer p?


(1) \(x = 5\)
(2) \(a = 0\)

Hi Bunuel

Why is the answer B? Nothing is mentioned about the possibility of x not being 0. How can statement 2 alone be sufficient?

I read in the below article that
Quote:
Any nonzero number to the power of 0 is 1.

https://gmatclub.com/forum/exponents-an ... 74993.html

We are given that p is a positive integer, while if x = 0, it won't be so. Hence x cannot be 0.
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