For all numbers p and q (where p 0 ), the operation $ is defined as

Question

For all numbers p and q (where p ≠ 0 ), the operation $ is defined as  p \$ q = (p-1)^{(q-1)} . What is the value of  (p +1) \$ (q +1) ?

(1) p =1

(2) q = 0

gmatophobia · Answer

Given  :

p $ q =  (p-1)^{(q-1)}

Statement 1

(1)  p =1

p + 1 = 1 + 1 = 2

(p +1) $ (q +1)  = (2 - 1) ^{(q + 1 - 1)} = 1^{q} = 1

The statement alone is sufficient. We can eliminate B, C, and E.

Statement 2

(2)  q = 0

q + 1 = 0 + 1 = 1

(p +1) $ (q +1)  = (p + 1 - 1) ^{(1 - 1)} = p^{0} = 1

The statement alone is sufficient.

Option D

Mathewseldo · Answer

In statement 1, you don't know the value of q so how can we assume 1^q is 1? same for statement 2, only after we know both values together can we determine the value right?

Which is why I got C as the answer.

Please correct me if I am wrong.

Bunuel · Answer

You do not need to know the exponent when the base is 1, because 1 raised to any power is 1.

Likewise, once the exponent is 0, the expression is fixed at 1, as long as the base is nonzero.

For all numbers p and q (where p 0 ), the operation $ is defined as

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