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

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Senior Manager
Joined: 18 Feb 2019
Posts: 386
Location: India
GMAT 1: 460 Q42 V13
GPA: 3.6
For all numbers p and q (where p ≠ 0 ), the operation \$ is defined  [#permalink]

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10 Apr 2019, 11:12
1
00:00

Difficulty:

65% (hard)

Question Stats:

25% (01:20) correct 75% (01:00) wrong based on 12 sessions

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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)

I. p =1

II. q = 0
Intern
Joined: 26 Mar 2019
Posts: 6
Re: For all numbers p and q (where p ≠ 0 ), the operation \$ is defined  [#permalink]

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10 Apr 2019, 23:38
can someone explain this ?
Intern
Joined: 16 Jan 2019
Posts: 49
Location: India
Concentration: General Management, Strategy
WE: Sales (Other)
For all numbers p and q (where p ≠ 0 ), the operation \$ is defined  [#permalink]

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11 Apr 2019, 00:01
can someone explain this ?

You can redefine the question stem as p\$q = (1st term - 1) raised to the power (2nd term - 1)

(p+1)\$(q+1) = ((p+1) - 1)^((q+1)-1) which simplifies to p^q

So, (p+1)\$(q+1) = p^q is what we need to find

Statement (1) says p=1, so the stem becomes 1^q

We know that 1 raised to any power is 1, so this is sufficient

Statement (2) says q=0, so the stem becomes p^0

We know that any number raised to the power 0 is 1, so this is sufficient

(1) and (2) are independently sufficient

(D)
For all numbers p and q (where p ≠ 0 ), the operation \$ is defined   [#permalink] 11 Apr 2019, 00:01
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# For all numbers p and q (where p ≠ 0 ), the operation \$ is defined

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