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# What is the value of p? (1) 5p-20 = 5 (2) (p-10)^2 = p^2 The

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Intern
Joined: 11 Oct 2005
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What is the value of p? (1) 5p-20 = 5 (2) (p-10)^2 = p^2 The [#permalink]  10 Nov 2005, 13:12
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What is the value of p?

(1) 5p-20 = 5
(2) (p-10)^2 = p^2

The answer to this DS is D. However, I got A. For statement 2, I took the squareroot of both sides and the p's canceled out. What did I do wrong? I feel like i"m missing something totally obvious.
Manager
Joined: 04 Oct 2005
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you cannot take the squarroot in this case, because the p is inside the fraction and the value depends on another number (10). This number could make the value positive or negative. since you do not know that, you cant take the sqrt.
you have to work the problem

(p-10)^2 = p^2
p^2 - 20p + 100 = p^2
100 = 20p
p = 5
Director
Joined: 24 Oct 2005
Posts: 576
Location: NYC
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Re: Algebra DS [#permalink]  10 Nov 2005, 14:07
crmasic wrote:
What is the value of p?

(1) 5p-20 = 5
(2) (p-10)^2 = p^2

The answer to this DS is D. However, I got A. For statement 2, I took the squareroot of both sides and the p's canceled out. What did I do wrong? I feel like i"m missing something totally obvious.

you cant just take the square root. try this

p^2 -20p + 100 = p^2

(a-b)^ 2 = a^2 -2ab +b^2
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SVP
Joined: 28 May 2005
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Location: Dhaka
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I got D

statement 1 is sufficient.
5p -20 =5
so p = 5

statement 2 is also sufficient

(p-10)^2 = p^2

we can write it as (p-10)^2 - p^2 = 0

or (p-10-P)(p-10+p) = 0

from here we can find p = 5

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hey ya......

SVP
Joined: 16 Oct 2003
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D is correct. Got with the same solution
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