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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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What is the value of p? (1) 5p-20 = 5 (2) (p-10)^2 = p^2 The [#permalink] New post 10 Nov 2005, 13:12
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D
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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.
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 [#permalink] New post 10 Nov 2005, 14:05
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
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Re: Algebra DS [#permalink] New post 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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 [#permalink] New post 10 Nov 2005, 14:31
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

so D is the answer.
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 [#permalink] New post 10 Nov 2005, 18:02
D is correct. Got with the same solution
  [#permalink] 10 Nov 2005, 18:02
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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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