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# What is the value of |x + 7|? (1) |x + 3| = 14 (2) (x + 2)^2

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What is the value of |x + 7|? (1) |x + 3| = 14 (2) (x + 2)^2 [#permalink]  04 Dec 2008, 13:33
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What is the value of |x + 7|?

(1) |x + 3| = 14
(2) (x + 2)^2 = 169
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Re: DS: abs [#permalink]  04 Dec 2008, 14:28
twilight wrote:
What is the value of |x + 7|?

(1) |x + 3| = 14
(2) (x + 2)^2 = 169

(1) |x + 3| = 14

if x is +ve: x + 3 = 14. x = 11
if x is -ve: x + 3 = - 14. x = -17

(2) (x + 2)^2 = 169
if x is +ve: (x + 2)^2 = 169
x + 2 = -13 or 13
x = either -15 or 11.

C: 1&2: x = 11.
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Re: DS: abs [#permalink]  05 Dec 2008, 11:10
whenever u have to slove abs and quadratics, you'll always have to expect two answers. there is a very similar problem up today on [url]gmatgremath.blogspot.com[/url]

jsut remember to set up two diff eqs whenyou have the abs value question
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Re: DS: abs [#permalink]  05 Dec 2008, 20:59
the Q is |x + 7| right,

why cant the answer be D

1) |x + 3| = 14 so |x + 7| = 18
2) (x + 2)^2 = 169 can be written as |x + 2| = 13 so |x + 7| = 18

Senior Manager
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Re: DS: abs [#permalink]  05 Dec 2008, 21:23
Answer is C. Here is my explanation.

Clue 1: |x+3| = 14 ==> x+3 = 14 or -(x+3) = 14. By solving them, x can be 11 or -17. So clue is INSUFFICIENT.
Clue 2: (x+2)^2 = 169. Taking the square root, x+2 = 13 or x+2 = -13. By Solving them, x can be 11 or -15. Clue alone is INSIFFICIENT.

Now considering clues 1 and 2, the common value of x is 11. We need to both the clues to solve the answer.
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Re: DS: abs [#permalink]  06 Dec 2008, 02:03
selvae wrote:
the Q is |x + 7| right,

why cant the answer be D

1) |x + 3| = 14 so |x + 7| = 18
2) (x + 2)^2 = 169 can be written as |x + 2| = 13 so |x + 7| = 18

|x + 7| = |x + 3 + 4|

|x+3| + 4 = 14 + 4 = 18

As |x + 3 + 4| <> |x+3| + 4 so |x + 7| <>18

Note that |x + y| <> |x| + |y| (you can prove by squaring both sides)
Re: DS: abs   [#permalink] 06 Dec 2008, 02:03
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