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Set 6 Q7 - DS

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Source: — Data Sufficiency |
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by josephcho77 » Thu Oct 04, 2007 7:39 am
I will also go with "C"

1) x can be 11 or -17

2) x can be 11 or -15

When combined, X=11
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by arocks » Thu Oct 04, 2007 7:43 am
Agreed. I had the same solution. So the OA is incorrect in this case. Thnx.
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by athakkar » Thu Oct 04, 2007 9:40 pm
I think the answer D is correct.

For first condition to be independently sufficient:

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

But according to (1), |x+3| = 14;

Therefore,
|x+7|=14 +4
= 18

For second condition to be independently sufficient:
According to (2),
(x+2) = 13
i.e x= 11

Therefore,
|x+7| = |11+7| = 18

Thus, the answer is D.
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by arocks » Fri Oct 05, 2007 8:40 am
athakkar,

This is how i think it works:
(1) │x + 3│= 14
So, x+3=14 OR x+3=-14
>x=11 OR x=-17

Hence, two values for x from stmt.1 > insuff.

(2) (x + 2)^2 = 169
>(x+2)^2=13^2
>x+2=+-13
>x=11 OR x=-15

Hence, two values for x from stmt.2 > insuff.
But if we combine stmt1 and stmt2, we get x=11.

So the answer is C.
Just two things to be kept in mind here, the treatment for lMODl in stmt.1 and that stmt.2 has a square power, so there will be two solutions - negative and positive.

Hope this clarifies!
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