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# If |d-9| = 2d, then d= (A) -9 (B) -3 (C) 1 (D) 3 (E) 9

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If |d-9| = 2d, then d= (A) -9 (B) -3 (C) 1 (D) 3 (E) 9 [#permalink]  18 Jan 2007, 07:49
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70% (01:41) correct 29% (00:54) wrong based on 48 sessions
If |d-9| = 2d, then d=
(A) -9
(B) -3
(C) 1
(D) 3
(E) 9
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Re: PS: Modulus D [#permalink]  18 Jan 2007, 09:42
1
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if d>9, on solving eqn u get d = -9 which is impossible since d>9.
if d<9, on sloving u get d = 3. Hence D is correct answer
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d has two solutions : d=3 and d=-9
D !
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IMO

2d is an absolute value, so d can't be negative.

Out of the +ve nos. d can be only 3
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Sumithra wrote:
IMO

2d is an absolute value, so d can't be negative.

Out of the +ve nos. d can be only 3

Same approach : it's better to plug 2 values mentally with the respect of abs always positive (or 0) than to solve the original equation (saving energy... 4 hours is long)
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Re: PS: Modulus D [#permalink]  10 Sep 2009, 11:40
1
KUDOS
If |d-9| = 2d, then d=
(A) -9
(B) -3
(C) 1
(D) 3
(E) 9

now the two eqs are
1) when (d-9) > 0
d-9 = 2d
d = -9

2) when (d-9) < 0
9-d = 2d
d = 3

the two initial solns are d = -9 and 3

but when we substitute d = -9 in original equation we get
|-9-9| = 2 * -9
18 = -18
which is not possible

Hence only solution is d = 3.
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Re: PS: Modulus D [#permalink]  23 Jan 2011, 07:51
vsaxenaGMAT wrote:
if d>9, on solving eqn u get d = -9 which is impossible since d>9.
if d<9, on sloving u get d = 3. Hence D is correct answer

GOOD EXPLANATION
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Re: PS: Modulus D [#permalink]  23 Jan 2011, 08:37
aurobindo wrote:
If |d-9| = 2d, then d=
(A) -9
(B) -3
(C) 1
(D) 3
(E) 9

You can approach this problem in several ways. For example: given |d-9| = 2d --> as LHS (|d-9|) is an absolute value then it's non-negative so RHS (2d or simply d) must also be non-negative thus answer choices A and B are out. Next you can quickly substitute the values to see that d=3 satisfies given inequality: |3-9|=|-6|=6=2*3.

Or you can try algebraic approach and expand |d-9| for 2 ranges:
If 0\leq{d}\leq{9} then -(d-9)=2d --> d=3 --> you have an answer D right away;
Just to check the second range: If {d}>9 then d-9=2d --> d=-9 --> not a valid solution as d can not be negative (also this value is not in the range we are considering).

Hope it's clear.
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Re: PS: Modulus D   [#permalink] 23 Jan 2011, 08:37
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