If |d - 9| = 2d, then d = A. -9 B. -3 C. 1 D. 3 E. 9 : PS Archive
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# If |d - 9| = 2d, then d = A. -9 B. -3 C. 1 D. 3 E. 9

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VP
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If |d - 9| = 2d, then d = A. -9 B. -3 C. 1 D. 3 E. 9 [#permalink]

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08 Jul 2008, 18:19
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If |d - 9| = 2d, then d =
A. -9
B. -3
C. 1
D. 3
E. 9
Director
Joined: 23 Sep 2007
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Kudos [?]: 185 [1] , given: 0

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08 Jul 2008, 18:37
1
KUDOS
D

Eliminate A/B, because d must be positive

The rest is just an easy number plugin.
VP
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08 Jul 2008, 20:15
I got D too. But the OA given for the Q is E.
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08 Jul 2008, 20:44
goalsnr wrote:
I got D too. But the OA given for the Q is E.

This questions seems ackward. I get two scenarios - 9 and 3 and since d has to be +ve D (3) is the answer

If E is the answer then |d -9| = |9 -9| = 0 which is not possible.

What is the source of your question?
Director
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08 Jul 2008, 21:14
D should be the correct answer.
VP
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Kudos [?]: 612 [0], given: 10

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08 Jul 2008, 22:28
x97agarwal wrote:
goalsnr wrote:
I got D too. But the OA given for the Q is E.

This questions seems ackward. I get two scenarios - 9 and 3 and since d has to be +ve D (3) is the answer

If E is the answer then |d -9| = |9 -9| = 0 which is not possible.

What is the source of your question?

This is how I solved :
|d-9|=2d

=> d> 0, d-9 =2d ->d = -9
but this contradicts the assumption d>0

=> d<0 , -(d-9 ) = 2d
-d+9=2d
3d =9 =>d=3
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09 Jul 2008, 01:00
goalsnr wrote:
x97agarwal wrote:
goalsnr wrote:
I got D too. But the OA given for the Q is E.

This questions seems ackward. I get two scenarios - 9 and 3 and since d has to be +ve D (3) is the answer

If E is the answer then |d -9| = |9 -9| = 0 which is not possible.

What is the source of your question?

This is how I solved :
|d-9|=2d

=> d> 0, d-9 =2d ->d = -9
but this contradicts the assumption d>0

=> d<0 , -(d-9 ) = 2d
-d+9=2d
3d =9 =>d=3

You're almost there, except you made a small confusion that led to a contradiction by the end of your proof too: you conclude with d=3 when you assumed d<0 to make the calculations

This is because |d-9| = d-9 when d $$\ge$$ 9 and 9-d when d $$\le$$ 9 (and not 0)
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 147 [0], given: 2

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09 Jul 2008, 03:50
I think I made the same mistake ... I came up with two answers, neither of which seemed to work:

for d>0 case, i got d=-9, which is inadmissable

for d<0 case, i got d=3, which also seems inadmissable

Where did I go wrong ?
Current Student
Joined: 12 Jun 2008
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09 Jul 2008, 03:56
1
KUDOS
pmenon wrote:
I think I made the same mistake ... I came up with two answers, neither of which seemed to work:

for d>0 case, i got d=-9, which is inadmissable

for d<0 case, i got d=3, which also seems inadmissable

Where did I go wrong ?

Because what separates the cases is not d>0 or d<0!

You want to know the sign of |d -9|, so what makes the separation is d-9$$\le$$0 or d-9$$\ge$$0, i.e. d$$\le$$9 or d$$\ge$$9

You then get:

- If d$$\ge$$9, solution is d=-9: this is a contradiction

- If d$$\le$$9, solution is d=3: this is perfectly correct
SVP
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Kudos [?]: 147 [0], given: 2

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09 Jul 2008, 04:07
Oski wrote:
pmenon wrote:
I think I made the same mistake ... I came up with two answers, neither of which seemed to work:

for d>0 case, i got d=-9, which is inadmissable

for d<0 case, i got d=3, which also seems inadmissable

Where did I go wrong ?

Because what separates the cases is not d>0 or d<0!

You want to know the sign of |d -9|, so what makes the separation is d-9$$\le$$0 or d-9$$\ge$$0, i.e. d$$\le$$9 or d$$\ge$$9

You then get:

- If d$$\ge$$9, solution is d=-9: this is a contradiction

- If d$$\le$$9, solution is d=3: this is perfectly correct

ah, c'est bon ! merci !! plus un !
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09 Jul 2008, 04:15
1
KUDOS
pmenon wrote:
ah, c'est bon ! merci !! plus un !

VP
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09 Jul 2008, 17:12
This is how I solved :
|d-9|=2d

=> d> 0, d-9 =2d ->d = -9
but this contradicts the assumption d>0

=> d<0 , -(d-9 ) = 2d
-d+9=2d
3d =9 =>d=3[/quote]
You're almost there, except you made a small confusion that led to a contradiction by the end of your proof too: you conclude with d=3 when you assumed d<0 to make the calculations

This is because |d-9| = d-9 when d $$\ge$$ 9 and 9-d when d $$\le$$ 9 (and not 0)[/quote]

THanks Oski for pointing out the error.Kudos!
SVP
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09 Jul 2008, 18:46
So is the correct answer (D) 3 even though the OA says it is E (which does not make any sense).
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Current Student
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09 Jul 2008, 23:58
Re: PS -mod   [#permalink] 09 Jul 2008, 23:58
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