Problem Solving

If |d - 9| = 2d, then d =

A. -9
B. -3
C. 1
D. 3
E. 9

IMO=>
|d - 9| =+(d-9) or -(d-9)
when +(d-9) =2d => d=-9(A)

when -(d-9) =2d => 3d=9 =>d=3([spoiler]answer-D)[/spoiler]

I certainly feel that this one is an ambiguous question. I am posting it just for opinions and to conform whether I missed anything or not.

since 2d=|d - 9|
=>2d must be >0 or 0 (2d cannot be negative)
=> d must be >0 or 0

of the 2 values of d you worked out, only d=3 satisfies the above condition
hence, D

scoobydooby wrote:since 2d=|d - 9|
=>2d must be >0 or 0 (2d cannot be negative)
=> d must be >0 or 0

of the 2 values of d you worked out, only d=3 satisfies the above condition
hence, D

Scooby, I think you are thinking 2d as d^2 which is indeed wrong.

2d = 2*d (it can be + or -)

here it is, 2*(-1) = -2(negative)

However, what I think you wanted to say was d^2= d*d = always positive(I agree )
You are mixing up both the things my friend

i beleive the expression within the modulus sign can have positive /negative value or 0, but the value outside the modulus can only be positive or 0. it can never be negative

|+2| or |-2|=+2 always.

in the question, if one takes d<0 gives 2d<0=>|d-9|<0 not possible. d must be >0

scoobydooby wrote:i beleive the expression within the modulus sign can have positive /negative value or 0, but the value outside the modulus can only be positive or 0. it can never be negative

|+2| or |-2|=+2 always.

in the question, if one takes d<0 gives 2d<0=>|d-9|<0 not possible. d must be >0

Let me get to the basics.

|d-9|= +(d-9) only when d>9

but when d<9 => |d-9| = -(d-9)

this - sign is used because we need the value after taking the " modulus||"

lets say when d<9
take d=4
so, |d- 9| = |4-9| = |-5|

As you said, the value after taking mod has to be positive, so when we remove mod, we need to place an additional (-) sign before the mod to make it positive.

|-5| => replace mod=> -(-5) => 5

gmat740,
thanks for the refresher, buddy. i only wanted to say that the question is not really ambiguous.

even you agree once we take the absolute value, whatever be the expression inside becomes positive.

let d-9 =y
given |y|=2d
we know |y|>0 or =0
so 2d must be >0 or =0
=>d>0 or d=0

y can be >0 or <0
=>d-9>0 or d-9<0
=>d=-9 or d=3 (as you found out)
of the two values of d, only d=3 works (being >0)
we get just one answer: D

You can solve the absolute value equation for d:

|d-9|=2d

d-9=2d---->d=-9

-and-

-d+9=2d---->d=3

Now test out both these solutions in the original equation:

|-9-9|=2(-9)---> 18 does not equal -18 so -9 is not a correct option. This leaves us with d being equal to 3.

Gmat740, when you came up with your solutions of 3 and -9, why didn't you just test them like I did? This problem really isn't that ambiguous at all.

x = |y|
x = y^2

x cannot be negative in both cases

truplayer256 wrote:You can solve the absolute value equation for d:

|d-9|=2d

d-9=2d---->d=-9

-and-

-d+9=2d---->d=3

Now test out both these solutions in the original equation:

|-9-9|=2(-9)---> 18 does not equal -18 so -9 is not a correct option. This leaves us with d being equal to 3.

Gmat740, when you came up with your solutions of 3 and -9, why didn't you just test them like I did? This problem really isn't that ambiguous at all.

Thanks a ton.
I forgot to check back again. I would keep in mind.

Scooby Thanks a lot buddy for discussion.