#-7# = a*-7*3 + b*-7 - 1 = 3

=> -21a-7b-1=3
=> 21a+7b=-4 -----(1)

#7# = a*7*3 + b*7 - 1 => 21a+7b-1 ---(2)

Replacing value from (1) to (2) we get -5

What does the pound sign represent though... I've never seen that before (or at least i've forgotten). I don't even know what i'm looking at.

If #p# = ap3+ bp – 1 where a and b are constants, and #-7# = 3, what is the value of #7#?

A)5
B)0
C)-2
D)-3
E)-5


Answer is E.

Explanation:

#-7# = 3

=> #-7# = a(-7)3+ b(-7) – 1 = 3

#7#?

#7# = a(7)3+ b(7) – 1
This can be written as
- (a(-7)3+ b(-7) – 1) - 2

from above a(-7)3+ b(-7) – 1 = 3

there fore - (a(-7)3+ b(-7) – 1) - 2 = -3 -2 = -5

Thanks,
Kishore

AFAIK, The pound sign does not represent anything in math world.

The Q defined a fictitious operation based on those symbols. They did not use +, * or any thing that is standard to avoid confusion.

Is this what you are looking at?

ohh... so this is really just the same thing as asking

f(p)=ap3+ bp – 1
f(-7) = 3

what is f(7)?

In that case, this is pretty simple! Thanks guys!

egybs wrote:ohh... so this is really just the same thing as asking

f(p)=ap3+ bp – 1
f(-7) = 3

what is f(7)?

In that case, this is pretty simple! Thanks guys!

Yah.. "wacky symbol" questions are really function questions. It seems that in recent versions of the GMAT the symbols are becoming rarer and function questions more common.

aatech wrote:#-7# = a*-7*3 + b*-7 - 1 = 3

=> -21a-7b-1=3
=> 21a+7b=-4 -----(1)

#7# = a*7*3 + b*7 - 1 => 21a+7b-1 ---(2)

Replacing value from (1) to (2) we get -5

I know this sounds silly but I'm missing the last step?
What exactly happens when you replace terms?

You ended up with two equations:
first: -21a - 7b = 4

and
second: +21a +7b -1 = x

rewrite this as
+21a +7b = x+1

since the left hand side of the second equation is simply -1 times the left had side of the first equation, you can set x+1 = -4 and therefore x = -5