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Pound sign?

by egybs » Fri May 30, 2008 12:39 pm
Hey folks - I've never seen this kind of notation before. What the heck does the pound mean? Now that I think about it, I think I remember seeing the same kind of thing while preparing for the SATs... Thanks!

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

OA:E
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by aatech » Fri May 30, 2008 12:52 pm
#-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
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by egybs » Fri May 30, 2008 12:56 pm
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.
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by kishore » Fri May 30, 2008 1:02 pm
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
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by chidcguy » Fri May 30, 2008 1:02 pm
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?
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by egybs » Fri May 30, 2008 1:06 pm
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!
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by Stuart@KaplanGMAT » Fri May 30, 2008 8:16 pm
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.
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by Strongt » Mon May 16, 2011 12:05 pm
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?
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by Tani » Mon May 16, 2011 1:09 pm
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
Tani Wolff
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