GMAT Math ?

[GoingUp]

For all integers a and b, the operation # is defined as a # b = (?a + b)(b + a). If a = 2 and b = 5, then b # a =


?25

?21 <--answer

?7

21

25


I guessed 21. I dont understand how to interpret what b#a means. It could be + - / or *.

Someone clue me in please.

 

[Anonemous]

? just plug in the numbers and chug.

b#a = (-b+a)(a+b)

if b =5 and a =2

(-5 +2)(2+5) = -3(7) = -21

"#" is not just a +,-,/,* it's a combination of those operations.

 

[cchen]

5 #2 = (-5 + 2)(5 + 2) = (-3)(7) = -21

are you seriously considering b-school??

 

[DingDingDao]

:laugh:

*pulls up chair and grabs popcorn*

The ensuing flamefest should be good for this one...

 

[Toastedlightly]

math != hard.

 

[saltedeggman]

Originally posted by: Toastedlightly
math != hard.

calculus is

 

Originally posted by: Anonemous
? just plug in the numbers and chug.

b#a = (-b+a)(a+b)

if b =5 and a =2

(-5 +2)(2+5) = -3(7) = -21

"#" is not just a +,-,/,* it's a combination of those operations.

Yeah, that! This is an application of abstract algebra. You use the definition given and replace the letters.

P.S. Couldn't you have just continued your questions in your already existing thread?

P.P.S. Let me explain a bit: So if they asked you for the value of g#x, you would go g#x = (-g+x)(x+g). Of course technically they would be required to define the set of values (i.e., field of numbers) we are working with.

 

[mitaiwan82]

Originally posted by: Gobadgrs
For all integers a and b, the operation # is defined as a # b = (?a + b)(b + a). If a = 2 and b = 5, then b # a =


?25

?21 <--answer

?7

21

25


I guessed 21. I dont understand how to interpret what b#a means. It could be + - / or *.

Someone clue me in please.

damn trick question...the definition is a # b, but they ask for b # a :frown:

 

[Lyfer]

Thats 6th grade math.

 

[iversonyin]

haha...I wonder which B School is going to accept this guy

 

[thereds]

Originally posted by: iversonyin
haha...I wonder which B School is going to accept this guy

DeVry?

 

[maziwanka]

Originally posted by: mitaiwan82
damn trick question...the definition is a # b, but they ask for b # a :frown:

hahaha i read too fast too

 

[GoingUp]

im quite capable of calculating (-2+5)(5+2) = 21. I was confused if they wanted us to flip the whole equation around or just change what was being substituted. All I thought the equation would change to would be (5+2)(-2+5)

 

[amoeba]

Its illustrating the use of different things as operator.

In this instance, the # is defined as an operator with the function of (a+b) (-a + b)

if you really want to be fancy you can change (a+b) (-a + b) to (b-a) (b+ a ) = b^2 -a^2

 

[Kyteland]

Originally posted by: Gobadgrs
For all integers a and b, the operation # is defined as a # b = (?a + b)(b + a). If x= 5 and y =2, then x # y = ?

Edited to remove trickyness.

Originally posted by: Gobadgrs
im quite capable of calculating (-2+5)(5+2) = 21. I was confused if they wanted us to flip the whole equation around or just change what was being substituted. All I thought the equation would change to would be (5+2)(-2+5)

You seem to be confused by that fact that they are using the same variable names to mean different things.

1) a # b = (?a + b)(b + a).
2) If a = 2 and b = 5, then b # a =

The variables a&b in 1) are not the same as the variables a&b in 2). a1=b2 and b1=a2.

2) is asking you to calculate b#a = 5#2 = (-5+2)(2+5) = (-3)(7) = -21

 

Originally posted by: Gobadgrs
im quite capable of calculating (-2+5)(5+2) = 21. I was confused if they wanted us to flip the whole equation around or just change what was being substituted. All I thought the equation would change to would be (5+2)(-2+5)

Think of it this way, if they told you:

A#B = A/B

What would you think if they asked you what B#A would be? Would you immediately think of division? Like A=2 and B =3, so 2#3 = 2/3; hence 3#2 = 3/2? No?