If the question is (-3x)^n the value of X is sufficient
If the question is -3* (x^n) we need both the values of x and n
So it is C for me
GMAT PRP QUESTION
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HSPA
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consider x = -2 and n = 3 then x^n = -8
1) -3* (x^n) = -3*-8 is +ve
2) (-3x)^n = (-3*-2)^n ... irrespective of value of n the value is positive
consider x= -2 and n=2
1) -3*4 = -ve
1) -3* (x^n) = -3*-8 is +ve
2) (-3x)^n = (-3*-2)^n ... irrespective of value of n the value is positive
consider x= -2 and n=2
1) -3*4 = -ve
Thanks for your reply! but look at it this way:
so I get what you are saying but shoudnt you be doing the exponent first ??? By RULE PEMDAS: it should be this order
-3(x)^n so we have to do (x)^n first but we have no parentheses.
Mathematics state that if the X has no parenthesis the sign is not powered so take x= - 2 we would get - (4) and then multiply -3*-4 = always positive..
On the other hand if the Q was written like this -3*(x)^n it would def be C bcos the sign is powered to N...
There is a convention tho that we always take it as if it X (or any number) had the () and then the number and then the sign were both powered to N... Maybe here is where I am going wrong: Once I see X, when I substitute it's both, the number and sign that are powered..
Do you see where I am coming from??? THANK YOU!!
so I get what you are saying but shoudnt you be doing the exponent first ??? By RULE PEMDAS: it should be this order
-3(x)^n so we have to do (x)^n first but we have no parentheses.
Mathematics state that if the X has no parenthesis the sign is not powered so take x= - 2 we would get - (4) and then multiply -3*-4 = always positive..
On the other hand if the Q was written like this -3*(x)^n it would def be C bcos the sign is powered to N...
There is a convention tho that we always take it as if it X (or any number) had the () and then the number and then the sign were both powered to N... Maybe here is where I am going wrong: Once I see X, when I substitute it's both, the number and sign that are powered..
Do you see where I am coming from??? THANK YOU!!
well, as far as I know that is a basic Maths rule..
But I think I know where I am going wrong here when I do the problem I write it like this:
-3*x^n and then that rule stands (again I think.. clearly not an expert) In this case same as -2^2 =-4 you only power the number not the sign..
but what I should do is when I substitute the X I should power THE X( NUMBER AND SIGN) and thus write it like this:
-3*(-2)^n and then when number and sign are powered to N I do need to know N's value ... C
Thanks anayway! If there is an expert around would appreciate insight on the norm stated!!!
THANKS!
But I think I know where I am going wrong here when I do the problem I write it like this:
-3*x^n and then that rule stands (again I think.. clearly not an expert) In this case same as -2^2 =-4 you only power the number not the sign..
but what I should do is when I substitute the X I should power THE X( NUMBER AND SIGN) and thus write it like this:
-3*(-2)^n and then when number and sign are powered to N I do need to know N's value ... C
Thanks anayway! If there is an expert around would appreciate insight on the norm stated!!!
THANKS!
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GMATGuruNY
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-3x^n = (-3)(x^n).lolap4 wrote:If n is an integer is -3x^n positive ?
(1) x is negative
(2) n is odd
My answer was A bcos it doesnt have parentheses so x would keep the negative sign no matter what n was. Thus -3*-x = would al ways yield > 0 RIGHTT???
(-3x)^n = (-3)^n * x^n.
Statement 1: x<0
If x = -1 and n=2, -3(x^n) = -3*(-1)² = -3*1 = -3.
If x = -1 and n=3, -3(x^n) = -3*(-1)³ = -3*-1 = 3.
Insufficient.
Statement 2: n is odd
If x = -1 and n=3, -3(x^n) = -3*(-1)³ = -3*-1 = 3.
If x = 1 and n=3, -3(x^n) = -3*(1³) = -3*1 = -3.
Insufficient.
Statements 1 and 2 together:
If x<0 and n is odd, then x^n < 0. (A negative number raised to an odd power stays negative.)
Thus, -3(x^n) = negative * negative = positive.
Sufficient.
The correct answer is C.
To confirm the notation used on the GMAT (and in all of math):
-2² = -(2²) = -(2*2) = -4.
(-2)² = (-2)*(-2) = 4.
Last edited by GMATGuruNY on Tue Mar 15, 2011 6:32 am, edited 1 time in total.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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