ook 4,medium question set,question no:10

[hasibul212003]

Hi,Manhattan math guide book 4,medium question set,question no:10 at the end of the book : x,y,r and t are integers such that x^y is negative and r^t is positive. If t is not a multiple of 2 which of the followings statements must be true?
A.xr>0
B.x-r<0
C.y is a multiple of 2
D.x^r
E.None of the above.
I selected ans E,but saw it sud be B. What is the explanation-I am not sure why B must be true.

[n00bpron00bpron00b]

Hey,

X,Y,R,T are integers

#1) x^y is negative
which means,

value of X is negative (can be even or odd ; does not matter)
value of Y is Odd

pick any random integer values satisfying the above conditions,

if x = -3 & y = 3
(-3)^3 = -27
if x= -2 & y = 1
(-2)^1 = -2

Y cannot be "even" in this case (negative base with an even exponent will make it positive) and X cannot be positive

#2) r^t is positive and t is not a multiple of 2
t is not a multiple of 2 => value of "t" is odd
if "t" is odd then the base (i.e. "r") has to be positive for the end result (r^t) to be positive

Because a negative base raised to and odd exponent will give us a negative end result (-3)^3 = -27

pick any random integer values satisfying the above conditions,

r = 2 (can be even or odd, must strictly positive)
t = 3
r^t = (2)^3 = 8

Let's see the answer choices using the following info -

#) X is negative and Y is ODD
#) R is positive and T is ODD

1) xr > 0
from above we know,
x is negative and r is positive
so "x * r" will be negative

2) x - r < 0
x is negative
r is positive

if x = -1 and r = 5 then,
(-1) - (+5) = -1-5 = -6 (which is less than 0)

3) Y is a multiple of 2
from 1st condition we know, "Y" is odd (so cannot be multiple of 2)

4) x^t > 0
x is negative
t is odd
Negative base raised to an odd exponent will always be odd
(-3)^3 = -27

So answer B

hope this helps

[hasibul212003]

Thanks n00bpron00bpron00b. Got it now.

[tommywallach]

Fantastic response, Noob. Thanks!

-t