beginners'bin#4

IMO B

Choose B
It will always yeild 1 as the result.

Fastest way would be to take examples. But, we needd to ensure that we have considered all cases.

Take (-1, -2, -3) & (-2,-3,-4).

Only B fits the bill.

thats correct, whenever solving a question by keeping examples in place of variables (x,y,z) it is important that you take more than 1 example..
Here, for example, if you take -1,-2,-3 even D would give a positive, Odd Number... So, its better you take one more example -2,-3,-4 to arrive at conclusion !

yup its B

as x, y and z are consecutive -ve integer

x-y = 1 and y-z = 1

(x-y)(y-z) = 1

Ans B

An alternate approach :

We know x > y >z . And x , y and z are consecutive integers , we also know consecutive integers alternate between odd and even integers .

But we do not know which variable is odd / even .

So there can be only 2 cases .

Case 1 : y = even .
So x = odd and z = odd .

Lets look at the answer options .

(A) x y z -> Odd x Even x Odd = Even , eliminate

(B) (x - y) (y - z) -> Odd - Even = Odd , Even - Odd = Odd .
Odd x Odd = Odd .
(C) x - y z -> Odd -Even = Odd .

(D) x (y + z) -> Odd x Odd = Odd .

(E) x + y + z -> Odd + Even + Odd = Even.

So far we have B , C , D as contenders .

Lets look at Case 2 :

y = odd .

x= even and z = even .

Lets check answer options now.

(A) x y z -> even x odd x even = Even , eliminate

(B) (x - y) (y - z) -> (even-odd)(odd - even)
odd x odd = odd .
(C) x - y z -> even - Even = even .

(D) x (y + z) -> even x Odd =even.

(E) x + y + z -> Even + Odd + Even = Odd .


Only option B gives the Odd result in both the cases .

Plugging in will be much faster but some times you may forget some values .

rockeyb wrote:An alternate approach :

We know x > y >z . And x , y and z are consecutive integers , we also know consecutive integers alternate between odd and even integers .

But we do not know which variable is odd / even .

So there can be only 2 cases .

Case 1 : y = even .
So x = odd and z = odd .

Lets look at the answer options .

(A) x y z -> Odd x Even x Odd = Even , eliminate

(B) (x - y) (y - z) -> Odd - Even = Odd , Even - Odd = Odd .
Odd x Odd = Odd .
(C) x - y z -> Odd -Even = Odd .

(D) x (y + z) -> Odd x Odd = Odd .

(E) x + y + z -> Odd + Even + Odd = Even.

So far we have B , C , D as contenders .

Lets look at Case 2 :

y = odd .

x= even and z = even .

Lets check answer options now.

(A) x y z -> even x odd x even = Even , eliminate

(B) (x - y) (y - z) -> (even-odd)(odd - even)
odd x odd = odd .
(C) x - y z -> even - Even = even .

(D) x (y + z) -> even x Odd =even.

(E) x + y + z -> Even + Odd + Even = Odd .


Only option B gives the Odd result in both the cases .

Plugging in will be much faster but some times you may forget some values .

@rockeyb

Rightly said!! I would say always try to plug in more distinct set of values to reach "conclusion".This hold true for DS also.
My inputs are 3 distinct sets.Even though it may take a extra 30-40 secs..u will certainly nail it!

Plugging in numbers is easier and faster on this kind of probelms...

IMO B