if you take x=-2 and y=-3, the equation does not satisfy so y has to be greater than = to 0

next if you take x=1 and y=0, still the equation does not satisfy so y has to be grater than 0

try x=2 and y=1, the equation satisfies hence y can be anything greater than 0

so answer is B

But why are you even taking -1 or -2 or 0 #s to plug in the first place , when they are already saying that they are consecutive positie integers????

oops, sorry thanks for correcting me. then you can just take from 1,2 but why you couldn't solve using 2,3. it does satisfy the equation.

bhumika.k.shah wrote:If x and y are consecutive positive integers, and x > y , x^2-1 > y^2 -4y + x -1

Then which of the following represents all the possible values of y?

A. y>= 0
B. y>0
C. y>1
D. y >7
E. y > 8

OA B

Since x>y and they are consecutive positive integers can i take x=3 and y=2 ?
Also if i do take it then its not coming correct!

Help ! [/spoiler]

x^2-1>y^2-4y+x_1
and x=y+1(bcoz x>y and bth are consecutive int)

x^2-1-y^2+4y-x+1>0
substituting x=y+1
we get 5y>0
hence y>0
and therfore B

o ok!
but can u explain by pluggin #s???

thephoenix wrote:

bhumika.k.shah wrote:If x and y are consecutive positive integers, and x > y , x^2-1 > y^2 -4y + x -1

Then which of the following represents all the possible values of y?

A. y>= 0
B. y>0
C. y>1
D. y >7
E. y > 8

OA B

Since x>y and they are consecutive positive integers can i take x=3 and y=2 ?
Also if i do take it then its not coming correct!

Help ! [/spoiler]

x^2-1>y^2-4y+x_1
and x=y+1(bcoz x>y and bth are consecutive int)

x^2-1-y^2+4y-x+1>0
substituting x=y+1
we get 5y>0
hence y>0
and therfore B

Also what made u use x=y+1
coz generally when its given they r consecutive positive integers , the first thing that would strike us is taking #s....

You use x=y+1 because it says that x & y are consecutive but x>y...

So for example, let's say that two consecutive integers we choose are 4 and 5. In this scenario, y=4 and x=5 (by the assumptions given), so we can also say that x = 5 = (4+1) = (y+1), thus x=y+1

This is a very important concept to understand especially for more advanced problems in the GMAT. 3 consecutive integers can be thought of as x, (x+1), (x+2)... OR, another way to think about 3 consecutive integers is (x-1), x, (x+1)


With regards to this specific question, you can even simplify the original equation with substitution to get..

(y+1)(y+2) > y(y-4), which is saying that the product of the next 2 larger consecutive integers (of y) is greater than the product of y and a number 4 units smaller. Thus this is true for any number, y, greater than 0.

B.

Wow ! thats some heavy dose of math for me !!!

bigmonkey31 wrote:You use x=y+1 because it says that x & y are consecutive but x>y...

So for example, let's say that two consecutive integers we choose are 4 and 5. In this scenario, y=4 and x=5 (by the assumptions given), so we can also say that x = 5 = (4+1) = (y+1), thus x=y+1

This is a very important concept to understand especially for more advanced problems in the GMAT. 3 consecutive integers can be thought of as x, (x+1), (x+2)... OR, another way to think about 3 consecutive integers is (x-1), x, (x+1)


With regards to this specific question, you can even simplify the original equation with substitution to get..

(y+1)(y+2) > y(y-4), which is saying that the product of the next 2 larger consecutive integers (of y) is greater than the product of y and a number 4 units smaller. Thus this is true for any number, y, greater than 0.

B.

bhumika.k.shah wrote:o ok!
but can u explain by pluggin #s???

plugging is time taking job as one has to plug no's for each choices
a)
y=0--->x=1(to satisfy x>y and x,y are consecutive)
check second case
x^2-1 > y^2 -4y + x -1
0>0( not true)

B)
let y=2 ---> x=3-----> satisfy first condition x>y and x and y are consecutive int
put in second case
x^2-1 > y^2 -4y + x -1
8>-6(true)
but to make it sure we need to check by plugging -ve int
y=-3---->x=-2

second case
3>18( which is not true)
hence y>0 must be true

like wise u can plug for others too....