If x and y are consecutive integers, x y, and x^2 - 1 y^2 - 4y + x

Question

If x and y are consecutive integers, x > y, and x^2 - 1 > y^2 - 4y + x - 1, then which of the following must be true?

A. x < 0
B. x ≠ 2
C. y > -1
D. y ≠ 1
E. x + y > 3

Bunuel · Accepted Answer

Question 37 - Official Solution: If x and y are consecutive integers, x > y , and x^2 - 1 > y^2 - 4y + x - 1 , which of the following must be true? A. x < 0 B. x ≠ 2 C. y > -1 D. y ≠ 1 E. x + y > 3 Given that x and y are consecutive integers, and x > y , we can say that x = y + 1 . Substituting this into the given inequality, we get: (y + 1)^2 - 1 > y^2 - 4y + (y + 1) - 1 ; y^2 + 2y + 1 - 1 > y^2 - 4y + y ; 2y > - 3y ; 5y > 0 ; y > 0 . y can be 1, 2, 3, 4, and so on. Since y = x - 1 , we can rewrite this inequality in terms of x : x - 1 > 0 , which gives x > 1 . x can be 2, 3, 4, 5, and so on. Since y > 0 , then y must also be greater than -1, making option C correct. Even though we have already determined that option C is correct, let's elaborate on the other options for the sake of understanding why they are not necessarily true: A. x < 0 . We found that x > 1 , so this option is not true. Discard. B. x ≠ 2 . Since we found that x > 1 , this option could be true, but it's not necessarily true. Discard. D. y ≠ 1 . Since we found that y > 0 , this option could be true, but it's not necessarily true. Discard. E. x + y > 3 . Since we found that y > 0 and x > 1 , the smallest possible value for x + y is 1 + 2 = 3 . Therefore, this option is not necessarily true. Discard. Answer: C

chetan2u · Answer

Since x and y are consecutive integers and x>y, we can say that x=y+1.
Substitute this value in equation. 
 x^2 - 1 > y^2 - 4y + x - 1 
  (y+1)^2 - 1 > y^2 - 4y + (y+1) - 1 
  y^2 +2y+1- 1 > y^2 - 4y + y+1- 1 
 5y>0…..y>0 
Also  y>0>-1

Thus,  y>-1  must be always true.

C

gmatophobia · Answer

Hi chetan2u , Bunuel Apologies if I am asking a silly question, but I didn't catch the last part of the working well Does this mean, if y is always greater than zero, its always greater than -1 ? I attempted the question using a similar approach as you did, however I also concluded that option A will also be true. Sharing an snippet of my working (in the attached image). Can you let me know where did I go wrong ? Explanation of my analysis - y > 0 We know that, y = x -1 x - 1 > 0 x > 1 Reasoning : If x is greater than 1, its always greater than 0. Therefore I concluded that A is also true.

Can you please let me know what I am missing in here ?

bellbell · Answer

I answered correctly but I also want official explanation because it seems like A&C is kinda the same?
Thank you.

Bunuel · Answer

Option A should have been x < 0, instead of x > 0. Edited. Thank you for noticing. +1. If x and y are consecutive integers, x > y, and x^2 - 1 > y^2 - 4y + x - 1, then which of the following must be true? A. x < 0 B. x ≠ 2 C. y > -1 D. y ≠ 1 E. x + y > 3 x and y are consecutive integers, and x > y means that x = y + 1 (y = x - 1). Substitute x = y + 1 ( or y = x - 1, it does not really matter which one ) into x^2 - 1 > y^2 - 4y + x - 1: x^2 - 1 > y^2 - 4y + x - 1 ; (y + 1)^2 - 1 > y^2 - 4y + (y + 1) - 1 ; y > 0 . Now, since y > 0 (y can be 1, 2, 3, 4, ...), the y will for sure be more than -1. So, option C must be correct. Answer: C. Other options: So, we got that y > 0 (y can be 1, 2, 3, 4, ...) and we know that x = y + 1, so x - 1 > 0 and x > 1 (x can be 2, 3, 4, 5, ...). A. x < 0. Not true because x > 1 (x can be 2, 3, 4, 5, ...). B. x ≠ 2. Not necessarily true because x could be 2 (x can be 2, 3, 4, 5, ...). C. y > -1. Must be true because y > 0. D. y ≠ 1. Not necessarily true because y could be 1 (y can be 1, 2, 3, 4, ...). E. x + y > 3. Not necessarily true. If y = 1 (the least possible value of y) then x = y + 1 = 2 (the least possible value of x), and in this case x + y = 3. Hope it helps.

bumpbot · Answer

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If x and y are consecutive integers, x > y, and x^2 - 1 > y^2 - 4y + x

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