M02-08

Question

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

A.  y \le 0  
B.  y \gt 0  
C.  y \gt 1  
D.  y \gt 7  
E.  y \gt 8

Bunuel · Accepted Answer

Official Solution:

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

A.  y \le 0  
B.  y \gt 0  
C.  y \gt 1  
D.  y \gt 7  
E.  y \gt 8

Given that  x  and  y  are consecutive integers and  x > y , then  x=y+1 . Substituting  x  with  y+1  in the provided equation, we get:
 
 (y+1)^2-1 \gt y^2-4y+(y+1)-1 ;
 
This simplifies to  5y \gt 0 ;
 
Therefore, it must be that  y \gt 0 .

Answer: B

AndrewN · Answer

Another way to approach the question is with  easy numbers  and then test the answers to see what you can eliminate. As long as you pick  consecutive integers  and the original inequality holds true, then you will have created a valid system of inputs. In this case, I arbitrarily chose y = 1 and x = 2  because I am lazy , and I would rather not do a bunch of time-consuming squaring and all that if I walk away with a better sense of  what must be true  from such simple inputs. We need to test in the original inequality:

(2)^2-1 \gt (1)^2-4(1)+(2)-1

4-1 \gt 1-4+2-1

3 \gt -2

This is true, 3 is greater than -2. Thus, we have created a valid system of inputs for our variables. Now check the answers against what we know to be true:

A.  y \le 0 
No, y is not less than 0. We chose 1, and the inequality held up.  Red light. 
 
B.  y \gt 0  
Yes, y is a positive integer, 1. Keep this for now.  Yellow light.

C.  y \gt 1  
No, our input of 1 for y is  not  greater than 1.  Red light.

D.  y \gt 7  
This is clearly not going to work. We just proved that y can equal 1, and 1 is not greater than 7.  Red light.

E.  y \gt 8 
This will not work either, for the same reason as in (D) above.  Red light.

With just one answer choice that checked out, the winner must be (B). For you skeptics out there who might be wondering what if a different combination of inputs had been selected, you might simply have to refine your numbers until one answer emerged, but I always advise keeping such numbers simple for ease of access. Why choose a negative if a positive would do? (More people make mistakes with signs when dealing with negatives.) Would y = 0 and x = 1 work? No, but you would get 0 on both sides--0 = 0, NOT 0 > 0--which would reveal that you had reached a crucial point, and that the answer was just around the corner.

Good luck with your studies.

- Andrew

AnirudhaS · Answer

x^2-1>y^2-4y+x-1 
or,  x^2>y^2-4y+x 
or,  x(x-1)>y(y-4) 
Substituting x=y+1, since they are consecutive integers and x>y
or,  (y+1)y>y(y-4) 
or,  y(y+1-y+4)>0 
or,   5y>0 
or,  y>0

Answer:  B

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

shreetamjena2000 · Answer

If Y > 0, then it's obvious that Y > 1,2,3,4,5,6,7,8... as well. Then why can't we consider the other options as correct? Are these options valid?

Bunuel · Answer

What if y = 1. Would any of the other options be correct?

vn049636 · Answer

I think this is a  high-quality  question and I agree with explanation.

Sanskar115 · Answer

I think this is a  high-quality  question.

Backoftheclass · Answer