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0 < x < y, and x and y are consecutive integers. If the difference bet
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27 Feb 2017, 01:27
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64% (02:03) correct 36% (02:05) wrong based on 141 sessions
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0 < x < y, and x and y are consecutive integers. If the difference between x^2 and y^2 is 12,201, then what is the value of x? (A) 6,100 (B) 6,101 (C) 12,200 (D) 12,201 (E) 24,402
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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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27 Feb 2017, 21:57
Hi vikasp99, This question can be solved in a couple of different ways. It's also built around some specific Number Properties, so if you're not sure how to approach the question, you can take advantage of those patterns and take a great guess. We're told that X and Y are CONSECUTIVE INTEGERS (0 < X < Y) and that Y^2  X^2 = 12,201. We're asked for the value of X. To start, it's interesting that the DIFFERENCE ends in a 1. When subtracting the squares of consecutive integers, there are only a couple of ways that this can occur: Y ends in a 1 and X ends in a 0 Y ends in a 6 and X ends in a 5 From the answer choices, we can clearly see that it's the first option  and that the correct answer is either Answer A or Answer C. If you're not sure what to do next, then you could guess and move on. However, if you want to continue working, you don't actually have to square two big numbers to find the solution to this question. Since the two variables are consecutive, we can rewrite Y as (X + 1), so the original equation can be written as... (X+1)^2  X^2 = 12,201 X^2 + 2X + 1  X^2 = 12,201 2X + 1 = 12,201 2X = 12,200 X = 6,100 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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27 Feb 2017, 01:34
Answer: (A) Since x and y are consecutive integers you can take y=x+1. So y^2x^2=2x+1=12201 2x=12200 x=6100. Sent from my Nexus 5 using GMAT Club Forum mobile app



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0 < x < y, and x and y are consecutive integers. If the difference bet
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Updated on: 20 Mar 2018, 11:34
EMPOWERgmatRichC wrote: Hi vikasp99, This question can be solved in a couple of different ways. It's also built around some specific Number Properties, so if you're not sure how to approach the question, you can take advantage of those patterns and take a great guess. We're told that X and Y are CONSECUTIVE INTEGERS (0 < X < Y) and that Y^2  X^2 = 12,201. We're asked for the value of X. To start, it's interesting that the DIFFERENCE ends in a 1. When subtracting the squares of consecutive integers, there are only a couple of ways that this can occur: Y ends in a 1 and X ends in a 0 Y ends in a 6 and X ends in a 5 From the answer choices, we can clearly see that it's the first option  and that the correct answer is either Answer A or Answer C. If you're not sure what to do next, then you could guess and move on. However, if you want to continue working, you don't actually have to square two big numbers to find the solution to this question. Since the two variables are consecutive, we can rewrite Y as (X + 1), so the original equation can be written as... (X+1)^2  X^2 = 12,201 X^2 + 2X + 1  X^2 = 12,201 2X + 1 = 12,201 2X = 12,200 X = 6,100 Final Answer: GMAT assassins aren't born, they're made, Rich Hi, EMPOWERgmatRichC When I first did the problem, I didn't actually come up with an answer, but did end up with a 61 while trying to figure out an approach. So I figured out that it had to be y^2  x^2 = 12,201. So, I simplified 12,201 to 12,200 to make it more manageable (without doing the algebra). I then divided 12,200 by 2 which gave me 6,100 (just because I needed to try something). I thought this was a trick or some flaw in my thinking so I actually eliminated that choice. Was I just lucky in getting the 6,100 or is there an avenue to the solution somewhere in that train of thought without realizing the substitution method of y=x+1?
Originally posted by rnz on 20 Mar 2018, 11:30.
Last edited by rnz on 20 Mar 2018, 11:34, edited 1 time in total.



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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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20 Mar 2018, 11:33
arkbatman wrote: Answer: (A) Since x and y are consecutive integers you can take y=x+1. So y^2x^2=2x+1=12201 2x=12200 x=6100. Sent from my Nexus 5 using GMAT Club Forum mobile appAnother way of factoring could be.. y^2  x^2 = ( y + x ) (y  x) Since y = x+1 This becomes 2x + 1 = 12201 2x = 12200 x = 6100. Best, Gladi Posted from my mobile device
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0 < x < y, and x and y are consecutive integers. If the difference bet
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26 Apr 2018, 04:18
or another way: (xy)*(x+y)= 12 201 (xy) = 12 201  which is impossible because xy= 1 (consecutive integers) and x<y (x+y) = 12 201  it is possible. Hence we must look at answer choices where C, D, E we can easily eliminate
We have A (6 100) and B (6 101), well it is not a big deal to understand that only A (6 100) suits to our problem.
A is the answer.



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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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27 Apr 2018, 09:55
vikasp99 wrote: 0 < x < y, and x and y are consecutive integers. If the difference between x^2 and y^2 is 12,201, then what is the value of x?
(A) 6,100
(B) 6,101
(C) 12,200
(D) 12,201
(E) 24,402 Since x and y are consecutive integers, we can let y = x + 1; thus: (x + 1)^2  x^2 = 12,201 x^2 + 2x + 1  x^2 = 12,201 2x + 1 = 12,201 2x = 12,200 x = 6,100 Answer: A
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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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28 Apr 2018, 08:57
Sum of 2 consecutive numbers is always equal to the difference of their Squares. This is always true.
So in this case sum should be equal to 12,201. out of the options, all the options are greater than 12,201 except for a,b and c. and since the numbers are consecutive, the numbers have to be 6,100 and 6,101
Since it is given that x<y, therefore, x=6,100
Ans  A



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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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02 Sep 2018, 14:30
EMPOWERgmatRichC wrote: Hi vikasp99, This question can be solved in a couple of different ways. It's also built around some specific Number Properties, so if you're not sure how to approach the question, you can take advantage of those patterns and take a great guess. We're told that X and Y are CONSECUTIVE INTEGERS (0 < X < Y) and that Y^2  X^2 = 12,201. We're asked for the value of X. To start, it's interesting that the DIFFERENCE ends in a 1. When subtracting the squares of consecutive integers, there are only a couple of ways that this can occur: Y ends in a 1 and X ends in a 0 Y ends in a 6 and X ends in a 5 From the answer choices, we can clearly see that it's the first option  and that the correct answer is either Answer A or Answer C. If you're not sure what to do next, then you could guess and move on. However, if you want to continue working, you don't actually have to square two big numbers to find the solution to this question. Since the two variables are consecutive, we can rewrite Y as (X + 1), so the original equation can be written as... (X+1)^2  X^2 = 12,201 X^2 + 2X + 1  X^2 = 12,201 2X + 1 = 12,201 2X = 12,200 X = 6,100 Final Answer: GMAT assassins aren't born, they're made, Rich Hi rich, hope u are well. I am very glad to have your explanationyour explanation is always fantastic! May I have another shortcut way, please? Thanks...
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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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02 Sep 2018, 14:38
ScottTargetTestPrep wrote: vikasp99 wrote: 0 < x < y, and x and y are consecutive integers. If the difference between x^2 and y^2 is 12,201, then what is the value of x?
(A) 6,100
(B) 6,101
(C) 12,200
(D) 12,201
(E) 24,402 Since x and y are consecutive integers, we can let y = x + 1; thus: (x + 1)^2  x^2 = 12,201 x^2 + 2x + 1  x^2 = 12,201 2x + 1 = 12,201 2x = 12,200 x = 6,100 Answer: A i'm a bit confused about the bold part above! if x and y are consecutive integers, we can also let x = y + 1, right? Thanks__
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Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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02 Sep 2018, 21:19
AsadAbu wrote: ScottTargetTestPrep wrote: vikasp99 wrote: 0 < x < y, and x and y are consecutive integers. If the difference between x^2 and y^2 is 12,201, then what is the value of x?
(A) 6,100
(B) 6,101
(C) 12,200
(D) 12,201
(E) 24,402 Since x and y are consecutive integers, we can let y = x + 1; thus: (x + 1)^2  x^2 = 12,201 x^2 + 2x + 1  x^2 = 12,201 2x + 1 = 12,201 2x = 12,200 x = 6,100 Answer: A i'm a bit confused about the bold part above! if x and y are consecutive integers, we can also let x = y + 1, right? Thanks__ Notice that we are also told that 0 < x < y, so y = x + 1.




Re: 0 < x < y, and x and y are consecutive integers. If the difference bet
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