[phpBB Debug] PHP Notice: in file /includes/viewtopic_mods/timer.php on line 168: array_key_exists() expects parameter 2 to be array, null given [phpBB Debug] PHP Notice: in file /includes/viewtopic_mods/timer.php on line 169: array_key_exists() expects parameter 2 to be array, null given [phpBB Debug] PHP Notice: in file /includes/viewtopic_mods/timer.php on line 170: array_key_exists() expects parameter 2 to be array, null given
m02#8 : Retired Discussions [Locked]

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

since y is the smaller positive integer than x, the above is always true. so y can be any number larger than zero (option E is a subset of option D; option D is a subset of option C; option C is a subset of option B).

an alternative way to solve this is as follows:

since y is smaller than x, but larger than zero, option A can be deleted.

then since the question asks for all possible numbers of y, it could be a smaller one rather than a larger value. we pick from option C as a start. if C fails, we move to B. if C works ok, we move to D.

The first thing I did was change the 2nd equation to look like this so that all of the variables are isolated on each side.

(x^2) - x - 1 > (y^2) - 4y - 1

Then I picked the numbers in the answer boxes to represent Y. Since it is asking for all of the possible solutions I started off with the smallest value for Y.

If Y = 0; then x has to equal 1 to satisfy the first equation x>y and is consecutive. This did not work because after plugging the numbers into the 2nd equation I got -1>0 which is not true. Cross out A.

Then I tried Y = 1; x = 2. This worked so all of the other answers will work too. I did try Y = 7; x = 8 just to be 100% sure though. _________________

I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

The first thing I did was change the 2nd equation to look like this so that all of the variables are isolated on each side.

(x^2) - x - 1 > (y^2) - 4y - 1

Then I picked the numbers in the answer boxes to represent Y. Since it is asking for all of the possible solutions I started off with the smallest value for Y.

If Y = 0; then x has to equal 1 to satisfy the first equation x>y and is consecutive. This did not work because after plugging the numbers into the 2nd equation I got -1>0 which is not true. Cross out A.

Then I tried Y = 1; x = 2. This worked so all of the other answers will work too. I did try Y = 7; x = 8 just to be 100% sure though.

it's a quicker way, but i think you don't even have to try y=0 because y is said to be a positive interger, which means that y is at least +1.

X and Y are consecutive integers. Therefore if Y = 0 then X has to equal 1. So if Y = 0 how could the second statement be true? O is not greater than 1. Therefore Y must be > 0. B.

Can somebody explain why im not getting results? below is my solution :

\left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}

initially i am adding 2 inequalities

x2 -1+x-x+1>y2-4y+y x2>y2-3y x=y+1 y2+2y+1>y2-3y

is it wrong to add inequalities above?

Anybody? why i cent get final answer with above? thx for feedback

Interesting. I am not too sure about my explanation, but just to add my thinking.

x is a greater value than y; i.e. 1 more; if you add a greater value to the LHS and a smaller value to RHS, the inequality will still be correct but the range of the inequality changes.

e.g. x>10 5>2 x+5>10+2 x>7

Thus, there is nothing wrong with the inequality but as we can see, the range changed. We started with; x>10 Now, x>7. Similar thing is happening in your example.

Well, the interesting thing to note is that; whether y>(-1/5) or y>0; the least value for y is going to be 1. So far; y>=1, we have the entire range. _________________

B. Simplify to x (X - 1) > y (y - 4) and plug in numbers for Y from answers with the restriction that X>Y and x and y are consecutive positive integers.

The OA states that "since x and y are consecutive positive integers (...)" but this is not stated by the question, it is actually our expected conclusion. In my view the OA should remove "positive" to avoid confusion

The OA states that "since x and y are consecutive positive integers (...)" but this is not stated by the question, it is actually our expected conclusion. In my view the OA should remove "positive" to avoid confusion

Actually it is: "If x and y are consecutive positive integers, and ..." _________________

BELOW IS REVISED VERSION OF THIS 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. y\geq{0} B. y>0 C. y>1 D. y>7 E. y>8

Since x and y are consecutive positive integers and x>y then x=y+1. Substitute x with y+1 in the given equation: (y+1)^2-1>y^2-4y+(y+1)-1 --> 5y>0 --> y>0.

i feel like they forgot to mention that they are positive integers. Wait actually that wouldn't matter because only dividing and multiplying with negative changes the inequality.