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Question Stats:
52% (01:58) correct
47% (01:30) wrong based on 73 sessions
If x and y are consecutive positive integers, and: \left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}which of the following represents all the possible values of y ? (A) y \ge 0(B) y \gt 0(C) y \gt 1(D) y \gt 7(E) y \gt 8Source: GMAT Club Tests - hardest GMAT questions
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Although the inequality looks complex, if we replace x with (y+1) in the inequality then we get y > 0.
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we know x>0 and y>0 and x,y are consecutive we know x>y ..... this means x=y+1 ( x,y consecutive) x² -1 > y²-4y+x-1 (y+1)² -1 > y²-4y+ (y+1) -1 -------- [ here i have substituted x=y+1] y²+2y+1 -1 > y²-4y +y+1-1 y²+2y > y²-4y +y y²+2y > y²-3y 2y > -3y 5y >0 y>0 SO B Hope this helps
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x^2-1=y^2-4y+x-1 => x^2-x=y^2-4y => x^2-x+\frac{1}{4}-\frac{1}{4}=y^2-4y+4-4 => (x-\frac{1}{2})^2-\frac{1}{4}=(y-2)^2-4 => (x-\frac{1}{2})^2>(y-2)^2-\frac{15}{4}
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.
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bigfernhead wrote: If x and y are consecutive positive integers, and: \left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}which of the following represents all the possible values of y ? (A) y \ge 0(B) y \gt 0(C) y \gt 1(D) y \gt 7(E) y \gt 8Source: GMAT Club Tests - hardest GMAT questions I picked numbers to solve this equation. 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.
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sonnco wrote: I picked numbers to solve this equation.
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.
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Has to be B. Here is the most simple reason why.
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.
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As x > y and x,y are consecutive x = y+1 ans - y > 0
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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?
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tinki wrote: tinki wrote: 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.
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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.
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if we replace x by Y+1 in inequality equation then we get 5y>0, hence Y>0.
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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
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Substituting x=y+1 into the second equations renders 5y>0, which gives y>0 Therefore B.
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bigfernhead wrote: If x and y are consecutive positive integers, and: \left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}which of the following represents all the possible values of y ? (A) y \ge 0(B) y \gt 0(C) y \gt 1(D) y \gt 7(E) y \gt 8Source: GMAT Club Tests - hardest GMAT questions 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. Answer: B.
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