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D01-43

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New post 29 Sep 2018, 00:08
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New post 14 Nov 2018, 03:01
Statement 1 : xy = 6
\(y = 6/x\)

Substitute above in main equation -
\(5x=y+7\)
\(\frac{30}{y} = y + 7\)
\(30 = y^2 + 7y\)
\(y^2 + 7y - 30 = 0\)
Solving we get \(y = 3\) and \(y = -10\)
You can stop here because we are getting two different values of y so obviously we have different values of x.
Hence statement 1 is insufficient.
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New post 04 Dec 2018, 02:21
If we solve the question stem by replacing the value of y initially, it gives us x - y >0 if x < 7/4. That approach doesn't work for statement2. any thoughts ?
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New post 11 Feb 2019, 16:05
Bunuel wrote:
Official Solution:


Statement 1: \(xy = 6\) is not sufficient to answer the question because solving the equation gives \(y = 3\) when \(x\) is 2 and \(y\) is -10 when \(x\) is \(-\frac{3}{5}\).

\(5x = y + 7\)

\(x = \frac{y + 7}{5}\) ................i

\(xy = 6\) ........................ ii

Replacing the value of \(x\) on eq ii: \(\frac{y(y + 7)}{5} = 6\)

\(y^2 + 7y = 30\)

\(y^2 + 10y - 3y - 30= 0\)

\(y (y + 10) - 3 (y + 10) = 0\)

\((y + 10) (y - 3) = 0\)

\(y = - 10\) or \(3\)

If \(y = - 10\), \(x = -\frac{3}{5}\). In this case, \(x - y = -\frac{3}{5} - (-10) = \frac{47}{5}\). Yes.

If \(y = 3\), \(x = 2\). In this case, \(x - y = 2 - 3 = -1\). No.

In each case \((x-y) \gt 0\) and \(\lt 0\). Hence statement 1 is not sufficient.

Statement 2: If \(x\) and \(y\) are consecutive integers, \(x = 2\) and \(y = 3\).

When \(x\) and \(y\) are consecutive integers, either \(x = y+1\) or \(y = x+1\) is possible.

(i) If \(x = y+1\)

\(5x = y + 7\)

\(5(y+1) = y + 7\)

\(4y = 2\)

\(y = \frac{1}{2}\). Then \(x = \frac{3}{2}\). However this is not possible because \(x\) and \(y\) are even not integers. So this option is not valid.

(ii) If \(y = x+1\)

\(5x = y + 7\)

\(5x = x + 1 + 7\)

\(4x = 8\)

\(x = 2\)

Then, \(y = 3\). This is valid because \(x\) and \(y\) are consecutive integers.

So \((x-y) = 2-3 = -1\). Hence statement 2 is sufficient.


Answer: B



Hi, Thank you for the detailed explanation.

The question is asking : is (x-y)>0?

Let us go straight to the last line of your explanation, when y
=x+1, x-y= -1, which is <0. Therefore (2) alone is not sufficient

Please advise.

Thanks
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New post 11 Feb 2019, 22:09
Cottonwood wrote:
Bunuel wrote:
Official Solution:


Statement 1: \(xy = 6\) is not sufficient to answer the question because solving the equation gives \(y = 3\) when \(x\) is 2 and \(y\) is -10 when \(x\) is \(-\frac{3}{5}\).

\(5x = y + 7\)

\(x = \frac{y + 7}{5}\) ................i

\(xy = 6\) ........................ ii

Replacing the value of \(x\) on eq ii: \(\frac{y(y + 7)}{5} = 6\)

\(y^2 + 7y = 30\)

\(y^2 + 10y - 3y - 30= 0\)

\(y (y + 10) - 3 (y + 10) = 0\)

\((y + 10) (y - 3) = 0\)

\(y = - 10\) or \(3\)

If \(y = - 10\), \(x = -\frac{3}{5}\). In this case, \(x - y = -\frac{3}{5} - (-10) = \frac{47}{5}\). Yes.

If \(y = 3\), \(x = 2\). In this case, \(x - y = 2 - 3 = -1\). No.

In each case \((x-y) \gt 0\) and \(\lt 0\). Hence statement 1 is not sufficient.

Statement 2: If \(x\) and \(y\) are consecutive integers, \(x = 2\) and \(y = 3\).

When \(x\) and \(y\) are consecutive integers, either \(x = y+1\) or \(y = x+1\) is possible.

(i) If \(x = y+1\)

\(5x = y + 7\)

\(5(y+1) = y + 7\)

\(4y = 2\)

\(y = \frac{1}{2}\). Then \(x = \frac{3}{2}\). However this is not possible because \(x\) and \(y\) are even not integers. So this option is not valid.

(ii) If \(y = x+1\)

\(5x = y + 7\)

\(5x = x + 1 + 7\)

\(4x = 8\)

\(x = 2\)

Then, \(y = 3\). This is valid because \(x\) and \(y\) are consecutive integers.

So \((x-y) = 2-3 = -1\). Hence statement 2 is sufficient.


Answer: B



Hi, Thank you for the detailed explanation.

The question is asking : is (x-y)>0?

Let us go straight to the last line of your explanation, when y
=x+1, x-y= -1, which is <0. Therefore (2) alone is not sufficient

Please advise.

Thanks


(2) gives a definite NO to the question, which means that the statement is sufficient.

There are two types of data sufficiency questions:

1. YES/NO DS Questions:

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no”while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".


2. VALUE DS QUESTIONS:

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.


Strategies and Tactics for DS Section




For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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New post 13 Feb 2019, 09:01
abastie wrote:
why does it half to be an integer? isnt this an assumption


We don't know they are integers from the base equation. In solving Statement 1, there are two solutions including \(x=-3/5\) and y = \(-10\). So a non-integer solution is possible here and is part of why S1 is not sufficient. If we knew \(xy=6\) AND that x and y were both integers, then S1 would have been sufficient.

Statement 2 tells us that x and y are consecutive integers. If we know for a fact that they are both integers and consecutive, there is only one possible solution. Therefore, S2 is sufficient information to answer the question.

Does that make sense?
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New post 13 Feb 2019, 14:39
The way I solved:
Changed original equation to 5x-y=7

(1) xy=6
x=3, y=2 (or vice versa)
x=1, y=6 (or vice versa)
plugging into original equation:
5(3)-2=13 NO
5(1)-6=-1 NO
5(6)-1=29 NO
5(2)-3=7 YES
NS bc we have yes and no

(2) We know from one that 2 and 3 are the only solutions, and they are both positive consecutive integers. Thus, 2 is sufficient.
Bunuel is this correct logic?
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New post 13 Feb 2019, 15:30
therman wrote:
The way I solved:
Changed original equation to 5x-y=7

(1) xy=6
x=3, y=2 (or vice versa)
x=1, y=6 (or vice versa)
plugging into original equation:
5(3)-2=13 NO
5(1)-6=-1 NO
5(6)-1=29 NO
5(2)-3=7 YES
NS bc we have yes and no

(2) We know from one that 2 and 3 are the only solutions, and they are both positive consecutive integers. Thus, 2 is sufficient.
Bunuel is this correct logic?


You have misread the question. You are correct that this is a Yes/No DS question, but the question to answer is whether \((x-y) > 0\)

In solving statement 1, you have assumed that x and y are both integers. However, there are many more solutions to \(xy = 6\). For statement 1, the exercise is to solve two equations in two variables. We know that both \(xy = 6\) AND \(5x=y+7\). Combining, you will need to solve a quadratic equation and find two solutions. One solution, x = 2 and y = 3 gives a NO answer to the questions \((x-y) > 0\). The other answer, x = -3/5 and y = -10 gives a yes answer. For this reason, statement 1 is not sufficient.
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New post 14 Feb 2019, 02:59
Bunuel would you please explain why for Statement 2 we have not considered x=-1 and y=-2 (2 consecutive negative integers)? If they are 2 valid values then (x-y)=1>0 and statement 2 alone would not be sufficient. Therefore I have answered C because we need Statement 1 to say x=2 and y=3.
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New post 14 Feb 2019, 03:20
frabzz wrote:
Bunuel would you please explain why for Statement 2 we have not considered x=-1 and y=-2 (2 consecutive negative integers)? If they are 2 valid values then (x-y)=1>0 and statement 2 alone would not be sufficient. Therefore I have answered C because we need Statement 1 to say x=2 and y=3.


x = -1 and y = -2 does not satisfy 5x = y + 7 (given in the stem).

5x = 5*(-1) = -5 while y + 7 = -2 + 7 = 5.
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New post 27 Mar 2019, 09:24
why cant a solution be -2,-3 since this satisfies the second statement and would make x-y=1
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New post 27 Mar 2019, 09:42
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Re: D01-43   [#permalink] 27 Mar 2019, 09:42

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