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Re D0143
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15 Sep 2014, 23:13
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 \((xy) \gt 0\) 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 \((xy) = 23 = 1\). Hence statement 2 is sufficient. Answer: B
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Re: D0143
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14 Nov 2014, 08:56
How did we get to y(y+10)3(y+10) from the step just before this.
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Re: D0143
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14 Nov 2014, 09:01



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Re: D0143
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14 Nov 2014, 10:20
Got it! thanks!



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Re: D0143
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19 Jan 2015, 05:23
if x and y are consecutive then x=y+1 and y=x+1 both can be true either ways both the numbers are consecutive my solution was like this given 5x=7+y xy=74x therefore, xy>0 when 74x>0 so when x is negative xy is postive



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How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.
Therefore I thought about it that way: 5x = y + 7 /*5 /y So xy = (4y+7)/5
Question: (4y+7)/5 > 0 ? Rephrased: y < 7/4 ?
1) Not sufficient: many ways y < or > 7/4
2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. > Sufficient
Is that right?
I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.



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Re: D0143
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19 Feb 2016, 11:26
The way I thought of it:
I rearranged the question to say 0= y+75x
Looking at the statements I skipped straight to B. Since they have to be consecutive integers Y=3 and X = 2 works, . Sufficient. With statement A) x could be 3 and y could be 2 or vice versa so not sufficient.



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Re: D0143
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17 Mar 2016, 12:00
If 5x=y+7, is (x−y)>0?
(1) xy=6 x(5x7)=6 => 5x^27x6=0 => x=(7+/ Sqrt (494*5*(6)))/10 => (7+/13)/10 => 2 or 3/5 x=2,y=3, xy=1<0 x=3/5, y=10, xy= 47/5=9.4 >0 Insufficient
(2) x and y are consecutive integers with the same sign
Let x=y+1 5x=y+7 => 4y=2 => y=0.5 not integer
Let y=x+1 5x=y+7 => 4x=8 =>x=2 , y=3 xy= 1<0 SUFFICIENT
Hence B
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14 Jul 2016, 12:52
I think this is a highquality question and I agree with explanation.



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rephrase as x y =7 4x ( eq.1) thus .. we can say... is 7  4x > 0.. i.e. x < 7/4.. x < 1.75? 1. x = 2 NO... X = 1.5 YES 2 * 3 = 6 1.4 * 4 =6 INSUFFICIENT 2. x = y + 1 substitu in eq. 1 x  x + 1 = 7  4x => x= 3/2= 1.5 so y = 0.5 ... NOT AN INTEGER... reject this case y = x+1 substiute in eq.1 x  x 1 = 7  4x => x = 2 ... so y = 3.. BOTH ARE CONSECUTIVE INTGERS.... Definite NO .. for x < 1.75? (2) is sufficient so B



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Re: D0143
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09 Sep 2016, 04:28
I like this question. This is, for sure, a great question. Thanks team at GC!



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Re: D0143
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01 Nov 2016, 20:57
from statement 1 we have xy=6 then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2 .if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?



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Re: D0143
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02 Nov 2016, 01:19
phanikrishna wrote: from statement 1 we have xy=6 then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2 .if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent? Please read the solution carefully. There are TWO sets of (x,y) given which satisfy the equations: x=2 and y=3 AND x=5/3 and y=10. Also, the red part above is wrong. xy=6 has infinitely many solutions for x and y. You considered only positive integers there but x and y can be negative and fractions too.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Joined: 17 Aug 2016
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madmax1000 wrote: How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.
Therefore I thought about it that way: 5x = y + 7 /*5 /y So xy = (4y+7)/5
Question: (4y+7)/5 > 0 ? Rephrased: y < 7/4 ?
1) Not sufficient: many ways y < or > 7/4
2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. > Sufficient
Is that right?
I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long. Bunuel, could you please suggest me if the highlight is correct? That "many ways" is without any proof, and as a matter of fact there are only two solutions to the system of equations XY=6 and 5X=Y7, which in the specific case are one with Y>7/4 and one with Y<7/4. But with the above approach how could I say that the first stm is not suff? Thank you in advance



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Re: D0143
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24 Jan 2017, 01:27
Is there a shortcut for this question ?



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Re: D0143
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24 Jan 2017, 01:35



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Re: D0143
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03 Jun 2017, 19:53
Statement 2 gives , both integers negative, both are positive, 0/+ and /0. Then we have so many options to pick, how will it be answer B then?



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04 Jun 2017, 05:35



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Re: D0143
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28 Sep 2018, 20:34
why does it half to be an integer? isnt this an assumption







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