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If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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15 Jan 2010, 09:32
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If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is less than 2
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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15 Jan 2010, 13:47
Hussain15 wrote: If (x/y)>2, is 3x+2y<18?
(1) xy is less than 2 (2) yx is less than 2
It will be great to see how do you guys approach this lethal one. Given x > 2y. Have to substantiate if 3x + 2y < 18. Stmt1: x < 2 + y. Keep substituting different values for y, we get ranges for x based on the stimulus condition and this statement, substitute these different values and we notice that certain values are applicable while many others aren't applicable to substantiate the posed question. Therefore, NS. Stmt2: can be rephrased as x > y  2. Do the same method as above, same situation, no definitive answer. Therefore, NS. combining both the statements, still substituting all possible values for y and deriving ranges for x, we can't really substantiate the given equation. My answer is E. I wonder if there is a simpler way of solving problems of this kind. I used the brute force approach of substituting valid numbers for y and ended up getting wierder ranges for x and again, choose something which accidentally would substantiate the equation and mostly certain other numbers that do not. Took me more than a 10 mins handling work simultaneously, and if such questions appear on the real deal, I might as well give up on GMAT and pursue a PhD in Pure Math.
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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15 Jan 2010, 14:02
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Hussain15 wrote: If (x/y)>2, is 3x+2y<18?
(1) xy is less than 2 (2) yx is less than 2
It will be great to see how do you guys approach this lethal one. I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well. \(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A.
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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15 Jan 2010, 22:35
Bunuel wrote: Hussain15 wrote: If (x/y)>2, is 3x+2y<18?
(1) xy is less than 2 (2) yx is less than 2
It will be great to see how do you guys approach this lethal one. I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well. x/y>2 tells us that x and y are either both positive or both negative, which means that all points (x,y) satisfying given inequality are in I or III quadrants. When they are both negative (in III quadrant) inequality 3x+2y<18 is always true, so we should check only for I quadrant. In I quadrant x and y are both positive and we can rewrite x/y>2 as x>2y>0 (remember x>0 and y>0). (1) xy<2. Subtract inequalities x>2y and xy<2 (we can do this as signs are in opposite direction) > x(xy)>2y2 > y<2. Now add inequalities xy<2 and y<2 (we can do this as signs are in the same direction) > xy+y<2+2 > x<4. We got y<2 and x<4. If we take maximum values x=4 and y=2 and substitute in 3x+2y<18, we'll get 12+4=12<18. Sufficient. (2) yx<2 and x>2y: x=3 and y=1 > 3x+2y=11<18 true. x=11 and y=5 > 3x+2y=43<18 false. Not sufficient. Answer: A. OA is "A". Thanks for detailed answer. You have plugged the numbers in option 2, can it be done algeberically??
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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I spent 5 min for this question with incorrect ans .. There was no way I could have solved this question .. Very nice explanation Brunel ..
But I failed to understand the theory of addition and substraction for equalities with same sign and opposite signs respective .. Can you pls throw some light ..



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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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21 May 2010, 09:28
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sandeepuc wrote: I spent 5 min for this question with incorrect ans .. There was no way I could have solved this question .. Very nice explanation Brunel ..
But I failed to understand the theory of addition and substraction for equalities with same sign and opposite signs respective .. Can you pls throw some light .. You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it helps.
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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01 Jun 2010, 15:26
Bunuel wrote: Hussain15 wrote: If (x/y)>2, is 3x+2y<18?
(1) xy is less than 2 (2) yx is less than 2
It will be great to see how do you guys approach this lethal one. I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well. \(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive. In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)). So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)? (1) \(xy<2\). Subtract inequalities \(x>2y\) and \(xy<2\) (we can do this as signs are in opposite direction) > \(x(xy)>2y2\) > \(y<2\). Now add inequalities \(xy<2\) and \(y<2\) (we can do this as signs are in the same direction) > \(xy+y<2+2\) > \(x<4\). We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\). Sufficient. (2) \(yx<2\) and \(x>2y\): \(x=3\) and \(y=1\) > \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) > \(3x+2y=43<18\) false. Not sufficient. Answer: A. +1 already for a great explanation. Followup question: Would you mind detailing a graphical approach to this problem? I haven't taken a math course in 7 years so am a little rusty. Knowing how to solve such problems with a graph seems like it would be very useful.
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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01 Jun 2010, 15:47



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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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28 Jun 2010, 00:19
Given x/y > 2. i. xy < 2: for this to be possible x and y have to be negative. Now since x and y are both negative, the equation in question will always result in a negative number. Hence, SUFFICIENT.
ii. yx <2: For this to be possible x and y have to be positive. Now since x and y both are positive and xy>2, multiple solutions exist. Hence, NOT SUFFICIENT.
Therefore, answer is A.



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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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28 Jun 2010, 05:54
sunnyarora wrote: Given x/y > 2. i. xy < 2: for this to be possible x and y have to be negative. Now since x and y are both negative, the equation in question will always result in a negative number. Hence, SUFFICIENT.
ii. yx <2: For this to be possible x and y have to be positive. Now since x and y both are positive and xy>2, multiple solutions exist. Hence, NOT SUFFICIENT.
Therefore, answer is A. OA for this question is A, but your reasoning is not correct: For (1) \(x=2>0\) and \(y=0.5>0\) satisfy both \(\frac{x}{y}>2\) and \(xy<2\), so x and y can be positive as well. For (2) \(x=2<0\) and \(y=0.5>0\) satisfy both \(\frac{x}{y}>2\) and \(yx<2\), so x and y can be negative as well. Solution for this problem is in earlier posts.
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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19 Jun 2014, 10:19
Hi Bunuel,
Thank you for the great solution. With regards to using graphs to solving the problem, do we get a grid kind of pad to be able to plot accurately and with ease?



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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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20 Jun 2014, 05:48



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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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30 Jun 2015, 09:07
Mo2men wrote: Hi Bunuel,
In statement 1, you got y<2 and x<4 but when x=2 & y=1 or even your point x=4 & y=2 so x/y>2 is not satisfied because 2/1 or 4/2 is not bigger 2. How come still statement 1 sufficient?
Thanks
Hi, responding to your Pm... Since you have asked here only the above doubt... y<2 and x<4.... you cannot take them as y=2 and x=4 as it is given both are less than these quantities... so if y=1.9 .. statement 1 says xy<2 or x1.9<2 or x<3.9, so it satisfies x<4.. however if we take the values of x and y slightly more than the max possible(x<4).. x=4and (x<2)..y=2, we find value of eq <18.. so suff
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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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19 Mar 2017, 20:26
If (x/y)>2, is 3x+2y<18?
x/y>2 means that both x & y have same sign.
(1) xy < 2
Let x=3 & y=5/4....... check x/y=12/5=2.4>2
Check statement xy<2.....3 1.25=1.75 <2 ......So (3*3) + (2*5/4)<18..........(Note that you can choose positive numbers with narrow range such 2 & 5/6 and will achieve same answer Yes )
Let x=10 & y=1......check x/y=10/1=10>2
Check statement xy<2.....10+1=9 <2............So (10*3)+ (1*2)<18.............Yes (Note that any negative numbers that satisfy both x/y> and fact 1 will always answer question with Yes)
Sufficient
(2) yx < 2
Let x=3 & y=5/4....... check x/y=12/5=2.4>2
Check statement yx<2.....1.25  3=1.75 <2 ......So (3*3) + (2*5/4)<18.....Answer is Yes
Let x=10 & y=1.......check x/y=10>2
Check statement yx<2.......110=9<2................So (3*10) + (2*1)<18.....Answer is NO
Insufficient
Answer: A
This question needs high skills to spot the strategic numbers an spot pattern also.



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Re: If x/y > 2, is 3x + 2y < 18? (1) x  y is less than 2 (2) y  x is les [#permalink]
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07 Apr 2018, 11:29
Hi All, We're told X/Y > 2 which means that X and Y are either BOTH positive or BOTH negative. The question asks if (3X + 2Y) < 18. This is a YES/NO question. This DS question has some useful Number Properties in it and can be solved by TESTing VALUES. To start, there are some patterns worth noting: If both X and Y are NEGATIVE, then the answer to the question is NO (and there'd be no reason to even do the math). If both X and Y are POSITIVE, then the math IS required because the answer to the question COULD be YES or NO (depending on how big X and Y are). 1) (X  Y) is less than 2 This tells us that X and Y must be relatively "close" to one another, BUT we also know that X/Y > 2, which means that X is MORE THAN TWICE Y. These 2 Facts severely LIMIT the possibilities... X = 2, Y = 1/2....3(2) + 2(1/2) IS < 18 The answer is YES X = 3, Y = 1.1....3(3) + 2(1.1) IS < 18 The answer is YES X CAN'T = 4 (or larger) because there's no value for Y that "fits" both Facts If X and Y are negative, then we get another YES Fact1 is SUFFICIENT 2) (Y  X) is less than 2 Here, we can use any of our TESTs from Fact 1 X = 2, Y = 1/2 ....The answer is YES But we also need to consider any other possibilities... X = 100, Y = 1....3(100 + 2(1) is NOT < 18 and the answer is NO Fact 2 is INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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