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Re: The Discreet Charm of the DS
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16 May 2012, 01:39
piyushksharma wrote: Bunuel wrote: 9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3 > \(3x<2y1\). \(x\) can be some very small number for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x<y\) BUT if \(x=2\) and \(y=2.1\) then the answer would be NO, \(x>y\). Not sufficient.
(2) 2x  3 < 3y  4 > \(x<1.5y\frac{1}{2}\) > \(x<y+(0.5y\frac{1}{2})=y+negative\) > \(x<y\) (as y+negative is "more negative" than y). Sufficient.
Answer: B. Hi bunuel, Did not got how u solved option 2.Could you please explain in detail. thanks. (2) 2x  3 < 3y  4 > \(x<1.5y\frac{1}{2}\) > \(x<y+(0.5y\frac{1}{2})\). Now, since \(y\) is a negative number then \(0.5y\frac{1}{2}=negative\) so, we have that: \(x<y+negative\). \(y+negative\) is less then \(y\) and if \(x\) is less than \(y+negative\) then it must also be less than \(y\) itself: \(x<y\). Hope it's clear.
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Re: The Discreet Charm of the DS
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25 May 2012, 11:58
Bunuel wrote: 3. If a, b and c are integers, is abc an even integer?
In order the product of the integers to be even at leas on of them must be even
(1) b is halfway between a and c > on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as \(b=\frac{a+c}{2}\). Now, does that mean that at leas on of them is be even? Not necessarily: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) and \(b=7\). Not sufficient.
(2) a = b  c > \(a+c=b\). Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.
Answer: B. What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.



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Re: The Discreet Charm of the DS
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25 May 2012, 12:29
avinash2603 wrote: Bunuel wrote: 3. If a, b and c are integers, is abc an even integer?
In order the product of the integers to be even at leas on of them must be even
(1) b is halfway between a and c > on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as \(b=\frac{a+c}{2}\). Now, does that mean that at leas on of them is be even? Not necessarily: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) and \(b=7\). Not sufficient.
(2) a = b  c > \(a+c=b\). Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.
Answer: B. What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'. Welcome to GMAT Club. Below is an answer to your question. Notice that zero is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. Since 0/2=0=integer then zero is even. For more on this subject please check Number Theory chapter of Math Book: mathnumbertheory88376.htmlHope it helps.
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Re: The Discreet Charm of the DS
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04 Jun 2012, 21:39
Bunuel wrote: 5. What is the value of integer x?
(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient.
\((x\frac{3}{2})(x3)<0\) I agree one solution of this inequality is \((x\frac{3}{2})\) > 0 , (x3)<0 => \((\frac{3}{2})\) < x < 3 However, Don't u think this can also resort to \((x\frac{3}{2})\) < 0 , (x3)>0 => x < \((\frac{3}{2})\) , x > 3 and in that case x can have infinite values. and if that is the case Stmt1 alone would not be sufficient.
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Re: The Discreet Charm of the DS
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05 Jun 2012, 00:51
anordinaryguy wrote: Bunuel wrote: 5. What is the value of integer x?
(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient.
\((x\frac{3}{2})(x3)<0\) I agree one solution of this inequality is \((x\frac{3}{2})\) > 0 , (x3)<0 => \((\frac{3}{2})\) < x < 3 However, Don't u think this can also resort to
\((x\frac{3}{2})\) < 0 , (x3)>0 => x < \((\frac{3}{2})\) , x > 3 and in that case x can have infinite values. and if that is the case Stmt1 alone would not be sufficient. The second case of \((x\frac{3}{2})<0\) and \((x3)>0\) is not possible. This condition leads to \(x<\frac{3}{2}\) and \(x>3\), but \(x\) can not be simultaneously less than \(\frac{3}{2}\) (to make \(x\frac{3}{2}\) negative) and more than 3 (to make \(x3\) positive). For more on how to solve such kind of inequalities check: x24x94661.html#p731476 (Check this first) inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps.
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Re: The Discreet Charm of the DS
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27 Jun 2012, 08:03
Hi Bunuel, Can we solve this question using graphical approach. Since (xa)^2+(yb)^2 = r^2 (A) Equation 1 represents equation of circle with radius 1. Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not. Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this. Thanks Bunuel wrote: Please read the thread. Solutions to ALL the questions are given on the previous pages.
2. Is xy<=1/2?
(1) x^2+y^2=1. Recall that \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) > \(x^22xy+y^2\geq{0}\) > since \(x^2+y^2=1\) then: \(12xy\geq{0}\) > \(xy\leq{\frac{1}{2}}\). Sufficient.
(2) x^2y^2=0 > \(x=y\). Clearly insufficient.
Answer: A.



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Re: The Discreet Charm of the DS
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27 Jun 2012, 08:17
imhimanshu wrote: Hi Bunuel, Can we solve this question using graphical approach. Since (xa)^2+(yb)^2 = r^2 (A) Equation 1 represents equation of circle with radius 1. Now, if we can maximize the value of both x and y, we can get the maximum value of xy and that could let us know whether statement (1) is sufficient or not. Here, I require your help. Intuitively, I find that x=y will maximize the equation x^2+y^2=1. But I am not sure about my reasoning as well as the Mathematics behind. Could you please look into this. Thanks Bunuel wrote: Please read the thread. Solutions to ALL the questions are given on the previous pages.
2. Is xy<=1/2?
(1) x^2+y^2=1. Recall that \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) > \(x^22xy+y^2\geq{0}\) > since \(x^2+y^2=1\) then: \(12xy\geq{0}\) > \(xy\leq{\frac{1}{2}}\). Sufficient.
(2) x^2y^2=0 > \(x=y\). Clearly insufficient.
Answer: A.
Useful property: For given sum of two numbers, their product is maximized when they are equal.(1) says that \(x^2+y^2=1\). So, \(x^2y^2\) will be maximized when \(x^2=y^2\): \(x^2+x^2=1\) > \(x^2=\frac{1}{2}\) > the maximum value of \(x^2y^2\) thus is \(\frac{1}{4}\) and the maixmum value of \(xy\) is \(\frac{1}{2}\). Hope it's clear.
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Re: The Discreet Charm of the DS
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28 Jul 2012, 07:36
Bunuel wrote: 2. Is xy<=1/2?
(1) x^2+y^2=1. Recall that \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) > \(x^22xy+y^2\geq{0}\) > since \(x^2+y^2=1\) then: \(12xy\geq{0}\) > \(xy\leq{\frac{1}{2}}\). Sufficient.
(2) x^2y^2=0 > \(x=y\). Clearly insufficient.
Answer: A. Why we recalled \((xy)^2\geq{0}\) ?
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Re: The Discreet Charm of the DS
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06 Dec 2012, 23:19
Bunuel wrote: 1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?
(1) x^2+y^2<12
(2) Bonnie and Clyde complete the painting of the car at 10:30am Responding to a pm: Time taken by Bonnie to complete one work = x hrs Time taken by Clyde to complete one work = y hrs x and y are odd integers i.e. they could take values such as 1/3/5/7/9/11... Question: Is x = y? i.e. is the time taken by Bonnie equal to time taken by Clyde? i.e. is the speed of Bonnie equal to the speed of Clyde? (1) x^2+y^2<12 This info is not related to work concepts. It's just number properties. x and y are odd integers. If x = y = 1, this inequality is satisfied. If x = 1 and y = 3, this inequality is satisfied. This means x may or may not be equal to y. Not sufficient. (2) Bonnie and Clyde complete the painting of the car at 10:30 am.r Together, they take 45 mins to complete the painting of the car. This means, if their rate of work were the same, each one of them would have taken 1.5 hrs working alone. But their time taken is an integer value. We can say that they do not take the same time i.e. x is not equal to y. Hence this statement is sufficient to say \(x \neq y\) Answer (B)
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Re: The Discreet Charm of the DS
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14 Feb 2013, 00:31
Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. Bunuel, Here's how I did . I don't know why it is wrong. Please advise : 6a=3b=7c => 12a=6b=14c From 1) ac=6b => ac=12a => c=12.. a can be cancelled because it is not inequality. . so c =12.. however, ac12a=0 > a(c12)=0 so a=0 OR c=12 .. Is this the reason why my soln is wrong?
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Re: The Discreet Charm of the DS
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14 Feb 2013, 01:24
Bunuel wrote: SOLUTIONS:
1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?
Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) > \(T=\frac{xy}{x+y}\)). Now, if \(x=y\) then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....
(1) x^2+y^2<12 > it's possible \(x\) and \(y\) to be odd and equal to each other if \(x=y=1\) but it's also possible that \(x=1\) and \(y=3\) (or viseversa). Not sufficient.
(2) Bonnie and Clyde complete the painting of the car at 10:30am > they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then \(x\) and \(y\) are not equal. Sufficient.
Answer: B. Bunuel, I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ?



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Re: The Discreet Charm of the DS
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14 Feb 2013, 02:08
Sachin9 wrote: Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. Bunuel, Here's how I did . I don't know why it is wrong. Please advise : 6a=3b=7c => 12a=6b=14c From 1) ac=6b => ac=12a => c=12.. a can be cancelled because it is not inequality. .so c =12.. however, ac12a=0 > a(c12)=0 so a=0 OR c=12 .. Is this the reason why my soln is wrong? Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.So, if you divide (reduce) ac=12a by a, you assume, with no ground for it, that a does not equal to zero thus exclude a possible solution (notice that both a=0 AND c12=0 satisfy the equation). Hope it's clear.
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Re: The Discreet Charm of the DS
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14 Feb 2013, 02:16
thinktank wrote: Bunuel wrote: SOLUTIONS:
1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?
Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) > \(T=\frac{xy}{x+y}\)). Now, if \(x=y\) then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....
(1) x^2+y^2<12 > it's possible \(x\) and \(y\) to be odd and equal to each other if \(x=y=1\) but it's also possible that \(x=1\) and \(y=3\) (or viseversa). Not sufficient.
(2) Bonnie and Clyde complete the painting of the car at 10:30am > they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then \(x\) and \(y\) are not equal. Sufficient.
Answer: B. Bunuel, I did not understand the reasoning behind statement 2 being sufficient. Line marked in red is my doubt. what does it mean ? From the stem we got that if \(x=y\) then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd then this time would be odd/2: 0.5 hours (1/2 hours = 0.5 hours), 1.5 hours (3/2 hours = 1.5 hours), 2.5 hours (5/2 hours = 2.5 hours), 3.5 hours (7/2 hours = 3.5 hours), 4.5 hours (9/2 hours = 4.5 hours), .... Now, from the second statement we got that they complete the job in 0.75 hours, since the total time (0.75 hours) is NOT odd/2 (0.5 hours, 1.5 hours, 2.5 hours, 3.5 hours, 4.5 hours, ....), then \(x\) and \(y\) are not equal. Hope it's clear.
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Re: The Discreet Charm of the DS
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22 Jul 2013, 23:53
Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. Hi Bunuel, In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c. Now substituting 12a=ac=14c I get 12a=ac > c = 12 and similarly a = 14 and hence b= 28. Can you please point where am I going wrong? Regards



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The Discreet Charm of the DS
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22 Jul 2013, 23:56
Spaniard wrote: Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. Hi Bunuel, In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c. Now substituting 12a=ac=14c I get 12a=ac > c = 12 and similarly a = 14 and hence b= 28. Can you please point where am I going wrong? Regards ac=12a (here you cannot reduce by a and write c=12 as you exclude possibility of a=0) > a(c12)=0 > either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12. Does this make sense?
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Re: The Discreet Charm of the DS
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11 Aug 2013, 00:27
Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. I followed a different approach but I am getting that (1) also answers the question. From 3b=7c => 6b=14c. So ac=6b  equation 1) 14c=6b  equation 2) Dividing 1) by 2) =>a/14= 1 So a=14!!!. Where am I committing a mistake in this logic? Bunuel, Please help!!



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Re: The Discreet Charm of the DS
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11 Aug 2013, 00:33
pjagadish27 wrote: Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. I followed a different approach but I am getting that (1) also answers the question. From 3b=7c => 6b=14c. So ac=6b  equation 1) 14c=6b  equation 2) Dividing 1) by 2) =>a/14= 1 So a=14!!!. Where am I committing a mistake in this logic? Bunuel, Please help!! You cannot divide ac by 14c because c could be zero and division by zero is not allowed. The same applies to division of 6b by 6b. Does this make sense?
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Re: The Discreet Charm of the DS
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08 Sep 2013, 08:55
Bunuel wrote: 9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3 > \(3x<2y1\). \(x\) can be some very small number for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x<y\) BUT if \(x=2\) and \(y=2.1\) then the answer would be NO, \(x>y\). Not sufficient.
(2) 2x  3 < 3y  4 > \(x<1.5y\frac{1}{2}\) > \(x<y+(0.5y\frac{1}{2})=y+negative\) > \(x<y\) (as y+negative is "more negative" than y). Sufficient.
Answer: B. How did you decide which approach to use in A, one could have solved like the way you did in B 3X < 2y  1 or x < .67y  0.33 , so x < y + negative , so x is more negative, so x < Y. Is this is a 700 question ?



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Re: The Discreet Charm of the DS
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08 Sep 2013, 11:02
ygdrasil24 wrote: Bunuel wrote: 9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3 > \(3x<2y1\). \(x\) can be some very small number for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x<y\) BUT if \(x=2\) and \(y=2.1\) then the answer would be NO, \(x>y\). Not sufficient.
(2) 2x  3 < 3y  4 > \(x<1.5y\frac{1}{2}\) > \(x<y+(0.5y\frac{1}{2})=y+negative\) > \(x<y\) (as y+negative is "more negative" than y). Sufficient.
Answer: B. How did you decide which approach to use in A, one could have solved like the way you did in B 3X < 2y  1 or x < .67y  0.33 , so x < y + negative , so x is more negative, so x < Y. Is this is a 700 question ? The approach used for the second statement does not work for the first: \(3x + 4 < 2y + 3\) > \(x<\frac{2y}{3}\frac{1}{3}\) > \(x<y\frac{y}{3}\frac{1}{3}\) > \(x<y+(\frac{y}{3}\frac{1}{3})\) > we don't know whether \(\frac{y}{3}\frac{1}{3}=positive+negative\) is negative. Hope it's clear.
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Re: The Discreet Charm of the DS &nbs
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