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Math Revolution GMAT Instructor
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27 Oct 2019, 19:51
MathRevolution wrote: [GMAT math practice question]
(equation) What are the values of \(x+y\) and \(xy\)?
\(1) x + y + xy = 2\)
\(2) (\frac{1}{x}) + (\frac{1}{y}) = 1\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since \(\frac{1}{x} + \frac{1}{y} = 1\) from condition 2), we have \(y + x = xy\) by multiplying both sides of the equation by \(xy\), which rearranges to get \(xy – (x+y) = 0.\) Since \(xy + (x+y) = 2\) from condition 1), we have \(xy  (x+y) + xy + (x+y) = 0 + 2\) by adding the two equations. Then \(2xy = 2\) or \(xy = 1\). Then we have \(x+y=1.\) Since both conditions together yield a unique solution, they are sufficient. Therefore, C is the answer. Answer: C Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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28 Oct 2019, 01:34
[GMAT math practice question] (algebra) What is the value of \(\frac{x}{(x+y)} + \frac{y}{(xy)}\)? \(1) (x+y):y = 3:1\) \(2) x + y = 8\)
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29 Oct 2019, 01:52
[GMAT math practice question] (number properties) \(x\) and \(y\) are positive integers. What is the value of \(x*y\)? \(1) x^{x+y} = y^4\) \(2) y^{x+y} = x^4\)
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30 Oct 2019, 00:40
MathRevolution wrote: [GMAT math practice question]
(algebra) What is the value of \(\frac{x}{(x+y)} + \frac{y}{(xy)}\)?
\(1) (x+y):y = 3:1\)
\(2) x + y = 8\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. The question \(\frac{x}{(x+y)} + \frac{y}{(xy)}\) is equivalent to \(\frac{(x^2+y^2)}{(x^2y^2)}\) for the following reason \(\frac{x}{(x+y)} + \frac{y}{(xy)}\) \(=> \frac{x(xy)}{(x+y)(xy)} + \frac{y(x+y)}{(x+y)(xy)}\) \(=> \frac{(x^2xy+xy+y^2)}{(x^2y^2)}\) \(=> \frac{(x^2+y^2)}{(x2y^2)}\) \(=> \frac{(x^2}{y^2+1)}/\frac{(x^2}{y^21)}\) by dividing the top and bottom by \(y^2\) \(=> [(\frac{x}{y})^2+1)/[(\frac{x}{y})^21]\) When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient. This tells us that A is most likely to be the answer to this question. Condition 1) The condition \((x+y):y = 3:1\) is equivalent to \(x = 2y\) since \(x + y = 3y\) from \((x+y):y = 3:1.\) Then \(\frac{(x^2+y^2)}{(x^2y^2)} = \frac{((2y)^2+y^2)}{((2y)^2y^2)} = \frac{(4y^2+y^2)}{(4y^2y^2)} = \frac{5y^2}{3y^2} = \frac{5}{3}.\) Since condition 1) yields a unique solution, it is sufficient. Condition 2) If \(x = 5\) and \(y = 3\), then we have \(\frac{x}{(x+y)} + \frac{y}{(xy)} = \frac{5}{8} + \frac{3}{2} = \frac{5}{8} + \frac{12}{8} = \frac{17}{8}.\) If \(x = 6\) and \(y = 2\), then we have \(\frac{x}{(x+y)} + \frac{y}{(xy)} = \frac{6}{8} + \frac{2}{4} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}.\) Since condition 2) does not yield a unique solution, it is not sufficient. Therefore, A is the answer. Answer: A
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Math Revolution DS Expert  Ask Me Anything about GMAT DS
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30 Oct 2019, 00:41
[GMAT math practice question] (algebra) max{\(x, y\)} denotes the maximum of \(x\) and \(y\), and min{\(x, y\)} denotes the minimum of \(x\) and \(y\). What is the value of \(x + y\)? 1) max\({x, y} = x + y\) 2) min\({x, y} = 2x + y  2\)
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31 Oct 2019, 01:20
MathRevolution wrote: [GMAT math practice question]
(number properties) \(x\) and \(y\) are positive integers. What is the value of \(x*y\)?
\(1) x^{x+y} = y^4\)
\(2) y^{x+y} = x^4\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) \(x = 1\) and \(y = 1\) satisfies both conditions 1) & 2) together and we have \(x*y = 1\). \(x = 2\) and \(y = 2\) satisfies both conditions 1) & 2) together and we have \(x*y = 4.\) Since both conditions together do not yield a unique solution, they are not sufficient. Therefore, E is the answer. Answer: E Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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31 Oct 2019, 01:21
[GMAT math practice question] (equation) What is the value of \(a + b\)? \(1) ax + by = 2(ax  by)  3 = x + y + 7\) \(2) x = 3, y = 1\)
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03 Nov 2019, 19:19
MathRevolution wrote: [GMAT math practice question]
(algebra) max{\(x, y\)} denotes the maximum of \(x\) and \(y\), and min{\(x, y\)} denotes the minimum of \(x\) and \(y\). What is the value of \(x + y\)?
1) max\({x, y} = x + y\)
2) min\({x, y} = 2x + y  2\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Case 1: \(x ≥ y\) Since max(\(x,y\)) = \(x\) and max(\(x,y\)) = \(x + y,\) we have \(x = x + y\) or \(y = 0.\) Since min(\(x,y\)) = \(y\) and min(\(x,y\)) = \(2x + y  2\), we have \(y = 2x + y  2, 0 = 2x  2, 2x = 2,\) or \(x = 1.\) Then we have \(x + y = 0 + 1 = 1.\) Case 2: \(x < y\) Since max(\(x,y\)) = \(y\) and max(\(x,y\)) = \(x + y\), we have \(y = x + y\) or \(x = 0.\) Since min(\(x,y\)) = \(x\), min(\(x,y\)) = \(2x + y  2\) and \(x = 0\), we have \(x = 2x + y  2\) or \(y = 2.\) Then we have \(x + y = 0 + 2 = 2.\) Since both conditions together do not yield a unique solution, they are not sufficient. Therefore, E is the answer. Answer: E Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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03 Nov 2019, 19:19
[GMAT math practice question] (number properties) \(x, y\), and \(z\) are positive integers. What is the value of \(xyz\)? \(1) xy + yz = 24\) \(2) xz + yz = 13\)
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03 Nov 2019, 19:22
MathRevolution wrote: [GMAT math practice question]
(equation) What is the value of \(a + b\)?
\(1) ax + by = 2(ax  by)  3 = x + y + 7\)
\(2) x = 3, y = 1\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. We have \(4\) variables (\(a, b, x\) and \(y\)). However, since both conditions have \(4\) equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since \(x = 3\) and \(y = 1\), we have \(3a + b = 2(3a  b)  3 = 3+1+7 = 11.\) Then we have \(3a + b = 11\) and \(6a  2b = 14\) or \(3a – b = 7.\) When we add those equations we have \(3a + b + 3a  b = 11 + 7, 6a = 18\) or \(a = 3\). Then we have \(3(3) + b = 11, 9 + b = 11\) or \(b = 2.\) Since both conditions together yield a unique solution, they are sufficient. Therefore, C is the answer. Answer: C Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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03 Nov 2019, 19:24
MathRevolution wrote: [GMAT math practice question]
(number properties) \(x, y\), and \(z\) are positive integers. What is the value of \(xyz\)?
\(1) xy + yz = 24\)
\(2) xz + yz = 13\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. Since we have \(3\) variables (\(x, y,\) and \(z\)) and \(0\) equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since we have \((x+y)z = 13\) from condition 2), we have \(x+y=13\) and \(z = 1.\) We have \((x+z)y = (x+1)y = 24\) from condition 1) since \(z = 1.\) If \(x = 1, y = 12, z = 1\), then \(xyz = 12.\) If \(x = 11, y = 2, z = 1\), then \(xyz = 22.\) Since both conditions together do not yield a unique solution, they are not sufficient. Therefore, E is the answer. Answer: E In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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03 Nov 2019, 23:44
Is x > 0? (1)3  x< x + 5 (2) 3  2x < x  1 A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B)Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D)EACH statement ALONE is sufficient. E)Statements (1) and (2) TOGETHER are NOT sufficient.
I am not able to solve this kind of absolute value questions. I need an algebraic approach to solve such kind of questions..
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Updated on: 06 Nov 2019, 02:03
[GMAT math practice question] (number properties) \(b = (1)+(1)^2+(1)^3+….+(1)^a\). What is the value of \(b\)? 1) \(a = 2019\) 2) \(a\) is an odd number.
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05 Nov 2019, 02:01
[GMAT math practice question] (algebra) What is the value of \(\frac{1}{(2x1)} + \frac{1}{(2y1)}\)? 1) \(\frac{1}{(2x+1)} + \frac{1}{(2y+1)} = 1\) 2) \(x = 4y = 1\)
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06 Nov 2019, 02:06
MathRevolution wrote: [GMAT math practice question]
(number properties) \(b = (1)+(1)^2+(1)^3+….+(1)^a\). What is the value of \(b\)?
1) \(a = 2019\)
2) \(a\) is an odd number. => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have \(2\) variables (\(a\) and \(b\)) and \(1\) equation, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) We have \((1) = (1)^3 = (1)^5 = … =(1)^{2019} = 1\) and \((1)^2 = (1)^4 = (1)^6 = … = (1)^{2018} = 1. (1) + (1)^2 + (1)^3+….+(1)^a. = ((1)+(1)^2) + ((1)^3+(1)^4) + … + ((1)^{2017} +(1)^{2018}) + (1)^{2019} = 0 + 0 + … + 0 + (1) = 1\) Since condition 1) yields a unique solution, it is sufficient. Condition 2) We have \((1) = (1)^3 = (1)^5 = … =(1)^a = 1\) and \((1)^2 = (1)^4 = (1)^6 = … = (1)^{a1} = 1. (1) + (1)^2 + (1)^3+….+(1)^a. = ((1)+(1)^2) + ((1)^3+(1)^4) + … + ((1)^{a2} +(1)^{a1}) + (1)^a = 0 + 0 + … + 0 + (1) = 1\) Since condition 2) yields a unique solution, it is sufficient. Therefore, D is the answer. Answer: D If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C or E.
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06 Nov 2019, 02:07
[GMAT math practice question] (algebra) What is the value of \(\frac{(4x3xy+4y)}{(3x+3y})\)? 1) \(x = y\) 2) \((\frac{1}{x}) + (\frac{1}{y}) = 3\)
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07 Nov 2019, 00:55
MathRevolution wrote: [GMAT math practice question]
(algebra) What is the value of \(\frac{1}{(2x1)} + \frac{1}{(2y1)}\)?
1) \(\frac{1}{(2x+1)} + \frac{1}{(2y+1)} = 1\)
2) \(x = 4y = 1\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. The question \(\frac{1}{(2x1)} + \frac{1}{(2y1)}\) is equivalent to \(\frac{(2x+2y2)}{(4xy2x2y+1)}\) for the following reason. \(\frac{1}{(2x1)} + \frac{1}{(2y1)}\) \(= \frac{(2y1)}{[(2x1)(2y1)]} +\frac{(2x1)}{[(2x1)(2y1)]}\) \(= \frac{(2x+2y2)}{(4xy2x2y+1)}\) Condition 2) Since \(x = 4y = 1\), we have \(4xy = 1\) and \(\frac{(2x+2y2)}{(4xy2x2y+1)} = \frac{(2x+2y2)}{(12x2y+1)} = \frac{(2x+2y2)}{(2x2y+2)} = \frac{{2(x+y1)}}{{(2)(x+y1)}} = 1.\) Since condition 2) yields a unique solution, it is sufficient. Condition 1) \(\frac{1}{(2x+1)} + \frac{1}{(2y+1)} = 1\) => \((2y+1)/[(2x+1)(2y+1)] + \frac{(2x+1)}{[(2x+1)(2y+1)}] = 1\) => \(\frac{(2x+1+2y+1)}{[(2x+1)(2y+1)]} = 1\) => \(\frac{(2x+2y+2)}{(4xy+2x+2y+1)} = 1\) => \(2x+2y+2 = 4xy+2x+2y+1\) => \(4xy = 1.\) Then we have \(\frac{(2x+2y2)}{(4xy2x2y+1)} = \frac{(2x+2y2)}{(12x2y+1)} = \frac{(2x+2y2)}{(2x2y+2)} = \frac{{2(x+y1)}}{{(2)(x+y1)}} = 1.\) Since condition 1) yields a unique solution, it is sufficient. Therefore, D is the answer. Answer: D Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2). This question is a CMT4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT4(B) questions, D is most likely to be the answer.
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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07 Nov 2019, 00:57
[GMAT math practice question] (algebra) \(a≠b\) and \(b≠c\) are given. Is it true that \(a=c\)? 1) \((ab)(bc)(ca) = 0\) 2) \(\frac{(a^2+3a)}{(a+1)} = \frac{(b^2+3b)}{(b+1)} = \frac{(c^2+3c)}{(c+1)}\)
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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07 Nov 2019, 22:12
Please let me know of good materials for DS. I am stuck



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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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08 Nov 2019, 00:31
MathRevolution wrote: [GMAT math practice question]
(algebra) What is the value of \(\frac{(4x3xy+4y)}{(3x+3y})\)?
1) \(x = y\)
2) \((\frac{1}{x}) + (\frac{1}{y}) = 3\) => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. The question asks the value of \((\frac{4}{3}) – \frac{xy}{(x+y)}\) for the following reason \(\frac{(4x3xy+4y)}{(3x+3y)}\) \(= \frac{[4(x+y)3xy]}{[3(x+y)]}\) \(= \frac{[4(x+y)]}{[3(x+y)]}  \frac{[3xy]}{[3(x+y)]}\) \(= (\frac{4}{3}) – \frac{xy}{(x+y)}\) Since we have \(\frac{(x+y)}{xy} = 3\) from condition 2),) for the following reason \((\frac{1}{x})+(\frac{1}{y}) = 3\) \((\frac{y}{xy}) + (\frac{x}{xy}) = 3\) \(\frac{(x+y)}{xy} = 3\) Then we have \(\frac{xy}{(x+y)} = \frac{1}{3}.\) Then \((\frac{4}{3}) – \frac{xy}{(x+y)} = (\frac{4}{3}) – (\frac{1}{3}) = 1.\) Since condition 2) yields a unique solution, it is sufficient. Condition 1) Since we have \(x=y, (\frac{4}{3}) – \frac{xy}{(x+y)} = (\frac{4}{3}) – \frac{x^2}{2x} = (\frac{4}{3})\frac{x}{2}.\) If \(x = y = 1,\) then \((\frac{4}{3})  \frac{x}{2} = \frac{5}{6}.\) If \(x = y = 2\), then \((\frac{4}{3})  \frac{x}{2} = \frac{1}{3}.\) Since condition 1) does not yield a unique solution, it is not sufficient. Therefore, B is the answer. Answer: B
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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