Ask me Anything about GMAT Data Sufficiency Hello, I am Max Lee, Founder and Lead Math Tutor at Math Revolution. We have over 9,000 posts and over 10,000 Kudos. This topic is a new feature on the DS Forum and a way for you to directly interact with our math team and ask anything about the DS, e.g. if you want a certain concept explained or have a particular question you want to us address, this is the place to post a link to it or your question. I intend to have this thread be as an "Everything You Need to Know about DS" type of thread. I will keep updating this post with links and resources that are helpful for the DS. Meanwhile, you can ask me anything Other discussions you may be interested in: How to Improve GMAT Quant from Q49 to Q51 - the Ultimate Q51 Quide Overview of GMAT Math Question Types and Patterns on the GMAT We will be posting new questions in this topic regularly - check it out for some new material! Thank you all - good luck on the GMAT and look forward to seeing you in the PS forum! - Max Lee.

=> 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}{(x-y)} is equivalent to \frac{(x^2+y^2)}{(x^2-y^2)} for the following reason \frac{x}{(x+y)} + \frac{y}{(x-y)} => \frac{x(x-y)}{(x+y)(x-y)} + \frac{y(x+y)}{(x+y)(x-y)} => \frac{(x^2-xy+xy+y^2)}{(x^2-y^2)} => \frac{(x^2+y^2)}{(x2-y^2)} => \frac{(x^2}{y^2+1)}/\frac{(x^2}{y^2-1)} by dividing the top and bottom by y^2 => 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^2-y^2)} = \frac{((2y)^2+y^2)}{((2y)^2-y^2)} = \frac{(4y^2+y^2)}{(4y^2-y^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}{(x-y)} = \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}{(x-y)} = \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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MathRevolution

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Updated on: 01 Aug 2021, 02:59

Originally posted by MathRevolution on 27 Oct 2019, 19:51.
Last edited by MathRevolution on 01 Aug 2021, 02:59, edited 1 time in total.

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[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.

MathRevolution

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28 Oct 2019, 01:34

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[GMAT math practice question]

(algebra) What is the value of $\frac{x}{(x+y)} + \frac{y}{(x-y)}$?

$1) (x+y):y = 3:1$

$2) x + y = 8$

MathRevolution

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29 Oct 2019, 01:52

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[GMAT math practice question]

(number properties) $x$ and $y$ are positive integers. What is the value of $x*y$?

$1) x^{x+y} = y^4$

$2) y^{x+y} = x^4$

MathRevolution

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Updated on: 01 Aug 2021, 03:00

Originally posted by MathRevolution on 30 Oct 2019, 00:40.
Last edited by MathRevolution on 01 Aug 2021, 03:00, edited 1 time in total.

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MathRevolution

[GMAT math practice question]

(algebra) What is the value of $\frac{x}{(x+y)} + \frac{y}{(x-y)}$?

$1) (x+y):y = 3:1$

$2) x + y = 8$

=>

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 $\frac{x}{(x+y)} + \frac{y}{(x-y)}$ is equivalent to $\frac{(x^2+y^2)}{(x^2-y^2)}$ for the following reason
$\frac{x}{(x+y)} + \frac{y}{(x-y)}$

$=> \frac{x(x-y)}{(x+y)(x-y)} + \frac{y(x+y)}{(x+y)(x-y)}$

$=> \frac{(x^2-xy+xy+y^2)}{(x^2-y^2)}$

$=> \frac{(x^2+y^2)}{(x2-y^2)}$

$=> \frac{(x^2}{y^2+1)}/\frac{(x^2}{y^2-1)}$ by dividing the top and bottom by $y^2$

$=> [(\frac{x}{y})^2+1)/[(\frac{x}{y})^2-1]$

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 $\frac{(x^2+y^2)}{(x^2-y^2)} = \frac{((2y)^2+y^2)}{((2y)^2-y^2)} = \frac{(4y^2+y^2)}{(4y^2-y^2)} = \frac{5y^2}{3y^2} = \frac{5}{3}.$

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
If $x = 5$ and $y = 3$, then we have $\frac{x}{(x+y)} + \frac{y}{(x-y)} = \frac{5}{8} + \frac{3}{2} = \frac{5}{8} + \frac{12}{8} = \frac{17}{8}.$

If $x = 6$ and $y = 2$, then we have $\frac{x}{(x+y)} + \frac{y}{(x-y)} = \frac{6}{8} + \frac{2}{4} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}.$

Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.
Answer: A

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30 Oct 2019, 00:41

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[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${x, y} = 2x + y - 2$

MathRevolution

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Updated on: 18 Dec 2021, 11:26

Originally posted by MathRevolution on 31 Oct 2019, 01:20.
Last edited by MathRevolution on 18 Dec 2021, 11:26, edited 1 time in total.

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MathRevolution

[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.

MathRevolution

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31 Oct 2019, 01:21

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[GMAT math practice question]

(equation) What is the value of $a + b$?

$1) ax + by = 2(ax - by) - 3 = x + y + 7$

$2) x = 3, y = 1$

MathRevolution

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Updated on: 18 Dec 2021, 11:33

Originally posted by MathRevolution on 03 Nov 2019, 19:19.
Last edited by MathRevolution on 18 Dec 2021, 11:33, edited 1 time in total.

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=>

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.

MathRevolution

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03 Nov 2019, 19:19

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[GMAT math practice question]

(number properties) $x, y$, and $z$ are positive integers. What is the value of $xyz$?

$1) xy + yz = 24$

$2) xz + yz = 13$

MathRevolution

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Updated on: 21 Dec 2021, 03:02

Originally posted by MathRevolution on 03 Nov 2019, 19:22.
Last edited by MathRevolution on 21 Dec 2021, 03:02, edited 1 time in total.

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[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.

MathRevolution

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[1]

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[1]

Updated on: 21 Dec 2021, 03:03

Originally posted by MathRevolution on 03 Nov 2019, 19:24.
Last edited by MathRevolution on 21 Dec 2021, 03:03, edited 1 time in total.

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[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

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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..

Posted from my mobile device

MathRevolution

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Updated on: 06 Nov 2019, 02:03

Originally posted by MathRevolution on 04 Nov 2019, 01:17.
Last edited by MathRevolution on 06 Nov 2019, 02:03, edited 1 time in total.

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[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.

MathRevolution

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05 Nov 2019, 02:01

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[GMAT math practice question]

(algebra) What is the value of $\frac{1}{(2x-1)} + \frac{1}{(2y-1)}$?

1) $\frac{1}{(2x+1)} + \frac{1}{(2y+1)} = 1$

2) $x = 4y = 1$

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06 Nov 2019, 02:06

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[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 $(-1)^2 = (-1)^4 = (-1)^6 = … = (-1)^{a-1} = 1. (-1) + (-1)^2 + (-1)^3+….+(-1)^a. = ((-1)+(-1)^2) + ((-1)^3+(-1)^4) + … + ((-1)^{a-2} +(-1)^{a-1}) + (-1)^a = 0 + 0 + … + 0 + (-1) = -1$

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

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[GMAT math practice question]

(algebra) What is the value of $\frac{(4x-3xy+4y)}{(3x+3y})$?

1) $x = y$

2) $(\frac{1}{x}) + (\frac{1}{y}) = 3$

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07 Nov 2019, 00:55

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[GMAT math practice question]

(algebra) What is the value of $\frac{1}{(2x-1)} + \frac{1}{(2y-1)}$?

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 $\frac{1}{(2x-1)} + \frac{1}{(2y-1)}$ is equivalent to $\frac{(2x+2y-2)}{(4xy-2x-2y+1)}$ for the following reason.

$\frac{1}{(2x-1)} + \frac{1}{(2y-1)}$

$= \frac{(2y-1)}{[(2x-1)(2y-1)]} +\frac{(2x-1)}{[(2x-1)(2y-1)]}$

$= \frac{(2x+2y-2)}{(4xy-2x-2y+1)}$

Condition 2)
Since $x = 4y = 1$, we have $4xy = 1$ and $\frac{(2x+2y-2)}{(4xy-2x-2y+1)} = \frac{(2x+2y-2)}{(1-2x-2y+1)} = \frac{(2x+2y-2)}{(-2x-2y+2)} = \frac{{2(x+y-1)}}{{(-2)(x+y-1)}} = -1.$

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 $\frac{(2x+2y-2)}{(4xy-2x-2y+1)} = \frac{(2x+2y-2)}{(1-2x-2y+1)} = \frac{(2x+2y-2)}{(-2x-2y+2)} = \frac{{2(x+y-1)}}{{(-2)(x+y-1)}} = -1.$

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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07 Nov 2019, 00:57

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[GMAT math practice question]

(algebra) $a≠b$ and $b≠c$ are given. Is it true that $a=c$?

1) $(a-b)(b-c)(c-a) = 0$

2) $\frac{(a^2+3a)}{(a+1)} = \frac{(b^2+3b)}{(b+1)} = \frac{(c^2+3c)}{(c+1)}$

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Please let me know of good materials for DS. I am stuck

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08 Nov 2019, 00:31

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[GMAT math practice question]

(algebra) What is the value of $\frac{(4x-3xy+4y)}{(3x+3y})$?

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

$\frac{(4x-3xy+4y)}{(3x+3y)}$

$= \frac{[4(x+y)-3xy]}{[3(x+y)]}$

$= \frac{[4(x+y)]}{[3(x+y)]} - \frac{[3xy]}{[3(x+y)]}$

$= (\frac{4}{3}) – \frac{xy}{(x+y)}$

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