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Math Revolution GMAT Instructor
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22 Nov 2019, 01:41
[GMAT math practice question] (Speed) The distance between Jane’s home and her school is \(24km\). It takes \(4\) hours and \(50\) minutes for Jane to walk from home to school and it takes \(5\) hours to come back. The road consists of an uphill section, a downhill section and a flat section. How long is the flat section? 1) The speed on the uphill section is \(4km/hr\), on the downhill section is \(6km/hr\) and on the flat section is \(5km/h.\) 2) The speed on the flat section is the arithmetic average of the speeds on the uphill and downhill sections.
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24 Nov 2019, 19:16
MathRevolution wrote: [GMAT math practice question]
(Algebra) What is the difference between \(x\) and \(y\)?
1) \(x  y\) is the first odd prime number
2) \(x\) and \(y\) are positive integers such that \(3x + 5y = 23\) => 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 \(x  y.\) Condition 1) Since the first odd prime number is \(3,\) we have \(x  y = 3.\) Since condition 1) yields a unique solution, it is sufficient. Condition 2) \(3x = 23 – 5y\) If \(y = 1\), then we have \(3x = 23 – 5 = 18\) or \(x = 6\) and \(x  y = 5.\) If \(y = 2\), then we have \(3x = 23 – 10 = 13\) and we don’t have an integer solution. If \(y = 3,\) then we have \(3x = 23 – 15 = 8\) and we don’t have an integer solution. If \(y = 4\), then we have \(3x = 23 – 20 = 3\) or \(x = 1\) and \(x  y = 3\) If \(y = 5,\) then we have \(3x = 23 – 25 = 2\) and we start to have negative numbers and we can stop this substitution process. Since condition 2) does not yield a unique solution, it is not sufficient. Therefore, A is the answer. Answer: A
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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24 Nov 2019, 19:20
MathRevolution wrote: [GMAT math practice question]
(Speed) The distance between Jane’s home and her school is \(24km\). It takes \(4\) hours and \(50\) minutes for Jane to walk from home to school and it takes \(5\) hours to come back. The road consists of an uphill section, a downhill section and a flat section. How long is the flat section?
1) The speed on the uphill section is \(4km/hr\), on the downhill section is \(6km/hr\) and on the flat section is \(5km/h.\)
2) The speed on the flat section is the arithmetic average of the speeds on the uphill and downhill sections. => 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. Assume \(f, u\) and \(d\) are distances on the flat sections, uphill sections and downhill sections when she travels from home to school. Then we have\( f + u + d = 24.\) Even though we have \(3\) variables and \(1\) equation, condition 1) has \(3\) equations. So we should check condition 1) first. Condition 1) The time she travels from home to school is \(\frac{f}{5} + \frac{u}{4} + \frac{d}{6} = 4(\frac{50}{60}) = \frac{29}{6}.\) The time she travels from school to home is \(\frac{f}{5} + \frac{u}{6} + \frac{d}{4} = 5\) since uphill sections becomes downhill sections and downhill sections becomes uphill when she travels back. Adding the equations together, we have \((\frac{2}{5})f + \frac{(5u + 5d)}{12} = (\frac{2}{5})f + (\frac{5}{12})(u+d) = \frac{29}{6} + 5 = \frac{59}{6}.\) Substituting in \(u + d = 24 – f,\) we have \((\frac{2}{5})f +(\frac{5}{12})(24  f) = \frac{59}{6}, (\frac{2}{5})f + 10  (\frac{5}{12})f = \frac{59}{6},\) or \((\frac{24}{60})f  (\frac{25}{60})f = \frac{59}{6}  \frac{60}{6}.\) Then we have \((\frac{1}{60})f = \frac{1}{6}\), or \(f = 10.\) Since condition 1) yields a unique solution, it is sufficient. Condition 2) Since the speed on the flat section is the arithmetic average of the speeds on the uphill and downhill sections, we can assume that s is the speed on the flat sections, \(s – a\) is the speed on the uphill sections and \(s + a\) is the speed on the downhill sections. Then we have the time she takes when she travels from home to school, \(\frac{f}{s} + \frac{u}{(sa)} + \frac{d}{(s+a)} = \frac{29}{6}\) and we have the time she takes when she travels from home to school, \(\frac{f}{s} + \frac{u}{(s+a)} + \frac{d}{(sa)} = 5.\) When we add those equations, we have \((\frac{2f}{s}) + (u+d)(\frac{1}{(sa)}+\frac{1}{(sb)}) = \frac{59}{6}.\) We can notice that there must be many possibilities for solutions. Since condition 2) does not yield a unique solution, it is not sufficient. Therefore, A is the answer. Answer: A 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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25 Nov 2019, 00:54
[GMAT math practice question] (functions) Is \(ax + 2y  3 = 4x + by + 5\) an equation of a line on the xyplane? 1) \(a ≠ 4.\) 2) \(b ≠ 2.\)
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26 Nov 2019, 06:09
[GMAT math practice question] (number properties) There are \(6\) bags. Each bag contains \(18, 19, 21, 23, 25\), and \(34\) beads, respectively. All the beads in one bag are broken, and no other bags have any broken beads. How many beads are broken? 1) Adam has \(3\) bags, and Betty has \(2\) bags, and no one has the bag of broken beads. 2) Adam has twice as many beads as Betty has.
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27 Nov 2019, 04:26
MathRevolution wrote: [GMAT math practice question]
(functions) Is \(ax + 2y  3 = 4x + by + 5\) an equation of a line on the xyplane?
1) \(a ≠ 4.\)
2) \(b ≠ 2.\) => 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. 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. \(ax + 2y  3 = 4x + by + 5\) \(⇔ (a  4)x + (2  b)y – 8 = 0.\) If we have \(a = 4\) and \(b = 2\), then the equation is equivalent to \(8 = 0\), which is not an equation of a line. So, each condition alone is sufficient. Therefore, D is the answer. Answer: D
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27 Nov 2019, 04:27
[GMAT math practice question] (arithmetic) \(a\) is a constant positive number. What is the maximum value of \(ax  y\)? 1) \(x ≥ 1\) and \(y ≥ 1.\) 2) \(x + y = 3.\)
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28 Nov 2019, 01:17
MathRevolution wrote: [GMAT math practice question]
(number properties) There are \(6\) bags. Each bag contains \(18, 19, 21, 23, 25\), and \(34\) beads, respectively. All the beads in one bag are broken, and no other bags have any broken beads. How many beads are broken?
1) Adam has \(3\) bags, and Betty has \(2\) bags, and no one has the bag of broken beads.
2) Adam has twice as many beads as Betty has. => 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 there are \(6\) bags, we have many variables and \(0\) equations, and 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) Assume \(a\) and \(b\) are the numbers of beads that Adam and Betty have, respectively. Since Adam chooses \(3\) bags and Betty chooses \(2\) bags, one bag is not chosen by Adam or Betty. Since we have \(a = 2b, a + b = 2b + b = 3b\), then \(a + b\) is a multiple of \(3.\) \(18 + 19 + 21 + 23 + 25 + 34 = 140\) has a remainder \(2\) when \(140\) is divided by \(3\). When we subtract the number of beads in \(1\) of the \(6\) bags from \(140\), it must be a multiple of \(3\) and only \(23\) satisfies this condition, since \(140 – 23 = 117\), which is a multiple of \(3.\) \(23\) is the number of broken beads. Since both conditions together yield a unique solution, they are sufficient. Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) There are \(6\) possibilities for the bag with broken beads. Since condition 1) does not yield a unique solution, it is not sufficient. Condition 2) Since condition 2) doesn’t give us any information about the number of broken beads, it is obviously not sufficient. Therefore, C is the answer. Answer: C 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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28 Nov 2019, 01:18
[GMAT math practice question] (inequalities) Which one of \(\frac{(a + c)}{(b + d)}\) and \(\frac{ac}{bd}\) is greater than the other one? 1) \(d > c\) and \(b > a.\) 2) \(a, b, c,\) and \(d\) are positive.
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29 Nov 2019, 01:36
MathRevolution wrote: [GMAT math practice question]
(arithmetic) \(a\) is a constant positive number. What is the maximum value of \(ax  y\)?
1) \(x ≥ 1\) and \(y ≥ 1.\)
2) \(x + y = 3.\) => 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. 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) In order to have a maximum value of \(ax – y, x\) must be the maximum value, and \(y\) is the minimum, which means \(x = 2\) and \(y = 1\). The answer is \(ax  y = 2a  1.\) 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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29 Nov 2019, 01:37
[GMAT math practice question] (inequalities) Which one of \(p + q\) and \(pq + 1\) greater than the other one? 1) \(1 < p < 1. \) 2) \(1 < q < 1.\)
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Math Revolution DS Expert  Ask Me Anything about GMAT DS
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01 Dec 2019, 21:02
MathRevolution wrote: [GMAT math practice question]
(inequalities) Which one of \(\frac{(a + c)}{(b + d)}\) and \(\frac{ac}{bd}\) is greater than the other one?
1) \(d > c\) and \(b > a.\)
2) \(a, b, c,\) and \(d\) are positive. => 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. \(\frac{(a + c)}{(b + d)} – \frac{ac}{bd}\) = \(\frac{{(a + c)bd – ac(b+d)}}{{bd(b+d)}}\) (by adding fractions with a common denominator) = \(\frac{{abd + cbd – acd – acd}}{{bd(b+d)}}\) (by multiplying through the brackets in the numerator) = \(\frac{{ab(d  c) + cd(b  a)}}{{bd(b + d)}}\) (by taking out common factors) Since we have \(4\) variables (\(a, b, c\), and \(d\)) 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 \(a, b, c\) and \(d\) is positive and we have \(d > c\) and \(b > a\), we have \(\frac{{ab(d  c) + cd(b  a)}}{{bd(b + d)}} > 0.\) It means we have \(\frac{(a + c)}{(b + d)} > \frac{ac}{bd}.\) Since both conditions together yield a unique solution, they are sufficient. Therefore, C is the answer. Answer: C 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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01 Dec 2019, 21:05
MathRevolution wrote: [GMAT math practice question]
(inequalities) Which one of \(p + q\) and \(pq + 1\) greater than the other one?
1) \(1 < p < 1. \)
2) \(1 < q < 1.\) => 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. \(pq + 1 – ( p + q )\) \(= pq – p – q + 1\) \(= (p1)(q1)\) The question asks if \((p1)(q1)\) is positive or negative. If we have \(p>1, q>1\) or \(p<1, q<1\), then \((p1)(q1)\) is positive. Thus, both conditions together are sufficient, since they tell \(p < 1\) and \(q < 1.\) Therefore, C is the answer. Answer: C
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02 Dec 2019, 01:52
[GMAT math practice question] (Inequality) What are the values of \(x\) and \(y\)? 1) \(x\) and \(y\) are numbers such that \(3x  2y = 6(x  1). \) 2) \(x\) and \(y\) are integers with \(3 < x ≤ 3,\)
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03 Dec 2019, 01:44
[GMAT math practice question] (algebra) \(A\) and \(B\) are polynomials. What polynomial is \(3A + B\)? 1) \(A + B = 3x^2  3xy + 4y^2\). 2) \(A – B = x^2 + xy  6y^2.\)
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04 Dec 2019, 02:14
MathRevolution wrote: [GMAT math practice question]
(Inequality) What are the values of \(x\) and \(y\)?
1) \(x\) and \(y\) are numbers such that \(3x  2y = 6(x  1). \)
2) \(x\) and \(y\) are integers with \(3 < x ≤ 3,\) => 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. 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) \(3x  2y = 6(x  1)\) => \(3x  2y = 6x  6\) => \(2y = 3x  6 \) => \(2y = 3(x  2)\) => \(2y = 3(x  2)\) => \(2y = 3(2  x)\) => \(y = (\frac{3}{2})(2  x)\) Since \(x\) and \(y\) are integers from condition 2) and \(x  2\) is an even number, \(x\) must be an even number. Also, \(y\) is a multiple of \(3.\) Since we have \(3 < x ≤ 3\) from condition \(2\), we have \(2 ≤ x ≤ 3\). The possible values of \(x\) are \(2, 0\) and \(2.\) Substituting these values into \(y = (\frac{3}{2})(2  x)\) gives the possible pairs of \((x, y)\), which are \((2, 6), (0, 3)\) and \((2, 6)\). 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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04 Dec 2019, 02:16
[GMAT math practice question] (number properties) \(x\) is a positive even integer, and \(y\) is a number. Which one of \(\frac{1}{2} ⅹ \frac{3}{4} ⅹ \frac{5}{6} ⅹ…..ⅹ \frac{x1}{x}\) and \(\frac{1}{y}\) is greater? 1) \(x = y^2.\) 2) \(y\) is positive.
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05 Dec 2019, 01:24
MathRevolution wrote: [GMAT math practice question]
(algebra) \(A\) and \(B\) are polynomials. What polynomial is \(3A + B\)?
1) \(A + B = 3x^2  3xy + 4y^2\).
2) \(A – B = x^2 + xy  6y^2.\) => 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 (\(A\) and \(B\)) 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) \(3A + B \) => \(= 2(A + B) + (A  B)\) (rearranging the equation to suit the conditions) => \(= 2(3x^2  3xy + 4y^2) + (x^2 + xy  6y^2)\) (substituting in the conditions) => \(= 6x^2  6xy + 8y^2 + x^2 + xy  6y^2\) (multiplying \(2\) through the bracket) => \(= 7x^2  5xy + 2y^2\) (adding like terms) 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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05 Dec 2019, 01:25
[GMAT math practice question] (number properties) \(N\) is a \(3\)digit positive integer \(s\) with the expression, \(abc\). What is the value of \(N\)? 1) \(b > 2a + c.\) 2) \(a + c > 12.\)
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06 Dec 2019, 01:23
MathRevolution wrote: [GMAT math practice question]
(number properties) \(x\) is a positive even integer, and \(y\) is a number. Which one of \(\frac{1}{2} ⅹ \frac{3}{4} ⅹ \frac{5}{6} ⅹ…..ⅹ \frac{x1}{x}\) and \(\frac{1}{y}\) is greater?
1) \(x = y^2.\)
2) \(y\) is positive. => 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. We have \(\frac{1}{2} < \frac{2}{3}, \frac{3}{4} < \frac{4}{5}, …, \frac{99}{100} < \frac{100}{101}, … , \frac{(x1)}{x} < \frac{x}{(x+1)}.\) Assume \(S\) is the multiple of the lefthand sides, \((\frac{1}{2})(\frac{3}{4})…\frac{(x1)}{x}\) and \(T\) is the multiple of the righthand sides, \((\frac{2}{3})(\frac{4}{5})…(\frac{x}{(x+1)})\). Then, we have \(S^2 < ST = (\frac{1}{2})(\frac{2}{3})…(\frac{(x1)}{x})(\frac{x}{(x+1)}) = \frac{1}{(x+1)}\) or \(S < (\frac{1}{2})(\frac{3}{4})…(\frac{x}{(x+1)}) < \frac{1}{ √(x+1)}\) 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) \(S < \frac{1}{ √(x+1)} < \frac{1}{√x} = \frac{1}{y}\) since \(y = √x\) because \(y > 0\) and \(x = y^2\) so \(y = √x.\) Then we have \(\frac{1}{2} ⅹ \frac{3}{4} ⅹ \frac{5}{6} ⅹ…..ⅹ \frac{x1}{x} < \frac{1}{y}.\) 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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