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14 Mar 2019, 18:13
[GMAT math practice question] If 0.02<x<0.04 and 100<y<250, which of the following could be the value of (yx)/(xy)? A. 10 B. 20 C. 30 D. 50 E. 100 => Note that (yx)/(xy) = 1/x – 1/y. 0.02 < x < 0.04 => 1/0.02 > 1/x > 1/0.04 => 50 > 1/x > 25 => 25 < 1/x < 50 100 < y < 250 => 1/100 > 1/y > 1/250 => 0.01 > 1/y > 0.004 => 0.004 < 1/y < 0.01 So, 25 – 0.01 < 1/x – 1/y < 50 – 0.004 24.99 < (yx)/(xy) < 49.996 The only possible value of (yx)/(xy) listed is 30. Therefore, the answer is C. Answer: C
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Re: The Ultimate Q51 Guide [Expert Level]
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17 Mar 2019, 18:20
[GMAT math practice question] x and y are integers. If y≠3, is x=4? 1) x+y=7 2) x^2+y^2=25 => 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. Since we have 2 variables (x and y) and 0 equations, C is most likely to be 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 y = 7 – x, x^2+y^2=25 is equivalent to x^2+(7x)^2=25 or 2x^2 – 14x + 24 = 0. This factors as 2(x^2 – 7x + 12) = 0 or 2(x3)(x4) = 0. So, x = 3 and y = 4, or x = 4 and y = 3. Since y ≠3, we must have x = 3 and x can’t be 4. Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient, when used together. Since this question is an integer 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) Since y = 7 – x ≠ 3, x can’t be 4. So, we have a unique answer, which is “no”. Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient. Condition 2) If x = 4 and y = 3, then x^2+y^2 = 25, and the answer is “yes”. If x = 3 and y = 4, then x^2+y^2 = 25, and the answer is “no”. Thus, condition 2) is not sufficient, since it does not yield a unique solution. Therefore, A is the answer. Answer: A 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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Re: The Ultimate Q51 Guide [Expert Level]
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18 Mar 2019, 18:22
[GMAT math practice question] If xyz≠0, is x^3y^4z^5>0? 1) xz>0 2) xyz>0 => 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Asking if x^3y^4z^5>0 is equivalent to asking if xz > 0 since we can ignore even exponents in inequalities. Thus, condition 1) is sufficient. Condition 2) If x = 1, y = 1 and z = 1, then x^3y^4z^5>0, and the answer is “yes”. If x = 1, y = 1 and z = 1, then x^3y^4z^5<0, and the answer is “no”. Condition 2) is not sufficient since it does not yield a unique answer. Therefore, A is the answer. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level]
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19 Mar 2019, 18:14
[GMAT math practice question] If f(x)=ax^2+bx+c, where a, b and c are integers, is b=0? 1) f(49)=f(49)=0 2) f(0)f(49)=0 => 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. In order to have b =0, f(x) must be symmetric about the yaxis. Thus, condition 1) is sufficient. By the factor theorem, condition 1) tells us that f(x) = a(x49)(x+49) = a(x^2 – 49^2) = ax^2 – 49^2a and b = 0. It is sufficient. Condition 2) If f(0)f(49) = 0 then either f(0)= 0 or f(49) = 0. Note that f(0) = c. If a = 1, b = 0 and c = 0, then f(0) = 0, so f(0)f(49) = 0, and the answer is ‘yes’. If a = a, b = 1 and c = 0, then f(0) = 0, so f(0)f(49) = 0, and the answer is ‘no’. Thus, condition 2) is not sufficient, since it does not yield a unique solution. Therefore, A is the answer. Answer: A
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20 Mar 2019, 18:14
[GMAT math practice question] Alice, Bob, Cindy, Darren, Eddie, Fabian sit on six chairs around a large round table. Alice must sit opposite Bob and Cindy must sit opposite Darren. How many seating arrangements are possible? A. 6 B. 8 C. 24 D. 120 E. 720 => Note: Because the table is round, Alice’s choice of chair does not change the number of possible arrangements. Once Alice is seated, Bob’s position is determined. Cindy has 4 remaining possible seats. Cindy’s choice determines Darren’s position. Eddie then has 2 remaining possible seats, and Fabian must sit in the remaining chair. Thus, the number of possible arrangements is 4*2 = 8. Therefore, B is the answer. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level]
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21 Mar 2019, 18:27
[GMAT math practice question] Which of the following could be the number of diagonals of an npolygon? A. 3 B. 6 C. 8 D. 12 E. 20 => The number of diagonals of an npolygon is nC 2 – n = n(n1)/2 – n=n(n3)/2. If n = 4, then the number of diagonals is 4C 2 – 4 = 6 – 4 = 2. If n = 5, then the number of diagonals is 5C 2 – 5 = 10 – 5 = 5. If n = 6, then the number of diagonals is 6C 2 – 6 = 15 – 6 = 9. If n = 7, then the number of diagonals is 7C 2 – 7 = 21 – 7 = 14. If n = 8, then the number of diagonals is 8C 2 – 8 = 28 – 8 = 20. Therefore, E is the answer. Answer: E
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Re: The Ultimate Q51 Guide [Expert Level]
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24 Mar 2019, 18:25
[GMAT math practice question] Is n an odd number? 1) n and n+2 are prime numbers 2) n+5 is a multiple of 5 => 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. If n and n+2 are prime numbers from condition 1), then both n and n + 2 are odd numbers. Note that 2 is the only even prime number, and 2 + 2 = 4 is not prime. Thus, condition 1) is sufficient. Condition 2) If n + 5 = 10, then n = 5 is an odd number and the answer is ‘yes’. If n + 5 = 15, then n = 10 is an even number and the answer is ‘no’. Thus, condition 2) is not sufficient, since it does not yield a unique solution. Therefore, A is the answer. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level]
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25 Mar 2019, 18:18
[GMAT math practice question] If m and n are positive integers, is m+n divisible by 15? 1) m is divisible by 9 and n is divisible by 15. 2) mn is divisible by 225. => 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. Since we have 2 variables (m and n) and 0 equations, C is most likely to be 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) If m = 45 and n = 45, then m + n = 90 is divisible by 15 and the answer is ‘yes’. If m = 9 and n = 225, then m + n = 234 is not divisible by 15 and the answer is ‘no’. Thus, both conditions together are not sufficient, since they do not yield a unique solution. 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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26 Mar 2019, 18:10
[GMAT math practice question] A box contains 3 red balls, 4 green balls, 5 yellow balls, 6 blue balls and 7 white balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 5 balls of a single color will be drawn? A. 10 B. 12 C. 15 D. 18 E. 21 => The maximum number of draws we can make without drawing 5 balls of a single color is 3 + 4 + 4 + 4 + 4 = 14. This occurs when we draw 3 red balls, 4 green balls, 4 yellow balls, 4 blue balls and 4 white balls. If we draw one more ball, then we will have drawn 5 balls of a single color. Thus, to guarantee that we have drawn at least 5 balls of a single color, we must draw 14+1 = 15 balls. Therefore, C is the answer. Answer: C
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Re: The Ultimate Q51 Guide [Expert Level]
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27 Mar 2019, 18:19
[GMAT math practice question] The number 1000 lies between two consecutive perfect squares. Which one of these square integers is closest to 1000? A. 29^2 B. 30^2 C. 31^2 D. 32^2 E. 33^2 => 31^2 = 961 32^2 = 1024 1024 is the perfect square closest to 1000. Therefore, D is the answer. Answer: D
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Re: The Ultimate Q51 Guide [Expert Level]
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28 Mar 2019, 18:33
[GMAT math practice question] What is the smallest positive integer n such that n / 420 can be expressed as a terminating decimal? A. 18 B. 21 C. 24 D. 30 E. 42 => n / 420 = n / {2^2*3^1*5^1*7^1} The fraction will only be a terminating decimal if both 3 and 7 can be canceled out. Thus, the smallest possible value of n is 3*7 = 21. Therefore, the answer is B. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level]
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29 Mar 2019, 05:14
not all but you have to get to hard questions and to do that you have to answer 1520 questions right in a row.



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Re: The Ultimate Q51 Guide [Expert Level]
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31 Mar 2019, 18:37
[GMAT math practice question] m, n and 2/m+3/n are positive integers. What is the value of mn? 1) m and n are prime numbers 2) The greatest common divisor of m and n is 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. The possible pairs (m,n) are (1,1),(1,3),(2,1),(2,3) and (5,5). Since we have 2 variables (m and n) and 0 equations, C is most likely to be 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) There is a unique pair of integers which satisfies both conditions. This is m = 2 and n = 3. So, mn = 6. Conditions 1) & 2) are sufficient, when applied together. Since this question is an integer 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) If m = 2 and n = 3, then mn = 6. If m = 5 and n = 5, then mn = 25. Condition 1) is not sufficient since it does not yield a unique solution. Condition 2) If m = 2 and n = 3, then mn = 6. If m = 1 and n = 1, then mn = 1. Condition 2) is not sufficient since it does not yield a unique solution. 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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Re: The Ultimate Q51 Guide [Expert Level]
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02 Apr 2019, 07:47
[GMAT math practice question] Is the fourdigit positive integer a,bc6 divisible by 4? 1) ac is an odd number 2) bc is an even 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. The number a,bc6 is divisible by 4 precisely when c is an odd number. Condition 1) tells us that both a and c are odd numbers. Thus, condition 1) is sufficient. Condition 2) If a = 1, b = 2, c = 1, then 1216 is divisible by 4 and the answer is ‘yes’. If a = 1, b = 1, c = 2, then 1126 is not divisible by 4 and the answer is ‘no’. Thus, condition 2) is not sufficient, since it does not yield a unique solution. Therefore, A is the answer. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level]
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03 Apr 2019, 18:36
[GMAT math practice question] What is the median of the three numbers a, b and 24? 1) a – 24 = 24  b 2) (a24)(b24) < 0 => 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Condition 1) means the average of a, b and 24 is (a + b + 24 ) / 3 = 24 since it implies that ( a + b ) / 2 = 24. Since one of the three data points is their average, it is also their median. Condition 1) is sufficient. Condition 2) If a > 24 and b < 24, the median of a, b and 24 is 24. If a < 24 and b > 24, the median of a, b and 24 is 24. Thus, we have a unique value for the median, and condition 2) is sufficient. Therefore, D is the answer. Answer: D Note: Condition 1) is easy to understand and condition 2) is hard to understand. Thus, this question is a CMT4(B) question. When one condition is easy to understand, and the other is hard to understand, D is most likely to be the answer.
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04 Apr 2019, 18:28
[GMAT math practice question] When f(x)=x^3+1/x^3, which of the following is equal to f(1/x)? A. f(x) B. f(x) C. 1/f(x) D. 1f(x) E. (f(x))^3 => f(1/x) = (1/x)^3 + 1/{1/(1/x)^3} = 1/x^3 – x^3 = (x^3+1/x^3) = f(x). Therefore, B is the answer. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level]
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07 Apr 2019, 18:44
[GMAT math practice question] Is x<y<z ? 1) x+1<y<z+1 2) x1<y<z1 => 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. Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be 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) By condition 1), x < y since x < x + 1 ≤  x + 1  < y. By condition 2), y < z since y < z – 1 < z. Therefore, x < y < z. Thus, both conditions 1) & 2) together are sufficient. Since this question is an absolute value 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) If x = 1, y = 3, and z = 5, then the answer is ‘yes’. If x = 1, y = 3, and z = 2.9, then the answer is ‘no’ since z < y. Thus, condition 1) is not sufficient, since it does not yield a unique solution. Condition 2) If x = 1, y = 3, and z = 5, then the answer is ‘yes’. If x = 3.1, y = 3, and z = 5, then the answer is ‘no’ since x > y. Thus, condition 2) is not sufficient, since it does not yield a unique solution. 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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Re: The Ultimate Q51 Guide [Expert Level]
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08 Apr 2019, 18:16
[GMAT math practice question] m and n are positive integers. Are m and n consecutive integers? 1) m^2+n^2 = 5 2) mn = 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Condition 1) The only positive integers satisfying m^2+n^2 = 5 are m = 1 and n = 2, or m=2 and n = 1. These are consecutive integers, so condition 1) is sufficient. Condition 2) Since the difference between m and n is 1, they are consecutive integers. Condition 2) is sufficient. Therefore, D is the answer. Answer: D
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Re: The Ultimate Q51 Guide [Expert Level]
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09 Apr 2019, 18:47
[GMAT math practice question] m and n are integers. Is m + n an odd number? 1) m – n = 2 2) m^2n^2 = 225 => 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Condition 2) Since m^2n^2 = 225 is an odd number, both m and n must be odd numbers, and m + n is an even number. Thus, the answer is ‘no’, and condition 2) is sufficient by CMT (Common Mistake Type) 1. Condition 1) Since m – n = 2, both m and n are odd numbers or even numbers. Since m and n have the same parity, m + n is an even number. Thus, the answer is ‘no’, and condition 1) is sufficient by CMT (Common Mistake Type) 1. Therefore, D is the answer. Answer: D
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10 Apr 2019, 18:13
[GMAT math practice question] What is the number of roots of the equation (x^25x+5)^{(x^25x6){=1? A. 2 B. 3 C. 4 D. 5 E. 6 => When we have an equation of the form a^b = 1, we need to consider the following three cases: Case 1: a = 1 Case 2: a = 1 and b is an even number Case 3: b = 0. Case 1: x^25x+5 = 1 => x^25x+4 = 0 => (x1)(x4) = 0 => x = 1 or x = 4 Thus, 1 and 4 are roots of the equation. Case 2: x^25x+5 = 1 => x^25x+6 = 0 => (x2)(x3) = 0 => x = 2 or x = 3 If x = 2, then the exponent b = x^25x6 = 4 – 10  6 = 12 is an even number. If x = 3, then the exponent b = x^25x6 = 9 – 15  6 = 12 is an even number. Thus, 2 and 3 are roots of the equation. Case 3: x^25x6 = 0 => (x+1)(x6) = 0 => x = 1 or x = 6. Thus 1 and 6 are roots of the equation. Hence, 1, 2, 3, 4, 6 and 1 are roots of the equation. Therefore, E is the answer. Answer: E
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