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Math Revolution GMAT Instructor
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT
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24 Oct 2018, 17:23
[ Math Revolution GMAT math practice question] If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is its width? A. 1 B. a+b C. ab D. 1/(a+b) E. 1/ab => Let x be the width of the rectangle. Then x(a+b) = 1/a + 1/b = (a+b)/ab. Thus, x = 1/ab. Therefore, the answer is E. Answer: E
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT
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25 Oct 2018, 17:18
[ Math Revolution GMAT math practice question] In the xyplane, does the graph of y=ax^2+c intersect the xaxis? 1) a>0 2) c>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. The question “does the graph of y=ax^2+c intersect the xaxis” is equivalent to asking “does the equation ax^2+c = 0 have a root”. Note that the statement “ax^2 + bx + c = 0 has a root” is equivalent to b^24ac ≥ 0. Thus, the question asks if 4ac ≥ 0, or ac ≤ 0, since b = 0 in this problem. When we consider both conditions together, we obtain ac > 0 and the answer is “no”, since a > 0 and c > 0. Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions together are sufficient. Note: Neither condition on its own provides enough information for us to determine whether ac ≤ 0. Therefore, C is the answer. Answer: C
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT
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28 Oct 2018, 18:20
[ Math Revolution GMAT math practice question] n is an integer. Is n(n+2) a multiple of 8? 1) n is an even integer 2) n is a multiple of 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Note that the product of two consecutive even integers is a multiple of 8 since one of them is a multiple of 4 and the other is an even integer. Thus, each of conditions is sufficient since each implies that n and n+2 are two consecutive even integers. Therefore, D is the answer. Answer: D
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29 Oct 2018, 17:25
[ Math Revolution GMAT math practice question] The range of set A is 24 and the range of set B is 20. What is the smallest possible range of sets A and B, combined? A. 20 B. 24 C. 40 D. 44 E. 48 => Note that the range of the combined set can’t be less than the maximum of the ranges of the two sets. The maximum of the ranges of the two sets is 24. So, the range of the combined set must be greater than or equal to 24. For example, if A = {0, 24} and B = { 0, 20 }, the combined set { 0, 20, 24 }, has range 24. 24 is the smallest possible range of the set A ⋃ B. Therefore, B is the answer. Answer: B
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30 Oct 2018, 17:29
[ Math Revolution GMAT math practice question] If x6=2x, then x=? A. 6 B. 4 C. 0 D. 2 E. 6 => x6=2x => x6 = ±2x => 6 = x ±2x => 6 = x or 6 = 3x => x = 6 or x = 2 However, 2x = x6 ≥ 0. Thus, we have the unique solution x=2. Therefore, the answer is D. Answer: D
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01 Nov 2018, 17:59
[ Math Revolution GMAT math practice question] The terms of a sequence are defined by a n=a n2+3. Is 411 a term of the sequence? 1) a 1=111 2) a 2=112 => 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 formula a n=a n2+3 tells us that alternate terms have the same remainder when they are divided by 3. Since a 1 = 111 = 3*37 is a multiple of three, all multiples of three greater than 111 can be obtained as oddnumbered terms. Therefore, 411= 3*137 is one of the oddnumbered terms, and 411 is in the sequence. Condition 1) is sufficient. a 2 = 112 = 3*37 + 1 and all evennumbered terms have a remainder of 1 when they are divided by 3. Since 411 = 3*137, it is not an evennumbered term. Since we don’t know any of the oddnumbered terms, condition 2) is not sufficient. Therefore, A is the answer. Answer: A
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04 Nov 2018, 19:31
[ Math Revolution GMAT math practice question] When a positive integer n is divided by 19, what is the remainder? 1) n17 is a multiple of 19 2) n19 is a multiple of 17 => 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 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on its own first. Condition 1) n – 17 = 19*k for some integer k. So, n = 19*k + 17, which means that n has a remainder of 17 when it is divided by 19. Condition 1) is sufficient. Condition 2) Now, n – 19 = 17*m or n = 17*m + 19. When m = 1, n = 36, and so n = 19*1 + 17 has a remainder of 17 when it is divided by 19. When m = 2, n = 53, and so n = 19*2 + 15 has a remainder of 15 when it is divided by 19. Since we don’t have a unique remainder, condition 2) 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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06 Nov 2018, 00:51
[ Math Revolution GMAT math practice question] A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 children with no pencils left over. What is the smallest possible number of pencils in the set? A. 6 B. 12 C. 15 D. 30 E. 60 => The smallest possible number of pencils must be the least common multiple of 2, 3, 4, 5 and 6. 2 = 2^1, 3 = 3^1, 4 = 2^2, 5 = 5^1 and 6 = 2^1*3^1. We multiply the maximum powers of each base to obtain the least common multiple 2^2*3^1*5^1 = 60. Therefore, the answer is E. Answer: E
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08 Nov 2018, 19:12
[ Math Revolution GMAT math practice question] If 1/3 = 1/m + 1/n, where m and n are different positive integers, what is the value of m+n? A. 9 B. 12 C. 16 D. 18 E. 20 => 1/3 = 1/m + 1/n => mn = 3n + 3m => mn – 3m – 3n = 0 => mn – 3m – 3n + 9 = 9 => (m – 3)(n – 3) = 9 The possible pairs of values (m3, n3) giving a product of 9 are (1,9), (9,1) and (3,3). Case 1: m – 3 = 1, n – 3 = 9. We have m = 4, n = 12 and m + n = 16 Case 2: m – 3 = 9, n – 3 = 1. We have m = 12, n = 4 and m + n = 16. Case 3: m – 3 = 3, n – 3 = 3 We have m = 6, n = 6. Since m = n, this case does not satisfy the original condition. Thus, m + n = 16. Therefore, the answer is C. Answer: C
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11 Nov 2018, 22:10
[ Math Revolution GMAT math practice question] What is the value of A? 1) The fourdigit number A77A is a multiple of 4. 2) The fourdigit number A77A is a multiple of 9. => 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 1 variable (A) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first. Condition 1) If the last two digit is a multiple of 4, the original number is a multiple of 4. Thus, A is 2 or 6 since 72 and 76 are multiples of 4. Condition 1) is not sufficient, because we don’t have a unique answer Condition 2) If the sum of all digits is a multiple of 9, the original number is a multiple of 9. A + 7 + 7 + A = 2A + 14 If 2A + 14 = 18, we have 2A = 4 or A = 2. We don’t have any digit integer A such that 2A = 9, 27 or 36. Thus A = 2 is the unique integer. Condition 2) is sufficient. Answer: C 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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12 Nov 2018, 19:28
[ Math Revolution GMAT math practice question] What is the units digit of a positive integer n? 1) n is a common multiple of 13 and 14. 2) n is a common multiple of 13 and 15. => 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 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first. Condition 1) Since n is a common multiple of 13 and 14, n is a multiple of lcm(13,14) = 182. Thus units digits of multiples of n are 0, 2, 4, 6, 8 since the units digit of 182 is 2. Condition 1) is not sufficient. Condition 2) Since n is a common multiple of 13 and 15, n is a multiple of lcm(13,15) = 195. Thus units digits of multiples of n are 0, 5 since the units digit of 182 is 2. Condition 2) is not sufficient. Conditions 1) & 2) The common units digit of multiples of 182 and 195 is 0 only. Both conditions together are sufficient. 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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14 Nov 2018, 19:22
[ Math Revolution GMAT math practice question] [x] is the greatest integer less than or equal x, what is the value of [√30]+[√40]+[√50]? A. 12 B. 15 C. 16 D. 18 E. 20 => Since 5 = √25 < √30 < √36 = 6, we have [√30] = 5. Since 6 = √36 < √40 < √49 = 7, we have [√40] = 6 Since 7 = √49 < √50 < √64 = 8, we have [√50] = 7 Thus [√30]+[√40]+[√50] = 18. Therefore, D is the answer. Answer: D
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16 Nov 2018, 02:05
[ Math Revolution GMAT math practice question] Which of the following are roots of an equation 10x^{2}+x^{1}21=0 A. 2/3 or 5/7 B. 2/3 or 7/5 C. 2/3 or 5/7 D. 2/3 or 7/5 E. 3/2 or 5/7 => 10x^{2}+x^{1}21=0 => 10+x21x^2=0 by multiplying x^2 => 21x^2x10=0 by multiplying (1) => (3x+2)(7x5)=0 => x = 2/3 or x = 5/7 Therefore, the answer is A. Answer: A
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT
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18 Nov 2018, 17:19
[ Math Revolution GMAT math practice question] When m and n are positive integers, is m!*n! an integer squared? 1) m = n + 1 2) m is an integer squared => 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) By condition 1), m!*n! = (n+1)!*n! = (n+1)(n!)(n!) = (n+1)(n!)^2 = m(n!)^2. Since (n!)^2 is an integer squared, and m is an integer squared by condition 2), m!*n! is an integer squared. Thus, both conditions together 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) If m = 4 and n = 3, then 4!*3! = 4*3!*3! = (2(3!))^2 and the answer is ‘yes’. If m = 3 and n = 2, then 3!*2! = 6*2 = 12, and the answer is ‘no’. Since it does not give a unique answer, condition 1) is not sufficient on its own. Condition 2) If m = 4 and n = 3, then 4!*3! = 4*3!*3! = (2(3!))^2 and the answer is ‘yes’. If m = 4 and n = 2, then 4!*2! = 24*2 = 48 and the answer is ‘no’. Since it does not give a unique answer, condition 2) is not sufficient on its own. Therefore, the answer is C. 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: Overview of GMAT Math Question Types and Patterns on the GMAT
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18 Nov 2018, 22:41
Hello.. Wanted to know if the same statistic is followed till now i.e November 2018 pr is there any revision in the topic and number of questions as the pattern has changed a bit with less number of questions.



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21 Nov 2018, 17:13
[ Math Revolution GMAT math practice question] What is the area of a regular octagon with sidelength 2? A. 4 + 4 √2. B. 8 + 8 √2. C. 16 + 16 √2. D. 16 E. 32 => Attachment:
11.21.png [ 11.62 KiB  Viewed 167 times ]
The above figure shows that the octagon can be divided up into one square of sidelength 2, 4 rectangles of sidelengths 2 and √2, and 4 rightangled isosceles triangles with equal sides of length √2. Thus its area is the sum of the areas of the square (2*2), 4 rectangles (2*√2) and 4 triangles ((1/2) (√2)( √2)). The area of the octagon is 2^2 + 4*2 √2 + 4*1 = 8 + 8 √2. Therefore, the answer is B. Answer: B
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT
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21 Nov 2018, 17:18
This picture is not that of an octagon
Posted from my mobile device



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25 Nov 2018, 17:40
[ Math Revolution GMAT math practice question] If p, q and r are prime, with p<q<r, p=? 1) (pq)^3=216 2) (pr)^3=1000 => 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 (p, q and r) 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) (pq)^3=216 => p^3q^3=2^33^3 => p = 2 and q = 3, since p and q are prime numbers with p < q. (pr)^3=1000 => p^3r^3=2^35^3 => p = 2 and r = 5, since p and r are prime numbers with p < r. While we have checked both conditions together, we have shown that conditions 1) and 2) are equivalent to each other in terms of p. So, each condition is sufficient by Tip 1). FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information. Therefore, the answer is D. Answer: D 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: Overview of GMAT Math Question Types and Patterns on the GMAT
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27 Nov 2018, 17:11
[ Math Revolution GMAT math practice question] A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer. What is the value of a + b? A. 5 B. 6 C. 7 D. 8 E. 9 => Since a234b is a multiple of 72, a234b is a multiple of both 8 and 9. The last three digits 34b of a234b form a multiple of 8. So, we must have b = 4 since 344 is the only 3digit multiple of 8 beginning with the digits, 34. Since a234b is a multiple of 9, a + 2 + 3 + 4 + b = a + 2 + 3 + 4 + 4 = a + 13 is a multiple of 9. This implies that a = 5, and a + b = 9. Therefore, E is the answer. Answer: E
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT
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29 Nov 2018, 17:33
[ Math Revolution GMAT math practice question] What is the sum of the prime factors of 2^81? A. 8 B. 16 C. 19 D. 22 E. 25 => 2^81 = (2^4+1)(2^41) = (2^4+1)(2^2+1)(2^21) = (2^4+1)(2^2+1)(2+1)(21) = 17*5*3*1. Thus, the sum of prime factors of 2^81 is 17 + 5 + 3 = 25. Therefore, the answer is E. Answer: E
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