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Math Revolution GMAT Instructor
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03 Jul 2019, 00:52
MathRevolution wrote: [GMAT math practice question]
(number properties) If \(k, m\) and \(n\) are positive integers, what is the value of \(kmn\)?
\(1) k^4mn = 240\)
\(2) m = 3\) => 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. 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) If \(k = 2, m = 3\) and \(n = 5,\) then \(kmn = 30.\) If \(k = 1, m = 3\) and \(n = 80\), then \(kmn = 240.\) Conditions 1) & 2) together are not sufficient, when applied together, since they don’t 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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03 Jul 2019, 00:53
[GMAT math practice question] (statistics) \(k, m\) and \(n\) are positive integers. Is their average equal to their median? 1) The median of \(k, m\) and \(n\) is \(11\). 2) The range of \(k, m\) and \(n\) is \(13\)
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04 Jul 2019, 01:08
MathRevolution wrote: [GMAT math practice question]
(number properties) \(p\) and \(q\) are different positive integers. What is the remainder when \(p^2 + q^2\) is divided by \(4\)?
1) \(p\) and \(q\) are prime numbers.
2) \(p\) and \(q\) are not consecutive integers. => 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 (\(p\) and \(q\)) 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 \(p\) and \(q\) are prime numbers, which are not consecutive integers, \(p\) and \(q\) are odd integers. So, both \(p^2\) and \(q^2\) have remainder \(1\) when they are divided by \(4\). Thus, \(p^2 + q^2\) has remainder \(2\) when it is divided by \(4\). Since conditions 1) & 2) yield a unique solution, when they are applied together, they are sufficient. 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 \(p = 2\) and \(q = 3\), then \(p^2 + q^2 = 4 + 9 = 13\), which has remainder \(1\) when it is divided by \(4\). If \(p = 3\) and \(q = 5\), then \(p^2 + q^2 = 9 + 25 = 34\), which has remainder \(2\) when it is divided by \(4\). Condition 1) is not sufficient since it doesn’t yield a unique solution. Condition 2) If \(p = 3\) and \(q = 5\), then \(p^2 + q^2 = 9 + 25 = 34\), which has remainder \(2\) when it is divided by \(4\). If \(p = 3\) and \(q = 6\), then \(p^2 + q^2 = 9 + 36 = 45\), which has remainder \(1\) when it is divided by \(4\). Condition 2) is not sufficient since it doesn’t 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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04 Jul 2019, 01:09
[GMAT math practice question] (number properties) \(m\) and \(n\) are positive integers. Is \(m^2 + n^2\) is divisible by \(3\)? \(1) m = 1234\) \(2) n = 4321\)
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05 Jul 2019, 00:49
MathRevolution wrote: [GMAT math practice question]
(statistics) \(k, m\) and \(n\) are positive integers. Is their average equal to their median?
1) The median of \(k, m\) and \(n\) is \(11\).
2) The range of \(k, m\) and \(n\) is \(13\) => 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. 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 the conditions, if necessary. Without loss of generality, we may assume \(k ≤ m ≤ n.\) If their average and their median are equal, then \(\frac{( k + m + n )}{3} = m\) or \(k + m + n = 3m\), so \(n + k = 2m\) and \(nk=2m2k\). So, the range of the numbers must be even. Thus, condition 2) yields the unique answer ‘no’, and is sufficient by CMT 1). Condition 1) If \(k = 10, m = 11\) and \(n = 12\), then their average and median are equal, and the answer is ‘yes’. If \(k = 10, m = 11\) and \(n = 13,\) then their average and their median are not equal, and the answer is ‘no’. Condition 1) is not sufficient since it doesn’t yield a unique solution. Therefore, B is the answer. Answer: B
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05 Jul 2019, 00:49
[GMAT math practice question] (absolute value) Is \(x > 1\)? \(1) x^2 > xx\) \(2) xx > x\)
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06 Jul 2019, 05:15
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?
(1) The vegetarians attended the party at a rate of 2 students to every 3 non–students, half the rate for non–vegetarians.
(2) 30% of the guests were vegetarian non–students.



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06 Jul 2019, 05:16
55 people live in an apartment complex with three fitness clubs (A, B, and C). Of the 55 residents, 40 residents are members of exactly one of the three fitness clubs in the complex. Are any of the 55 residents members of both fitness clubs A and C but not members of fitness club B?
(1) 2 of the 55 residents are members of all three of the fitness clubs in the apartment complex.
(2) 8 of the 55 residents are members of fitness club B and exactly one other fitness club in the apartment complex.



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06 Jul 2019, 07:10
the1aspirant wrote: Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?
(1) The vegetarians attended the party at a rate of 2 students to every 3 non–students, half the rate for non–vegetarians.
(2) 30% of the guests were vegetarian non–students. Discussed here: https://gmatclub.com/forum/guestsata ... 04547.htmlPlease follow our rules when posting a question: https://gmatclub.com/forum/rulesforpo ... 33935.html
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06 Jul 2019, 07:11
the1aspirant wrote: 55 people live in an apartment complex with three fitness clubs (A, B, and C). Of the 55 residents, 40 residents are members of exactly one of the three fitness clubs in the complex. Are any of the 55 residents members of both fitness clubs A and C but not members of fitness club B?
(1) 2 of the 55 residents are members of all three of the fitness clubs in the apartment complex.
(2) 8 of the 55 residents are members of fitness club B and exactly one other fitness club in the apartment complex. Discussed here: https://gmatclub.com/forum/55peopleli ... 50894.htmlPlease follow our rules when posting a question: https://gmatclub.com/forum/rulesforpo ... 33935.html
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07 Jul 2019, 18:21
MathRevolution wrote: [GMAT math practice question]
(number properties) \(m\) and \(n\) are positive integers. Is \(m^2 + n^2\) is divisible by \(3\)?
\(1) m = 1234\)
\(2) n = 4321\) => 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. We should simplify the conditions if necessary. Recall that the remainder when an integer is divided by \(3\) is the same as the remainder when the sum of all of its digits is divided by \(3\). The square of an integer \(k\) has remainder \(0\) or \(1\) when it is divided by \(3\). If \(k\) is a multiple of \(3\), then \(k = 3a\) for some integer \(a\), and \(k^2 = (3a)^2 = 3(3a^2)\) has remainder \(0\) when it is divided by \(3\). If \(k\) has remainder \(1\) when it is divided by \(3\), then \(k = 3a + 1\) for some integer \(a\), and \(k^2 = (3a+1)^2 = 9a^2 +6a + 1 = 3(3a^2 +2a) + 1\) has remainder \(1\) when it is divided by \(3\). If \(k\) has remainder \(2\) when it is divided by \(3\), then \(k = 3a + 2\) for some integer \(a\), and \(k^2 = (3a+2)^2 = 9a^2 +12a + 4 = 3(3a^2 +4a+1) + 1\) has remainder \(1\) when it is divided by \(3\). Since \(m=1234\) has remainder \(1\) when it is divided by \(3\), \(m^2\) has remainder \(1\) when it is divided by \(3.\) Since \(n^2\) could have remainder \(0\) or \(1\) when it is divided by \(3, m^2 + n^2\) is never divisible by \(3\), regardless of the value of \(n\). Condition 1) is sufficient, since it yields the unique answer, ‘no’. Since \(n=4321\) has remainder \(1\) when it is divided by \(3, n^2\) has remainder \(1\) when it is divided by \(3\). Since \(m^2\) could have remainder \(0\) or \(1\) when it is divided by \(3, m^2 + n^2\) is never divisible by \(3\), regardless of the value of \(m\). Condition 2) is sufficient, since it yields the unique answer, ‘no’. Therefore, D is the answer. Answer: D Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient.
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07 Jul 2019, 18:23
MathRevolution wrote: [GMAT math practice question]
(absolute value) Is \(x > 1\)?
\(1) x^2 > xx\)
\(2) xx > x\) => 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. We should simplify the conditions, if necessary. Condition 1) \(x^2 > xx\) \(=> x^2 > xx\) \(=> x^2  xx > 0\) \(=> x(x  x) > 0\) \(=> (x  x) > 0\) \(=> x > x\) \(=> x < 0\) Thus, condition 1) is sufficient since it yields the unique answer, ‘no’. Condition 2) \(xx > x\) Assume \(x ≥ 0. xx > x ⇔ x^2 > x ⇔ x^2 – x > 0 ⇔ x(x1) > 0 ⇔ x < 0\) or \(x > 1.\) We must have \(x > 1\) by the assumption. Assume \(x < 0. xx > x ⇔ x^2 > x ⇔ x^2 + x < 0 ⇔ x(x+1) < 0 ⇔ 1 < x < 0.\) In this case, \(1 < 0 < 1.\) So, condition 2) tells us that \(1 < x < 0\) or \(x > 1.\) In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient The solution set of the question “\(x > 1\)” doesn’t include the solution set of the condition 2) “\(1 < x < 0\) or \(x > 1\)”. Condition 2) is not sufficient. Therefore, A is the answer. Answer: A
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08 Jul 2019, 00:36
[GMAT math practice question] (number properties) \(x, y\) are two different positive integers greater than \(6\). What are \(x\) and \(y\)? \(1) x * y =1008\) 2) the greatest common divisor of \(x\) and \(y\) is \(6\)
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08 Jul 2019, 03:34
MathRevolution wrote: [GMAT math practice question]
(number properties) \(x, y\) are two different positive integers greater than \(6\). What are \(x\) and \(y\)?
\(1) x * y =1008\)
2) the greatest common divisor of \(x\) and \(y\) is \(6\) Combining 1 and 2 yields more than one combination of numbers for x and y i.e. 24,42; 18,56; 12,84. Hence both statements are clearly insufficient. Answer is E. Posted from my mobile device



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09 Jul 2019, 00:39
[GMAT math practice question] (algebra) \(x, y, z\) are \(3\) consecutive positive integers. What is the value of \(x\)? \(1) x<y<z\) \(2) 12+13+14+15=x+y+z\)
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Math Revolution DS Expert  Ask Me Anything about GMAT DS
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09 Jul 2019, 01:06
MathRevolution wrote: [GMAT math practice question]
(algebra) \(x, y, z\) are \(3\) consecutive positive integers. What is the value of \(x\)?
\(1) x<y<z\)
\(2) 12+13+14+15=x+y+z\) Combining statement 1 and 2: 12+13+14+15=x+y+y=54 Mean of x,y,z = midian of x,y,z =54/3 = 18 Hence the consecutive numbers are 17,18,19. From statement 1, x<y<z. Hence x is 17. Statement 1 only gives the order of the numbers that x is the first number. Hence 1 on its own is insufficient. Statement 2 helps us to get the midian which in turn facilitated getting the other numbers. Statement 2 does not give any idea about the order of the numbers, neither does the preamble of the question. Hence x can be first number, second number or third number. Statement 2 on its own is also insufficient. The correct answer is C. Posted from my mobile device



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09 Jul 2019, 06:52
MathRevolution wrote: [GMAT math practice question]
(algebra) \(x, y, z\) are \(3\) consecutive positive integers. What is the value of \(x\)?
\(1) x<y<z\)
\(2) 12+13+14+15=x+y+z\) Approach 1: Clearly individual statement are insufficients. Now combining both the statements : Sum= 12+13+14+15 = 54. We have to make sum of three numbers 54 that is even ..So possible solution will be odd+even + odd..Note we cant have first term as even ..otherwise even+odd+even will be equal to odd but we need sum as even.(54) Now lets try sum of three consecutive odd term to end in 4. 5+6+7=18 7+8+9=24.(ends in 4) Now 17+18+19= 54 . Approach 2:For consecutive terms Mean=Median so Y will be the average of terms and the median of term Therefore since we know sum is 54 so Mean= median = 54/3 = 18. Now the terms cane be easily found as 17,18,19 Press Kudos if it helps!!



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10 Jul 2019, 00:29
MathRevolution wrote: [GMAT math practice question]
(number properties) \(x, y\) are two different positive integers greater than \(6\). What are \(x\) and \(y\)?
\(1) x * y =1008\)
2) the greatest common divisor of \(x\) and \(y\) is \(6\) => 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) Assume \(x = 6a\) and \(y = 6b\) where \(a\) and \(b\) are relatively prime. We have \(x*y = (6a)*(6b)= 1008\) or \(ab = 28\). Then we have four pairs of \((a,b)\) which are \((1,28), (4,7), (7,4)\) and \((28,1).\) Thus we have four pairs of \((x,y)\) which are \((6, 168), (24, 42), (42, 24)\) AND \((6,168)\) and we don’t have a unique answer. 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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10 Jul 2019, 00:30
[GMAT math practice question] (inequality) Five consecutive integers satisfies \(a<b<c<d<e.\) what is the maximum of \(a+e\)? 1) the summation of five integers is negative 2) \(e\) is positive
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11 Jul 2019, 01:07
MathRevolution wrote: [GMAT math practice question]
(algebra) \(x, y, z\) are \(3\) consecutive positive integers. What is the value of \(x\)?
\(1) x<y<z\)
\(2) 12+13+14+15=x+y+z\) => 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. Consecutive integers have two variables for the first number and the number of integers. Since the number of integers is \(3\), we need one more equation and D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) If \(x = 1, y = 2\) and \(z = 3,\) then we have \(x = 1\). If \(x = 2, y = 3\) and \(z = 4\), then we have \(x = 2.\) Since condition 1) doesn’t yield a unique solution, it is not sufficient. Condition 2) Since \(12 + 13 + 14 + 15 = x + y + z,\) we have \(x + y + z = 54.\) If \(x = 17, y = 18, z = 19,\) then we have \(x = 17.\) If \(x = 19, y = 18, z = 17,\) then we have \(x = 19.\) Since condition 2) doesn’t yield a unique solution, it is not sufficient. Conditions 1) & 2) Since \(y = x + 1\) and \(z = x + 2\) from condition 1), we have \(12 + 13 + 14 + 15 = 54 = x + y + z = x + x + 1 + x + 2 = 3x + 3 or 3x = 51.\) Thus we have \(x = 17.\) Since both conditions together yield a unique solution, they are sufficient. 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. Therefore, C is the answer. 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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