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Math Revolution DS Expert  Ask Me Anything about GMAT DS
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27 Feb 2019, 05:43
MathRevolution wrote: n is a positive integer. What is the remainder when n is divided by 3?
1) n^2 has remainder 1 when it is divided by 3 2) n has remainder 7 when it is divided by 9 We need to find r when n=3q+r Statement 1: \(n^2\)=3q+1 By plugging in numbers, \(n^2\)= 1, 7, 10, 13, 16....25... => n= 1, \(\sqrt{7}\), \(\sqrt{10}\), \(\sqrt{13}\), 4.....5.... Plug in any value of n in n=3q+r. We get more than one value for r. NOT SUFFICIENT Statement 2: n=9q+7 n= 7, 25, 34, 43.......... By plugging in any value of n in the eq. n=3q+r, you'll get r=1. SUFFICIENT IMO the answer is option B.



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28 Feb 2019, 07:17
MathRevolution wrote: If n is an integer, is (n+1)(n+2)(n+3) divisible by 12?
1) n is an even number. 2) n is a multiple of 4.
Answer: B 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. Since n+1, n+2 and n+3 are three consecutive integers, (n+1)(n+2)(n+3) is a multiple of 3. Condition 2) tells us that n+1 and n+3 are odd integers, and n+2 is an even number which is not a multiple of 4. Thus, (n+1)(n+2)(n+3) is not a multiple of 4. CMT(Common Mistake Type 1) states “no” is also an answer and a condition giving rise to the unique answer “no” is sufficient. Thus, condition 2) is sufficient. Condition 1) If n = 2, then (n+1)(n+2)(n+3) = 3*4*5 = 60 is a multiple of 12 and the answer is “yes”. If n = 4, then (n+1)(n+2)(n+3) = 5*6*7 = 210 is not a multiple of 12 and the answer is “no”. Thus, condition 1) is not sufficient, since it does not yield a unique solution. Therefore, B is the answer.
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28 Feb 2019, 07:23
Is x = y? 1) x ≤ y 2) x ≥ y Answer: E
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01 Mar 2019, 00:55
MathRevolution wrote: n is a positive integer. What is the remainder when n is divided by 3?
1) n^2 has remainder 1 when it is divided by 3 2) n has remainder 7 when it is divided by 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 (\(n\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) \(n\) could be any of the integers \(1, 2, 4, 5, 7, 8, …\) If \(n\) is one of \(1, 4, 7\), then \(n\) has a remainder \(1\) when it is divided by \(3\). If \(n\) is one of \(2, 5, 8\), then \(n\) has a remainder \(2\) when it is divided by \(3\). Thus, condition 1) is not sufficient, since it does not yield a unique solution. Condition 2) \(n = 9k +7\) can be expressed as \(n = 9k + 7 = 9k + 6 + 1 = 3(3k+2)+1\). Therefore, \(n\) has remainder \(1\) when it is divided by \(3\). Thus, condition 2) is sufficient. Therefore, B is the answer. Answer: B
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01 Mar 2019, 00:56
[GMAT math practice question] (number properties) If \(m\) and \(n\) are prime numbers, is \(m^2 + n^2\) an even number? \(1) m > 10\) \(2) n > 20\)
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01 Mar 2019, 06:53
MathRevolution wrote: [GMAT math practice question]
(number properties) If \(m\) and \(n\) are prime numbers, is \(m^2 + n^2\) an even number?
\(1) m > 10\) \(2) n > 20\) It is given that both m and n are prime numbers. We need to find whether \(m^2 + n^2\) an even number or not. To get an even number, both m and n must be either odd or even. Statement 1: m>10 Prime numbers greater than 10 is always odd. By plugging in numbers.... If m=11 and n=13, \(m^2 + n^2\) is even.....yes But, if m=11 and n=2, \(m^2 + n^2\) is odd.....no Different answers. Insufficient. Statement 2: n>20 Prime numbers greater than 20 is always odd. By plugging in numbers.... If m=11 and n=23, \(m^2 + n^2\) is even.....yes But, if m=2 and n=23, \(m^2 + n^2\) is odd.....no Different answers. Insufficient. Combined 1 & 2: m>10 & n>20 All prime numbers in this range is odd. So, \((odd)^2\)+\((odd)^2\) will ALWAYS be even. Hence, sufficient. IMO answer is option C.



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02 Mar 2019, 04:33
MathRevolution wrote: If n is an integer, is (n+1)(n+2)(n+3) divisible by 12?
1) n is an even number. 2) n is a multiple of 4.
Answer: B #1 for n = 2 ; we get yes sufficient n=4 . we get not sufficient #2 n=4,8,12 for all such values of n . the expression is not divisible by 12 sufficinet IMO B



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02 Mar 2019, 04:40
MathRevolution wrote: n is a positive integer. What is the remainder when n is divided by 3?
1) n^2 has remainder 1 when it is divided by 3 2) n has remainder 7 when it is divided by 9 #1 n=2,5,8 we get remainder as 2 when divided by 3 and n=4,7. we get remainder as 1 when divided by 3 in sufficient #2 n= 11,34 remainder as 1 sufficient IMO B



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02 Mar 2019, 04:48
MathRevolution wrote: Is x = y?
1) x ≤ y 2) x ≥ y
Answer: E #1 y=1, x=1 yes sufficeint y=2 ,x=1, in sufficient #2 x & y can be of same or opposite signs in sufficient IMO E



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02 Mar 2019, 04:51
MathRevolution wrote: [GMAT math practice question]
(number properties) If \(m\) and \(n\) are prime numbers, is \(m^2 + n^2\) an even number?
\(1) m > 10\) \(2) n > 20\) #1 m=11,13,17.. n=2,3,5,7,11.. so m^2+n^2 = even and odd both as n value not given in sufficeint #2 same as #1 value of m not sure . in sufficient as at m=2 eqn would be odd and else even from 1 & 2 m>10 and n>20 we have odd integers so addition of odd + odd= even ; sufficient IMO C



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03 Mar 2019, 18:31
MathRevolution wrote: Is x = y?
1) x ≤ y 2) x ≥ y
Answer: E => 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) If \(x =1\), and \(y = 1\), then \(x\) and \(y\) satisfy both conditions, and the answer is “yes” since \(x = y\). If \(x = 2\), and \(y = 1\), then \(x\) and \(y\) satisfy both conditions, but the answer is “no” since \(x ≠ y\). Thus, both conditions together are not sufficient, since they do not yield a unique solution. Therefore, E is the answer. Answer: E
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03 Mar 2019, 18:34
MathRevolution wrote: [GMAT math practice question]
(number properties) If \(m\) and \(n\) are prime numbers, is \(m^2 + n^2\) an even number?
\(1) m > 10\) \(2) n > 20\) => 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. In order for \(m^2 + n^2\) to be even, either both \(m\) and \(n\) must be even or both \(m\) and \(n\) must be odd. Since \(m\) and \(n\) are primes, we must have both \(m\) and \(n\) odd numbers as they are both greater than \(10\). Thus, C is the answer. 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 \(m = 11\) and \(n = 3\), then \(m^2+n^2 = 11^2+3^2 = 121 + 9 = 130\) is an even number, and the answer is “yes”. If \(m = 11\) and \(n = 2\), then \(m^2+n^2 = 11^2+2^2 = 121 + 4 = 125\) is an odd number, and the answer is “no”. Thus, condition 1) is not sufficient since it does not yield a unique solution. Condition 2) If \(m = 3\) and \(n = 23\), then \(m^2+n^2 = 3^2+23^2 = 9 + 529 = 538\) is an even number, and the answer is “yes”. If \(m = 2\) and \(n = 23\), then \(m^2+n^2 = 2^2+23^2 = 4 + 529 = 533\) is an odd number, and the answer is “no”. Thus, 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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04 Mar 2019, 02:01
[GMAT math practice question] (number properties) \(k\) is a positive integer. When \(2^{\frac{k}{2}} ≤ \frac{1}{1},024, k=?\) 1) \(k\) is an integer less than \(21\) 2) \(k\) is an integer greater than \(12\)
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05 Mar 2019, 01:16
[GMAT math practice question] (number properties) When \(32\) is divided by \(k\), the remainder is \(k3\). What is the value of \(k\)? \(1) k>20\) \(2) k<40\)
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05 Mar 2019, 21:15
MathRevolution wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] (number property) If \(n\) is an integer greater than \(1\), what is the value of \(n\)? 1) \(n\) is a prime number 2) \(\frac{(n+2)}{n}\) is an integer => 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 there are many prime numbers, condition 1) is not sufficient. Condition 2) If \(n = 1\), then \(\frac{(n+2)}{n} = 3\) is an integer. If \(n = 2,\) then \(\frac{(n+2)}{n} = 2\) is an integer. Since we don’t have a unique solution, condition 2) is not sufficient. Conditions 1) & 2) If \(n = 2\), then \(\frac{(n+2)}{n} = 2\) is an integer. If \(n = 3\), then \(\frac{(n+2)}{n} = \frac{5}{2}\) is not integer. If \(n\) is a prime number bigger than \(2\), \(\frac{(n+2)}{n}\) is not an integer. Thus \(n = 2\) is the unique solution and both conditions together are sufficient. 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. The answer should B as it is already mentioned in the question that n is greater than 1 (n>1)



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06 Mar 2019, 01:59
MathRevolution wrote: [GMAT math practice question]
(number properties) \(k\) is a positive integer. When \(2^{\frac{k}{2}} ≤ \frac{1}{1},024, k=?\)
1) \(k\) is an integer less than \(21\) 2) \(k\) is an integer greater than \(12\) => 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. \(2^{\frac{k}{2}}≤\frac{1}{1,024}\) \(=> 2^{\frac{k}{2}}≥1,024\) \(=> 2^{\frac{k}{2}}≥1,024=2^{10}\) \(=> \frac{k}{2} ≥ 10\) \(=> k ≥ 20\) We have the simplified original condition \(k ≥ 20\). Condition 1) “\(k < 21\)” yields the unique solution \(k = 20\), since \(k\) is an integer and \(k ≥ 20\). Condition 1) is sufficient. Condition 2) We have \(k > 12\) from condition 2) and \(k ≥ 20\) from the original condition. The integer \(k\) has many possible values. Since it does not yield a unique solution, condition 2) is not sufficient. Therefore, A is the answer. Answer: A
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06 Mar 2019, 02:00
[GMAT math practice question] (geometry) Is a triangle, with one side of length \(12\), inscribed in a circle a right triangle? 1) The area of the circle is \(36π\). 2) The circumference of the circle is \(12π\).
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07 Mar 2019, 01:13
MathRevolution wrote: [GMAT math practice question] (number properties) When \(32\) is divided by \(k\), the remainder is \(k3\). What is the value of \(k\)?
\(1) k>20\) \(2) k<40\) => 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 original condition says that \(32 = k*q + ( k – 3 )\) or \(35 = k*q + k = k(q+1).\) That is, \(k\) is a factor of \(35\). So, \(k = 1, 5, 7\) or \(35.\) Since condition 1) “\(k>20\)” yields the unique solution “\(k=35\)”, condition 1) is sufficient. Condition 2) yields \(k = 1, 5, 7\) or \(35\). Since it does not yield a unique solution, condition 2) is not sufficient. Therefore, A is the answer. Answer: A
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07 Mar 2019, 01:14
[GMAT math practice question] (statistics) Is the average (arithmetic mean) of \(x, y,\) and \(z\) equal to their median? 1) \(y\) is the average (arithmetic mean) of \(x, y,\) and \(z\) 2) \(z = 2y  x\)
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08 Mar 2019, 02:36
MathRevolution wrote: [GMAT math practice question]
(geometry) Is a triangle, with one side of length \(12\), inscribed in a circle a right triangle?
1) The area of the circle is \(36π\). 2) The circumference of the circle is \(12π\). => 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. Attachment:
3.8.png [ 7.42 KiB  Viewed 158 times ]
If a side of a triangle is the diameter of its circumscribed circle, then the triangle is a right triangle. Condition 1) A circle with area \(36π\) has radius \(6\) and diameter \(12\). So, the answer is ‘yes’ and condition 1) is sufficient. Condition 2) A circle with circumference \(12π\) has diameter \(12\). So, the answer is ‘yes’ and condition 2) is sufficient. Therefore, D is the answer. Answer: D Note that if conditions 1) and 2) yield the same information, Tip 1 of the VA method tells us that D is most likely to be the answer.
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