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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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30 Oct 2018, 18:06
MathRevolution wrote: [ Math Revolution GMAT math practice question] (algebra) \((x^23x+2)(y^25y+6)=?\) \(1) x=1\) \(2) y=1\) From statement 1: If x = 1. Then the expression becomes zero. Sufficient. From statement 2: If y = 1. then expression reduces to 2x^26x+4 or (x2)(2x+2). Since the RHS side is unknown. It's insufficient. A is the answer.



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31 Oct 2018, 01:34
MathRevolution wrote: [ Math Revolution GMAT math practice question] (algebra) \((x^23x+2)(y^25y+6)=?\) \(1) x=1\) \(2) y=1\) => 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. Factoring the question gives \((x^23x+2)(y^25y+6) = (x1)(x2)(y2)(y3).\) Condition 1) If \(x = 1\), then \((x1)(x2)(y2)(y3) = 0.\) Condition 1) is sufficient. Condition 2) If \(y = 1\), then \((x1)(x2)(12)(13) = 2(x1)(x2).\) Since we don’t know the value of \(x\), condition 2) is not sufficient. Note: Tip 4) of VA (Variable Approach) method states that if both conditions are trivial, they are not usually sufficient. Thus, conditions 1) & 2) are not sufficient by Tip 4). Therefore, A is the answer. Answer: A
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31 Oct 2018, 01:34
[ Math Revolution GMAT math practice question] (number property) If \(m\) and \(n\) are integers, is \(m+m^2n\) an even number? 1) \(m\) is an even number 2) \(n\) is an even number
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31 Oct 2018, 01:38
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number property) If \(m\) and \(n\) are integers, is \(m+m^2n\) an even number? 1) \(m\) is an even number 2) \(n\) is an even number \(m+m^2\) = m(m+1) = always even. So B is sufficient. B is the answer.



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31 Oct 2018, 04:36
MathRevolution wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] (inequality) Is \(1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1x)}\)? \(1) x>0\) \(2) x<1\) => 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 question \(1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1x)}\) is equivalent to \(0 < x < 1\) as shown below: For \(x ≠1\), =>\(1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1x)}\) \(=> (1+x+x^2+x^3+x^4+x^5+x^6)(1x)^2< (1x)\) \(=> (1  x^7)(1  x) < 1 – x\) \(=> 1  x^7 – x +x^8 < 1  x\) \(=>  x^7 + x^8 < 0\) \(=> x^7( x – 1 ) < 0\) \(=> x( x – 1 ) < 0\) \(=> 0 < x < 1\) Since both conditions must be applied together to obtain this inequality, both conditions 1) & 2) are sufficient, when applied together. Therefore, C is the answer. Answer: C Can you please explain, how did you reach to (1x^7)(1x)<1x step, I didn't get 1x^7 Regards



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31 Oct 2018, 05:00
R2S wrote: Afc0892 wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] (sequence) 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 From statement 1: \(a_1\) = 111. \(a_3 = a_1\) + 3 = 114 \(a_5 = a_3\) + 3 = 117. The series becomes 111, 114, 117.... the terms are 1, 3, 5, 7.... The common difference between the terms 1,3,5... is 3. Here the common difference is 2; 31=2; 53=2Let's consider 411 as the last term. then 411 = 111 +(n1)3. Then 411 becomes 101th term. Now the formula is, a = a1+(n1)*d 411 = 111+(n1)*2 411111=(n1)*2 300=(n1)*2 150=n1 151=n This shows that 411 is the 151st term in the sequence. Is it correct?Based on either, the answer choice will still remain the same i.e. A
A is sufficient. From statement 2: We get \(a_2\), \(a_4\) and \(a_6\) and so on.. But first term is unknown. B is sinsufficient. A is the answer. Thanks R2S. Missed it in urgency.



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31 Oct 2018, 05:16
Yes, it is pretty clear now Thanx Afc0892



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31 Oct 2018, 06:28
MathRevolution wrote: [ Math Revolution GMAT math practice question] (sequence) 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 Let me present the proof of the insufficiency of statement (2) alone, for the benefit of the mostrigorous students! \(S\,\,{\rm{sequence:}}\,\,\left\{ \matrix{ {{\rm{a}}_{\rm{1}}},{a_2}, \ldots \hfill \cr {a_n} = {a_{n  2}} + 3\,\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,n \ge 3 \hfill \cr} \right.\,\,\,\,\,\left( * \right)\) \(411\,\,\mathop \in \limits^? \,\,\,S\) \(\left( 2 \right)\,\,{a_2} = 112\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{a_4} = 112 + 3\,\,\,\, \Rightarrow \,\,\,{a_6} = 112 + 2 \cdot 3\,\,\, \Rightarrow \,\,\,\, \ldots\) \(411 \ne 112 + k \cdot 3\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,k\,\, \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{ \,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,{{\rm{a}}_{\rm{1}}}{\rm{ = 111}}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \,\rm{Take}\,\,{a_1} = 112\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\) Obs.: once a_1 and a_2 are given, the sequence S is uniquely defined. This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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31 Oct 2018, 23:33
MathRevolution wrote: [ Math Revolution GMAT math practice question] (sequence) 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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31 Oct 2018, 23:34
[ Math Revolution GMAT math practice question] (number property) 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 \(17\)
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01 Nov 2018, 22:26
MathRevolution wrote: Ask me Anything about GMAT Data SufficiencyHello, I am Max Lee, Founder and Lead Math Tutor at Math Revolution.
I have over 6,000 posts and almost 5,000 Kudos. This topic is a new feature on the DS Forum and a way for you to directly interact with me and ask anything about the DS, e.g. if you want a certain concept explained or have a particular you question you want me addressed, this is the place to post a link to it or your question. I intend to have this thread be as a "Everything You Need to Know about DS" type of thread. I will keep updating this post with links and resources that are helpful for the DS. Meanwhile, you can ask me anything
My other discussions you may be interested in:
Thank you all  good luck on the GMAT and look forward to seeing you in the DS forum!  Max Lee. Please review my logic to this question. Is z even? (1) 5z is even (2) 3z is even Solution (1) 5z is even z=2, 5z is even yes z=2/5 , 5z is even no Insufficient. (2) 3z is even z=2, 3z is even yes z=2/3 , 3z is even no Insufficient. Combining (1) and (2) since even + even =even Adding both (1) and (2) 5z+3z= even or 8z=even hence z could be odd or even. hence answer should be E



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02 Nov 2018, 00:29
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number property) If \(m\) and \(n\) are integers, is \(m+m^2n\) an even number? 1) \(m\) is an even number 2) \(n\) 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. \(m+m^2n = m(m+1) – n\). Now, m(m+1) is an even number since \(m(m+1)\) is the product of two consecutive integers. Thus, the parity of \(m+m^2n\) depends on the parity of \(n\) only. Thus, condition 2) is sufficient. Therefore, B is the answer. Answer: B
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02 Nov 2018, 00:29
[ Math Revolution GMAT math practice question] (statistics) If the average (arithmetic mean) of \(p, q\), and \(r\) is \(6\), what is the value of \(r\)? \(1) p=r\) \(2) p=q\)
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03 Nov 2018, 03:05
MathRevolution wrote: [ Math Revolution GMAT math practice question] (statistics) If the average (arithmetic mean) of \(p, q\), and \(r\) is \(6\), what is the value of \(r\)? \(1) p=r\) \(2) p=q\) The question states (p+q+r)/3 = 6 p+q+r = 6*3 If we know the value of p and q, we can find the value of 'r' Condition 1: p = r r+q+r = 18 (+ and  r will cancel out each other) q = 18 insufficient; 'cuz it provides value of q instead of r. Condition 2: p = q q+q+r = 18 r = 18 sufficient. The correct choice is B.



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04 Nov 2018, 19:05
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number property) 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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04 Nov 2018, 19:11
MathRevolution wrote: [ Math Revolution GMAT math practice question] (statistics) If the average (arithmetic mean) of \(p, q\), and \(r\) is \(6\), what is the value of \(r\)? \(1) p=r\) \(2) p=q\) => 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 asks for the value of \(r\), where \(r = 18 – ( p + q )\). If we know the value of \(p + q\), then we can determine the value or \(r\). Condition 2) implies that \(p + q = 0\). Therefore, \(r = 0 + r = ( p + q ) + r = 18,\) and condition 2) is sufficient. Condition 1) If \(p = 1, q =18\), and \(r = 1\), then \(p + q + r = 18\) and \(r = 1.\) If \(p = 2, q =18\), and \(r = 2\), then \(p + q + r = 18\) and \(r = 2.\) Since it doesn’t give a unique value of r, condition 1) is not sufficient. Therefore, B is the answer. Answer: B
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05 Nov 2018, 01:35
[ Math Revolution GMAT math practice question] (inequality) Is \(x^2>xy?\) \(1) x>y\) \(2) y>0\)
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05 Nov 2018, 22:53
R2S wrote: MathRevolution wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] (inequality) Is \(1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1x)}\)? \(1) x>0\) \(2) x<1\) => 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 question \(1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1x)}\) is equivalent to \(0 < x < 1\) as shown below: For \(x ≠1\), =>\(1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1x)}\) \(=> (1+x+x^2+x^3+x^4+x^5+x^6)(1x)^2< (1x)\) \(=> (1  x^7)(1  x) < 1 – x\) \(=> 1  x^7 – x +x^8 < 1  x\) \(=>  x^7 + x^8 < 0\) \(=> x^7( x – 1 ) < 0\) \(=> x( x – 1 ) < 0\) \(=> 0 < x < 1\) Since both conditions must be applied together to obtain this inequality, both conditions 1) & 2) are sufficient, when applied together. Therefore, C is the answer. Answer: C Can you please explain, how did you reach to (1x^7)(1x)<1x step, I didn't get 1x^7 Regards \((1+x+x^2+x^3+x^4+x^5+x^6)(1x)^2< (1x)\) \(⇔(1+x+x^2+x^3+x^4+x^5+x^6)(1x)(1x)< (1x)\) \(⇔(1+x+x^2+x^3+x^4+x^5+x^6xx^2x^3x^4x^5x^6x^7)(1x)< (1x)\) \(⇔(1x^7)(1x)< (1x)\)
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06 Nov 2018, 00:47
[ Math Revolution GMAT math practice question] (number property) 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 \(9\)
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07 Nov 2018, 00:46
MathRevolution wrote: [ Math Revolution GMAT math practice question] (inequality) Is \(x^2>xy?\) \(1) x>y\) \(2) y>0\) => 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. \(x^2>xy\) \(=> x^2xy > 0\) \(=> x(xy) > 0\) \(=> x>0, x>y\) or \(x<0, x<y\) 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 we have \(x > y\) and \(x > 0\) from \(x > y > 0\) which are a combined inequality of both conditions, both conditions together are sufficient. Since this question is an inequality 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) \(x > y\) If \(x = 2, y = 1\), then we have \(x^2>xy\) and the answer is ‘yes’. If \(x = 1, y = 2,\) then we have \(x^2<xy\) and the answer is ‘no’. Condition 2) \(y > 0\) If \(x = 2, y = 1\), then we have \(x^2>xy\) and the answer is ‘yes’. If \(x = 1, y = 2,\) then we have \(x^2<xy\) and the answer is ‘no’. 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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