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Math Revolution GMAT Instructor
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17 Jan 2019, 02:06
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number properties) If \(x, y\) are integers, is \(x^2+x+y\) an odd integer? 1) \(x\) is an odd integer 2) \(y\) is an odd 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. The parity of \(x^2+x+y = x(x+1) + y\) is same as the parity of \(y\), since \(x^2+x = x(x+1)\) is the product of two consecutive integers and so it is always an even integer. Thus, asking whether \(x^2+x+y = x(x+1) + y\) is odd is equivalent to asking whether \(y\) is odd. Therefore, B is the answer. Answer: B
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17 Jan 2019, 02:07
[ Math Revolution GMAT math practice question] (number properties) A sequence A n satisfies A n+3=A n+8. What is the remainder when A 99 is divided by 8? 1) A 1=1 2) A 2=2
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18 Jan 2019, 05:07
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number properties) If \(x\) and \(y\) are integers, is \(x^2y^2\) an even integer? 1) \(x^3y^3\) is an even integer 2) \(x+y\) is an even 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Asking whether \(x^2y^2\) is even is equivalent to asking whether \((x+y)(xy)\) is an even integer. Condition 2) is sufficient since \(x^2y^2 = (x+y)(xy)\) is an even integer if \(x + y\) is an even integer. Condition 1) In order for \(x^3y^3\) to be an even integer, \(x\) and \(y\) must both have the same parity. There are only two cases to consider: both \(x\) and \(y\) are even integers or both are odd integers. Since \(x\) and \(y\) have the same parity, \(x – y\) is always an even integer. Thus, \(x^2y^2\) is an even integer. Condition 1) is sufficient. Therefore, D is the answer. Answer: D
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18 Jan 2019, 05:38
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute value) Is \(ab<0\)? \(1) a+b =  ( a + b )\) \(2) a+b + 1 = a + b\) for ab<0 either of a or b has to be ve #1: a+b =  ( a + b ) LHS would always be +ve and RHS also has to be +ve for RHS to be +ve the values of ( a+b) has to be ve which would possible when both a & b are ve or when either of a & b are ve and the ve value is greater than +ve value in that case ab<0 wont be sufficeint #2: a+b + 1 = a + b[/m] this can be written as: ab+1 = a+b 1= 2 ( a+b) a+b= 1/2 this would be possible when either of values of a & b are either + & ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient combining 1 & 2 :  ( a + b )+1 = a + b or say 1/2 = a+b again this would be possible when either of values of a & b are either + & ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient IMO E



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18 Jan 2019, 15:09
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute value) Is \(ab<0\)? \(1) a+b =  ( a + b )\) \(2) a+b + 1 = a + b\) Excellent problem, Max. Congrats (and kudos)! \(ab\mathop {\,\, < }\limits^? \,\,0\) \(\left( 1 \right)\,\,\,\left {a + b} \right =  \left( {a + b} \right)\,\,\,\, \Leftrightarrow \,\,\,\,\,a + b \le 0\) \(\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {  2, 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,\left {a + b} \right + 1 = \left a \right + \left b \right\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,ab < 0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\) \(\left( * \right)\,\,ab \ge 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left {a + b} \right = \left a \right + \left b \right\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left {a + b} \right + 1 \ne \left a \right + \left b \right\,\,\,\,,\,\,\,{\rm{impossible}}\) We follow the notations and rationale taught in the GMATH method. Regards, Fabio.
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20 Jan 2019, 18:38
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number properties) A sequence A n satisfies A n+3=A n+8. What is the remainder when A 99 is divided by 8? 1) A 1=1 2) A 2=2 => 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 initial condition, A n+3=A n+8, tells us that every third term has the same remainder when it is divided by 8. By condition 1), A 1 = 1, A 4 = 9, A 7 = 17, … , A 97 = 257, A 100 = 265 …. Each of these terms has remainder 1 when it is divided by 8. By condition 2), A 2 = 2, A 5 = 10, A 8 = 18, … , A 98 = 258, A 101 = 266 …. Each of these terms has remainder 2 when it is divided by 8. A 99 does not appear in either of these lists. Thus, both conditions together do not provide enough information to find the remainder when A 99 is divided by 8. Therefore, E is the answer. Note: in order to find the remainder when A 99 is divided by 8, we need the value of A 3. Thus, we need the values of 3 variables, and we are given only 2, so E should be the answer. Therefore, the correct answer is E. Answer: E
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Re: Math Revolution DS Expert  Ask Me Anything about GMAT DS
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20 Jan 2019, 18:39
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute value) Is \(ab<0\)? \(1) a+b =  ( a + b )\) \(2) a+b + 1 = a + b\) => 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. You should remember that the inequality \(x+y < x + y\) is equivalent to the inequality \(xy < 0.\) Condition 2) tells us that \(a+b + 1 = a + b.\) Thus, \(a + b < a + b\) and \(ab < 0\). Thus, condition 2) is sufficient. Condition 1) If \(a = 2\) and \(b = 1\), then the answer is ‘yes’. If \(a = 1\) and \(b = 1\), then the answer is ‘no’. Since it does not yield a unique solution, condition 1) is not sufficient. Therefore, the answer is B. Answer: B
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21 Jan 2019, 01:14
[ Math Revolution GMAT math practice question] (number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)? 1) \(a+b\) is an even integer 2) \(ab\) is an odd integer.
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21 Jan 2019, 05:24
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)? 1) \(a+b\) is an even integer 2) \(ab\) is an odd integer. \(n = \left\langle {ab6} \right\rangle \,\,\,\, \Rightarrow \left\{ \matrix{ \,n > 0\,\,\,\left( {{\rm{implicitly}}} \right) \hfill \cr \,a \in \left\{ {1,2,3, \ldots ,9} \right\} \hfill \cr \,b \in \left\{ {0,1,2,3, \ldots ,9} \right\} \hfill \cr} \right.\) \({{\left\langle {ab6} \right\rangle } \over 4}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}}\) \(\left( 1 \right)\,\,\,a + b = {\rm{even}}\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {2,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,ab = {\rm{odd}}\,\,\, \Rightarrow \,\,\,b = {\rm{odd}}\,\,\, \Rightarrow \,\,\,\left\langle {b6} \right\rangle \in \left\{ {16,36,56,76,96} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\) The correct answer is therefore (B). We follow the notations and rationale taught in the GMATH method. Regards, Fabio.
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21 Jan 2019, 07:44
MathRevolution wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute value) Is \(ab<0\)? \(1) a+b =  ( a + b )\) \(2) a+b + 1 = a + b\) => 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. You should remember that the inequality \(x+y < x + y\) is equivalent to the inequality \(xy < 0.\) Condition 2) tells us that \(a+b + 1 = a + b.\) Thus, \(a + b < a + b\) and \(ab < 0\). Thus, condition 2) is sufficient. Condition 1) If \(a = 2\) and \(b = 1\), then the answer is ‘yes’. If \(a = 1\) and \(b = 1\), then the answer is ‘no’. Since it does not yield a unique solution, condition 1) is not sufficient. Therefore, the answer is B. Answer: B MathRevolutionQUOTING YOUR LINE : You should remember that the inequality \(x+y < x + y\) is equivalent to the inequality \(xy < 0.\)
BUT HERE #2 \(2) a+b + 1 = a + b\)ITS NOT > BUT = .. SO HOW IS THAT RELATION VALID? IMO #2: a+b + 1 = a + b[/m] this can be written as: ab+1 = a+b 1= 2 ( a+b) a+b= 1/2 this would be possible when either of values of a & b are either + & ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient combining 1 & 2 :  ( a + b )+1 = a + b or say 1/2 = a+b again this would be possible when either of values of a & b are either + & ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient IMO E



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22 Jan 2019, 00:29
[ Math Revolution GMAT math practice question] (absolute values) If \(y=x1+x+1,\) then \(y=\)? \(1) x>1\) \(2) x<1\)
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22 Jan 2019, 03:42
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)? 1) \(a+b\) is an even integer 2) \(ab\) is an odd integer. given n= ab6 is it divisible by 4 any no which is two times divisible by 2 would be divisible by 4 #1 a+b is an even integer here a,b can both be odd or even so in sufficient #2 ab is an odd integer means both ab are odd integers so ab6 is not divisible by 4 sufficient IM O B



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22 Jan 2019, 03:47
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute values) If \(y=x1+x+1,\) then \(y=\)? \(1) x>1\) \(2) x<1\) IMO e would be sufficeint #1: x>1 ; x can be any no from 1 to infinity #2 x<1 : x can be any value <1 to infinity from 1 & 2 range of 1>x>1 ; x can be 0 ,0.5 , 0.5 .. combining both we get answer different values of y IMO E



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22 Jan 2019, 07:14
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute values) If \(y=x1+x+1,\) then \(y=\)? \(1) x>1\) \(2) x<1\) \(y = \left {x  1} \right + \left {x + 1} \right\) \(? = y\) \(\left( 1 \right)\,\,x >  1\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,? = 2\, \hfill \cr \,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,? = 4\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,x < 1\,\,\,\left\{ \matrix{ \,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,? = 2\, \hfill \cr \,{\rm{Take}}\,\,x =  2\,\,\,\, \Rightarrow \,\,\,? = 4\, \hfill \cr} \right.\) \(\left( {1 + 2} \right)  1 < x < 1\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{ \,\left {x  1} \right = 1  x \hfill \cr \,\left {x + 1} \right = x + 1 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2\) We follow the notations and rationale taught in the GMATH method. Regards, Fabio.
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22 Jan 2019, 07:19
Archit3110 wrote: from 1 & 2 range of 1>x>1 ; x can be 0 ,0.5 , 0.5 ..
combining both we get answer different values of y
Hi, Archit3110 ! Please read my solution above, and take into account that 1<x<1 implies: (a) 1<x hence x+1 > 0 , hence x+1 = x+1 (b) x<1 hence x1 < 0 , hence x1 = (x1) = 1x Then y = (x+1) + (1x) = 2 , therefore x will have infinite possible values (between 1 and 1), but y will be always 2 (when 1<x<1). I hope I could help you! Regards and success in your studies! Fabio.
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23 Jan 2019, 01:06
MathRevolution wrote: [ Math Revolution GMAT math practice question] (number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)? 1) \(a+b\) is an even integer 2) \(ab\) is an odd 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We can determine whether \(a\) number is divisible by \(4\) from its final two digits. Numbers with the final digits \(16, 36, 56, 76\) and \(96\) are divisible by \(4\) and those with final digits \(06, 26, 46, 66\) and \(88\) are not divisible by \(4\). Thus, asking whether \(n\) is divisible by \(4\) is equivalent to asking whether \(b\) is odd. Since it implies that both \(a\) and \(b\) are odd integers, condition 2) is sufficient. Condition 1) There are two cases to consider. If \(a\) is an even integer and \(b\) is an odd integer, the answer is ‘yes’. If \(a\) is an odd integer and \(b\) is an even integer, the answer is ‘no’. Since it does not yield a unique solution, condition 1) is not sufficient. Therefore, B is the answer. Answer: B
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23 Jan 2019, 01:08
[ Math Revolution GMAT math practice question] (algebra) \(x=\)? \(1) x^3x=0\) \(2) x=x\)
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24 Jan 2019, 06:24
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute values) If \(y=x1+x+1,\) then \(y=\)? \(1) x>1\) \(2) x<1\) => 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. There are three ranges of values of x to consider. If \(x > 1\), then \(y =  x – 1  +  x + 1  = x – 1 + x + 1 = 2x\) and we don’t have a unique value of \(y\). If \(1 ≤ x ≤ 1\), then \(y =  x – 1  +  x + 1  =  ( x – 1 ) + x + 1 = 2\) and we have a unique value of \(y\). If \(x < 1\), then \(y =  x – 1  +  x + 1  = ( x – 1 ) – ( x + 1 ) = 2x\) and we don’t have a unique value of \(y\). Asking for the value of \(y\) is equivalent asking if \(1 ≤ x ≤ 1\). Both conditions yield the inequality \(1 < x < 1\), when applied together. Therefore, both conditions are sufficient, when taken together. 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. Therefore, C is the answer. Answer: C
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24 Jan 2019, 06:27
[ Math Revolution GMAT math practice question] If \(x\) and \(y\) are nonzero numbers and \(x≠±y\), then \(\frac{( x^2 + y^2 )}{( x^2  y^2 )}=?\) \(1) \frac{x}{y}=\frac{1}{3}\) \(2) y=3x\)
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25 Jan 2019, 02:28
MathRevolution wrote: [ Math Revolution GMAT math practice question] (algebra) \(x=\)? \(1) x^3x=0\) \(2) x=x\) => 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. Since we have \(1\) variable (\(x\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) \(x^3x = 0\) \(=> x(x^21) = 0\) \(=> x(x+1)(x1) = 0\) \(=> x = 0, x = 1\) or \(x = 1.\) Since it does not yield a unique solution, condition 1) is not sufficient. Condition 2) \(x = x\) \(=> 2x = 0\) \(=> x = 0.\) Since it gives a unique solution, condition 2) is sufficient. Therefore, B is the answer. Answer: B
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