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11 Jan 2019, 07:28
MathRevolution wrote: [ Math Revolution GMAT math practice question] (function) In the \(xy\)plane, a circle has center \((0,0)\) and radius \(5\). Is the point \((r,s)\) inside or on the circle? \(1) 3 < r < 3\) \(2) 4 < s < 4\) \(\left( {r,s} \right)\,\,\,\mathop \in \limits^? \,\,\,\left\{ {\,\left( {x,y} \right)\,\,:\,\,{{\left( {x  0} \right)}^2} + {{\left( {y  0} \right)}^2} \leqslant {5^2}\,} \right\}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,\,{r^2} + {s^2}\,\,\,\mathop \leqslant \limits^? \,\,\,25\,}\,\,\) \(\left( 1 \right)\,\,\,\,\left r \right < 3\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,6} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,\,\left s \right < 4\,\,\,\,\left\{ \matrix{ \,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {6,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( {1 + 2} \right)\,\,\,\,\left\{ \matrix{ \,{r^2} = {\left r \right^2} < {3^2} \hfill \cr \,{s^2} = {\left s \right^2} < {4^2} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{r^2} + {s^2} < 25\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Math Revolution DS Expert  Ask Me Anything about GMAT DS
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11 Jan 2019, 08:09
MathRevolution wrote: [ Math Revolution GMAT math practice question] (geometry) \(x, y\) and \(z\) are the sides of the triangle shown and \(h\) is its height. Is the perimeter, \(x + y + z\) of the triangle greater than \(1\)? \(1) h = \frac{1}{2}\) \(2) x = y = \frac{1}{3}\) \(x + y + z\,\,\mathop > \limits^? \,\,\,1\) \(\left( 1 \right)\,\,\,\left\{ \matrix{ \,x\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr \,z\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x + y + z > 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\) \(\left( 2 \right)\,\,\left\{ \matrix{ \,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{left}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\, < \,\,\,\,3\left( {{1 \over 3}} \right)\,\, = \,\,1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr \,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{right}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\,\mathop < \limits^{{\rm{as}}\,{\rm{near}}\,\,{\rm{as}}\,\,{\rm{desired}}!} \,\,\,\,{2 \over 3} + {1 \over 3}\left( {\sqrt 2 } \right)\,\, = \,\,{{\sqrt 2 + 2} \over 3}\,\,\,\,\,\left[ { > \,\,\,1} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\) The correct answer is therefore (A). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. P.S.: it is interesting to study the case in which D does not belong to the side BC!
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13 Jan 2019, 17:52
MathRevolution wrote: [ Math Revolution GMAT math practice question] (geometry) \(x, y\) and \(z\) are the sides of the triangle shown and \(h\) is its height. Is the perimeter, \(x + y + z\) of the triangle greater than \(1\)? \(1) h = \frac{1}{2}\) \(2) x = y = \frac{1}{3}\) Attachment: 1.10.png => 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. In a triangle, the sum of the lengths of two sides is always greater than the length of the other side. Thus, from triangle \(ABD\), we must have \(AB + BD > AD = h\), and from triangle \(ABD\), we must have \(AC + CD > AD = h\). If \(h = \frac{1}{2}\), then \(AB + BC + CA = AB + BD + DC + CA > \frac{1}{2} + \frac{1}{2} = 1.\) Condition 1) is sufficient. Condition 2): If \(AC = \frac{1}{2}\), then \(AB + BC + CA = \frac{1}{3} + \frac{1}{3} + \frac{1}{2} = \frac{7}{6} > 1\) and the answer is ‘yes’. If \(AC = \frac{1}{4}\), then \(AB + BC + CA = \frac{1}{3} + \frac{1}{3} + \frac{1}{4} = \frac{11}{12} < 1\) and the answer is ‘no’. Since it does not yield a unique solution, condition 2) is not sufficient. Therefore, the correct answer is A. Answer: A
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13 Jan 2019, 18:00
Attachment:
1.14.png [ 22.82 KiB  Viewed 235 times ]
MathRevolution wrote: [ Math Revolution GMAT math practice question] (function) In the \(xy\)plane, a circle has center \((0,0)\) and radius \(5\). Is the point \((r,s)\) inside or on the circle? \(1) 3 < r < 3\) \(2) 4 < s < 4\) => 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. The inequality satisfied by points inside or on the circle is \(r^2+s^2≤5^2=25\). Since we have \(2\) variables (\(r\) and \(s\)) and \(0\) equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. Conditions 1) & 2): Since \(3<r<3\) and \(4<s<4\), we have \(0≤r^2<3^2=9\) and \(0≤s^2<4^2=16\). Thus, \(0≤r^2+s^2<9+16=25\) and both conditions together are sufficient. Attachment:
1.14.png [ 22.82 KiB  Viewed 235 times ]
Therefore, the answer is C. Answer: C
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14 Jan 2019, 01:04
[ Math Revolution GMAT math practice question] (absolute value, geometry) Is \(a > b  c\)? \(1) cb < a\) \(2) a, b\), and \(c\) are the lengths of the \(3\) sides of a triangle.
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14 Jan 2019, 05:51
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute value, geometry) Is \(a > b  c\)? \(1) cb < a\) \(2) a, b\), and \(c\) are the lengths of the \(3\) sides of a triangle. \(a\,\,\mathop > \limits^? \,\,b  c\) \(\left( 1 \right)\,\,\,a\,\, > \,\,\left {c  b} \right\,\,\, = \,\,\,\left {b  c} \right\,\,\,\, \ge \,\,\,b  c\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\) \(\left( 2 \right)\,\, \Rightarrow \,\,\left( 1 \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\) The correct answer is therefore (D). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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15 Jan 2019, 00:58
[ 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
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15 Jan 2019, 06:51
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 \(x,y\,\,{\text{ints}}\,\,\,\left( * \right)\) \({x^{\text{2}}} + x + y\,\, = \,\,\underbrace {x\left( {x + 1} \right)}_{\left( * \right)\,\,{\text{even}}} + y\,\,\mathop = \limits^? \,\,\,{\text{odd}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,y\,\,\mathop = \limits^? \,\,\,{\text{odd}}\,\,}\) \(\left( 1 \right)\,\,x\,\,{\rm{odd}}\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,y\,\,{\text{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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16 Jan 2019, 01:14
MathRevolution wrote: [ Math Revolution GMAT math practice question] (absolute value, geometry) Is \(a > b  c\)? \(1) cb < a\) \(2) a, b\), and \(c\) are the lengths of the \(3\) sides of a triangle. => 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. Condition 1) is sufficient since it yields \(b – c ≤  b  c  =  c – b  < a.\) Condition 2) Since the sum of the lengths of two sides of a triangle is greater than the length of the third side, we must have \(a + c > b\). Thus, condition 2) is sufficient. Therefore, D is the answer. Answer: D 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.
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16 Jan 2019, 01:15
[ 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
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Updated on: 16 Jan 2019, 09:02
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 VERY beautiful problem, Max. Congrats (and kudos)! \(x,y\,\,{\rm{ints}}\,\,\,\,\left( * \right)\) \(\left( {x + y} \right)\left( {x  y} \right) = {x^{\text{2}}}  {y^2}\,\,\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\boxed{\,\,x + y\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,x  y\mathop = \limits^? \,\,{\text{even}}\,\,}\) \(\left( {**} \right)\,\,\,\left\{ \matrix{ \,x + y = {\rm{even}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,x  y = x + y  2y = {\rm{even}} \hfill \cr \,x  y = {\rm{even}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,x + y = x  y + 2y = {\rm{even}} \hfill \cr} \right.\) \(\left( 1 \right)\,\,\,{x^3}  {y^3} = {\rm{even}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\{ \matrix{ \,x\,\,{\rm{even}}\,\,,y\,\,{\rm{even}} \hfill \cr \,{\rm{OR}} \hfill \cr \,x\,\,{\rm{odd}}\,\,,y\,\,{\rm{odd}} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x + y = {\rm{even}}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\) \(\left( 2 \right)\,\,x + y = {\rm{even}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Originally posted by fskilnik on 16 Jan 2019, 05:59.
Last edited by fskilnik on 16 Jan 2019, 09:02, edited 1 time in total.



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16 Jan 2019, 06:20
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 \(x^2+x+y\) can be written as x(x+1) +y so in this irrespective of value of x ; y has to be an odd integer to get \(x^2+x+y\) = odd #1: value of y not know in sufficient #2: yes sufficient IMO B
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16 Jan 2019, 06:29
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 \(x^2y^2\) an even integer would be valid for values when both x & y are either odd or even #1: \(x^3y^3\) is an even integer for both x & y being either even or odd ; it would be valid and sufficient #2: \(x+y\) is an even integer again when both x & y are either even or odd then relation would be valid and sufficinet for all values we can test with integers : even 2,4 and odd 1,3 IMO D
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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, 05:09
[ Math Revolution GMAT math practice question] (absolute value) Is \(ab<0\)? \(1) a+b =  ( a + b )\) \(2) a+b + 1 = a + b\)
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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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