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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]
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04 Jan 2017, 05:12
What is the median of the 5 consecutive integers? 1) The average (arithmetic mean) of the 5 consecutive numbers is 5 2) The average (arithmetic mean) of the 5 consecutive numbers is an odd number ==> In the original condition, there is 1 variable (and, n+1, n+2, n+3, n+4), and in order to match the number of variables to the number of equations, there must be 1 equation as well. Therefore, D is most likely to be the answer. For con 1), in case of consecutive numbers, you get median=mean, and thus median=5 and it is sufficient. For con 2), median=1,3,5,.. and hence it is not unique and not sufficient. Therefore, it is a question where A becomes the answer, but you must be aware of CMT 3. Con 2) and con 1) must be independent from each other, and you can’t assume that they are the same after solving con 1). The answer is A. Answer: A
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Re: Math Revolution Approach (DS) [#permalink]
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05 Jan 2017, 05:39
If x and y are positive integers, are they consecutive? 1) x+y=3 2) xy=1 ==> In the original condition, there are 2 variables (x, y), and in order to match the number of variables to the number of equations, there must be 2 equations as well. Therefore, C is most likely to be the answer. By solving con 1) and con 2), For con 2), it is always yes, hence it is sufficient. For con 1), only (x,y)=(1,2), (2,1) satisfies, hence it is yes. This question is also related to CMT 4(B). The answer is D. Answer: D
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Re: Math Revolution Approach (DS) [#permalink]
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06 Jan 2017, 02:39
If r and s are positive integers, is the least common multiple of r and s equal to the product of r and s? 1) r and s are consecutive odd numbers 2) r=s+2 ==> In the original condition, there are 2 variables (r, s), and therefore C is most likely to be the answer. By solving con 1) and con 2), it is yes, hence sufficient. Therefore, C is the answer. However, since it is an integer question, one of the key questions, if you apply CMT 4(A) on con 1), you get yes and hence it is sufficient. Therefore, the answer is A. Answer: A
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]
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08 Jan 2017, 18:08
Is (a1)(b+1)=0? 1) \((a1)^2=0\) 2) \(a1+(b+1)^2=0\) ==> In the original condition, there are 2 variables (a, b), so there must be 2 more equations. Therefore, C is most likely the answer. By solving con 1) and con 2), you get con 1) = con 2), and a=1 and b=1, hence it is always yes. The answer is D. Answer: D
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]
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08 Jan 2017, 18:09
\(2^a2^b<4\)? 1) a+b<1. 2) a+b>0. ==> If you modify the original condition and the question, from \(2^a^+^b<4=2^2\), you get a+b<2? Then, for con 1), the range of the question includes the range of the condition, and therefore it is sufficient. For con 2), you get a+b=1 yes, but a+b=3 no, and therefore it is not sufficient. Answer A
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Re: Math Revolution Approach (DS) [#permalink]
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10 Jan 2017, 19:36
If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12? 1) a=even 2) b=odd ==> If you modify the original condition and the question, since (ab+2)(ab+3)(ab+4) is the product of 3 consecutive integers, so it is always divisible by 3. Then, since (ab+2)(ab+3)(ab+4) is divisible by 3, in order for it to be divisible by 12, it needs to be divisible by 4, so you must get ab+2=even. For con 1), if a=even, you get ab=even, then you get ab+2=even and ab+4=even, so it is always divisible by 4, and hence it is yes. The answer is A. It is a CMT 4(A) question. Answer: A
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Re: Math Revolution Approach (DS) [#permalink]
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12 Jan 2017, 18:14
Is n an even? 1) 3n=even 2) 5n=even ==> In the original condition, there is 1 variable (n), and therefore D is most likely to be the answer. For con 1) n=2 yes and n=2/3 no, so it is not sufficient. For con 2), n=2 yes and n=2/5 no, so it is not sufficient. By solving con 1) and con 2), you always get n=even, and hence it is sufficient. Therefore, the answer is C. Answer: C
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Re: Math Revolution Approach (DS) [#permalink]
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13 Jan 2017, 01:47
When a positive integer n has 4 different factors, n=? 1) n has only 1 prime factor 2) n<10 ==> In the original condition, there is 1 variable(n), which should match with the number of equations. Then, 1 equation is needed as well. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), \(n=2^3, 3^3\),…, which is not sufficient. For 2), only \(n=2^3\) is possible, which is unique and sufficient. Hence, the answer is B. Answer: B
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Re: Math Revolution Approach (DS) [#permalink]
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15 Jan 2017, 18:37
What is the remainder when \(2^k\) is divided by 10? 1) k is a positive multiple of 10 2) k is a positive multiple of 4 ==> Modify the original condition and the question. The remainder dividing \(2^k\) by \(10 is ~2^1=~2, ~2^2=~4, ~2^3=~8, ~2^4=~6.\) Thus, when it comes to ones, 2>4>8>6>2… are repeated, which means they have a cycle of four. Hence, the answer is 2) and B. Answer: B
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Re: Math Revolution Approach (DS) [#permalink]
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15 Jan 2017, 18:42
Is x+y=x+y? 1) xy>0 2) x<0 ==> When you modify the original condition and the question, \(x+y^2=(x+y)^2?, x^2+2xy+y^2=x^2+2xy+y^2\)? is derived. If you delete \(x^2 y^2\) from the both equations, 2xy=2xy?, xy=xy? is derived, which becomes xy≥0?. Then, in a case of 1), it is yes and sufficient. Answer: A
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Re: Math Revolution Approach (DS) [#permalink]
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16 Jan 2017, 10:14
MathRevolution wrote: A circle has the same area as a rectangle. What is the circumference of the circle? 1) The perimeter of the rectangle is 18 2) The area of the rectangle is 20
ANS: We can modify the original condition and the question. The area of the circle and the area of the rectangle are same. Hence, since it is asking for the circumference of the circle, we only have to know the area of the circle (which is same as the area of the rectangle). So, the correct answer is B. sir, if we apply your variable approach then there are 3 variable r, l,b and 3 eqn are needed one is already given so 2 are needed. statement 1 gives 1 and stmnt 2 gives 1 so answer should be C as per your you tube demo.. pls explain this deviation and when not to use this approach.



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Re: Math Revolution Approach (DS) [#permalink]
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17 Jan 2017, 01:15
If x and y are positive, is \(x+y>2√xy\)? 1) \(x>y\) 2) \((xy)^2>0\) ==> Modify the original condition and the question, and square the both equations. Then, \((x+y)^2>(2√xy)^2? X^2+2xy+y^2>4xy?, x^22xy+y^2>0?, (xy)^2?0?\) is derived. From \((xy)^2?0?\), x≠y? is derived. Then, 1)=2) and it is yes and sufficient. Hence, the answer is D. Answer: D
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Re: Math Revolution Approach (DS) [#permalink]
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17 Jan 2017, 04:33
@mathrevolution pls answer my query Sent from my Redmi Note 3 using GMAT Club Forum mobile app



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Re: Math Revolution Approach (DS) [#permalink]
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19 Jan 2017, 01:28
In the xy plane, what is the slope of line K? 1) Line K intersects with line equation y=3x 2) Line K is parallel to line equation y=2x ==> In case of 1), if it meets y=x, the slope of line K cannot be decided in a unique way, which is not sufficient. In case of 2), if they are parallel, the slope of line K is always 2, which is unique and sufficient. Hence, the answer is B. Answer: B
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Re: Math Revolution Approach (DS) [#permalink]
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20 Jan 2017, 01:14
For a positive integer x, what is the remainder when \(3^x+1\) is divided by 10? 1) x=4n+2, n is a positive integer 2) x>4 ==> If you modify the original condition and the question, for the remainder question when divided by 10, the exponent has the period of 4. In other words, you get \(~3^1=~3, ~3^2=~9, ~3^3=~7, ~3^4=~1\), so the ones digits has the period of 3>9>7>1>3>…. Thus, for con 1), from x=4n+2, the exponent always gets a period of 4, and since you get \(3^4^n^+^2+1=(~3^2)+1=(~9)+1=~0\), the remainder is always 0, hence it is sufficient. Therefore, the answer is A. Answer: A
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Re: Math Revolution Approach (DS) [#permalink]
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22 Jan 2017, 22:27
Is \(x^y>1\)? 1) x>1 2) y<1 ==> In the original condition, there are 2 variables (x, y), and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), you get (x,y)=(2,1/2) yes, but (x,y)=(2,2) no, NOT sufficient. Therefore, the answer is E. Answer: E
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Re: Math Revolution Approach (DS) [#permalink]
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22 Jan 2017, 22:36
In the xy plane what is the area of a certain circle? 1) The circle has a center (0,0) 2) The circle pass through (3,4) ==> If you look at the original condition, there are 3 variables (center a, b, and radius r) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), the center is (0,0) and the radius is \(√((30)^2+(40)^2)=5\), so the area of the circle becomes \(πr^2= π(5^2)=25π\), hence it is unique and sufficient. Therefore, the answer is C. Answer: C
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Re: Math Revolution Approach (DS) [#permalink]
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25 Jan 2017, 05:09
Is x^2  y^2 < 15? (1) x  y < 3 (2) x + y < 5 ==> If you modify the original condition and the question, you get \(15<x^2y^2<15?\), 15<(xy)(x+y)<15? There are 2 variables (x, y), and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from 3<xy<3 and 5<x+y<5 to 15<(xy)(x+y)<15, it is always yes, hence it is sufficient. Therefore, the answer is C. Answer: C
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Re: Math Revolution Approach (DS) [#permalink]
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26 Jan 2017, 00:55
In the xy plane, there is line K, (x/a)+(y/b)=1. What is the yintercept of line K? 1) a=b 2) b=5 ==> If you modify the original condition and the question, the yintercept is the value of y when x=0, so if you substitute x=0, from y/b=1, you get y=b, so you only need to know b. According to con 2), it is unique and sufficient. Therefore, the answer is B. Answer: B
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Re: Math Revolution Approach (DS) [#permalink]
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27 Jan 2017, 04:19
A certain set X contains the number 20. Is the range of the numbers greater than 12? 1) The maximum of the set is 50 2) The set contains the number 25 ==> In the original condition, from range=Maxmin, there are 3 variables (r,M,m) and 1 equation (r=Mn), and in order to match the number of variables to the number of equations, there must be 2 more equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. However, for con 1), if Max=20, set X already includes 20, so the range is at least 5020=30>12, hence it is always yes and sufficient. Therefore, the answer is A. Answer: A
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