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Math Revolution Approach (DS)

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New post 22 Jun 2017, 02:15
Is |a-b|-(a-b)>0?

1) a<b
2) a>0 and b>0

==> If you modify the original condition and the question, what satisfies |a-b|>a-b? must be a-b<0.

Therefore, the answer is A.
Answer: A
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New post 23 Jun 2017, 00:07
Is y^4<x^2y^2?

1) x<y
2) y<0

==> If you modify the original condition and the question, if y≠0 from y^4<x^2y^2?, y^4-x^2y^2<0?, y^2(y^2-x^2)<0?, you get y^2-x^2<0?. There are 2 variables, and in order to match the number of variables to the number of equations, there must be 2 equations as well. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) & con 2), from x<y<0, you get x^2-y^2>0, hence yes, it is always sufficient.

Therefore, the answer is C.
Answer: C
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New post 25 Jun 2017, 17:12
If x≠0, is x^2<|x|?

1) x<1
2) x>-1

==> If you modify the original condition and the question, from x^2<|x|?, |x|^2<|x|?, when you divide both sides by |x|, (Since |x|>0, the inequality sign doesn’t change after division) you get |x|<1?, and then -1<x<1?.

The answer is C.
Answer: C
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New post 25 Jun 2017, 17:15
|a-b|=?

1) |a+b|=10
2) |a^2-b^2|=20

==> In the original condition, there are 2 variables (a,b) 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 |a^2-b^2|=|a+b||a-b|=20, 10|a-b|=20, |a-b|=2.

The answer is C.
Answer: C
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New post 28 Jun 2017, 00:08
Is a<b?

1) a<0
2) |b|<1

==> In the original condition, there are 2 variables (a, b) 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 -1<b<1, if a=-0.1 and b=-0.2, no, but if b=0.1 and a=-0.1, yes.

Hence, the answer is E.
Answer: E
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New post 29 Jun 2017, 01:57
If x and y are integers, what are the values of x and y?

1) 2^x3^y=16/27
2) x+y=1

==> For con 1), you get 2^x3^y=16/27=2^43^-3, then x=4 and y=-3. Therefore, the answer is A. In other words, it is CMT 4(A), in which A and C are both the answers. In the original condition, there are 2 variables, so the answer is C, but since it is an integer question, you apply CMT 4(A) and get the final answer as A.

This is a 4950 level question.
Answer: A
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New post 30 Jun 2017, 00:01
If x, y, z are positive integers, is xyz divisible by 6?

1) x, y, z are consecutive.
2) x+y+z is a multiple of 3.

==> The product of 3 consecutive integers is always a multiple of 6. For con 1), it is yes and sufficient, and for con 2), (x,y,z)=(1,2,3) yes but (x,y,z)=(2,2,2) no, hence it is not sufficient.

Therefore, the answer is A.
Answer: A
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New post 02 Jul 2017, 17:09
3^-(x+y)/3^-(x-y)=?

1) x=3
2) y=2
==> If you modify the original condition and the question, you get 3^-(x+y)/3^-(x-y)=3^x-y+x-y=3^-2y, so you only need to know y.

The answer is B.
Answer: B
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New post 02 Jul 2017, 17:11
Is |x+y|<|x|+|y|?

1) x+y<0
2) x<y

==> If you modify the original condition and the question, for |x+y|<|x|+|y|?, you can square both sides, and if you expand from (|x+y|)^2<(|x|+|y|)^2? Or, (x+y)^2<(|x|+|y|)^2?, you get x^2+y^2+2xy<|x|^2+|y|^2+2|xy|?, or x^2+y^2+2xy<x^2+y^2+2|xy|?, 2xy<2|xy|?, xy<|xy|?. However, in con 1) and con 2), xy<0 is not mentioned, so the answer is E.

Answer: E
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New post 05 Jul 2017, 00:02
If x and y are positive integers and y(2^x)=24, x=?

1) x 2
2) y is even

==> In the original condition, there are 2 variables (a,b), and in order to match the number of variables to the number of equations there must be 2 equations as well. 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 24=6(2^2).

The answer is C.
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New post 06 Jul 2017, 00:22
When x+y is integer, is y an integer?

1) x is an integer.
2) x+2y is an integer.

==> In the original condition, there are 2 variables (x,y) and 1 equation (x+y=integer). In order to match the number of variables to the number of equations, there must be 1 more equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), from x=integer and x+y=integer, integer+y=integer, you get y=integer, which is yes and sufficient. For con 2), from x+2y=x+y+y=integer, integer+y=integer, you get y=integer, which is also yes and sufficient.

Therefore, the answer is D.
Answer: D
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New post 07 Jul 2017, 00:06
For x and y, what is the value of x?

1) (x-3)(x+y)=0
2) (x-3)(2x+y)=0

==> 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=3, which is unique and sufficient.

Therefore, the answer is C.
Answer: C
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New post 09 Jul 2017, 17:23
If mr≠0, m/r=?

1) (m+r)/r=3
2) r/(m+r)=1/3

==> In the original condition, there are 2 variables (m,r) 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 con 1) = con 2), so from m/r+1=3, you get m/r=2, hence it is unique and sufficient.

Therefore, the answer is D.
Answer: D
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New post 09 Jul 2017, 17:29
1-c-d=?

1) c+d=-1
2) c=-1-d

==> In the original condition, there are 2 variables (c,d) 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 con 1) = con 2), so both of them becomes c+d=-1, and 1-c-d=1-(c+d)=1-(-1)=2, hence it is unique and sufficient.

Therefore, the answer is D.
Answer: D
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New post 12 Jul 2017, 00:09
Is x^3y^2z<0?

1) x^2y<0
2) yz<0


==> If you modify the original condition and the question, when xyz≠0, you get xz<0?. There are 3 variables (x,y,z), 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), E is most likely to be the answer. By solving con 1) and con 2), you get x^2y<0, y<0 and yz<0, z>0. Since x is unknown, it is not sufficient.

Therefore, the answer is E.
Answer: E
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New post 13 Jul 2017, 01:05
Is a positive integer x an even number?

1) The smallest prime factor of x is 2.
2) The greatest prime factor of x is 13.

==> In the original condition, there is 1 variable (x) and in order to match the number of variables to the number of equations, there must be 1 equations. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), since x always has 2 as the factor, you get x=even, hence yes, it is sufficient. For con 2), you get x=2*13 yes but x=13 no, hence it is not sufficient.

The answer is A.
Answer: A
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New post 14 Jul 2017, 00:00
If x, y, and z are positive integers, is x+y divisible by 3?

1) x+z is divisible by 3
2) y+z is divisible by 3

==> In the original condition, there are 3 variables (x,y,z) 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), you need 1 more equation, so E is most likely to be the answer. By solving con 1) and con 2), if x=y=z=3 yes, but if z=2, x=y=1 no, hence it is not sufficient.

Therefore, the answer is E.
Answer: E
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New post 16 Jul 2017, 17:43
Is the standard deviation of set A greater than that of set B?

1) The median of set A is greater than that of set B
2) The average (arithmetic mean) of set A is greater than that of set B

==> In the original condition, more than 90% of the questions related to the the relationship between median, mean, and standard deviation have E as the answer.

The answer of this question is also E.
Answer: E
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New post 16 Jul 2017, 17:45
Is n>0?

1) n>-1
2) |1+n| > |n-3|

==> In the original condition, there is 1 variable (n), and in order to match the number of variables to the number of equations, there must be 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. Also, for inequality questions, if the range of the question includes the range of the condition, the condition is sufficient. Thus, for con 1), the range of the question does not include range of the condition, hence it is not sufficient. For con 2), if you square both sides, you get |1+n|^2 > |n-3|^2, (1+n)^2 > (n-3)^2, n^2+2n+1>n^2-6n+9, 8n>8, n>1, so the range of the question includes the range of the condition, hence it is sufficient.

Therefore, the answer is B.
Answer: B
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New post 19 Jul 2017, 00:58
Out of 50 people, there are 35 people with college degrees and 30 females. How many females with college degree are there?

1) 10 are males with college degree.
2) 5 are females without college degree.

==> If you modify the original condition and the question, it is a 2 by 2 question that appears most frequently on gmat math.

Attachment:
7.19.png
7.19.png [ 2.96 KiB | Viewed 188 times ]


According to the table above, there are 4 variables (w,x,y,z) and 3 equations (w+x+y+z=50, w+y=35, w+x=30). In order to match the number of variables to the number of equations, there must be 1 more equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. You get con 1) = con 2), and then you get w=25.

The answer is D.
Answer: D
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Re: Math Revolution Approach (DS)   [#permalink] 19 Jul 2017, 00:58

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