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Math Revolution Approach (DS)

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New post 24 May 2017, 00:36
When a positive integer x is divided by 12, what is the remainder?

1) x is divisible by 9
2) When x is divided by 4, the remainder is 1

==> In the original condition, there is 1 variable 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. For remainder questions, you find the first overlapping number after direct substitution, then you add the least common multiple of the dividing numbers.
For con 1), from x=9=12(0)+9, the remainder is 9.
For con 2), from x=4p+1=1,5,9…., the remainder of x=1=12(0)+1 is 1, and from x=5=12(0)+5, the remainder is 5, hence it is not unique and not sufficient.
By solving con 1) and con 2), the first overlapping number is 9 and the least common multiple of the dividing numbers is LCM(From 4,9+36, you get x=9, 9+36=45, 45+36=81,.., and the remainder when divided by 12 always becomes 9, hence it is unique and sufficient. Therefore, the answer is C.

Answer: C
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New post 25 May 2017, 01:28
r=?

1) r^3-r^2=0
2) r=-r

==> In the original condition, there is 1 variable (r), 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.
For con 1), from r^2(r-1)=0, you get r=0,1, hence it is not unique and not sufficient.
For con 2), from 2r=0, you get r=0, hence it is unique and sufficient. Therefore, the answer is B.

Answer: B
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New post 25 May 2017, 17:41
Is 12 a factor of x(y^2)?

1) X is a multiple of 24
2) Y is a multiple of 6

==> 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. However, for con 1), x has 12 as a factor from 24=2(12), hence yes, it is sufficient. For con 2), y^2 also has 36 as a factor, and from 36=3(12), y^2 has 12 as a factor as well, hence yes. The answer is D.

Answer: D
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New post 29 May 2017, 00:02
In the coordinate plane system, does line L pass through 3rd quadrant?

1) The slope of line L is positive.
2) Line L has x-intercept -2 and y-intercept 3.

==> In the original condition, there are 2 variables (m and n from y=mx+n) 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. However, con1)=con2), hence it is always yes and sufficient. Therefore, the answer is D.

Answer: D
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New post 31 May 2017, 00:56
If n=s^at^b, where r, a, b and s are integers, is √n an integer?

1) a+b is an even number
2) a is an even number

==>In the original condition, there are 5 variables (n,s,t,a,b) and 1 equation (n=s^at^b). In order to match the number of variables to the number of equations, there must be 5 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), s=t=2 and a=b=2 yes, but s=t=2 and a=b=-2 no, hence it is not sufficient. Therefore, the answer is E.

Answer: E
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New post 01 Jun 2017, 01:08
If m and n are positive integers, is mn a multiple of 9?

1) m+n is a multiple of 3
2) mn is a multiple of 3

==> In the original condition, there are 2 variables (m,n) 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 the multiple of mn=3, it becomes a multiple of m or n=3. According to con 1), you always get a multiple of m=n=3, hence yes, it is sufficient.

The answer is C.
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New post 02 Jun 2017, 00:59
If xy≠0, is \(\frac{x^3y^2}{x^2y^3}=1\)?

1) \(\frac{x^2}{y^2}=1\)
2) \(\frac{x^3}{y^3}=1\)

==> If you modify the original condition and the question, you get \(\frac{x^3y^2}{x^2y^3}=1\), \(\frac{x}{y}=1\)? For con 1), you get \(\frac{x}{y}=-1\), 1, hence it is not unique and not sufficient. But fo con 2), you only get \(\frac{x}{y}=1\), hence it is unique and sufficient.

Therefore, the answer is B.
Answer: B
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New post 04 Jun 2017, 17:53
Is x=y?

1) \(|x|+|y|=0\)
2) \(x^2+y^2=0\)

==> 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. However, since con 1)=con2), you get x=y=0, hence yes, it is sufficient.

The answer is D.
Answer: D
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New post 04 Jun 2017, 17:54
If a and b are positive integers, what is the number of the different prime factors of ab?

1) a has different 2 prime factors.
2) b has different 3 prime factors.

==> If you modify the original condition and the question, the number of prime factors doesn’t affect the exponent. In other words, the number of prime factors of ab is equal to the number of prime factors of a.

Therefore, the answer is A.
Answer: A
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New post 07 Jun 2017, 01:18
Is xyz>0?

1) xy>0
2) yz>0

==> 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), E is most likely to be the answer. By solving con 1) and con 2), if x=y=z=1, yes, but if x=y=x=-1, no, hence it is not sufficient.

Therefore, the answer is E.
Answer: E
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New post 08 Jun 2017, 00:53
There is a total of 400 students. 310 students participated in A, and 120 students participated in B. What is the number of students that participated in both A and B?

1) 24 students did not participate in either A or B.
2) 66 students participated only in B.

==> If you modify the original condition and the question, you get a 2 by 2 as shown in the table below.
Attachment:
6.8.png
6.8.png [ 3.11 KiB | Viewed 191 times ]

As shown above, from a+b+c+d=400, a+c=310, and a+b=120, there are 4 variables and 3 equations. 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 d=24, you get a=54, so con 1) = con 2), hence it is unique and sufficient.

The answer is D.
Answer: D
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New post 09 Jun 2017, 02:33
If x,y are positive integers, is \(x^y\) even?

1) x is even
2) y is even

==> In the original condition, there are 2 variables and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from \(x^y=even^{even}=even\), yes, hence the answer is C. However, this is an integer question, one of the key questions, so you apply CMT 4 (if you apply A, then 1). For con 1), from x=even, you get \(x^y=even^y=even\), hence yes and sufficient.

Therefore, the answer is A, not C.
Answer: A
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New post 11 Jun 2017, 18:23
What is the units digit of a positive integer n?

1) The units digit of \(n^2\) is 4
2) The units digit of \(n^3\) is 8

==> 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 as well. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), from \(n^2\)=~4,you get n=~2, ~8, hence it is not unique and not sufficient. For con 2), from \(n^3\)=~8, you get n=~2, hence it is unique and sufficient.

Therefore, the answer is B.
Answer: B
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New post 11 Jun 2017, 18:24
a=?

1) 4a=9+a
2) a*a=3a

==> In the original condition, there is 1 variable (a) 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 con 2), D is most likely to be the answer. For con 1), you get 4a=9+a, 3a=9, a=3, hence it is unique and sufficient.
For con 2), from \(a^2=3a, a^2-3a=0\), a(a-3)=0, you get a=0, 3, hence it is not unique and not sufficient.

Therefore, the answer is A.
Answer: A
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New post 14 Jun 2017, 01:26
If f(x)=ax²+bx+c, for all x is f(x)<0?

1) b²-4ac<0
2) a<0

==> In the original condition, there are 3 variables (a, b, c) 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), if discriminant =b^2-4ac<0, it doesn’t meet with the x-axis, and if a<0, you always get f(x)<0, hence yes, it is sufficient.

Therefore, the answer is C.
Answer: C
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New post 15 Jun 2017, 01:19
If x, y are integers, is x+y an odd number?

1) y=3x+5
2) y=2x-3

==> If you modify the original condition and the question, in order to get x+y=odd, you need to get (x,y)=(even,odd)or (odd,even). For con 1), you get (x,y)=(even,odd) or (odd, even), hence always yes and sufficient.

Therefore, the answer is A.
Answer: A
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New post 16 Jun 2017, 01:38
When a positive integer n is divided by 13, what is the remainder?

1) n+1 is divisible by 13
2) n+14 is divisible by 13

==> In the original condition, there is 1 variable, 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. By solving con 1) and con 2), you get con 1) = con 2), and you use direct substitution for the rest, so you get n=12,25,38,… When you divide by 13, the remainder always becomes 12, hence it is unique and sufficient.

The answer is D.
Answer: D
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New post 18 Jun 2017, 17:59
Is ab+cd>0?

1) ac+bd>0
2) bc+ad>0

==> In the original condition, there are 4 variables (a, b, c, d), and in order to match the number of variables to the number of equations, there must be 4 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 a=2, b=3, c=-1, d=2: yes BUT a=5,b=-1,c=1,d=2:, hence it is not sufficient.

Therefore, the answer is E.
Answer: E
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New post 18 Jun 2017, 18:00
If there are 11 numbers, what is the median of them?

1) The smallest 6 numbers are less than or equal to 10.
2) The largest 6 numbers are greater than or equal to 10.

==> In the original condition, there are 11 variables, and in order to match the number of variables to the number of equations, there must be 11 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 largest 6 numbers are greater than or equal to 10, and the smallest 6 numbers are less than or equal to 10, so the 6th number becomes 10, which is the median. Since median=10, it is unique and sufficient.

Therefore, the answer is C.
Answer: C
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New post 21 Jun 2017, 01:02
If average (arithmetic mean) score of 9 students is 90, what is the median score?

1) The 5 smallest scores are 84, 83, 82, 81, and 80
2) The 4 largest scores are 89, 88, 86, and 85

==> If you modify the original condition and the question, and list the 9 numbers in the ascending order, you get 5th=median. Thus, according to con 1), you get 5th=median=84.

Therefore, the answer is A.
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