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Which of the following points is reflect to y=-x at (-2,1)?

A. (-1, 2) B. (1,-2) C. (2,1) D. (2,-1) E. (1,2)

==> You can figure out a point reflecting to y=-x by substituting –y to x-coordinate and –x to y-coordinate. Then, (-2,1) --> (-1, -(-2))=(-1,2) is derived.

Hence, the answer is A. Answer: A
_________________

There are a bunch of buildings and their building numbers are all even numbers. If the numbers are 312 between 590 inclusive, what is the number of buildings? A. 139 B. 140 C. 141 D. 142 E. 143

==> The number of consecutive even numbers of odd numbers=(the last term+the first term) )/2 +1. That is, (590-312)/2 +1=139+1=140. That is, (590-312)/2 +1=139+1=140 and the answer is B.

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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In the x-y plane, there is line K, (x/a)+(y/b)=1. What is the y-intercept of line K? 1) a=b 2) b=5

==> If you modify the original condition and the question, the y-intercept 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
_________________

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=Max-min, there are 3 variables (r,M,m) and 1 equation (r=M-n), 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 50-20=30>12, hence it is always yes and sufficient.

Therefore, the answer is A. Answer: A
_________________

If m and n are positive integers, what is the value of m+n? 1) m/n=3/5 2) The greatest common divisor of m and n is 5

==>In the original condition, there are 2 variables (m, n) for the right triangle, 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 M=3*5=13 and n=5*5=25, so m+n=25+15=40, hence it is unique and sufficient.

Therefore, the answer is C. Answer: C
_________________

Overview of GMAT Math Question Types and Patterns on the GMAT [#permalink]

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13 Feb 2017, 20:59

MathRevolution wrote:

Below is the 5051 question.

Is \(x<xy<y\)? 1) \(x<y\) 2) \(0<x<1<y\)

==> In the original condition, there are 2 variables (x,y), so C is highly likely to be the answer. Through 1) & 2), 2) is true, and from 2) \(x<1\), both sides of the equation can be multiplied by y, and you get \(xy<y\). From \(1<y\), you multiply both sides of the equation by \(x\), you get \(x<xy\), then you get \(x<xy<y\), hence yes, and sufficient. B is the answer Answer: B

Dear MathRevolution & Bunuel, If \(x=0.5\) and \(y=1.5\), statement 1 still valid. Why D is not the answer?
_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

==> 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 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), it is always yes, hence it is sufficient. For con 2), x=0 yes, but x=-3.5 no, hence it is not sufficient.

Therefore, the answer is A. Answer: A
_________________

==> 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 (a,b,c)=(1,1,1), you get a+b+c=1+1+1=3=odd, so no, but if (a,b,c)=(2,2,2), you get a+b+c=2+2+2=6=even, so yes, hence it is not sufficient.

n=? 1) twice n equals to n+1 2) n times n equals to n ==> 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. For con 1), from 2n=n+1, you get n=1, hence sufficient. For con 2), from \(n^2=n\) and \(n^2-n=0\), n(n-1)=0, you get n=0,1, hence it is not unique and not sufficient.

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