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Math Revolution GMAT Instructor V
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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.

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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.

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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.
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If f(n)=n(n+1), which of following is equal to f(8)/f(2)?
A. f(3) B. f(4) C. f(5) D. f(6) E. f(7)

==>You get f(8)/f(2)=8*9/2*3=4*3=f(3), so the answer is A.
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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.

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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.

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When $$8^x=16$$, x=?

A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{3}{4}$$
D. $$\frac{4}{3}$$
E. $$\frac{4}{5}$$

==>From $$8^x=16$$ to$$2^3^x=2^4$$, you get $$3x=4, x=\frac{4}{3}$$. Therefore, the answer is D.
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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.

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

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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

Dear MathRevolution & Bunuel, If $$x=0.5$$ and $$y=1.5$$, statement 1 still valid. Why D is not the answer?
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Originally posted by BillyZ on 13 Feb 2017, 19:59.
Last edited by BillyZ on 17 Feb 2017, 22:14, edited 1 time in total.
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MathRevolution

Could you please provide me information about which type of question comes under
1. DATA interpretation
2. Common Mistake Type (CMT)
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What is the range of 30 consecutive even numbers?

A. 54 B. 56 C. 58 D. 60 E. 62

==> The number of consecutive numbers become (last-first/2)+1=(range/2)+1. In other words, from (range/2)+1=30, you get range/2=29, range=2(29)=58.

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Is -3<x<4?

1) -2<x<3
2) -4<x<4

==> 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.

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2x+3y=?

1) 2x+4y=3
2) 4x+6y=6

==> In the original condition, you get 2x+3y=(1/2)(4x+6y)=? From con 2), you get 4x+6y=6, hence it is unique and sufficient.

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10+8÷3×6-2=?

A. 24
B. 34
C. 22
D. 12
E. 11

==> When calculating the numbers, even if there are no brackets, multiplication and division come first.
You get 10+8÷3×6-2=10+(8/3)6-2=10+16-2=24.

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If $$x^2=2x+1, x^3$$=？

A. 4x
B. 5x+2
C. 5x-1
D. 3x+2
E. 3x-2

==> $$x^3=x(x^2)=x(2x+1)=2x^2+x=2(2x+1)+x=5x+2$$
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If a, b, and c are integers, is abc an even?

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

==> 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.

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Which of the following is equal to
$$3^m7^{m-1}$$?

$$A. 3(21^m)$$
$$B. 7(21^m)$$
$$C. 3(21^{m-1})$$
$$D. 7(21^{m-1})$$
$$E. 21^{m-1}$$

==> From $$3^m7^m^-^1=3(3^m^-^1)(7^m^-^1)=3(3*7)^m^-^1=3(21^m^-^1)$$, the answer is C.

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MathRevolution wrote:
Questions related to geometry are continuously increasing. Let’s have a look at the example of the recent trend.

(ex 4)
Attachment:
GEOMETRY.jpg

If n regular pentagons are tangent each other in points of a circle as above figure, n=?

A. 8
B. 9
C. 10
D. 11
E. 12

Questions like the above are increasing. Therefore, students preparing for GMAT should focus on geometry more intensively.

Can someone please explain the solution of this question?
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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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If x and y are integers greater than 1 and x>y, what are the values of x and y?
1) x+y=13
2) xy=22

==> 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=11 and y=2, hence it is unique and sufficient. The answer is c. However, this is an integer question, one of the key questions, so you apply CMT 4(A).
For con 1), from (x,y)=(11,2),(10,3), it is not unique and not sufficient.
For con 2), you only get (x,y)=(2,11), hence it is unique.

Therefore, the answer is B, not C.
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