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# Overview of GMAT Math Question Types and Patterns on the GMAT

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

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25 May 2017, 01:14
Is a positive integer n a multiple of 6?

1) n is the product of the 4 consecutive integers
2) n is a multiple of 12

==> 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 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), the product of consecutive 4 number is always the multiple of 24, hence yes, it is sufficient.
For con 2), if it is a multiple of 12, it is also a multiple of 6, hence yes, it is sufficient.

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28 May 2017, 23:51
Is x^2>y^2?

1) |x|>y
2) x>|y|

==> 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 variables. 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 x>|y|≥0, it is x>0, so from x>|y|, it is |x|>|y|, which becomes x^2>y^2, hence yes, it is sufficient. The answer is C. However, this is an absolute value question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), you get (x,y)=(2,1) yes (2,-3) no, hence not sufficient. For con 2), from x>|y|≥0, you get x>0, which becomes |x|>|y|, then x^2>y^2, it is always yes and sufficient. Therefore, the answer is B.

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01 Jun 2017, 18:00
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.

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04 Jun 2017, 18:00
We define the average (harmonic mean) as the reciprocal of the average (arithmetic mean) of reciprocals. What is the average (harmonic mean) of 2, 3, and 6?

A. 1/3
B. 1/2
C. 1
D. 2
E. 3

==> If you find the average of the reciprocals, you get $$\frac{++}{3}=13$$ . Since it is the reciprocal of that, the answer is E.

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08 Jun 2017, 18:45
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.

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11 Jun 2017, 18:31
Which of the following is the closest value of x such that $$x^{20}+x^2+0.0000019=0.09?$$

A. 0.1
B. 0.01
C. 0.3
D. 0.03
E. 0.003

==> From $$x^{20}+x^2+0.0000019=0.09, x^{20}$$ and 0.0000019 are very close to 0, so you can ignore them, and the equation becomes $$x^2=0.09=0.3^2$$. Thus, you get x=0.3.

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13 Jun 2017, 12:28
Hi, MathRevolution, just checking - do these patterns still hold true?

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14 Jun 2017, 18:14
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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 =b2-4ac<0, it doesn’t meet with the x-axis, and if a<0, you always get f(x)<0, hence yes, it is sufficient.

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18 Jun 2017, 18:05
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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.

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20 Jun 2017, 19:53
If the sum of the first 50 even numbers is 2,550, what is the sum of the first 50 odd numbers?

A. 1,275
B. 2,550
C. 2,500
D. 2,600
E. 3,000

==> You get 2+4+…..+98+100=2,550, then 1+3+….+97+99=?. If you compare each numbers, 1 is 1 less than 2, 3 is 1 less than 4, 97 is 1 less than 98, and 99 is 1 less than 100. Since there are 50 numbers in total, it is 50 less than 2,550.

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22 Jun 2017, 18:42
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.

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26 Jun 2017, 17:52
|a-b|=?

1) |a+b|=10
2) |a2-b2|=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.

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28 Jun 2017, 16:51
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.

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03 Jul 2017, 18:04
If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?

A. 15
B. 20
C. 25
D. 40
E. 47

==> You get x=4z and y=2z, and the average=(x+y+3z)/3=(4z+2z+3z)/3=3z, so it always needs to be the multiple of 3.

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04 Jul 2017, 18:08
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^x*3^y=16/27=2^4*3^-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.
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06 Jul 2017, 18:27
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).

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09 Jul 2017, 18:32
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.

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10 Jul 2017, 17:54
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.

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12 Jul 2017, 19:07
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.

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16 Jul 2017, 18:49
If x>0, y<0, and a>b>0, then which of the following is (are) positive?
Ⅰ. ax+by Ⅱ. ax-by Ⅲ. by-ax

A. Ⅰonly
B. Ⅱ only
C. Ⅲ only
D.Ⅰ& Ⅱ only
E. Ⅱ & Ⅲ only

==> You getⅠ. ax+by a=2, b=x=1, y=-10, which is negative. (X)
From Ⅱ. ax-by by<0, you get –by>0 and ax>0, ax-by>0, hence it is always positive. (O)
From Ⅲ. by-ax a=2, b=1, x=1, y=-10, it is negative. (X)

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Re: Overview of GMAT Math Question Types and Patterns on the GMAT   [#permalink] 16 Jul 2017, 18:49

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