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# The Ultimate Q51 Guide [Expert Level]

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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02 Mar 2017, 18:31
$$(x-y)^2=?$$

1) x and y are integers
2) xy=3

==> 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,y)=(1,3),(3,1),(-1,-3),(-3,-1), and all become $$(x-y)^2=4$$, hence it is unique and sufficient.

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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05 Mar 2017, 18:24
What is the scope including 1/11+1/12+1/13+......+1/20?

A 1/6~1/5
B. 1/5~1/4
C. 1/4~1/3
D. 1/3~1/2
E. 1/2~1

The sum of consecutive reciprocal number sequence is decided by the first number and the last number. Thus, from
10/20=1/11+1/12+.....+1/20<1/11+1/12+.....+1/20<1/11+1/12+.....+1/20=10/11, you get 1/2=10/20<1/11+1/12+.....+1/20<10/11<1, which becomes 1/2~1.

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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06 Mar 2017, 18:18
$$(x+y)^2-(x-y)^2=?$$

1) xy=5
2) x+y=6

==> If you modify the original condition and the question, $$(x+y)^2-(x-y)^2=?$$ becomes (x+y-x+y)(x+y+x-y)=?, and if you simplify this, you get (2y)(2x)=?, 4xy=?. Thus, for con 1), you get xy=5, hence it is unique and sufficient. The answer is A.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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08 Mar 2017, 18:26
Is a=b?

1) a^2=b^2
2) a=2

==> 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 a=2, you get b2=22=4, then b=±2. Thus, (a,b)=(2,2) yes but (a,b)=(2,-2) no, hence it is not sufficient. The answer is E.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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09 Mar 2017, 22:37
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What is the perimeter of a certain right triangle?

1) The hypotenuse’s length is 10
2) The triangle’s area is 24

==> In the original condition, for a right triangle, there are 2 variables (2 legs) 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 6:8:10 and the perimeter of the right triangle becomes 6+8+10=24, hence unique and sufficient.

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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12 Mar 2017, 18:51
$$Let A=6^2^0, B=2^6^0, and C=4^5^0.$$

Which of the following is true?

A. A<B<C
B. A<C<B
C. B<A<C
D. B<C<A
E. C<B<A

==> You can compare the big and small numbers by making the base or the exponent the same. From$$A=6^2^0, B=2^6^0=(2^3)^2^0=8^2^0$$,
you get 6<8, which becomes A<B, and from $$B=2^6^0=(2^2)^3^0=4^3^0, C=4^5^0,$$
you get 30<50, which becomes B<C.

Thus, you get A<B<C. The answer is A.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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13 Mar 2017, 18:28
$$\frac{5.09}{0.149}$$ is closest to which of the following?

A. 0.34
B. 3.4
C. 34
D. 340
E. 3,400

From $$\frac{5.09}{0.149} = \frac{5.10}{0.15} = \frac{510}{15} =34$$, the answer is C.

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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15 Mar 2017, 18:39
In the x-y plane, what is the slope between y-intercept and a negative x-intercept of $$y=x^2+x-6$$?

A. -2
B. -3
C. 2
D. 4
E. 6

==> From y=x2+x-6=(x+3)(x-2)=0, the x-intercept becomes (-3,0) or (2,0), and (0,-6). From these, the negative x-intercept is (-3,0) and the negative y-intercept is (0,-6). The slope of the straight line that passes through the two points becomes $$\frac{0-(-6)}{-3-0=-2}$$

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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16 Mar 2017, 18:21
If m and n are positive integers, what is the number of factors of $$3^m7^n$$?

A. mn+m+n
B. m-n+1
C. (m-1)(n-1)
D. (m+1)(n+1)
E. mn

==> The number of factors of $$3^m7^n$$ becomes (m+1)(n+1).

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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20 Mar 2017, 03:44
Is a positive integer n a multiple of 12
1) n is a multiple of 6
2) n is a multiple of 24

==> 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 remainder questions, you always use direct substitution.
For con 1), n=6 no, n=12 yes, hence not sufficient.
For con 2), n=24,48,…, hence it is always yes and sufficient.

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20 Mar 2017, 18:48
Is x>y>z?
1) x>y
2) y>z

==> 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), from x>y>z, it is always yes and sufficient.

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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22 Mar 2017, 18:33
Is a positive integer x a factor of 24?
1) 3x is a factor of 48
2) 2x is a factor of 24

==> 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), x is a factor of 16, so x=8 yes, x=16 no, hence not sufficient.
For con 2), x is a factor of 12, so it is always yes, hence sufficient.

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26 Mar 2017, 18:34
In the x-y plane there is a line K, (x/a)+(y/b)=1. What is the x-intercept of line K?

1) a=b
2) a=1

==> If you modify the original condition and the question, the x-intercept is the value of x when y=0, hence from (x/a)=1, you get x=a, so you only need to find a.

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27 Mar 2017, 18:05
If the average (arithmetic mean) of 5 consecutive multiples of 5 is 30, what is the smallest number of them?

A. 5
B. 10
C. 15
D. 20
E. 25

==> If the average of 5 consecutive multiples of 5 is 30, from 20,25,30,35,40, the smallest multiple of 5 becomes 20.

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29 Mar 2017, 18:52
If Tom goes y miles in x hours, how many miles does he go per minute, in terms of x and y?

A. 60x/y
B. y/60x
C. 60y/x
D. x/60y
E. 60xy

==>You get miles:hours=y(miles):x(hours)=y(miles):60x(minutes), and from y:60x=some:1, you get some=y/60x.

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30 Mar 2017, 18:33
m=?

1) 5 is a factor of m
2) m is a prime number

==> In the original condition, there is 1 variable (m) 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 m=5,10…, it is not unique and not sufficient.
For con 2), from m=5,7…, it is not unique and not sufficient.
By solving con 1) and con 2), you get m=5, hence it is unique and sufficient.

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02 Apr 2017, 18:25
If two times x is 5 greater than three times y, what is the value of y, in terms of x?

A. y=2x-5
B. y=6x-5
C. y=(x/2)-5
D. y=(x/3)-5
E. y=(2x-5)/3

==> According to the Ivy Approach, you get is:”=” and greater than:”+”, so you get 2x=5+3y. From 3y=2x-5, y=(2x-5)/3, the answer is E.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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04 Apr 2017, 06:49
Is x<1<y?

1) x<√x<y
2) x<√y<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 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 x<√x, you get 0<x<1, and from √y<y, you get y>1. Then, you get x<1<y, hence always yes.

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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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06 Apr 2017, 18:30
[m] is defined as the greatest integer less than or equal to m, what is the value of [m]?

1) 1<m<2
2) |m|<1

==> In the original condition, there is 1 variable (m) 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 [m]=1, it is unique and sufficient.
For con 2), from -1<m<1, you get [0]=0, but from [0.1]=-1, it is not unique and not sufficient.

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09 Apr 2017, 18:39
If 1 male, 2 females, and 1 child are to be selected at random from 8 males, 10 females, and 8 children, respectively, how many such cases are possible?

A. 980
B. 1,440
C. 1,880
D. 2,480
E. 2,880

==> From (8C1)(10C2)(8C1)=(8)(10*9/2!)(8)=2,880, the answer is E.
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Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 09 Apr 2017, 18:39

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# The Ultimate Q51 Guide [Expert Level]

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