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# The Ultimate Q51 Guide [Expert Level]

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Math Revolution GMAT Instructor
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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

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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]

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10 Apr 2017, 18:46
What is the range of 25 consecutive even numbers?

A. 44
B. 46
C. 48
D. 50
E. 52

==> The number of consecutive even numbers=(The last number-the first number)/2 +1=range/2+1, so from range/2+1=25, range/2=25-1=24, you get range=2(24)=48. Therefore, the answer is C.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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12 Apr 2017, 18:31
If x and y are positive integers, is x divisible by y?

1) x is divisible by 6
2) y is divisible by 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. By solving con 1) and con 2), (x,y)=(6,6) yes, but (x,y)=(6,12) no, hence it is not sufficient.

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

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13 Apr 2017, 18:22
If (4/9)÷n=3/5, n=?

A. 20/27
B. 27/20
C. 10/9
D. 9/10
E. 9/4

==>(4/9)÷n=3/5 -> (4/9)(1/n)=3/5. And 4/9n=3/5 -> n=(4/9)(5/3)=20/27.

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

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16 Apr 2017, 18:32
2x+3y=?

1) x=6
2) 4x+6y=18

==> In the original condition, you get 2x+3y=?. For con 2), if you divide both sides by 2, you get 2x+3y=9, hence it is unique and sufficient.

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

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17 Apr 2017, 18:16
If 3 juniors from 8 juniors are selected at random to make a committee, how many cases are possible?

A. 8
B. 28
C. 35
D. 42
E. 56

==> In general, you solve probability questions using nCr, a combination. In other words, from 8C3=56, the answer is E.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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18 Apr 2017, 18:48
If $$x^{-2}+x^{-1}=0$$, x=?

A. -1
B. 0
C. 1
D. 0, 1
E. -1, 0

==> If you multiply $$x^2$$ on both sides, you get 1+x=0, x=-1. The answer is A.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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19 Apr 2017, 18:31
If b≠0, |a/b|=a/b, is b>0?

1) a>0
2) a+b>0

==> If you modify the original condition and the question, in order to satisfy |a/b|=a/b, you must get a/b≥0, and if you multiply b^2 on both sides, from (b^2)a/b≥(b^2)0, you get ab≥0. Since the question is b>0? becomes a≥0?, con 1) is yes and sufficient. For con 2), using CMT 4 (B: if you get A or B too easily, consider D), you only get ab≥0 and a+b>0 when a>0 and b>0, hence yes, it is sufficient.

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

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20 Apr 2017, 18:28
When n and k are positive integers, what is the greatest common divisor of n+k and n?

1) n=2
2) k=1

==> 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 con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), you get n+k=2+1=3 and n=2, and GCD(3,2)=1, hence it is unique and sufficient. Therefore, the answer is C. However, this is an integer 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), k is unknown hence it is not sufficient.
For con 2), if k=1, n+k(=n+1) and n becomes 2 consecutive integers, so always GCD=1, hence it is unique and sufficient.

Therefore, the answer is B, not C.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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24 Apr 2017, 18:44
If x and y are positive, is (y/x)+(x/y)>2?

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

==> If you modify the original condition and the question, even if you multiply xy on both sides, you get xy>0, hence the inequality sign doesn’t change. Thus, you get is (y/x)+(x/y)>2?, or $$y^2+x^2>2xy?$$, or $$y^2+x^2-2xy>0$$?, or $$(x-y)^2>0$$?, or x≠y?. From con 1) = con 2), it is always yes and sufficient.

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

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25 Apr 2017, 18:10
When a certain coin is flipped, the probability that the coin will land on head or tail is 1/2 each. If the coin is flipped 4 times, what is the probability that it will land on tail at least twice on 4 flips?

A. 3/8
B. 1/16
C. 1/2
D. 5/8
E. 11/16

==> In general, you solve probability questions using nCr combination. In other words, from TTHH, there are (4!/2!/2!)=6 possibilities, so you get (1/2)(1/2)(1/2)(1/2)*6=3/8. From TTTH, there are (4!/3!)=4 possibilities, so you get (1/2)(1/2)(1/2)(1/2)*4=1/4. From TTTT, there is only 1 possibility, so you get (1/2)(1/2)(1/2)(1/2)=1/16 and (3/8)+(1/4)+(1/16)=(6+4+1)/16=11/16. The answer is E.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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26 Apr 2017, 17:58
There are a lot of products. Is the standard deviation of the prices of the products less than $60? 1) The median value of their price is$100
2) The range of their price is $110 ==> If you modify the original condition and the question, you get standard deviation(d)≤range/2. Then, from d≤range/2=$110/2=$55<$60, it is always yes and sufficient.

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

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27 Apr 2017, 19:13
When a/b=1.2, (a-b)/(a+b)=?

A. 1/15
B. 1/11
C. 1/9
D. 3/8
E. 1/3

==> From a/b=1.2=6/5, if you substitute a=6 and b=5, you get (a-b)/(a+b)=(6-5)/(6+5)=1/11.

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

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30 Apr 2017, 18:14
If nk≠0, n is what percent of k?

1) k=0.2n
2) (k+n)/n=1.2

==> If you modify the original condition and the question, from n=(some)(1/100)k, you get some=100(n/k). However, con 1) = con 2), so you get some=100(0.2)=20, hence it is unique and sufficient.

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

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01 May 2017, 18:47
If the sum of the annual salary of n persons is $x and the monthly salary per person is$y, what is the value of n in terms of x and y?

A. $x/12y B.$12x/y
C. $12xy D.$12y/x
E. \$xy/12

==> If the annual salary per person: s, you get ns=x, s=12y. If you substitue this, you get n(12y)=x, n=x/12y.

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

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03 May 2017, 18:43
If n is the product of 3 consecutive integers, which of the following must be true?

I. a multiple of 2 II. a multiple of 3 III. a multiple of 4

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

==> From n=m(m+1)(m+2) , where m=integer, the product of 3 consecutive integers is always a multiple of 6. Thus, I and II is always true. For III, if m=1, from n=(1)(2)(3)=6, it is not a multiple of 4, hence it is false.

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

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07 May 2017, 18:05
If the average (arithmetic mean) of set A is 10,000 and the average (arithmetic mean) of set B is 10,000, what is the range of set A and set B combined?

1) The range of set A is 6,000
2) The range of set B is 3,000

==> If you modify the original condition and the question, since there are 2 sets, set 1’s range=set 1’s Max-set 1’s min, and set 2’s range=set 2’s Max-set 2’s min. Thus, there are 6 variables and 2 equations, and in order to match the number of variables to the number of equations, there must be 4 more 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 max and the min when combined is unknown, hence it is not sufficient. Therefore, the answer is E.
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Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 07 May 2017, 18:05

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# The Ultimate Q51 Guide [Expert Level]

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