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# Math Revolution Approach (DS)

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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]

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26 Apr 2017, 01:15
Of the 100 people, 72% bought stocks and 65% bought bonds. What is the number of people who bought stocks but not bonds?

1) The number of people who bought both stocks and bonds is 45
2) The number of people who bought neither stocks nor bonds is 8

==> If you modify the original condition and the question, you get a 2 by 2, as shown on the table below.
Attachment:

4.26.png [ 2.5 KiB | Viewed 163 times ]

From a+b+c+d=100, a+c=100(72%)=72, and a+b=100(65%)=65, there are 4 variables and 3 equations. 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 a=45, you get 45+c=27, hence c=27 and it is unique and sufficient.
For con 2), from d=8, you get a+b+c+d=65+c+8=100, hence c=27 and it is unique and sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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27 Apr 2017, 01:49
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: Math Revolution Approach (DS) [#permalink]

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28 Apr 2017, 01:31
Attachment:

gc.png [ 1.21 KiB | Viewed 154 times ]

A larger playground has a rectangular-shaped track shaded shown as above figure such that the track has a uniform d as its width. What is the track’s area?

1) The perimeter of the smaller playground is 100.
2) The perimeter of the larger playground is 200.

==> In the original condition, if you set the width and the height of the playground as a and b, there are 3 variables (a,b,d). 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 2(a+b)=100 and 2(a+b)+8d=200, you cannot find a and b in a unique way, hence it is not sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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30 Apr 2017, 18:07
At least one of x, y, and z is 1?

1) Two of them are odd
2) x, y, and z are different integers

==> In the original condition, there are 3 variables (a,b,d) 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), (x,y,z)=(1,2,3) yes, but (x,y,z)=(2,3,5)no, hence it is not sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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30 Apr 2017, 18:08
n=?

1) 7 is a factor of n
2) n is a prime number

==> 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 2), n=7, 14, hence not unique and not sufficient. For con 2), n=7, 11, hence not unique and not sufficient. By solving con 1) and con 2), you get n=7, hence it is unique and sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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04 May 2017, 01:19
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.

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Re: Math Revolution Approach (DS) [#permalink]

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04 May 2017, 01:20
If a is an integer, is a-b an integer?

1) 100 is a factor of a
2) b is 37 percent of a

==> If you modify the original condition and the question and check the question again, from a-b=int?, int-b=int?, you get b=int-int=int?. Since there is 2 variables (a,b), 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 b=0.37a=0.37(100int)=37int=int, hence it is always yes and sufficient. Therefore, the answer is C.
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Re: Math Revolution Approach (DS) [#permalink]

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05 May 2017, 01:46
Set S has six numbers and their average (arithmetic mean) is 32. What is the median of the numbers?

1) The six numbers are greater than or equal to 31
2) There is “37” in set S

==> In the original condition, there are 6 variables and 1 equation. In order to match the number of variables to the number of equations, there must be 5 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), you get
31, 31, 31, 31, 31, 37, and so the median=31+310/2=31, hence it is unique and sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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07 May 2017, 17:58
Event E and event F are independent. Is the probability that both event E and event F will occur less than 0.3?

1) The probability that event E will occur is 0.25.
2) The probability that event F will not occur is 0.75.

==> In the original condition, there are 3 variables (P(E), P(F), P(E∩F)), 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), since
Event E and event F are independent, you get P(E∩F)=P(E)P(F). Also, con1)=con2), so you get P(E∩F)=P(E)P(F)=(0.25)P(F)=P(E)(1-0.75)=P(E)(0.25)<0.3, because P(E)=P(F)≤1, hence it is yes and sufficient.

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07 May 2017, 18:00
m=?

1) $$2^{2m+1}=32$$
2) $$3^{3m-1}=243$$

==> 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 $$2^{2m+1}=32=2^5, 2m+1=5, 2m=4$$, you get m=2, hence it is unique and sufficient. For con 2), you get $$3^{3m-1}=243=3^5$$, 3m-1=5, 3m=6, m=2, hence it is unique and sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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10 May 2017, 01:29
Attachment:

5.8.png [ 4.21 KiB | Viewed 101 times ]

What is the area of the region AOB shaded shown as above?

1) $$∠AOB=120^o$$
2) $$AB=50$$

==> In the original condition, you need to find the radius of the circle and the central angle, so there are 2 variable. 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), the answer is C.

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Re: Math Revolution Approach (DS) [#permalink]

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11 May 2017, 01:21
If the average (arithmetic mean) of a, b, and c is (a+2b)/3, what is the value of b?

1) a=1
2) c=2

==> If you modify the original condition and the question and check the question again, from is (a+2b)/3=(a+b+c)/3, you get a+2b=a+b+c, then b=c. In order to find b, you only need to know c.

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Re: Math Revolution Approach (DS) [#permalink]

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12 May 2017, 01:40
abcd=?

1) abc=1
2) bcd=1

==> In the original condition, there are 4 variables and in order to match the number of variables to the number of equations, there must be 4 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 a=1, b=1,c=1,d=1 and a=2, b=1/2, c=1, d=2, it is not unique and not sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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14 May 2017, 18:31
For positive integer n, is 12 a factor of n?

1) n is a factor of 48
2) n is a multiple of 24

==> In the original condition, there is 1 variable. 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), n=2 no, n=12 yes, hence it is not sufficient.
For con 2), n=24t(t=any positive integer), hence it is always yes and sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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14 May 2017, 18:32
Is 3-digit integer n a prime number?

1) The hundreds digit of n is equal to the tens digit of n
2) The units digit of n is 3

==> In the original condition, there are 3 variables (3-digit integer). 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 2), you get n=333 no and n=113, hence yes, it is not sufficient.

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Re: Math Revolution Approach (DS) [#permalink]

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17 May 2017, 01:30
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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Re: Math Revolution Approach (DS) [#permalink]

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18 May 2017, 01:06
If x and y are integers, is x+y an odd number?

1) y=3x+1
2) y=2x+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), from 3x+1=2x+3, you get x=2 y=7. Then, you get x+y=2+7=9=odd, hence yes, it is sufficient. 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), from y=3x+1, you get (x,y)=(even,odd) or (odd,even), which always becomes x+y=odd, hence yes, it is sufficient. For con 2), you get (x,y)=(1,5) no, (2,7) yes, it is not sufficient.

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19 May 2017, 01:50
Is kr<0?

1) $$k^2r^3<0$$
2) $$|k+r|<|k|+|r|$$

==> For con 1), you ignore the square, so t<0, and for con 2), you get kr<0, hence yes, it is sufficient. The reason is that from $$(|k+r|)^2<(|k|+|r|)^2$$, you get $$k^2+r^2+2kr<k^2+r^2+2|kr|$$, and if you get rid of $$k^2+r^2$$ from both sides, you get 2kr<2|kr|, then kr<|kr|, which becomes kr<0.

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Re: Math Revolution Approach (DS) [#permalink]

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21 May 2017, 17:59
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.

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21 May 2017, 17:59
If a and b are 1-digit positive integers, is 100a+10b+2 divisible by 4?

1) a=2.
2) b=3.

==> If you modify the original condition and the question, the remainder when an integer n is divided by 4 is equal to the remainder when only the units digit and the tens digit are divided by 4, and so you only need to find b. Thus, for con 2), you get b=3, which becomes 10(3)+2=32, and so it is always divisible by 4, hence yes, it is sufficient.

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Re: Math Revolution Approach (DS)   [#permalink] 21 May 2017, 17:59

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