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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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09 Nov 2016, 00:36
(ex 20) (function) In(on) the xy coordinate plane, what is the area of a circle K?
1) The circle has (2,1) as its center
2) The circle passes through a point (3,4)

==>In the original condition, since the circle is on the coordinate plane, there are 3 variables (the center (a,b)) and becomes the radius r), so E is highly likely to be the answer. Through 1) & 2), you get the radius r=√(〖(2-3)〗^2+〖(1-4)〗^2 )=√(1+9)=√10, and the area of the circle is π(√10)^2=10π. Since the answer is unique, it is sufficient. The answer is C.
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]

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10 Nov 2016, 00:18
There is an arithmetic sequence An where n is a positive integer such that An+1=An+3, what is the value of A10?
1) A1=3.
2) A3+A6=27

==>In the original condition, there are 2 variables (An+1, An), and 1 equation (An+1=An+3), so D is highly likely to be the answer. 1)=2), so you get A10=30, hence unique, and sufficient. The answer is D.

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11 Nov 2016, 00:06
If x/300=y/100, x=?
1) x+y=400
2) y=100

==>In the original condition, there are 2 variables (x,y) and 1 equation (x/300=y/100), so D is highly likely to be the answer. Hence 1)=2), the remainder of both is x=300, so it is unique, and sufficient. The answer is D.
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]

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14 Nov 2016, 00:32
Is |x-1|>|x-3|?
1) x>2
2) x>1

==>If you modify the original condition and the problem, the sign of inequality does not change even if you square both sides. In other words, |x-1|>|x-3|?,Is |x-1|^2>|x-3|^2?, (x-1)^2>(x-3)^2?, and if you expand this, you get x^2-2x+1>x2-6x+9?, 6x-2x>9-1?, 4x>8?. That is, you get x>2?. 1) is yes, and sufficient. The answer is A.
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]

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14 Nov 2016, 00:33
For all integers k, the function g is defined by g(k) = ak+4, where a is a constant. What is the value of g(1)?

(1) g(3) = 7

(2) g(-1) =3

If you modify the original condition and the problem, you get g(1)=a×1+4=a+4=?. Since you get 1 variable(a), and in order for it to match with the number of equation it needs one, hence 1 for 1), and 1 for 2). Thus, there is high chance that D is the answer.

In the case of 1), from g(3)=3a+4=7 you get a=1, therefore unique and sufficient.
In the case of 2), from g(-1)=a(-1)+5=-a+4=3 you get a=1, therefore unique and sufficient.

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

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17 Nov 2016, 03:16
m=?
1) 3^3m-1=9
2) 5^2m+1=125

==> In the original condition, there is 1 variable (m), and therefore D is most likely to be the answer.
In case of con 1), from 3^3m-1=9=3^2, you get 3m-1=2 and m=1, and hence it is sufficient.
In case of con 2), from 5^2m+1=125=5^3 and 2m+1=3, you get m=1, and hence it is sufficient.
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Re: Math Revolution Approach (DS) [#permalink]

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18 Nov 2016, 00:23
If x and y are integers greater than 1 and x>y, what are the values of x and y?
1) x+y=14
2) xy=33

==> In the original condition, there are 2 variables (x, y) and therefore C is most likely to be the answer. By solving con 1) and con 2), you get x=11 and y=3. However, since this question is an integer question, one of the key questions, if you apply CMT 4 (A), in case of con 1), from x=11, y=3 or x=10 y=4, it is not unique and not sufficient, and in case of con 2), you only get x=11 and y=3, and hence it is unique and sufficient. Therefore, the answer is B.
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21 Nov 2016, 00:59
Is ab+cd>0?
1) ac+db>0

==> In the original condition, there are 4 variables (a, b, c, d). Therefore, E is most likely to be the answer. By solving con 1) and con 2), (a, b, c, d) = (3, -4, -1, -1) is no, and (1,1,1,1) is yes. Therefore, the answer is E.

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Last edited by MathRevolution on 21 Nov 2016, 23:39, edited 1 time in total.
Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]

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21 Nov 2016, 01:01
What is the standard deviation of 3 products’ prices?

1) The sum of any two prices of these 3 products is \$1,000
2) At least one of them is 500

==> In the original condition, there are 3 variables. Therefore, E is most likely to be the answer. For con 1), the sum of any 2 prices of the products is always 1,000, so if you add the three equations a+b=1,000, b+c=1,000, and c+a=1,000 together, you get 2(a+b+c)=3,000 and a+b+c=1,500 each, and using substitution, a=b=c=500. Thus, the standard deviation is 0, hence it is sufficient. Therefore, the answer is A. This question is CMT 4 (A).

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

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23 Nov 2016, 00:19
If x is positive integer, is √2x an integer?

1) x/2 is an integer squared
2) x/200 is an integer squared

==> In the original condition, there is 1 variable (x). Therefore, D is most likely to be the answer.
For con 1), from x=2t^2, you get √2x = √(2*2t^2 ) =2t, hence yes, it is sufficient.
For con 2), from x=200s^2, you get √2x = √(2*200s^2 ) =20s, hence yes, it is sufficient.

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24 Nov 2016, 01:41
If a+b is even, is b an integer?
1) a-b=integer
2) a+3b=even

==> In the original condition, there are 2 variables (a, b) and 1 equation (a+b=even). Therefore, D is most likely to be the answer.
For con 1), a=b=1 is yes, and a=5/2, b=1/2 is no, hence it is not sufficient.
For con 2), a+3b=a+b+2b=even, even+2b=even, 2b=even, b=integer, hence yes, it is sufficient. Therefore, the answer is B.

This type of question is a 5051-level question related to CMT 4 (A).
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25 Nov 2016, 01:20
1
KUDOS
Expert's post
The question below is also a 5051-level question, a typical integer question.

When a positive integer n has 6 different factors, n=?
1) n has 2 prime factors
2) n<18

==> In the original condition, in case of standard deviation questions, there is 1(n) variable, and therefore D is most likely to become the answer.

In case of con 1), you get $$n=2^23^1, 2^13^2,$$ hence it is not unique and not sufficient.
In case of con 2), you get only $$n=2^23^1$$, hence it is unique and sufficient.

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28 Nov 2016, 00:34
Is x<0?
1) $$x^5$$<0
2)$$x^5$$+x+1=0

==> In the original condition, there is 1 variable, and therefore D is most likely to be the answer.
In case of con 1), from $$x^5$$<0, if you divide both sides by $$x^4$$, you get x<0, hence yes, it is sufficient.
In case of con 2), it cannot be factored, and this is when you use CMT 4 (B). In other words, if you get A and B too easily, then consider D. Since$$x^5$$+x=-1, x($$x^4$$+1)=-1, and x=-1/$$(x^4$$+1) always agree with x4+1>0, and hence yes, it is sufficient. Therefore, the answer is D.
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS) [#permalink]

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28 Nov 2016, 00:36
If $$a^3b^4c^5<0$$, is abc<0?
1) c<0
2) b>0

==> If you modify the original condition and the question, when you divide both sides of $$a^3b^4c^5<0$$ by $$a^2b^4c^4,$$ you get ac<0, and the question abc<0? becomes b>0?. From con 2) b>0, it becomes yes, hence it is sufficient.
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Math Revolution GMAT Instructor
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30 Nov 2016, 02:08
If m and n are positive integers, is $$n^m-n$$ divisible by 6?
1) m=3
2) n=2

==> 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. Therefore, C is most likely to be the answer. By solving con 1) and con 2), from$$2^3-2=6$$, you get yes, and hence it is sufficient. The answer is C. However, this question is an integer question, one of the key questions, so you need to apply CMT 4. For con 1), from $$n^3-n=(n-1)n(n+1)$$, it is the multiple of the three consecutive integers, which always becomes the multiple of 6, hence yes, it is sufficient. For con 2), from n=2 and m=3 yes, m=2 no, and hence it is not sufficient. Therefore, the answer is A.

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

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01 Dec 2016, 00:57
|a-b|=b-a?
1) a<b
2) ab<0 and $$a^2b$$=1

==> If you modify the original condition and question, to get |A|=-A, you need to get A≤0?, so it becomes |a-b|=b-a=-(a-b)?, a-b<0?, a<b?. Thus, for con 1), it is always yes, hence sufficient, and for con 2), if you apply CMT 4 (B), it also becomes a<b, hence yes, it is sufficient. Therefore, the answer is D.
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Re: Math Revolution Approach (DS) [#permalink]

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02 Dec 2016, 00:32
If x is a positive integer greater than 1, is 1/x a terminating decimal?

1) x has 3 as a factor
2) x is a factor of 81

==> If you modify the original condition and the question, to get 1/x as the terminating decimal, only 2 or 5 can be the prime factors of denominator x. However, for con 2), in order for x to become factors of 81=$$3^4$$, only 3 can be the prime factor of x, hence no, it is sufficient. Therefore, the answer is B.

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05 Dec 2016, 00:34
If 5 different positive integers have 3 as its median, is the average (arithmetic mean) of them greater than 5?
1) The greatest integer of them is 16
2) The smallest integer of them is 1

==> If you modify the original condition and the question, the sum of 5 integers>585=25?, and so there are 5 variables and 1 equation. Therefore, E is most likely to be the answer. However, if the question is “greater than”, you need to find the least value. By solving con 1) and con 2),
The least value of the sum becomes 1+2+3+4+16=26>25 yes, hence it is sufficient. The answer is C. However, this question is a key question, so you need to apply CMT 4 (A).
For con 1), the least value of the sum=1+2+3+4+16=26>25, hence yes, it is sufficient.
For con 2), 1+2+3+4+5=15<25 is no, 1+2+3+10+30=46>25 is yes, hence it is not sufficient. Therefore, the answer is A.
This question, related to CMT 4 (A), is 5051-level question in current GMAT.

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05 Dec 2016, 00:37
x=?
1) 1.1x=-1.2x
2) $$1.1x^2$$=-$$1.2x^2$$

==> 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 as well. Therefore, D is most likely to be the answer.
For con 1), you get from 2.3x=0 to x=0, hence it is unique and sufficient, and
For con 2), you get from 2.3x2=0 to x=0, hence it is also unique and sufficient. Therefore, the answer is D.

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07 Dec 2016, 00:33
What is the median of the 9 numbers?

1) The smallest 5 numbers of them are less than or equal to 10.
2) The largest 5 numbers of them are more than or equal to 10.

==> In the original condition, since there are 9 variables, E is likely to be the answer.
Through 1) & 2), in order of size, the first 5 numbers are smaller than 10 or the same. Also, the last 5 numbers are bigger than 10 or the same. Hence, the fifth number should be always 10. Then, the fifth number becomes median and median=10, which is unique and sufficient. Hence, the answer is C.
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Re: Math Revolution Approach (DS)   [#permalink] 07 Dec 2016, 00:33

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