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Intern  Joined: 26 Nov 2016
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Re: Math Revolution Approach (DS)  [#permalink]

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MathRevolution wrote:
At least 10 cars have tinting window and fog light. 40% cars which have tinting windows also have fog light, is the number of the cars with tinting windows larger than that of the cars with the fog light?
1) 80% cars which have fog light also have tinting windows
2) 52 cars have tinting windows or fog light or both

ANS: If we were to solve this question with our trick, we should observe if one of conditions include ratio. If one of conditions contains ratio, then the condition with ratio has 90% of chance to become the answer. Since the condition 1) has ratio, the correct answer is A.

Can you please explain the solution in detail

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Math Revolution GMAT Instructor V
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a, b, c, d, and e are integers. Is the median of the integers greater than the average (arithmetic mean) of the integers?

1) a<b<c<d<e
2) b-a=e-d

==> In the original condition, there are 5 variables, so E is most likely to be the answer.
By solving con 1) & con 2), you get (a,b,c,d,e)=(1,2,3,4,5), hence it is no, and you get (a,b,c,d,e)=(1,2,101,102,103),
hence it is yes, so it is not sufficient. Therefore, the answer is E.

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Is x<y<z?

1) x-2<y-1<z
2) x+2<y+1<z

==> In the original condition, there are 3 variables (x, y, z), so E is most likely to be the answer.
By solving con 1) & con 2), you get 2x<2y<2z. From x<y<z, it is always yes, hence it is sufficient.
However, if you look at con 2), from x+1<x+2<y+1<z<z+1, you get x+1<y+1<z+1, and x<y<z, so it is always yes, hence it is sufficient.

Therefore, the answer is B.
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Re: Math Revolution Approach (DS)  [#permalink]

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Hemanth2017 wrote:
MathRevolution wrote:
At least 10 cars have tinting window and fog light. 40% cars which have tinting windows also have fog light, is the number of the cars with tinting windows larger than that of the cars with the fog light?
1) 80% cars which have fog light also have tinting windows
2) 52 cars have tinting windows or fog light or both

ANS: If we were to solve this question with our trick, we should observe if one of conditions include ratio. If one of conditions contains ratio, then the condition with ratio has 90% of chance to become the answer. Since the condition 1) has ratio, the correct answer is A.

Can you please explain the solution in detail

Sent from my C6902 using GMAT Club Forum mobile app

Hi Hemanth2017,

In general, one con is % and if one con is number, % is likely to be an answer.
When it comes to 1), amongst cars with tinting windows, 40% of them have fog light.
For the same cars with fog light, 80% have tinting windows.
For instance, cars with only tinting window - six cars, both tinting window and fog light - four cars, only fog light - one car, which is 10>5 and yes.
Hence, the answer is A.

Happy Studying!
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If the ratio of a to b to c to d is 1 to 2 to 3 to 4, what is the range of a, b, c and d?

1) The average (arithmetic mean) of b and d is 9
2) The sum of a and d is 15

==> If you modify the original condition and the question and look at the question again, from a=k, b=2k, c=3k, d=4k, there are 5 variables (a, b, c, d, k), and 4 equations. Therefore, D is most likely to be the answer. Con 1) = con 2), so k=3, and from range=d-a=4k-k=3k=3(3) = 9, it is sufficient.

Therefore, D is the answer.

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What is the standard deviation (d) of m, n, p?

1) m, n, and p are different
2) $$d^2-2d=0$$

==> From the original condition, there are 3 variables, and therefore E is most likely to be the answer. By solving con 1) & con 2), from d(d-2)=0, you get d=0, 2. From con 1), m, n, and p are different and d≠0, it is always d=2 only, so it is unique and sufficient.

Therefore, the answer is C.

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What is the median of 5 numbers?

1) Each of the 3 numbers is 11
2) The 2 greatest numbers of them are 21 and 25

==> If you modify the original condition and the question, the median of 5 numbers become the third number.
In case of con 1), from 11,11,11,( ),( ) or ( ),11,11,11,( ) or ( ), ( ),11,11,11, it is always median=11, so it is unique and sufficient.

Therefore, the answer is A.
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If x≠0, |x|/x=?
1) x<0
2) x=-1

==> If you modify the original condition and the question, you get |x|=-x (x<0). Thus, if x<0, you get |x|/x=-x/x=-1, so con 1) = con 2), and therefore the answer is D.
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If the average (arithmetic mean) of 3 integers a, b, and c is 5, what is the value of a?

1) b=7
2) b=1-c

==> If you modify the original condition and the question and look at the question again, from a+b+c=15, you get a=15-(b+c), so you only need to know b+c. From con 2), you get b+c=1, so from a=15-(b+c)=15-1=14, it is unique and sufficient. Therefore, the answer is B. This question is related to mistake type 4(A).

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If x and y are prime numbers, what is the smallest prime factor of $$xy^3$$ ?

1) x=even
2) x+y=odd

==> In the original condition, there are only 2 variables (x, y), hence C is most likely to be the answer. For con 1), if x=even, an even number that is also a prime number is only “2”, so the smallest prime factor of $$xy^3$$ becomes 2, and hence it is sufficient.
For con 2), in order to get x+y=odd, since it is even+odd=odd, x or y always becomes “2”, so the smallest prime factor of $$xy^3$$ becomes “2”, and hence it is also sufficient.
This question is also related to mistake type 4(B), where con 1) is easy and con 2) is difficult. This type of question is a 5051-level question.

The answer is D.
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If $$a^3b^4c^5<0$$, is $$ab^2<0$$?

1) a<0
2) c>0

==> If you modify the original condition and the question and look at the question again, from If $$a^3b^4c^5<0$$, there are only odd variables, so you get If ac<0. Then, the que is $$ab^2<0$$? becomes a<0?, and hence con 1) is yes and sufficient. Also, from ac<0, you get c>0?, and hence con 2) is yes and sufficient.

Therefore, the answer is D.
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If ab=c, b=?
1) c≠0
2) |a-c|≤0

==> In the original condition, there are 3 variables (a, b, c) and 1 equation (ab=c). In order to match the number of variables to the number of equations, there must be 2 equations, and therefore C is most likely to be the answer.
By solving con 1) and con 2), from con 2), you get a=c, and if you substitute this into ab=x, you get ab=a. In order to divide a from both sides, it needs to be a≠0. From con 1), it is c=a≠0, and if you divide both sides by a, you get b=1, hence it is unique and sufficient.

Therefore, the answer is C.
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If x, y and z are different integers, what is the value of y?

1) The average (arithmetic mean) of x, y and z is 2
2) x<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, and therefore E is most likely to be the answer. By solving con 1) & con 2), you get (x+y+z)/3=2. Since it is x<y<z, you always get median=average=y=2. Hence it is unique and sufficient.

The answer is C.
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If r and s are positive integers, is r+s an odd number?
1) r and s are consecutive
2) r=s+1

==> In the original condition, there are 2 variables (r, s), and therefore C is most likely to be the answer. By solving con 1) & con 2), you get con 1) = con 2), so it becomes r+s=s+1+s=2s+1=odd, and hence it is yes and sufficient.

Therefore, the answer is D.
This is a 5051-level question related to mistake type 4(B).
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If x and y are positive integers, xy-2=?  [#permalink]

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If x and y are positive integers, $$x^y^-^2$$=?
1) $$x^2=1$$
2) $$y^2=4$$

==> In the original condition, there are 2 variables (x), and in order to match the number of variables to the number of equations, there must be 2 equations, and therefore C is most likely to be the answer. By solving con 1) & con 2), you get x=-1, 1 and y=-2, 2. Since x and y are positive integers, only x=1 and y=2 are possible, and con 1) becomes $$1^y^-^2=x^2^-^2=x^0=1$$.

Therefore, the answer is D.
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If a and b are integers, is ab an odd?

1) a=0
2) b=1-a

==> In the original condition, there are 2 variables (a, b), and therefore C is most likely to be the answer. By solving con 1) and con 2), from a=0 and b=1, you get ab0*1=0=even, hence no, it is sufficient. Therefore, C is the answer. However, this is an integer question, one of the key questions. Thus, if you apply CMT 4 (A, B), if 1) a=0, you get ab=0 and it is always even, hence no, it is sufficient. Also, for con 2), from a+b=1=odd, and (a, b)=(odd, even), (even, odd), you get ab=even, hence yes, it is sufficient. Therefore, the answer is D. This question is related to CMT 4(B). In other words, con 1) is easy and con 2) is difficult, so you apply CMT 4(B) (If you get A and B easily, consider B).
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Is n an even number?

1) n(n+1)/2 is an even number
2) n(n+2) is an even number

==> In the original condition, there is 1 variable (n), and therefore D is most likely to be the answer.
However, for con 1), you get n=3 no, n=8 yes, and hence it is not sufficient, and for con 2), in order to get n(n+2)=even, you need to get n=even, and hence yes, it is sufficient.

Therefore, the answer is B.
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Is 1/a>1/b?
1) a<b
2) a<b<0

==> If you modify the original condition and the question and check the question again, you get a<b<0--> 1/a>1/b. In other words, it does not become a<b --> 1/a>1/b.

Therefore, B is the answer.
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If ab=0, is |a-b|>0?
1) a=0
2) b<0

==> If you modify the original condition and the question, you get is a≠b? Also, there are 2 variables (a, b) and 1 equation (ab=0), in order to match the number of variables to the number of equations, there must be 1 equation as well, and therefore D is most likely to be the answer.
For con 1), you get a=b=0 no a=0 and b=3, hence yes, it is not sufficient.
For con 2), you get b<0 and a=0, hence yes, it is sufficient.

Therefore, the answer is B.
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If m and n are positive integers, is $$10^n$$+m divisible by 3?
1) n=1
2) m=2

==> If you modify the original condition and the question, “Is $$10^n$$+m=3t? (t=any positive integer)” is equal to “Is the sum of all the digits of $$10^n$$+m divisible by 3?” Then, as from con 2), if you know m=2, from $$10^n$$+2 => 12, 102, 1002…., the sum of the digits always become 3, which is divisible by 3, and hence yes, it is sufficient.

Therefore, the answer is B.
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