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The Ultimate Q51 Guide [Expert Level]

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New post 29 Nov 2016, 04:28
Also, integer questions constantly appear recently.

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.

Answer: A
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New post 02 Dec 2016, 02:01
|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.

Answer: D
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New post 05 Dec 2016, 19:14
) 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=34, only 3 can be the prime factor of x, hence no, it is sufficient. Therefore, the answer is B.
Answer: B
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New post 08 Dec 2016, 18:31
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 \(2.3x^2\)=0 to x=0, hence it is also unique and sufficient. Therefore, the answer is D.

Answer: D
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New post 12 Dec 2016, 02:16
As always, more advanced questions related to mistake type 3 and 4 are being released. Look at the below. This question, a 5051-level question, was released recently, and it also falls into the mistake type 4(A). You must strengthen your skills in these type of questions. You need to know the relationship between variable approach and mistake types.

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).

Answer: B
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New post 14 Dec 2016, 19:57
The integer question below is also a 5051-level question.

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.
Answer: D
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New post 18 Dec 2016, 23:49
You must be careful with inequality questions as below.

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.
Answer: D
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New post 21 Dec 2016, 19:09
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.
Answer: C
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New post 26 Dec 2016, 18:11
If m and n are positive integers, when \(3^4^n\)+m is divided by 5, what is the remainder?

1) n=1
2) m=4

==> If you modify the original condition and the question, you only need to know “m” from \(3^4^n+m= (34)^n+m=(81)n+m=(~1)+m\), so the answer is B.
This type of question is a 5051-level question related to mistake type 4(A).

Answer: B
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New post 31 Dec 2016, 01:58
1) As always, more developed questions combined with CMT 3 and 4 are being released. Look at the question below. This question, a 5051-level question including CMT 4(A), was released recently. To strengthen your skills for this type of questions, you need to know the relationship between variable approach and CMT.

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).
Answer: D
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New post 02 Jan 2017, 19:08
The question below is also a 5051-level question.

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.
Answer: C
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New post 03 Jan 2017, 18:38
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).

Answer: D
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New post 05 Jan 2017, 05:55
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.
Answer: B
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New post 08 Jan 2017, 18:14
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.

Answer: B
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New post 11 Jan 2017, 17:58
What is the median of the 5 consecutive integers?

1) The average (arithmetic mean) of the 5 consecutive numbers is 5
2) The average (arithmetic mean) of the 5 consecutive numbers is an odd number

==> In the original condition, there is 1 variable (and, n+1, n+2, n+3, n+4), 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), in case of consecutive numbers, you get median=mean, and thus median=5 and it is sufficient.
For con 2), median=1,3,5,.. and hence it is not unique and not sufficient. Therefore, it is a question where A becomes the answer, but you must be aware of CMT 3. Con 2) and con 1) must be independent from each other, and you can’t assume that they are the same after solving con 1).

The answer is A.
Answer: A
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New post 12 Jan 2017, 18:57
Is (a-1)(b+1)=0?
1) \((a-1)^2=0\)
2) \(|a-1|+(b+1)^2=0\)

==> In the original condition, there are 2 variables (a, b), so there must be 2 more equations. Therefore, C is most likely the answer. By solving con 1) and con 2), you get con 1) = con 2), and a=1 and b=-1, hence it is always yes. The answer is D.

Answer: D
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New post 13 Jan 2017, 04:15
Hello Math revolution

You mentioned in your post that integer questions are very important and if someone gets all integer questions correct, then he can score a good 49 or so.

What is the reason behind this ? Are the first 20 questions mainly integer questions?

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New post 15 Jan 2017, 18:47
\(2^a2^b<4\)?

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

==> If you modify the original condition and the question, from \(2^a^+^b<4=2^2\), you get a+b<2? Then, for con 1), the range of the question includes the range of the condition, and therefore it is sufficient.
For con 2), you get a+b=1 yes, but a+b=3 no, and therefore it is not sufficient.
Answer A
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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New post 15 Jan 2017, 21:14
MathRevolution wrote:
The question below is also a 5051-level question.

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.
Answer: C


Your solution is wrong.

You can let x = -2, y = 0, z = 8 and still meet all of the conditions. You are confounding the mean with the median.

E is the correct answer choice.
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New post 16 Jan 2017, 18:55
Is n an even?
1) 3n=even
2) 5n=even

==> In the original condition, there is 1 variable (n), and therefore D is most likely to be the answer. For con 1) n=2 yes and n=2/3 no, so it is not sufficient.
For con 2), n=2 yes and n=2/5 no, so it is not sufficient. By solving con 1) and con 2), you always get n=even, and hence it is sufficient.

Therefore, the answer is C.
Answer: C
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Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 16 Jan 2017, 18:55

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