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==> 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. Answer: C
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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.

Therefore, the answer is E. Answer: E
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==> 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.

Therefore, the answer is D. Answer: D
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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. Answer: B
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==> 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.

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. Answer: E
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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.

Therefore, the answer is B. Answer: B
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==> 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.

Therefore, the answer is D. Answer: D
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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.

Therefore, the answer is D. Answer: D
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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. Answer: E
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A. (1/6) and (1/5) B. (1/5) and (1/4) C. (1/4) and (1/3) D. (1/3) and (1/2) E. (1/2) and 1

==>The sum of consecutive reciprocal numbers is decided by the first and the last number. In other words, you get (1/9)+(1/9)+(1/9)< (1/7)+(1/8)+(1/9)< (1/7)+(1/7)+(1/7), and if you reorganize this, from =1/3=3/9=(1/9)+(1/9)+(1/9)<(1/7)+(1/8)+(1/9)<(1/7)+(1/7)+(1/7)=3/7<3/6=1/2, you get between (1/3) and (1/2).

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.