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20 Jul 2017, 01:01
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When n is divided by 4, what is the remainder?
1) When n is divided by 3, the remainder is 1
2) When n+1 is divided by 4, the remainder is 2
==> 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 remainder questions, you can directly substitute. Therefore, for con 1), from n=3p+1=1,4,…, the remainder when divided by 4 becomes 1=4(0)+1, which makes remainder=1, and from 4=4(1)+0, you get remainder=0, hence it is not unique and not sufficient. For con 2), from n+1=4q+2 and n=4q+1, the remainder when divided by 4 always becomes 1, hence it is unique and sufficient.
Therefore, the answer is B.
Answer: B
1) When n is divided by 3, the remainder is 1
2) When n+1 is divided by 4, the remainder is 2
==> 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 remainder questions, you can directly substitute. Therefore, for con 1), from n=3p+1=1,4,…, the remainder when divided by 4 becomes 1=4(0)+1, which makes remainder=1, and from 4=4(1)+0, you get remainder=0, hence it is not unique and not sufficient. For con 2), from n+1=4q+2 and n=4q+1, the remainder when divided by 4 always becomes 1, hence it is unique and sufficient.
Therefore, the answer is B.
Answer: B
21 Jul 2017, 01:03
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Is 3x greater than x-6?
1) x is greater than -4
2) x is greater than 0
==> If you modify the original condition and the question, you get 3x>x-6?, 2x>-6?, x>-3?. There is 1 variable (x) 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 x>-4, the range of the question doesn’t include the range of the condition, hence it is not sufficient.
For con 2), from x>0, the range of the question includes the range of the condition, hence it is sufficient.
Therefore, the answer is B.
Answer: B
1) x is greater than -4
2) x is greater than 0
==> If you modify the original condition and the question, you get 3x>x-6?, 2x>-6?, x>-3?. There is 1 variable (x) 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 x>-4, the range of the question doesn’t include the range of the condition, hence it is not sufficient.
For con 2), from x>0, the range of the question includes the range of the condition, hence it is sufficient.
Therefore, the answer is B.
Answer: B
23 Jul 2017, 18:27
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Is x^y=?
1) x=1
2) y=1
➔ For this type of questions, the answer is A.
==> In the original condition, there is 2 variable (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), you get x^y=1^1=1, hence it is unique and sufficient. The answer is C. However, according to tip4, a trivial condition cannot be the answer, so if you solve con 1) and con 2) separately, for con 1), you substitute x=1, and get x^y=1^y=1, which is unique and sufficient.
The answer is A.
Answer: A
1) x=1
2) y=1
➔ For this type of questions, the answer is A.
==> In the original condition, there is 2 variable (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), you get x^y=1^1=1, hence it is unique and sufficient. The answer is C. However, according to tip4, a trivial condition cannot be the answer, so if you solve con 1) and con 2) separately, for con 1), you substitute x=1, and get x^y=1^y=1, which is unique and sufficient.
The answer is A.
Answer: A
