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GMAT Inequalities is a high-frequency topic in GMAT Quant, but many students struggle because the concepts behave differently from standard algebra. Understanding the right rules, patterns, and edge cases can significantly improve both speed and accuracy.- May 06
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29 Nov 2016, 04:28
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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
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
02 Dec 2016, 02:01
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|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
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
05 Dec 2016, 19:14
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) 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
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
