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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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26 Feb 2017, 21:33
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Is a=b?
1) \(a^2=b^2\)
2) \(a=2\)
==> In the original condition, there are 2 variables (a,b) 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), from a=2, you get \(b^2=2^2=4\), then b=±2. Thus, (a,b)=(2,2) yes but (a,b)=(2,-2) no, hence it is not sufficient.
The answer is E.
Answer: E
1) \(a^2=b^2\)
2) \(a=2\)
==> In the original condition, there are 2 variables (a,b) 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), from a=2, you get \(b^2=2^2=4\), then b=±2. Thus, (a,b)=(2,2) yes but (a,b)=(2,-2) no, hence it is not sufficient.
The answer is E.
Answer: E
02 Mar 2017, 01:38
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If mn≠0, what is the value of \(\frac{m+1}{m}-\frac{n+1}{n}\)?
1) m=n
2) m=1
==> If you modify the original condition and the question, you get \(\frac{m+1}{m}-\frac{n+1}{n}=\frac{n(m+1)-m(n+1)}{mn}= \frac{nm+n-mn-m}{mn}=\frac{n-m}{mn}=?\). Then, for con 1), m=n, so n-mmn=0, hence it is unique and sufficient.
The answer is A.
Answer: A
1) m=n
2) m=1
==> If you modify the original condition and the question, you get \(\frac{m+1}{m}-\frac{n+1}{n}=\frac{n(m+1)-m(n+1)}{mn}= \frac{nm+n-mn-m}{mn}=\frac{n-m}{mn}=?\). Then, for con 1), m=n, so n-mmn=0, hence it is unique and sufficient.
The answer is A.
Answer: A
03 Mar 2017, 01:12
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If a, b, and c are integers, is abc an even?
1) a+b is an even
2) b+c is an even
==> In the original condition, there are 3 variables (a,b,c) and in order to match the number of variables to the number of equations, there must be 3 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), if (a,b,c)=(1,1,1), you get a+b+c=1+1+1=3=odd, so no, but if (a,b,c)=(2,2,2), you get a+b+c=2+2+2=6=even, so yes, hence it is not sufficient.
The answer is E.
Answer: E
1) a+b is an even
2) b+c is an even
==> In the original condition, there are 3 variables (a,b,c) and in order to match the number of variables to the number of equations, there must be 3 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), if (a,b,c)=(1,1,1), you get a+b+c=1+1+1=3=odd, so no, but if (a,b,c)=(2,2,2), you get a+b+c=2+2+2=6=even, so yes, hence it is not sufficient.
The answer is E.
Answer: E
