[Math Revolution GMAT math practice question]
Is 1/x>1/y?
1) x < y
2) x > 0
Is 1/x>1/y?
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- Max@Math Revolution
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- Brent@GMATPrepNow
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Target question: Is 1/x > 1/y?Max@Math Revolution wrote:
Is 1/x > 1/y?
1) x < y
2) x > 0
This is a good candidate for rephrasing the target question.
Take: 1/x > 1/y
Subtract 1/y from both sides to get: 1/x - 1/y > 0
Rewrite with common denominators: y/xy - x/xy > 0
Combine: (y - x)/xy > 0
REPHRASED target question: Is (y - x)/xy positive?
Statement 1: x < y
Let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 1 and y = 2. In this case, (y - x)/xy = 1/2. So, the answer to the REPHRASED target question is YES, (y - x)/xy IS positive
Case b: x = -1 and y = 2. In this case, (y - x)/xy = 3/-2 = -3/2. So, the answer to the REPHRASED target question is NO, (y - x)/xy is NOT positive
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x > 0
Since we have not information about y there's no way to answer the REPHRASED target question with certainty.
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that x < y
Statement 2 tells us that 0 < x
When we combine the statements, we get: 0 < x < y
If x < y, then y - x > 0. In other words, (y - x) is POSITIVE
Also, if x and y are both greater than 0, we know that the product xy is POSITIVE
So, (y - x)/xy = POSITIVE/POSITIVE = POSITIVE
So, the answer to the REPHRASED target question is YES, (y - x)/xy IS positive
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
- Max@Math Revolution
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=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The question may be modified as follows:
1/x > 1/y
=> xy^2 > x^2y by multiplication by x^2y^2
=> xy^2 - x^2y > 0
=> xy(y-x) > 0
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Applying both conditions together yields y > x > 0. So, x>0, y>0 and y-x>0.
It follows that the product xy(y - x) is positive. Therefore, both conditions are sufficient when considered together.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
If x = 2 and y = 3, then 1/x = 1/2, 1/y = 1/3, and the answer is 'yes'.
If x = -2 and y = 3, then 1/x = -1/2, 1/y = 1/3, and the answer is 'no'.
Condition 1) is not sufficient on its own.
Condition 2)
This condition provides us with no information about the variable y, so it is not sufficient.
Therefore, the answer is C.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provides an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The question may be modified as follows:
1/x > 1/y
=> xy^2 > x^2y by multiplication by x^2y^2
=> xy^2 - x^2y > 0
=> xy(y-x) > 0
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Applying both conditions together yields y > x > 0. So, x>0, y>0 and y-x>0.
It follows that the product xy(y - x) is positive. Therefore, both conditions are sufficient when considered together.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
If x = 2 and y = 3, then 1/x = 1/2, 1/y = 1/3, and the answer is 'yes'.
If x = -2 and y = 3, then 1/x = -1/2, 1/y = 1/3, and the answer is 'no'.
Condition 1) is not sufficient on its own.
Condition 2)
This condition provides us with no information about the variable y, so it is not sufficient.
Therefore, the answer is C.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provides an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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