MathRevolution wrote:
[GMAT math practice question]
(number properties) Given two different positive integers, what is the ratio of the larger number to the smaller one?
1) the sum of the two numbers is 1000 less than the product of the two numbers
2) one of the two numbers is a perfect square
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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.
Since we have \(2\) variables 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) and 2)
Suppose \(x\) and \(y\) are the integers and \(x\) is a perfect square.
Then \(xy = x + y + 1000\), and \(xy – x – y + 1 = 1001.\)
Thus, \((x-1)(y-1) = 1001 = 7*11*13.\)
Since \(x\) is a perfect square, only \(11*13 + 1 = 144\) is a perfect square out of all possible values \(7+1, 11+1, 13+1, 7*11+1, 7*13+1, 11*13+1, and 7*11*13+1.\)
Thus, \(x = 144\) and \(y = 8.\)
Therefore, \(x : y = 144:8.\)
Since both conditions together yield a unique solution, they are sufficient.
Since this question is an integer 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)
We have \(xy = x + y + 1000\) or \(xy – x – y + 1 = 1001.\)
Thus \((x-1)(y-1) = 1001 = 7*11*13.\)
We can find pairs of solutions \(x=2\) and \(y=1002\), and \(x=1002\) and \(y=2.\)
Since condition 1) doesn’t yield a unique solution, it is not sufficient.
Condition 2)
Since it provides no information about the second number, condition 2) 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) provide 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.