calreg11 wrote:
What is the distance between x and y on the number line?
(1) |x| – |y| = 5
(2) |x| + |y| = 11
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Condition (1)
In the case that \(x = 6\), \(y = 1\), the distance \(x\) and \(y\), \(| x - y | = 5\).
In the case that \(x = 6\), \(y = -1\), the distance \(x\) and \(y\), \(| x - y | = 7\).
Thus we don't have a unique solution.
Condition (2)
In the case that \(x = 6\), \(y = 5\), the distance \(x\) and \(y\), \(| x - y | = 1\).
In the case that \(x = 6\), \(y = -5\), the distance \(x\) and \(y\), \(| x - y | = 11\).
Thus we don't have a unique solution.
Condition (1) & (2)
If we add two equation, we have \(2|x| =16\) or \(|x| = 8\). Thus \(x = \pm 8\).
If we subtract the first equation from the second one, we have \(2|y| =6\) or \(|y| = 3\). Thus \(y = {\pm}3\).
In the case that \(x = 8\) and \(y = 3\), the distance \(x\) and \(y\), \(| x - y | = 5\).
In the case that \(x = 8\) and \(y = -3\), the distance \(x\) and \(y\), \(| x - y | = 11\).
Thus we don't have a unique solution.
Therefore the answer is E.
Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.