David and Stacey are riding bicycles on a flat road at a constant rate

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.

David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?

(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.


In the original condition, there are 2 variables(you need to figure out the speed of David and Stacey as you can figure out the time when they are 1 mile apart by knowing their speed). In order to match with the number of equations, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), for 1), they are 2 miles apart at 1 hour. So, to be 1 mile apart, 30 minutes should pass, which is sufficient.
For 2), 45 minutes:1.5 miles=x;1 mile, x=30 minutes is derived, which is unique and sufficient.
Since 1) = 2), the answer is D.


 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, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. 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 using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.