There are 3 different positive integers.

It is actually stated, but it's subtle. If the largest is twice the smallest, then the remaining number must by definition be in between the two. The problem is that if we just represent this as, say, 3x + y = 24, we end up removing that information from the setup. It might be helpful to start by making three blanks in a row and labeling them while this information is fresh.

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.

There are 3 different positive integers. If their average (the arithmetic mean) is 8, what are their values?
1) The largest integer is twice the smallest integer.
2) One of them is 9.


In the original condition, there are 3 variables(a<b<c) and 1 equation(a+b+c=24), which should match with the number of equations. So you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), 5, 9, and 10 are valid, which is unique. So, the answer is C. However, this is an integer and statistics question, one of the key questions, and apply the mistake type 4(A). For 1), only 5, 9, 10 are valid as well, because they are different positive integers. Thus, 6,6,12 -> impossible and 4,8,12 -> impossible. Therefore, the answer is A.


 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.

Great explanation! Thanks a lot.

Cuhmoon, did you miss that the third integer had to be in between the other two? If we overlook that, there are plenty of possible combinations. It's an easy restriction to miss!

Thanks Dmitry. But is that implied or should that be part of the question?

Thanks a lot for the explanation!

Great explanation! Thanks v much!!