How many cylindrical cans with a radius of 2 inches and a height of 6

good question for testing the common sense

only have to see that that there can fit no can of dia 4 inches in the box of length 3 inches ( what ever the other sides be)
so (2) is sufficient

and form one we don't get any idea about the fitting of cans or not
i.e. there can be cans or as in (2) there can be no one

so (1) is not sufficient

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.

How many cylindrical cans with a radius of 2 inches and a height of 6 inches can fit into a rectangular box?

(1) The volume of the box is 240 cubic inches.

(2) The length of the box is 3 inches.

For this question, the radius of the can is given to be 2 inches and the height is 6 inches. From condition 2, the box with length 3 inches cannot fit the cylindrical can, so the answer is 0 and (B), but the question does not come out this way.
Originally, this question has 3 variables (length, width, height) of the box, and we need 3 equations when only 2 equations are given from the conditions provided, so the answer becomes (E) as they are insufficient.

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% 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 E 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, C or D.

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