=>
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.
Visit
https://www.mathrevolution.com/gmat/lesson for details.
Since we have the standard deviations of two sets, we have 2 variables and 0 equations, C is most likely 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) & 2)
Assume A
1, A
2, … , A
5 are the “push-up” numbers of the employees in department A and B
1, B
2, … , B
6 are the “push-up” numbers of the employees in department B.
The combined average of departments A and B is
( 5*7 + 6*7 ) / 11 = 11*7/11 = 77/11 = 7.
The variances of A and B are squares of the standard deviations of A and B, respectively.
The variance of A is { (A1-7)2 + … + (A5-7)2 } / 5 = 1 and we have { (A1-7)2 + … + (A5-7)2 = 5.
The variance of B is { (B1-7)2 + … + (B6-7)2 } / 6 = 1 and we have { (B1-7)2 + … + (B6-7)2 = 6.
The combined variance of sets A and B is
{ (A
1-7)^2 + … + (A
5-7)^2 + (B
1-7)^2 + … + (B
6-7)^2 } / 11
= { 5 + 6 } / 11 = 11/11 = 1.
The standard deviation of the sets A and B is the square root of the combined variance equal to 1.
Since both conditions together yield a unique solution, they are sufficient.
Since this question is a statistics 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)
Since we don’t have any information about department B, it is not sufficient, obviously.
Condition 2)
Since we don’t have any information about department A, it is not sufficient, obviously.
Therefore, C is the answer.
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.
