vinay1983 wrote:I need a simple way of solving this problem.

Mary and Nancy can each perform a certain task in m and n hours, respectively. Is m<n?

1. Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is greater than m.

2. Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is less than n.

Many DS rate problems can be solved quickly by REASONING our way to the correct answer.

Statement 1: Twice the time it would take Mary and Nancy to perform the task together, each working at their respective rates, is greater than m.
In other words, if we double the time needed by Mary and Nancy together, the result is MORE time than Mary needs to complete the task alone.
This means that Mary is working FASTER than Nancy.
Thus, Mary needs LESS TIME than Nancy to complete the task alone, and m<n.
SUFFICIENT.

Statement 2: Twice the time it would take Mary and Nancy to perform the task together, each working at their respective constant rates, is less than n.
In other words, if we double the time needed by Mary and Nancy together, the result is LESS time than Nancy needs to complete the task alone.
This means that Nancy is working MORE SLOWLY than Mary.
Thus, Nancy needs MORE TIME than Mary to complete the task alone, and n>m.
SUFFICIENT.

The correct answer is D.

Algebraic approach:

Statement 1: Twice the time for Mary and Nancy working together is greater than m.
Rate for Mary = 1/m.
Rate for Nancy = 1/n.
Combined rate for Mary and Nancy = 1/m + 1/n.

Time for Mary and Nancy working together = 1/(1/m + 1/n) = 1/ [(m+n)/mn] = mn/(m+n).
Twice the time for Mary and Nancy working together = 2mn/(m+n).

Since twice the time for Mary and Nancy working together is greater than m:
2mn/(m+n) > m.
2n/(m+n) > 1.
2n > m+n.
n > m.
SUFFICIENT.

Statement 2: Twice the time for Mary and Nancy working is less than n.
2mn/(m+n) < n.
2m/(m+n) < 1.
2m < m+n.
m < n.
SUFFICIENT.