Hi Guys... Here is my attempt at writing a PS problem. A group of people, m, are divided into teams of n members. If all teams except one have an equal number of members, while the final incomplete team contains only 64% of the members of any complete team, then how many people are in the team containing the least number of people? Each team must contain less than 50 people. A - 64 B - 32 C - 25 D - 16 E - 12

Nycgrl wrote:Is ans D??? YES! Answer is D Way to go!

I used back solving method. it says that incomplete team consist of 64% of members of any team let number of peopel on complete team be X start with C= 25 64/100*X=25 X= 6.2..there can't be 6.2 people on team check D=16 64/100*X=16 X=25 If you try any other number you will not get X= integer except 16. There could be another method to solve this question....

Nycgrl wrote:There could be another method to solve this question.... Yes - there is! The other method I propose is to think of "the remainders" of imperfect divisions. Your method works, NYCgrl. I hope it was clearly expressed. It took me an hour to craft the wording of it. I hope it didn't take longer than 2 min to solve either. Really, this question could be solved in less than 20 sec. I'll write a full explanation soon..

Here's the official solution: m = total population n = number of people in a complete team t = number of complete teams, a whole number m = nt + 0.64n m = n*(t + 64/100) Since t is a whole number, 64/100 represents the remainder as a percentage of a complete team. (64/100) = (32/50) = (16/25) Since question states that teams must contain less than 50 members, the remaining people who are put into the final, incomplete team, equals 16. I hope it's clear... Let me know!