nidhis.1408 wrote:Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?

IMO : 12

what is the OA??

20 for division A
30/2=15 for division B
40/3= 13 for division C, 1 left

C is the limiting factor

for the group- 13 A (20-13)=7 left
2*13=26 B (30-26)= 4 left
1 left in C

therefore the smallest number of people who will not be able to join a group will be= 7+4+1= 12

Its a manhattan problem.

Here's how I do these types of problems:

I start off with division C and look at Division A as well. So if Division A has 15 people, Division C has to have 45 people. 15*3. Thats wrong. Even less is needed so we go down to 14 and realize that 14*3 is 42. Still too high but now were making progress and you get the feel of the pattern so you catch that 13 is the correct number. You quickly verify that B will have the correct amount of people and it doesn't go over and then you calculate how many people are not listed.

A- 13= 7 people are left out
B - 13*2= 26--> 4 people are left out
C- 13*3= 39--> 1 person is left out


7+4+1= 12

Thanks

It is more like

The number of groups = x

Total no. of people listed to form groups= x+2x+3x

Max 3x = 39, groups = 13

So,
13 -> 7 left
26 -> 4 left
39 -> 1 left

Total = 12

Could you explain for me how you got Max 3x = 39 please

thank you