Question 116422: Five girls- Fiorenza, Gladys, Helene, Jocelyn, and Kaitlin - and four boys- Abe, Bruce, Clive, and Doug ride to school each day in three seperate buses.
Abe and Fiorenza always ride together.
Gladys and Helen always ride together.
Jocelyn and Kaitlin never ride together.
Doug always rides in the bus with the fewest children.
Boys cannot outnumber girls in any bus.
The maximum number of children in any bus is four.
The bus in which Doug rides can hold how many children?
Bruce can ride with each of the following except
a.Abe
b.Clive
c.Doug
f.Fiorenza
e.Helene
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! First thing we know is that there are three buses and nine children. The situation where each bus carries three children is not possible because there are four boys, so one of the buses would have to carry two of the boys and on that bus the boys would outnumber the girls 2 to 1.
We are told that the most children on any bus is 4, so that gives us a high end to the range. Could any of the buses carry just 1 passenger? Since Doug rides on the bus with the fewest children, he would have to be that 1 passenger, but then he would outnumber the girls on that bus 1 to 0. So a bus with only 1 passenger is not possible, and the bus with the fewest passengers, and the one that Doug rides, has 2 children. The numbers of passengers have to be 4 for one bus, 3 for the second bus, and 2 for the remaining bus. (3, 3, and 3; 4, 4, and 1; 4, 3, and 2 are the only ways to divide 9 people into three groups where the largest group is 4)
Since Doug rides the bus with the fewest passengers, he must be on the bus with 2 passengers, and the "Boys cannot outnumber girls in any bus" rule means that the second passenger on this bus must be one of the girls. That means that the second passenger can't be Bruce, and one answer to the second part of the question is c: Doug.
However, let's make certain that the other four responses to the second part of the problem are ok with the rules -- in other words, is it possible to create a combination of passengers including Bruce and the listed person that doesn't violate any of the conditions?
a: Abe. Abe always rides with Fiorenza. If Bruce were with Abe, then they couldn't be on the 3 person bus because it would be boys 2, girls 1. So for Bruce and Abe to be together, they would have to be on the 4 person bus, and the fourth passenger would have to be a girl, perhaps Jocelyn or Kaitlin. So this is possible. That would mean that Clive, Gladys, and Helene were on the 3 person bus, and Doug would be with whichever of Jocelyn or Kaitlin was not on the 4 person bus.
b: Clive: Clive and Bruce together means that there would have to be two girls to ride with them and they would all be on the 4 person bus. But this is possible because the two girls could be Gladys and Helene. This would mean that Abe and Fiorenza were on the 3 person bus with either Jocelyn or Kaitlin, and Doug would be with Kaitlin or Jocelyn.
d: Fiorenza. This is the same analysis as Bruce and Abe, therefore possible.
e: Helene: Bruce could either be with Gladys and Helene on the 3 person bus, or with Gladys, Helene, and Clive on the 4 person bus. Note that Clive is the only possibility for this configuration because if the fourth person were a girl, you couldn't keep the "Boys cannot outnumber girls in any bus" rule on the 3 person bus, Doug is on the 2 person bus already, and Abe has to be with Fiorenza.
Bruce, Gladys, and Helene on the 3 person bus means that Abe, Fiorenza, Clive, and either Jocelyn or Kaitlin were on the 4 person bus, and again, Doug is riding with either Kaitlin or Jocelyn
Bruce, Gladys, Helene, and Clive on the 4
Abe, Fiorenza, and (Jocelyn or Kaitlin) on the 3
Doug, (Kaitlin or Jocelyn) on the 2
So, the only correct answer to the second part of the question is c: Doug.
Answer by Edwin McCravy(20064)  (Show Source):
You can put this solution on YOUR website!
Five girls- Fiorenza, Gladys, Helene, Jocelyn, and Kaitlin - and four boys- Abe, Bruce, Clive, and Doug ride to school each day in three seperate buses.
Abe and Fiorenza always ride together.
Gladys and Helen always ride together.
Jocelyn and Kaitlin never ride together.
Doug always rides in the bus with the fewest children.
Boys cannot outnumber girls in any bus.
The maximum number of children in any bus is four.
The bus in which Doug rides can hold how many children?
Bruce can ride with each of the following except
a.Abe
b.Clive
c.Doug
f.Fiorenza
e.Helene
To make it easier to read, I used initials instead of names:
Five girls- F, G, H, J, and K - and four boys- A, B, C, and D
ride to school each day in three separate buses.
1. A and F always ride together.
2. G and H always ride together.
3. J and K never ride together.
4. D always rides in the bus with the fewest children.
5. Boys cannot outnumber girls in any bus.
6. The maximum number of children in any bus is four.
The bus in which D rides can hold how many children?
Let's see if we can all the ways to seat all 9 children in the buses
X, Y, and Z following those rules.
Make this table:
Bus X Bus Y Bus Z
Girls
Boys
I'll start by putting A,F,G,H on Bus X, which fits 1 and 2
Bus X Bus Y Bus Z
Girls FGH
Boys A
Now because of 3, we have to put J and K on different buses.
And because of 6, neither can go on Bus X
So let's put J in Bus Y and K in Bus Z.
Bus X Bus Y Bus Z
Girls FGH J K
Boys A
That only leaves boys B, C, D.
I can only put one of those three boys on Bus Y because of 5.
So I'll put B on Bus Y and the other two boys on Bus Z. But
that won't do because there'd be no place to put D, since
D has to be on the bus with the fewest children. D can't go
on Bus Y and leave K alone for she would then be on the bus
with the least number of children. And it's the same thing
if you put him in Bus Z. So it's not possible for J and K
to be on buses Y and Z.
So it is impossible to seat all 9 when F,G,H,A are all
on the same bus, So let's start over, putting F,G on one bus
and H,A on another.
Bus X Bus Y Bus Z
Girls FG H
Boys A
Now we seat J and K. By 3, there are 3 ways to seat them. This is
the first way:
Bus X Bus Y Bus Z
Girls FGJ HK
Boys A
That leaves B, C, D, Then because of 4, D must go in Bus Z
Bus X Bus Y Bus Z
Girls FGJ HK
Boys A D
Now we have three ways to place B and C. This way:
Girls FGJ HK
Boys B AC D
********************************************************
I underlined it with stars to show it is a solution.
This way
Girls FGJ HK
Boys B A CD
********************************************************
And this way:
Girls FGJ HK
Boys AB CD
********************************************************
Now go back to where we putting F,G on one bus
and H,A on another.
Bus X Bus Y Bus Z
Girls FG H
Boys A
Now we seat J and K. By 3, there are 3 ways to seat them. This is
the second way:
Bus X Bus Y Bus Z
Girls FGJ H K
Boys A
This leaves boys B, C, D. We can't put any of these boys on Bus Y or
they'd outnumber girl H. So if we put D on Bus Z then D wouldn't be
on the bus with the fewest children, and we can't put D on Buses X or Y.
So we can't put J and K on Buses X and Z. So this is ruled out. So
that leaves only this way to place J and K on different buses:
Bus X Bus Y Bus Z
Girls FG HJ K
Boys A
This leaves boys B, C, D. D obviously has to go in Bus Z or D wouldn't be
on the bus with the fewest children.
So we have
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys A
Now one of B,C must go on Bus X for it to have more children than
the bus D is on.
So we have
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys B A
Now C can either go on Bus X or Y. Either this
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys BC A
**********************************************************
or
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys B AC
***********************************************************
That leaves B, C, D, Then because of 4, D must go in Bus Z
Bus X Bus Y Bus Z
Girls FGJ HK
Boys A D
Now we have three ways to place B and C. This way:
Girls FGJ HK
Boys B AC D
********************************************************
I underlined it with stars to show it is a solution.
This way
Girls FGJ HK
Boys B A CD
********************************************************
And this way:
Girls FGJ HK
Boys AB CD
******************************************************
Now we've found all possible solutions to the seating
arrangements on the busses.
The first question is:
The bus in which Doug rides can hold how many children?
From all the solutions starred, D is on bus Z either
by himself, with C, or with D. So at most 2 children
can be on the bus D rides on.
Bruce can ride with each of the following except ___
Look at all the solutions and you'll see there is no
solution that has B on the same bus with D.
Edwin
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