SOLUTION: Hi On a bus there were 54 passengers. There were 4 more boys than girls on the bus. 1/5 of the adult passengers and 1/6 of the children passengers were equal to 10. How many

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Question 1122763: Hi
On a bus there were 54 passengers. There were 4 more boys than girls on the bus. 1/5 of the adult passengers and 1/6 of the children passengers were equal to 10.
How many boys were on the bus.
thanks
Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Boys    g+4
Girls    g
Children  c
Adults    a
Total    54


54 total passengers, c is sum of (g+4) and g.

system%28%28g%2B4%29%2Bg=c%2Ca%2F5%2Bc%2F6=10%2Cc%2Ba=54%29

system%282g%2B4=c%2C6a%2B5c=300%2Ca%2Bc=54%29
.
.
.
system%28c=2g%2B4%2C3a%2B5g=140%2Ca%2B2g=50%29

highlight%28g=10%29
-
c=24
-
g%2B4=highlight%2814%29-------boys

Answer by MathTherapy(10553) About Me  (Show Source):
You can put this solution on YOUR website!
Hi
On a bus there were 54 passengers. There were 4 more boys than girls on the bus. 1/5 of the adult passengers and 1/6 of the children passengers were equal to 10.
How many boys were on the bus.
thanks
This is EXTREMELY easy and doesn't have to be as COMPLEX as some may make it out to be.

Let number of boys be B

Then number of girls = B - 4

Number of children: boys + girls = B + B - 4, or 2B - 4

Also, number of adults = 54 - (2B - 4) = 58 - 2B

We then get the following: 

Solve this for B, and you should get: highlight_green%28matrix%281%2C2%2C+14%2C+boys%29%29

That's ALL!! Nothing more, nothing less!

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

            I am trying to find more simple way to solve it . . . 


From the condition, you have these two equations for two unknowns


a  +  c = 54        (1)                    (a = # of adults;  c = # of children)

a%2F5 +  c%2F6 = 10       (2)


Multiply eq(2) by 6 (both sides).  Keep eq(1) as is.  You will get


a   +   c = 54      (3)                  

%286a%29%2F5 +  c = 60      (4)


From eq(4) subtract eq(3)  (both sides).  You will get


a%2F5 = 60-54 = 6  ====>  a = 6*5 = 30   ====>  c = 54 - 30 = 24.


Now we have this simple system


Boys + Girls = 24
Boys - Girls =  4.

---------------------------- Add the two equations. You will get

2*Boys = 24 + 4 = 28  ====>  Boys = 28%2F2 = 14.

Solved.

===============

Surely,  I could start from a single equation for the unknown  "C"  only,  instead of using two equations for  "A"  and  "C".

But the major idea is clear:  The most simple way is to solve the problem in two steps:

        first determine the number of children - and then  (or after that)  determine the number of boys.