Jurassic Zoo problem
- Thread starter Sarsath
- Start date
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- Nov 24, 2012
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Hello, and welcome to FMH! 
Let \(A\) be the number of adults and \(C\) be the number of children. Do we know the sum of these two values? Can you state the sum of the amount charged for all the adults and the amount charged for all the children, and equate this sum to the total bill?
If you complete these two equations, you will have enough to find a unique solution.
EDIT: Since the equations and solution have been handed to you in the post below, with no regard for the fact that I was trying to provide judicious help and engage you in the process so you might actually learn, I'll go ahead and just give you the full solution and at least offer some kind of explanation.
We know the number of adults plus the number of children must be equal to the total number of people who went on the trip, which we are told is 132. Thus we may state:
[MATH]A+C=132[/MATH]
We know for each adult, $7 was charged, so \(7A\) would be the charge for all the adults, and likewise \(3C\) would be the charge for all the children. Added together, we are told this is $520, and so we may write:
[MATH]7A+3C=520[/MATH]
Now, if we multiply the first equation by -3, and add it to the second equation, the variable \(C\) will be elminated:
[MATH]4A=124\implies A=31[/MATH]
Putting this value into the first equation, we get:
[MATH]31+C=132\implies C=101[/MATH]
And so we conclude that there were 101 children and 31 adults on the school trip.
Let \(A\) be the number of adults and \(C\) be the number of children. Do we know the sum of these two values? Can you state the sum of the amount charged for all the adults and the amount charged for all the children, and equate this sum to the total bill?
If you complete these two equations, you will have enough to find a unique solution.
EDIT: Since the equations and solution have been handed to you in the post below, with no regard for the fact that I was trying to provide judicious help and engage you in the process so you might actually learn, I'll go ahead and just give you the full solution and at least offer some kind of explanation.
We know the number of adults plus the number of children must be equal to the total number of people who went on the trip, which we are told is 132. Thus we may state:
[MATH]A+C=132[/MATH]
We know for each adult, $7 was charged, so \(7A\) would be the charge for all the adults, and likewise \(3C\) would be the charge for all the children. Added together, we are told this is $520, and so we may write:
[MATH]7A+3C=520[/MATH]
Now, if we multiply the first equation by -3, and add it to the second equation, the variable \(C\) will be elminated:
[MATH]4A=124\implies A=31[/MATH]
Putting this value into the first equation, we get:
[MATH]31+C=132\implies C=101[/MATH]
And so we conclude that there were 101 children and 31 adults on the school trip.
Last edited:
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
Hello, and welcome to FMH!
Let \(A\) be the number of adults and \(C\) be the number of children. Do we know the sum of these two values? Can you state the sum of the amount charged for all the adults and the amount charged for all the children, and equate this sum to the total bill?
If you complete these two equations, you will have enough to find a unique solution.
EDIT: Since the equations and solution have been handed to you in the post below, with no regard for the fact that I was trying to provide judicious help and engage you in the process so you might actually learn, I'll go ahead and just give you the full solution and at least offer some kind of explanation.
We know the number of adults plus the number of children must be equal to the total number of people who went on the trip, which we are told is 132. Thus we may state:
[MATH]A+C=132[/MATH]
We know for each adult, $7 was charged, so \(7A\) would be the charge for all the adults, and likewise \(3C\) would be the charge for all the children. Added together, we are told this is $520, and so we may write:
[MATH]7A+3C=520[/MATH]
Now, if we multiply the first equation by -3, and add it to the second equation, the variable \(C\) will be elminated:
[MATH]4A=124\implies A=31[/MATH]
Putting this value into the first equation, we get:
[MATH]31+C=132\implies C=101[/MATH]
And so we conclude that there were 101 children and 31 adults on the school trip.
Thank you, both of you.