The cost of admission to a exhibition for an adult is 3 times the cost of a child....

Math/Bio2018

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Hello,
I am doing a Wildlife Biology degree program and am in semester one with an Essential Scientific Maths exam next Tuesday. There is a question in past exam papers that I am finding difficulty in. At first I thought it was sim equations but that doesn't seem like it will work. Here is the question:
"The cost of admission to a exhibition for an adult is 3 times the cost of a child. Ayouth is charged more than a child but less than an adult. The total cost for an adult,a youth and a child is 28 euros. If 5 youths 2 children and 2 adults pay 80 euros, whatis the cost of admission for a youth?"

Thank You
 
Hello,
I am doing a Wildlife Biology degree program and am in semester one with an Essential Scientific Maths exam next Tuesday. There is a question in past exam papers that I am finding difficulty in. At first I thought it was sim equations but that doesn't seem like it will work. Here is the question:
"The cost of admission to a exhibition for an adult is 3 times the cost of a child. Ayouth is charged more than a child but less than an adult. The total cost for an adult,a youth and a child is 28 euros. If 5 youths 2 children and 2 adults pay 80 euros, whatis the cost of admission for a youth?"

Thank You

Rule #1 If you don't know what it is, give it a name so we can talk about it.

A = Cost of Adult Ticket
Y = Cost of Youth ticket
C = Cost of Child's Ticket

Translate!

"The cost of admission to a exhibition for an adult is 3 times the cost of a child."

1) A = 3*C

"A youth is charged more than a child but less than an adult."

C < Y < A

"The total cost for an adult,a youth and a child is 28 euros"

2) C + Y + A = 28

"If 5 youths 2 children and 2 adults pay 80 euros"

Give it a go. There IS enough information to find a unique solution. See if you actually need that inequality.
 
Hello,
I am doing a Wildlife Biology degree program and am in semester one with an Essential Scientific Maths exam next Tuesday. There is a question in past exam papers that I am finding difficulty in. At first I thought it was sim equations but that doesn't seem like it will work. Here is the question:
"The cost of admission to a exhibition for an adult is 3 times the cost of a child. A youth is charged more than a child but less than an adult. The total cost for an adult,a youth and a child is 28 euros. If 5 youths 2 children and 2 adults pay 80 euros, what is the cost of admission for a youth?"

Thank You

It can definitely be solved by simultaneous equations. You have three equations in three variables, along with an inequality (which isn't needed to solve the problem, but may serve as a check). I'm guessing that it is the inequality that is troubling you; just ignore it.

What are your equations? Can you show your work in trying to solve them?
 
a = adult, c = child, y = youth

> The cost of admission to a exhibition for an adult is 3 times the cost of a child.

a = 3c [1]

> A youth is charged more than a child but less than an adult.
Not required

> The total cost for an adult,a youth and a child is 28 euros.

a + c + y = 28 [2]

> If 5 youths 2 children and 2 adults pay 80 euros,what is the cost of admission for a youth?"

2a + 2c + 5y = 80 [3]

3 equations, 3 unknowns.
Let's see your stuff!
 
"The cost of admission to a exhibition for an adult is 3 times the cost of a child.
So one adult is equal to three children.

A youth is charged more than a child but less than an adult. The total cost for an adult,a youth and a child is 28 euros.
Since one adult is equal to three children, this may be restated as:

. . . . .4 children plus 1 youth cost 28

If 5 youths 2 children and 2 adults pay 80 euros, what is the cost of admission for a youth?"
Since one adult is equal to three children, this may be restated as:

. . . . .8 children plus 5 youths cost 80

The first "equation" above can be multiplied by two to give us:

. . . . .8 children plus 2 youths cost 56

Subtract one of the "equations" from the other to "cancel out" the children. Solve what's left (by dividing by three) to get the cost of one youth. ;)
 
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