Maths word equations. Linear relations and equations

Question

The registration fees for a mathematics competition are $1.20 for students aged 8–12
years and $2 for students 13 years and over. One hundred and twenty-five students
have already registered and an amount of $188.40 has been collected in fees. How
many students between the ages of 8 and 12 have registered for the competition?

Answer 1

This is a straightforward two-equation two-variable problem. Your information tells you how many total registrations there are (125) and how much money has been collected (188.40).

 

 

So you now have p = number of preteen registrations

and                   t = number of teenage registrations

 

p + t = 125 for the number of students.

 

1.2p + 2t = 188.40 for the money collected.

 

 

Spoiler space

 

 

 

 

 

 

 

 

 

 

Next step is to rearrange the first equation to t = 125 - p and substitute that value in for t in the second equation.

 

Thus: 1.2p + 2(125-p) = 188.4

 

1.2p + 250 -2p = 188.4

 

Combining terms, we get   -.8p = -61.6

 

and dividing by -0.8 the final answer of  p = 77.

 

There are other methods of doing the problem, but if you pick a substitution for t you get the desired answer directly.

Answer 2

Let x = the number of students ages 8-12

Let y = the number of students ages 13+

 

Using these variable, can create a system of equations.  One equation that represents the total number of students, and the other to represent the total cost.

 

     x + y = 125                           eq1

1.2x + 2y = 188.40                    eq2

 

 

Since we want to find the number of students ages 8-12 who attended, we only need to solve for x.  We can substitute eq1 into eq2 so that eq2 is in terms of x.

 

1.2x + 2(125 - x) = 188.40

 

 

Solve for x from this equation.  I'll let you take it from here.