Search 73,740 tutors
0 0

# I need help showing my work for a couple mathematical problems, can anyone help me?

I couldn't figure out how to solve these two problems and have to show ALL my work for both, if someone could help me I would appreciate it thanks!

1. The perimeter of a rectangular field is 300 meters. This field is to be fenced along three sides only. The fencing along it's two lengths and one width is to be done at \$50 per meter. If the total cost of the fencing is \$12,000, what are the field's dimensions?

2. The admission fee at a fair is \$2 for children and \$4 for adults. On a certain day, 2,000 people enter the fair and \$6,000 is collected. How many children and how many adults went to see the fair on that day?
Tutors, please sign in to answer this question.

### 1 Answer

1. The rectangular field's perimeter can be represented by an equation:
2x + 2y = 300m (perimeter). Since the fence will not span one side, we can say the length of the fence could be 2x + y = f (fence).
Since we know each meter of fence costs \$50 & the total cost is \$12,000, we can find the length of the fence, in meters:
12,000/50 = 240
Now substitute 240m for the f in the fence equation & subtract the length of the fence from the entire perimeter:
2x + y = 240
(2x-2y)-(2x-y)=y
300-240=y
y=60
Thus the "missing" side is 60 meters long & the "y" side of the fence is 60 meters long.
Next substitute 60 for the y in the fence equation & solve for x:
2x+60=240
2x=240-60
x=180/2
x=90
The field is 60x90.

2. In this one we'll use x to represent a number of people & we'll designate them as children & adults by the dollar amounts:
2x+4x=\$6000
Now solve for x:
6x=\$6000
x=6000/6
x=1000
Check your answer:
1000+1000=2000
There were 1000 adults & 1000 children.