Help with Grade 10 math question
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Help with Grade 10 math question

[From: ] [author: ] [Date: 11-12-09] [Hit: ]
5 sites per acre. Developed campgrounds have 4 sites per acre. The amount of land used for basic sites is twice the amount of land used for deveolped sites.a) determine the number of acres to be used for each type of campground if all 57 acres of land are used.b) How many campsites of each type will there be?-1)Ld +Lb = 57where Ive used Ld to represent the acreage for developed sites; obvious what Lb is then.......
Please help solve and show all of your steps, thank youuuuuu!!

Solve using TWO equations

A local nature club wants to convert 57 acres of land to campgrounds. Basic campgrounds have a density of 1.5 sites per acre. Developed campgrounds have 4 sites per acre. The amount of land used for basic sites is twice the amount of land used for deveolped sites.

a) determine the number of acres to be used for each type of campground if all 57 acres of land are used.
b) How many campsites of each type will there be?

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1) Ld +Lb = 57 where I've used Ld to represent the acreage for developed sites; obvious what Lb is then.

2) Lb = 2Ld So equation (1) becomes Ld + (2Ld) = 57; solve and get Ld = 19 acres, making Lb = 38.

There are to be 1.5 sites per acre of Lb land ==> 1.5(38) = 57 sites on the 38 acres of Lb.

On the Ld land you have 4(19) = 76 sites.

-
B=basic campground acreage
D=developed campground acreage

two equations:
B=2D
57=B+D

Substitute 2D for B in the second equation:
57=2D+D = 3D
D = 57/3 = 19

Substitute 19 for D in first equation:
2*19=B
B=38

a) Basic has 38 acres, Developed has 19 acres.
b) Basic campgrounds = 38*1.5 = 57. Developed campgrounds = 19*4 = 76

-
a)
x=y
2x+y=57
57/3=19
2x=19
x(basic site)=9.5 acres
y(developed site)=19 acres

b)
n(total sites for basic site)=9.5x1.5=13.25 (or 13 1/4 )
m(total sites for developed site)=19x4=76
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