Question

Question #de470

Answer 1

Kilos of rambutan: `8`
Kilos of duku: `6`
Kilos of langsat: `4`

(see below for solution method)

Explanation:

Part a
Let `R` be the number of kilos of rambutan,
`K` be the number of kilos of duku (I used `D` for Determinant, so I didn't want to use it for "duku" as well), and
`L` be the number of kilos of langsat.

We are told the total cost (and individual per kilo costs); so:
`color(white)("XXX")0.75R+0.90K+0.60L=13.80`
Also the total weight:
`color(white)("XXX")R+K+L=18`
And that the difference of the total cost of the duku and langsat minus the cost of the rambutan:
`color(white)("XXX")-0.75R+0.90K+0.60L=1.80`

Part b
Setting this up as an augmented matrix:

`color(white)("XXX")Rcolor(white)("XXX")Kcolor(white)("XXX")Lcolor(white)("XXX")"constants"`

#( (0.75,0.90,0.60," | ",13.80), (1,1,1," | ",18), (-0.75,0.90,0.60," | ",1.80) )#

and using the (hopefully) standard designations for the derived matrices:
`M, M_R, M_K, M_L`

Crammer's Rule tells us that (for example):
`color(white)("XXX")R=("Determinant"(M_R))/("Determinant"(M))`

Here it is as evaluated in a spreadsheet: