Jk=6r,kl=3r and jl = 27

2 answers

If kl = 3r then 2kl = 6r
but 6r = jk
then jk = 2kl
j = 2l

given: jl = 27
2l(l) = 27
l^2 = 27/2 or 54/4
l = √54/2 = (3/2)√6
j = 3√6

If 6r = jk
6r = 3k√6
r = k√6/2

I get the same result from 3r = kl

so k could be any real number c and
r = c√6/2

e.g let k = 5
then r = 5√6/2

check for l and j
jk = 6r
LS = 3√6(5) = 15√6
RS = 6(5√6)/2 = 15√6 = LS

kl = 3r
LS = 5(3/2)√6 = (15/2)√6
RS = 3(5√6/2) = (15/2)√6 = LS
and of course ...
jl = 3√6(3/2)√6 = (9/2)√36 = (9/2)(6) = 27 as it should

final conclusion:
j = 3√6
l = (3/2)√6
k = any number
r = (k/2)√6

the answer would be c, 3