In a class of 40 students,
That means:
(1) c+d+e+f+g+h+i+j = 40
17 have ridden an airplane,
(2) c+d+f+g = 17
28 have ridden a boat,
(3) d+e+g+h = 28
10 have ridden a train,
f+g+h+i = 10
12 have ridden both an airplane and a boat.
(4) d+g = 12
3 have ridden a train only
(5) i = 3
and 4 have ridden an airplane only.
(6) c = 4
Some students have not yet ridden any of
the three modes of transportation
(7) j > 0 (that is, j is not 0)
and an equal number have taken all the three.
(8) g = j but we'll need the letters on the left side
so we'll write it as:
(8) g-j = 0
How many students have used all the three
modes of transportation?
This asks for the value of g.
How many students have taken only the boat?
This asks for the value of e.
So we put all equations down and line up the unknowns
[we don't need (7)]
(1) c+d+e+f+g+h+i+j = 40
(2) c+d +f+g = 17
(3) d+e +g+h = 28
(4) d +g = 12
(5) i = 3
(6) c = 4
(8) g = j
Change the bottom equation to g-j = 0
(1) c+d+e+f+g+h+i+j = 40
(2) c+d +f+g = 17
(3) d+e +g+h = 28
(4) d +g = 12
(5) i = 3
(6) c = 4
(8) g -j = 0
We have the 7 x 9 augmented matrix:
Using row operations to reduce this matrix to
row reduced echelon form, we have:
Which describes this system:
c = 4
d = 8
e = 16
f = 1
g = 4
i = 3
j = 4
So we have the answers g = 4 and e = 16.
But let's complete the Venn diagram:
We have everything but h.
But we can find h by substituting in
(3) d+e +g+h = 28
8+16 +4+h = 28
28+h = 28
h = 0
So the complete Venn diagram is:
Edwin
