Algebra (Linear Algebra)

Question

A tourist has a collection of 33 coins consisting of Belgian francs ( 2 cents each), British shillings (14 cents), and German marks (25 cents each). The total value of the collection is $6.33. He has three times as many marks as shillings. How many coins of each kind has he?

Answer 1

Let f = number of francs, s = number of shillings and m = number of marks.

From the description:

f+s+m = 33

.02f+.14s+.25m= 6.33 --> 2f+14s+25m=633

m = 3s --> -3s + m = 0

Since this is linear algebra, put the equations into a matrix

[1 1 1 33]

[0 -3 1 0]

[2 14 25 63]

R2 = -R2/3

R3 = R3-2R1

[1 1 1 33]

[0 1 -1/3 0]

[0 12 23 633]

R3 = R3-12R2

[1 1 1 33]

[0 1 -1/3 0]

[0 0 27 567]

Back substituting

m = 567/27 = 21

s = m/3 = 21/3 = 7

f + 7 + 21 = 33 --> f = 5

The tourist has 21 francs, 7 shillings and 5 marks