A bus has 4 seats for passengers. Those willing to pay first class fares can take 60 kg of baggage each, but tourist class passengers are restricted to 20kg each. If 120 kg can be carried altogether. Find the number of passengers of each kind by using determinants.

A bus has 4 seats for passengers. Those willing to pay first class fares can take 60 kg of baggage each, but tourist class passengers are restricted to 20kg each. If 120 kg can be carried altogether. Find the number of passengers of each kind by using determinants.

Solution: Given:
No. of seats for Passengers = 4
First class fares can take 60kg
Tourists class passenger are restricted to 20kg each Bus can carry 120kg Baggage altogether. Let first class fares Passenger = x
Also, let tourist class passengers = y
x + y = 4
and, 60x + 20y = 120
Or, 3x + y = 6
Thus, two equations are:
x + y = 4……(i)
3x + y + 6……. (ii)
Writing above equation in matrix Format:
$\left[ {\begin{array}{*{20}{c}}1&1\\3&1\end{array}} \right]$
So, D = $\left| {\begin{array}{*{20}{c}}1&1\\3&1\end{array}} \right|$=1-3=-2
For x:
$D_x$=$\left| {\begin{array}{*{20}{c}}4&1\\6&1\end{array}} \right|$
Thus, x = $\frac{{{D_x}}}{D}$ 
=$\frac{{\left| {\begin{array}{*{20}{c}}4&1\\6&1\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}1&1\\3&1\end{array}} \right|}}$
=$\frac{{4 - 6}}{{ - 2}}$
=1
For y: 
$D_y$=$\left| {\begin{array}{*{20}{c}}1&4\\3&6\end{array}} \right|$
Thus, y=$\frac{{{D_y}}}{D}$
=$\frac{{\left| {\begin{array}{*{20}{c}}1&4\\3&6\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}1&1\\3&1\end{array}} \right|}}$
=$\frac{{6 - 12}}{{ - 2}}$ =3
Thus, Number of passengers in First class = 1
And Number of passengers in tourist class = 3