A journey of 192 km from …

Let speed of passenger train be x km/h
{tex}\therefore {/tex} speed of superfast train = (x + 16) km/h
By question, {tex}T _ { \text { passenger } } = \frac { 192 } { x }{/tex} and {tex}\mathrm { T } _ { \text { superfast } } = \frac { 192 } { ( x + 16 ) }{/tex} 
or, {tex}\frac { 192 } { x } - \frac { 192 } { x + 16 } = 2{/tex} 
or,  {tex}192 ( x + 16 ) - 192 x = 2 \left( x ^ { 2 } + 16 x \right){/tex} 
or, {tex}192 x + 192 \times 16 - 192 x = 2 \left( x ^ { 2 } + 16 x \right){/tex} 
192x + 3072 - 192x {tex}{/tex}{tex} = 2 \left( x ^ { 2 } + 16 x \right){/tex}( divide throughout by 2, we get,
96x + 1536 - 96x {tex} = \left( x ^ { 2 } + 16 x \right){/tex} 
or x(x + 48) - 32(x + 48) = 0
or, (x - 32) (x + 48) = 0
or, x = 32 or - 48
Since speed can't be negative, therefore - 48 is not possible.
{tex}\therefore {/tex} Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h