Question

cyclist travels from town A to town C through another town B. He travels A to B at a speed of 6 km/h and B to C at a speed of 9 km/h. The time taken for journey is 4 hrs. On his return journey, the travels at a speed of 6 km from C to B and at a speed of 9 km/h. The time taken for return journey is 4 hrs and 20 minutes. Find the distance AC and BC.

Answer 1

Let the distance from A to B =X km and from B to C =Y km.

On the going journey ;Time taken from A to B=$\frac{X} {6}$ hrs and B to C=$\frac{Y} {9}$ hrs.

Hence ; $\frac{x} {6}$ +$\frac{y} {9}$ = 4hrs ............ (i)

On the Return journey ;Time taken from C to B =$\frac{y} {6}$ hrs and B to A =$\frac{y} {9}$ hrs.

Hence ;$\frac{y} {6}$ +$\frac{y} {9}$ =$\frac{13} {3}$ hrs........... (ii)

Making y the subject of the formula in equation (i) ;

y =(4-$\frac{x} {6}$) 9=

y=36-$\frac{3} {2}$ x

Substituting y in the second equation ;

$(\frac{36-1.5x}{6})$ +$\frac{x} {9}$ =$\frac{13}{3}$

=108-4.5x+2x=78

=-2.5x=-30

x=12

y=36-(1.5×12)=18

Hence Distance AC=12+18=30km

Distance BC =18km