Question

The grid shows the travel graph for a train travelling from Audley to Cawley, stopping at Brookland.
Between which two towns is the train journey fastest ?
Give a reason for your answer.

Answer 1

1) Consider first part of the journey i.e. Audley to Brookland.

Distance between two stations is $$10\quad km$$

Train left Audley at $$9.00$$ and reached Brookland at $$9:20$$

Thus, time taken by train to reach from Audley to Brookland is $$20\quad minutes$$

Thus, average speed of the train between these two stations is given by,

$${ v }_{ 1 }=\frac { 10 }{ 20 } \quad \frac { km }{ min } $$

$$\therefore { v }_{ 1 }=\frac { 1 }{ 2 } \quad \frac { km }{ min } $$ (1)

2) Consider second part of the journey i.e. Brookland to Cawley.

Distance covered by train between these two stations is $$35-10=25\quad km$$

Train left Brookland at $$9:25$$ and reached Cawley at $$9:40$$

Thus, time taken by train to reach from Brookland to Cawley is $$15\quad minutes$$

Thus, average speed of the train between these two stations is given by,

$${ v }_{ 2 }=\frac { 25 }{ 15 } \quad \frac { km }{ min } $$

$$\therefore { v }_{ 2 }=\frac { 5 }{ 3 } \quad \frac { km }{ min } $$

Now, we have to compare two fractions i.e. $$\frac { 1 }{ 2 } $$ and $$\frac { 5 }{ 3 } $$

LCM of $$2$$ and $$3$$ is $$6$$

$$\frac { 1 }{ 2 } =\frac { 1\times 3 }{ 2\times 3 } =\frac { 3 }{ 6 } $$

$$\frac { 5 }{ 3 } =\frac { 5\times 2 }{ 3\times 2 } =\frac { 10 }{ 6 } $$

$$\frac { 10 }{ 6 } >\frac { 3 }{ 6 } $$

$$\therefore \frac { 5 }{ 3 } >\frac { 1 }{ 2 } $$

The train journey was fastest between Brookland and Cawley.