#### Question

The distance by road between two towns A and B is 216 km, and by rail, it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate:

(1) the time is taken by the car to reach town B from A, in terms of x;

(2) the time is taken by the train to reach town B from A, in terms of x.

(3) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.

(4) Hence, find the speed of the train.

#### Solution 1

Speed of car = x km/hr

Speed of train = (x + 16) km/hr

1) we know Time = `"Distance"/"Speed"`

Time is taken by the car to reach town B From A = `216/"x"` hrs

2) Time taken by the train to reach town B from A = `208/("x" + 16)` hrs

3) From the given information

`216/x - 208/(x + 16) = 2`

`(216x + 3456 - 208x)/(x(x + 16)) = 2`

`(8x + 3456)/(x(x + 16)) = 2`

`4x + 1728 = x^2 + 16x`

`x^2 + 12x - 1728 = 0`

`x^2 + 48x - 36x - 1728 = 0`

`x(x + 48)- 36(x + 48) = 0`

(x + 48)(x - 36) = 0

x = -48, 36

But speed cannnot be negative So x = 36

4) Speed of train = (36 + 16) km/ hr = 52 km/hr

#### Solution 2

The distance by road between A and B = 216km

and the distance by rail = 208km

speed of car = x km/hr

and speed of train = (x + 16)km/hr.

(i) Time taken by car = `(216)/x"hours"`

(ii) Time taaken by train = `(208)/(x + 16)"hours"`.

(iii) According to the condition,

`(216)/x - (208)/(x + 16)` = 2

⇒ `(216x + 3456 - 208x)/(x(x + 16)` = 2

⇒ `(8x + 3456)/(x^2 + 16x) = (2)/(1)`

⇒ 8x + 3456 = 2x^{2} + 32x

⇒ 2x^{2} + 32x - 8x - 3456 = 0

⇒ 2x^{2} + 24x - 3456 = 0

⇒ x^{2} + 12x - 1728 = 0 ...(Dividing by 2)

⇒ x^{2 }+ 48x - 36x - 1728 = 0

⇒ x(x + 48) -36(x - 48) = 0

⇒ (x + 48)(x -36) = 0

Either x + 48 = 0,

then x = -48,

which is not possible as it is negative.

or

x - 36 = 0,

then x = 36.

(iv) Speed of the train

= (x + 16)km/hr.

= (36 + 16)km/hr.

= 52km/hr.