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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/h, and the train travels at a speed which is 16 km/h faster than the car. Calculate :

(i) The time taken by car , to reach town B from A, in terms of x.

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

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

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

Quadratic Equations

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Answer

(i) Speed of car = x km/h

Distance between point A and B by road = 216 km

Time taken = DistanceSpeed=216x\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{216}{x} hours

(ii) Speed of train = (x + 16) km/h

Distance between point A and B by rail = 208 km

Time taken = DistanceSpeed=208x+16\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{208}{x + 16} hours

(iii) Given,

Train takes 2 hours less than car to reach town B

216x208x+16=2216(x+16)208xx(x+16)=2216x208x+3456=2x(x+16)8x+3456=2x2+32x2x2+32x8x3456=02x2+24x3456=02(x2+12x1728)=0x2+12x1728=0x2+48x36x1728=0x(x+48)36(x+48)=0(x36)(x+48)=0x36=0 or x+48=0x=36 or x=48\therefore \dfrac{216}{x} - \dfrac{208}{x + 16} = 2 \\[1em] \Rightarrow \dfrac{216(x + 16) - 208x}{x(x + 16)} = 2 \\[1em] \Rightarrow 216x -208x + 3456 = 2x(x + 16) \\[1em] \Rightarrow 8x + 3456 =2x^2 + 32x \\[1em] \Rightarrow 2x^2 + 32x -8x - 3456 = 0 \\[1em] \Rightarrow 2x^2 + 24x - 3456 = 0 \\[1em] \Rightarrow 2(x^2 + 12x - 1728) = 0 \\[1em] \Rightarrow x^2 + 12x - 1728 = 0 \\[1em] \Rightarrow x^2 + 48x - 36x - 1728 = 0 \\[1em] \Rightarrow x(x + 48) - 36(x + 48) = 0 \\[1em] \Rightarrow (x - 36)(x + 48) = 0 \\[1em] \Rightarrow x - 36 = 0 \text{ or } x + 48 = 0 \\[1em] \Rightarrow x = 36 \text{ or } x = -48

Since speed of car cannot be negative hence x ≠ -48

∴ x = 36

Equation : 216x208x+16=2\dfrac{216}{x} - \dfrac{208}{x + 16} = 2

(iv) Speed of train = (x + 16) = (36 + 16) = 52 km/h

Hence, speed of train is 52 km/h.

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