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Mathematics

Mr. and Mrs. Das were traveling by car from Delhi to Kasauli for a holiday. Distance between Delhi and Kasauli is approximately 350 km (via NH 152D). Due to heavy rain they had to slow down. The average speed of the car was reduced by 20 km/hr and time of the journey increased by 2 hours. Find:

(a) the original speed of the car.

(b) with the reduced speed, the number of hours they took to reach their destination.

Quadratic Equations

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Answer

(a) Let the original speed be x km/hr.

Distance = 350 km

In first case (original speed):

Time =DistanceSpeed=350x=\dfrac{\text{Distance}}{\text{Speed}}=\dfrac{350}{x} hrs

In second case (reduced speed):

Speed = (x - 20) km/hr

Time =350x20=\dfrac{350}{x-20} hrs

According to the question,

Time taken to reach destination increases by 2 hours in the second case.

350x20350x=2350x350(x20)x(x20)=2350x350x+7000x(x20)=27000x(x20)=2x(x20)=70002x220x=3500x220x3500=0x2+50x70x3500=0x(x+50)70(x+50)=0(x+50)(x70)=0(x+50)=0 or (x70)=0x=50 or x=70.\therefore \dfrac{350}{x-20}-\dfrac{350}{x}=2 \\[1em] \Rightarrow \dfrac{350x-350(x-20)}{x(x-20)}=2 \\[1em] \Rightarrow \dfrac{350x - 350x + 7000}{x(x - 20)} = 2 \\[1em] \Rightarrow \dfrac{7000}{x(x-20)}=2 \\[1em] \Rightarrow x(x-20) = \dfrac{7000}{2} \\[1em] \Rightarrow x^2 - 20x = 3500 \\[1em] \Rightarrow x^{2} - 20x - 3500 = 0 \\[1em] \Rightarrow x^{2} + 50x - 70x - 3500 = 0 \\[1em] \Rightarrow x(x + 50) - 70(x + 50) = 0 \\[1em] \Rightarrow (x + 50)(x - 70) = 0 \\[1em] \Rightarrow (x + 50) = 0 \text{ or } (x - 70) = 0 \\[1em] \Rightarrow x = -50 \text{ or } x = 70.

Since speed cannot be negative, x = 70 km/hr.

Hence, original speed of the car = 70 km/hr.

(b) Reduced Speed = x - 20 = 70 - 20 = 50 km/hr.

Time taken with reduced speed =35050= \dfrac{350}{50} = 7 hours.

Hence, with reduced speed the time taken to reach destination = 7 hours.

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