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Mathematics

Two cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then :

If car X used 4 litres of diesel more than car Y in the journey, then :

  1. 72x−3−12x=4\dfrac{72}{x - 3} - \dfrac{12}{x} = 4

  2. 72x+3−12x=4\dfrac{72}{x + 3} - \dfrac{12}{x} = 4

  3. 72x−72x+3=4\dfrac{72}{x} - \dfrac{72}{x + 3} = 4

  4. 72x−3−72x+3=4\dfrac{72}{x - 3} - \dfrac{72}{x + 3} = 4

Quadratic Equations

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Answer

Diesel used by car X = 72x\dfrac{72}{x} litres

Diesel used by car Y = 72x+3\dfrac{72}{x + 3} litres

Given,

Car X used 4 litres of diesel more than car Y in the journey.

∴72x−72x+3=4\therefore \dfrac{72}{x} - \dfrac{72}{x + 3} = 4

Hence, option 3 is the correct option.

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