By increasing the speed of a car by $10 k m / h r$ , the time of journey for a distance of $72 k m$ is reduced by $36$ minutes. Write an equation for the given information and check if it is a Quadratic Equation?

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Solution

let the speed be $x k m / h r$
$\frac{72}{x} - \frac{72}{x + 10} = \frac{36}{60} \frac{1}{x} - \frac{1}{x + 10} = \frac{1}{120} \frac{x + 10 - x}{x^{2} + 10 x} = \frac{1}{120} x^{2} + 10 x = 1200 x^{2} + 10 x - 1200 = 0 (x + 40) (x - 30) = 0$
$∴ x = 30 k m / h r$

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By increasing the speed of a car by 10km/hr, the time of journey for a distance of 72km is reduced by 36 minutes. Write an equation for the given information and check if it is a Quadratic Equation?

let the speed be xkm/hr72x−72x+10=36601x−1x+10=1120x+10−xx2+10x=1120x2+10x=1200x2+10x−1200=0(x+40)(x−30)=0 ∴x=30km/hr

By increasing the speed of a car by $10 k m / h r$ , the time of journey for a distance of $72 k m$ is reduced by $36$ minutes. Write an equation for the given information and check if it is a Quadratic Equation?

let the speed be $x k m / h r$
$\frac{72}{x} - \frac{72}{x + 10} = \frac{36}{60} \frac{1}{x} - \frac{1}{x + 10} = \frac{1}{120} \frac{x + 10 - x}{x^{2} + 10 x} = \frac{1}{120} x^{2} + 10 x = 1200 x^{2} + 10 x - 1200 = 0 (x + 40) (x - 30) = 0$
$∴ x = 30 k m / h r$