Question 10(i)A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed. If it takes 3 hours to complete the total journey then form a quadratic equation and solve it to find its original average speed.

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Accepted Answer

Given,A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed.Total time taken to complete the journey = 3 hours∴72x+81x+6=3⇒72(x+6)+81xx(x+6)=3⇒72(x+6)+81x=3x(x+6)⇒72x+432+81x=3x2+18x⇒3x2+18x−72x−81x−432=0⇒3x2−135x−432=0⇒3(x2−45x−144)=0⇒x2−45x−144=0⇒x2−48x+3x−144=0⇒x(x−48)+3(x−48)=0⇒(x+3)(x−48)=0⇒x+3=0 or x−48=0⇒x=−3 or x=48.\therefore \dfrac{72}{x} + \dfrac{81}{x + 6} = 3 \[1em] \Rightarrow \dfrac{72(x + 6) + 81x}{x(x + 6)} = 3 \[1em] \Rightarrow 72(x + 6) + 81x = 3x(x + 6) \[1em] \Rightarrow 72x + 432 + 81x = 3x^2 + 18x \[1em] \Rightarrow 3x^2 + 18x - 72x - 81x - 432 = 0 \[1em] \Rightarrow 3x^2 - 135x - 432 = 0 \[1em] \Rightarrow 3(x^2 - 45x - 144) = 0 \[1em] \Rightarrow x^2 - 45x - 144 = 0 \[1em] \Rightarrow x^2 - 48x + 3x - 144 = 0 \[1em] \Rightarrow x(x - 48) + 3(x - 48) = 0 \[1em] \Rightarrow (x + 3)(x - 48) = 0 \[1em] \Rightarrow x + 3 = 0 \text{ or } x - 48 = 0 \[1em] \Rightarrow x = -3 \text{ or } x = 48.∴x72​+x+681​=3⇒x(x+6)72(x+6)+81x​=3⇒72(x+6)+81x=3x(x+6)⇒72x+432+81x=3x2+18x⇒3x2+18x−72x−81x−432=0⇒3x2−135x−432=0⇒3(x2−45x−144)=0⇒x2−45x−144=0⇒x2−48x+3x−144=0⇒x(x−48)+3(x−48)=0⇒(x+3)(x−48)=0⇒x+3=0 or x−48=0⇒x=−3 or x=48.As speed cannot be negative in this case,Hence, the original speed = 48 km/hr.

Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 18

Question 10(i)