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(B) Let the original speed of the car be $x\,km/hr$.Time taken at original speed,$T_1 = \frac{150}{x}$ hours.New speed = $(x + 5)\,km/hr$.Time taken a

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$25$

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(B) Let the original speed of the car be $x\,km/hr$.Time taken at original speed,$T_1 = \frac{150}{x}$ hours.New speed = $(x + 5)\,km/hr$.Time taken at new speed,$T_2 = \frac{150}{x + 5}$ hours.Given that the car takes $60$ minutes ($1$ hour) less,so $T_1 - T_2 = 1$.$\frac{150}{x} - \frac{150}{x + 5} = 1$.$150(x + 5 - x) = x(x + 5)$.$150(5) = x^2 + 5x$.$x^2 + 5x - 750 = 0$.$(x + 30)(x - 25) = 0$.Since speed cannot be negative,$x = 25\,km/hr$.

$A$ car takes $60$ minutes less for a journey of $150\,km$,if its speed is increased by $5\,km/hr$ from its usual speed. Find the original speed of the car in $km/hr$.

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