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(C) Let the usual speed of the train be $x \,km/hr$.Time taken to cover $900 \,km$ at usual speed is $T_1 = \frac{900}{x} \,hours$.When the speed is d

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$50 \,km/hr$

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(C) Let the usual speed of the train be $x \,km/hr$.Time taken to cover $900 \,km$ at usual speed is $T_1 = \frac{900}{x} \,hours$.When the speed is decreased by $10 \,km/hr$,the new speed is $(x - 10) \,km/hr$.Time taken to cover $900 \,km$ at the new speed is $T_2 = \frac{900}{x - 10} \,hours$.According to the problem,$T_2 - T_1 = 4.5 \,hours$ (which is $9/2 \,hours$).So,$\frac{900}{x - 10} - \frac{900}{x} = \frac{9}{2}$.Dividing by $9$,we get $\frac{100}{x - 10} - \frac{100}{x} = \frac{1}{2}$.$100 \left( \frac{x - (x - 10)}{x(x - 10)} ight) = \frac{1}{2}$.$100 \left( \frac{10}{x^2 - 10x} ight) = \frac{1}{2}$.$1000 	imes 2 = x^2 - 10x$.$x^2 - 10x - 2000 = 0$.Solving the quadratic equation: $x^2 - 50x + 40x - 2000 = 0$.$x(x - 50) + 40(x - 50) = 0$.$(x - 50)(x + 40) = 0$.Since speed cannot be negative,$x = 50 \,km/hr$.

When there is a decrease of $10 \,km/hr$ in the usual speed of a train,it takes $4 \frac{1}{2}$ hours more to cover a $900 \,km$ distance. Find the usual speed of the train.

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