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(B) Let the original average speed of the train be $x \, km/h$.According to the problem,the time taken for the first part is $\frac{63}{x} \, hours$ a

Vedclass · Accepted Answer

$42$

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(B) Let the original average speed of the train be $x \, km/h$.According to the problem,the time taken for the first part is $\frac{63}{x} \, hours$ and for the second part is $\frac{72}{x+6} \, hours$.The total time taken is $3 \, hours$,so:$\frac{63}{x} + \frac{72}{x+6} = 3$Dividing the entire equation by $9$:$\frac{7}{x} + \frac{8}{x+6} = \frac{1}{3}$Multiplying by $3x(x+6)$ to clear the denominators:$21(x+6) + 24x = x(x+6)$$21x + 126 + 24x = x^2 + 6x$$45x + 126 = x^2 + 6x$$x^2 - 39x - 126 = 0$Factoring the quadratic equation:$(x - 42)(x + 3) = 0$This gives $x = 42$ or $x = -3$.Since speed cannot be negative,the original average speed is $42 \, km/h$.

$A$ train travels at a certain average speed for a distance of $63 \, km$ and then travels a distance of $72 \, km$ at an average speed of $6 \, km/h$ more than its original speed. If it takes $3 \, hours$ to complete the total journey,what is its original average speed in $km/h$?

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