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/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original speed of the train in km/hr?

Correct Option

Let the original speed of the train be $x$ km/hr.

• The distance traveled at the original speed is 63 km. The time taken for this segment is $\frac{63}{x}$ hours.
• The train then travels 72 km at an increased speed of $(x+6)$ km/hr. The time taken for this segment is $\frac{72}{x+6}$ hours.
• The total journey time is given as 3 hours.

Formulating the equation based on the total time:

$$\frac{63}{x}+\frac{72}{x+6}=3$$

To solve for $x$, multiply the entire equation by $x(x+6)$ to eliminate the denominators:

$$63(x+6)+72x=3x(x+6)$$

Expand and simplify the equation:

$$63x+378+72x=3x^2+18x$$

$$135x+378=3x^2+18x$$

Rearrange the terms to form a quadratic equation:

$$3x^2+18x-135x-378=0$$

$$3x^2-117x-378=0$$

Divide the entire equation by 3 to simplify:

$$x^2-39x-126=0$$

Factor the quadratic equation:

$$x^2-42x+3x-126=0$$

$$x(x-42)+3(x-42)=0$$

$$(x-42)(x+3)=0$$

This yields two possible values for $x$: $x=42$ or $x=-3$.

Since speed cannot be a negative value, $x=-3$ is rejected. Therefore, the original speed of the train is 42 km/hr.

Incorrect Options

Options 1 (24 km/hr), 2 (33 km/hr), and 4 (66 km/hr) are incorrect because substituting these values into the derived equation $$\frac{63}{x}+\frac{72}{x+6}=3$$ does not satisfy the condition that the total journey time is 3 hours.

• If $x=24$ km/hr: $$\frac{63}{24}+\frac{72}{24+6}=2.625+\frac{72}{30}=2.625+2.4=5.025\ne 3$$
• If $x=33$ km/hr: $$\frac{63}{33}+\frac{72}{33+6}\approx 1.909+\frac{72}{39}\approx 1.909+1.846\approx 3.755\ne 3$$
• If $x=66$ km/hr: $$\frac{63}{66}+\frac{72}{66+6}\approx 0.954+\frac{72}{72}=0.954+1=1.954\ne 3$$