A worker reaches his factory 3 minutes late if his speed from his house to the factory is 5 km/hr. If he walks at a speed of 6 km/hr., then he reaches the factory 7 mins early. The distance of the factory from his house is

Correct Option

Let the distance from the house to the factory be $x$ km and the scheduled time to reach the factory be $t$ hours.

According to the problem statement, two scenarios are given:

• When the worker walks at 5 km/hr, he is 3 minutes late. The time taken is $\frac{x}{5}$ hours. Converting 3 minutes to hours: $\frac{3}{60}$ hours.

So, the equation is: $\frac{x}{5}=t+\frac{3}{60}$ …(1)

• When the worker walks at 6 km/hr, he is 7 minutes early. The time taken is $\frac{x}{6}$ hours. Converting 7 minutes to hours: $\frac{7}{60}$ hours.

So, the equation is: $\frac{x}{6}=t-\frac{7}{60}$ …(2)

To find $x$, subtract equation (2) from equation (1):

$$\frac{x}{5}-\frac{x}{6}=(t+\frac{3}{60})-(t-\frac{7}{60})$$

Simplify the left side:

$$\frac{6x-5x}{30}=\frac{x}{30}$$

Simplify the right side:

$$t+\frac{3}{60}-t+\frac{7}{60}=\frac{3}{60}+\frac{7}{60}=\frac{10}{60}$$

Equating both sides:

$$\frac{x}{30}=\frac{10}{60}$$

$$\frac{x}{30}=\frac{1}{6}$$

Solve for $x$:

$$x=\frac{30}{6}$$

$$x=5\text{ km}$$

Thus, the distance of the factory from his house is 5 km.

Incorrect Options:

Options 1 (3 km), 2 (4 km), and 4 (6 km) are incorrect because they do not satisfy the conditions derived from the given speeds and time differences. The calculated distance based on the established equations is uniquely 5 km.