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…

Correct Option

Let the distance from the house to the factory be $x$x km and the scheduled time to reach the factory be $t$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}$5x​ hours. Converting 3 minutes to hours: $\frac{3}{60}$603​ hours.

So, the equation is: $\frac{x}{5} = t + \frac{3}{60}$5x​=t+603​ …(1)

• When the worker walks at 6 km/hr, he is 7 minutes early. The time taken is $\frac{x}{6}$6x​ hours. Converting 7 minutes to hours: $\frac{7}{60}$607​ hours.

So, the equation is: $\frac{x}{6} = t - \frac{7}{60}$6x​=t−607​ …(2)

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

$$\frac{x}{5} - \frac{x}{6} = \left( t + \frac{3}{60} \right) - \left( t - \frac{7}{60} \right)$$5x​−6x​=(t+603​)−(t−607​)

Simplify the left side:

$$\frac{6x - 5x}{30} = \frac{x}{30}$$306x−5x​=30x​

Simplify the right side:

$$t + \frac{3}{60} - t + \frac{7}{60} = \frac{3}{60} + \frac{7}{60} = \frac{10}{60}$$t+603​−t+607​=603​+607​=6010​

Equating both sides:

$$\frac{x}{30} = \frac{10}{60}$$30x​=6010​

$$\frac{x}{30} = \frac{1}{6}$$30x​=61​

Solve for $x$x:

$$x = \frac{30}{6}$$x=630​

$$x = 5 \text{ km}$$x=5 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.