A person goes to his office by scooter at the speed of 50 km/hr and reaches 10 minutes earlier. If he goes at the speed of 40 km/hr, then he reaches 20 minutes late. What will be the speed (in km/hr) of the scooter to reach on time?

The correct option is 46 ²/₁₃ km/hr (which corresponds to the option 46 2/13).

To find the speed required to reach the office exactly on time, let us break down the problem step-by-step.

Step 1: Understand the relation between speed, distance, and time
Let the distance to the office be $d$ km, and the scheduled time to reach the office on time be $t$ hours.
Recall the basic formula relating distance, speed, and time:
$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

Step 2: Set up equations for both scenarios
In the first scenario, the person travels at a speed of 50 km/hr and reaches 10 minutes earlier than the scheduled time $t$.
Since 10 minutes is equal to $\frac{10}{60}=\frac{1}{6}$ hours, the time taken is:
$\frac{d}{50}=t-\frac{1}{6}$              (Equation 1)

In the second scenario, the person travels at a speed of 40 km/hr and reaches 20 minutes late.
Since 20 minutes is equal to $\frac{20}{60}=\frac{1}{3}$ hours, the time taken is:
$\frac{d}{40}=t+\frac{1}{3}$              (Equation 2)

Step 3: Solve for the distance ($d$)
Subtract Equation 1 from Equation 2 to eliminate the scheduled time $t$:
$\frac{d}{40}-\frac{d}{50}=\left(t+\frac{1}{3}\right)-\left(t-\frac{1}{6}\right)$

Simplify both sides:
$\frac{5d-4d}{200}=\frac{1}{3}+\frac{1}{6}$

$\frac{d}{200}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$

Solving for $d$:
$d=200\times \frac{1}{2}=100\text{km}$

Step 4: Solve for the scheduled time ($t$)
Substitute $d=100$ into Equation 1:
$\frac{100}{50}=t-\frac{1}{6}$

$2=t-\frac{1}{6}$

$t=2+\frac{1}{6}=\frac{13}{6}\text{hours}$

Step 5: Find the required speed to be on time
The required speed $s$ to cover the distance $d$ in scheduled time $t$ is given by:
$s=\frac{d}{t}=\frac{100}{\left(\frac{13}{6}\right)}$

$s=100\times \frac{6}{13}=\frac{600}{13}\text{km/hr}$

Converting this improper fraction to a mixed fraction:
$600÷13=46$ with a remainder of $2$.
Therefore,
$s=46\frac{2}{13}\text{km/hr}$