MAH-CET 2026 — Mathematics PYQ

1. Given Data

• Speed of the bus excluding stoppages ($S_{\text{fast}}$) = $54\text{ kmph}$

• Speed of the bus including stoppages ($S_{\text{slow}}$) = $45\text{ kmph}$

2. Formula

The standard shortcut formula to find the stoppage time per hour is:

$$\text{Stoppage time per hour}=\frac{\text{Speed excluding stoppages}-\text{Speed including stoppages}}{\text{Speed excluding stoppages}}\times 60\text{ minutes}$$

$$\text{Stoppage time per hour}=\frac{S_{\text{fast}}-S_{\text{slow}}}{S_{\text{fast}}}\times 60$$

3. Calculation

Substitute the given values into the formula:

$$\text{Stoppage time per hour}=\frac{54-45}{54}\times 60$$

$$\text{Stoppage time per hour}=\frac{9}{54}\times 60$$

Simplify the fraction $\frac{9}{54}$:

$$\text{Stoppage time per hour}=\frac{1}{6}\times 60$$

$$\text{Stoppage time per hour}=10\text{ minutes}$$

4. Alternative Conceptual Method

• In $1\text{ hour}$ without stopping, the bus travels $54\text{ km}$.

• With stoppages, it travels only $45\text{ km}$ in $1\text{ hour}$.

• This means the bus loses a distance of $54-45=9\text{ km}$ due to stops.

• The time required to cover this lost distance of $9\text{ km}$ at its original speed ($54\text{ kmph}$) is:

$$\text{Time}=\frac{\text{Distance}}{\text{Speed}}=\frac{9}{54}\text{ hours}=\frac{1}{6}\text{ hours}$$

Convert hours into minutes:

$$\text{Time}=\frac{1}{6}\times 60=10\text{ minutes}$$

Correct Answer: (b) 10