Question

Abha starts for the school at 6.45 AM and comes back home at 2.15 PM. How long does she stay in school if 40 minutes are spent in going and coming?

Answer 1

Hint: Here, we are given starting time and ending time. For finding the time period between them, we will first change units of both times to the same units. After that, we will know the time period that she stays outside. From this time, we will subtract 40 minutes of going and coming time to find time spent in school.

Complete step-by-step answer:
We are given arriving time at 6.45 am that is 6 hours and 45 minutes. Converting it into hours, as we know, 1 hour = 60 minutes, therefore, 1 minute = $\dfrac{1}{60}$ hours. Therefore, changing 45 minutes to hours, we get $\Rightarrow 45\times \dfrac{1}{60}\text{hours}=\dfrac{3}{4}\text{hours}$.
Hence, 6 hours 45 minutes become \[6+\dfrac{3}{4}\text{hours}=\dfrac{27}{4}\text{hours}\cdots \cdots \cdots \cdots \left( i \right)\]
Time at which she comes back is given as 2.15 pm. To remove pm, we can convert the time to a 24 hour clock by adding 12 to the hour's side. Hence, we get time as 14 hours 15 minutes.
Converting 15 minutes into hours, we get:
\[15\text{ minutes}=15\times \dfrac{1}{60}\text{hours}=\dfrac{1}{4}\text{hours}\]
Therefore, 14 hours 15 minutes becomes \[14+\dfrac{1}{4}\text{hours}=\dfrac{57}{4}\text{hours}\cdots \cdots \cdots \cdots \left( ii \right)\]
Calculating time difference between them, by subtracting (i) from (ii) we get:
\[\dfrac{57}{4}-\dfrac{27}{4}=\dfrac{30}{4}\text{hours}=7\dfrac{1}{2}\text{hours}\]
We can write $7\dfrac{1}{2}\text{hours}$ by \[7+\dfrac{1}{2}\text{hours,}\dfrac{1}{2}\text{hour}=\dfrac{60}{2}\text{minutes}=30\text{minutes}\]
Hence, time difference is \[\text{7 hours 3}0\text{ minutes }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }\left( \text{iii} \right)\]
Time spent coming and going is 40 minutes. So, we will subtract this from (iii) to obtain the required answer.
1 hour = 60 minutes
Therefore, 7 hours 30 minutes \[\Rightarrow 7\times 60+30=420+30=450\text{ minutes}\]
Required time \[\Rightarrow 450-40\text{ minutes}=410\text{ minutes}\]
Converting minutes into hour we get:
\[410\text{ minutes}=\dfrac{410}{60}\text{hours}=6\dfrac{50}{60}\text{hours}\]
Which gives us
\[\begin{align}
  & 6+\dfrac{50}{60}\text{hours and }\dfrac{50}{60}\text{hours}=\dfrac{50}{60}\times 60\text{ minutes} \\
 & \Rightarrow \text{50 minutes} \\
\end{align}\]
Hence, required time is found to be 6 hours 50 minutes.

Note: Students can calculate time difference orally also. Students should carefully check if it is am or pm. If pm is given, it must be converted into 24 hours format before proceeding further. Also, try to convert time into hours or minutes completely to avoid complex calculations.