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Number Play - Clock and Calendar Numbers

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The 24-hour clock system format ranges from 00:0000:00 (midnight) to 23:5923:59. To convert PM time to 24-hour format, we add 1212 to the hours (except for 1212 PM).

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A leap year consists of 366366 days, where February has 2929 days. A non-leap year (ordinary year) has 365365 days.

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A year is a leap year if it is divisible by 44. For century years (years ending in 0000), the year must be divisible by 400400 to be a leap year.

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To find the duration of an event, we subtract the start time from the end time. If the minutes in the end time are less than the start time, we borrow 11 hour (6060 minutes).

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In calendar math, the concept of 'odd days' helps find the day of the week. Odd days are the remainder obtained when the total number of days is divided by 77.

📐Formulae

Time Duration=End Time−Start Time\text{Time Duration} = \text{End Time} - \text{Start Time}

Total Days in n weeks=7×n\text{Total Days in } n \text{ weeks} = 7 \times n

Leap Year Condition: (Year÷4=integer) and for centuries (Year÷400=integer)\text{Leap Year Condition: } (\text{Year} \div 4 = \text{integer}) \text{ and for centuries } (\text{Year} \div 400 = \text{integer})

💡Examples

Problem 1:

Convert 4:254:25 PM into 24-hour clock format.

Solution:

4:25 PM=4+12=16:25 hours4:25 \text{ PM} = 4 + 12 = 16:25 \text{ hours}

Explanation:

To convert PM time to 24-hour format, we add 1212 to the hours. Thus, 44 hours +12+ 12 hours =16= 16 hours.

Problem 2:

A train leaves a station at 21:4521:45 and reaches its destination at 02:1502:15 the next day. Find the duration of the journey.

Solution:

Time from 21:45 to midnight (24:00)=24:00−21:45=2 hours 15 minutes\text{Time from } 21:45 \text{ to midnight (24:00)} = 24:00 - 21:45 = 2 \text{ hours } 15 \text{ minutes} Time from midnight to 02:15=2 hours 15 minutes\text{Time from midnight to } 02:15 = 2 \text{ hours } 15 \text{ minutes} Total Duration=2 h 15 min+2 h 15 min=4 hours 30 minutes\text{Total Duration} = 2 \text{ h } 15 \text{ min} + 2 \text{ h } 15 \text{ min} = 4 \text{ hours } 30 \text{ minutes}

Explanation:

First, calculate the time remaining in the first day by subtracting 21:4521:45 from 24:0024:00. Then add the time elapsed in the second day.

Problem 3:

How many days are there from 25th25^{\text{th}} January 20242024 to 5th5^{\text{th}} March 20242024? (Include both dates)

Solution:

Days in Jan=31−25+1=7 days\text{Days in Jan} = 31 - 25 + 1 = 7 \text{ days} Days in Feb (2024 is leap year)=29 days\text{Days in Feb (2024 is leap year)} = 29 \text{ days} Days in Mar=5 days\text{Days in Mar} = 5 \text{ days} Total days=7+29+5=41 days\text{Total days} = 7 + 29 + 5 = 41 \text{ days}

Explanation:

20242024 is divisible by 44, so it is a leap year and February has 2929 days. We count the remaining days in January, all of February, and the first 55 days of March.

Problem 4:

Subtract 44 hours 4545 minutes from 99 hours 2020 minutes.

Solution:

9 h20 min−4 h45 min\begin{array}{r} 9 \text{ h} & 20 \text{ min} \\ -4 \text{ h} & 45 \text{ min} \\ \hline \end{array} Borrow 11 hour (6060 min) from 99 hours: 8 h80 min−4 h45 min4 h35 min\begin{array}{r} 8 \text{ h} & 80 \text{ min} \\ -4 \text{ h} & 45 \text{ min} \\ \hline 4 \text{ h} & 35 \text{ min} \end{array}

Explanation:

Since 2020 minutes is less than 4545 minutes, we borrow 11 hour from the 99 hours, making it 88 hours and adding 6060 minutes to 2020 to get 8080 minutes.

Clock and Calendar Numbers Class 6 Notes & Examples