krit.club logo

Time and Temperature - Time Intervals and Duration

Grade 5ICSE

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

🔑Concepts

•

Understanding the Clock Face: A clock is divided into 1212 main sectors representing hours. Each sector is further divided into 55 small markings representing minutes. Visualize the minute hand moving through all 6060 small markings to complete one full rotation, which equals 11 hour, while the hour hand moves from one number to the next.

•

12-hour and 24-hour Formats: In the 12-hour system, we use a.m. (ante meridiem) for the first 1212 hours of the day (midnight to noon) and p.m. (post meridiem) for the next 1212 hours (noon to midnight). In the 24-hour system, time is expressed as a four-digit number from 00000000 to 24002400. Visualize a timeline where 1:001:00 p.m. is represented as 13:0013:00 hours by adding 1212 to the p.m. hour.

•

Conversion of Units: To convert larger units to smaller units, we multiply (e.g., hours to minutes). To convert smaller units to larger units, we divide (e.g., seconds to minutes). Visualize a 'conversion ladder' where stepping down from hours to minutes requires multiplying by 6060, and stepping down from minutes to seconds requires multiplying by 6060 again.

•

Calculating Time Intervals (Duration): Duration is the amount of time that passes between a start time and an end time. To calculate this, imagine a jump method on a horizontal timeline: first jump from the start time to the next whole hour, then jump in hourly blocks to the target hour, and finally add the remaining minutes.

•

Borrowing in Time Subtraction: When subtracting time (e.g., 55 hours 2020 minutes minus 22 hours 4545 minutes), if the minutes in the subtrahend are greater than the minuend, we borrow 11 hour from the hours column. Importantly, visualize that 11 hour borrowed becomes 6060 minutes, not 100100. So, 2020 minutes becomes 20+60=8020 + 60 = 80 minutes.

•

Calculating Days between Dates: When counting days between two dates, identify the number of days left in the starting month, the full months in between, and the days in the final month. Visualize a calendar where you decide whether to include or exclude the start and end dates based on the problem statement.

•

Leap Years: A year is a leap year if it is divisible by 44. However, for century years (like 19001900 or 20002000), they must be divisible by 400400. In a leap year, February has 2929 days instead of 2828, making the total days in the year 366366 instead of 365365.

📐Formulae

1 hour=60 minutes1 \text{ hour} = 60 \text{ minutes}

1 minute=60 seconds1 \text{ minute} = 60 \text{ seconds}

1 day=24 hours1 \text{ day} = 24 \text{ hours}

1 week=7 days1 \text{ week} = 7 \text{ days}

1 ordinary year=365 days1 \text{ ordinary year} = 365 \text{ days}

1 leap year=366 days1 \text{ leap year} = 366 \text{ days}

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

End Time=Start Time+Duration\text{End Time} = \text{Start Time} + \text{Duration}

24-hour Time (for p.m.)=12-hour Time+12:00\text{24-hour Time (for p.m.)} = \text{12-hour Time} + 12:00

💡Examples

Problem 1:

A train departs from Station A at 09:4509:45 a.m. and reaches Station B at 03:2003:20 p.m. on the same day. Find the total duration of the journey.

Solution:

  1. Convert p.m. time to 24-hour format: 03:20 p.m.=3:20+12:00=15:2003:20 \text{ p.m.} = 3:20 + 12:00 = 15:20 hours.
  2. Set up the subtraction: 15 hours 20 minutes−9 hours 45 minutes15 \text{ hours } 20 \text{ minutes} - 9 \text{ hours } 45 \text{ minutes}.
  3. Since 20<4520 < 45, borrow 11 hour from 1515.
  4. 15 hours15 \text{ hours} becomes 14 hours14 \text{ hours}.
  5. 20 minutes20 \text{ minutes} becomes 20+60=80 minutes20 + 60 = 80 \text{ minutes}.
  6. Subtract minutes: 80−45=35 minutes80 - 45 = 35 \text{ minutes}.
  7. Subtract hours: 14−9=5 hours14 - 9 = 5 \text{ hours}.
  8. Total duration = 5 hours 35 minutes5 \text{ hours } 35 \text{ minutes}.

Explanation:

To find the duration, we subtract the start time from the end time. Converting to a 24-hour clock simplifies the calculation by avoiding the transition through noon. We use the regrouping method (borrowing 6060 minutes for 11 hour) because we cannot subtract 4545 from 2020 directly.

Problem 2:

Calculate the number of days from 12th April to 5th June (including both days).

Solution:

  1. Days in April: Total days in April is 3030. Days from 12th to 30th (including 12th) = 30−12+1=1930 - 12 + 1 = 19 days.
  2. Days in May: Full month = 3131 days.
  3. Days in June: Up to 5th (including 5th) = 55 days.
  4. Total days = 19+31+5=5519 + 31 + 5 = 55 days.

Explanation:

The problem requires counting the days across three different months. We calculate the remaining days in the first month, add the total days of the middle month(s), and add the specified days of the final month. Adding 11 in the first step ensures the start date is included in the count.