Examkin

Question

Difficulty: MediumClocks and Calendars

A city hall maintains an antique mechanical tower clock that has a constant drift, falling behind standard time by exactly 2.52.5 minutes every 2424 hours. The maintenance crew calibrates the clock to the precise standard time on January 14, 19001900, at exactly 12:00 PM (Noon). Calculate the total amount of time, in minutes, that the clock will have drifted behind standard time by exactly 12:00 PM (Noon) on March 15, 19001900.

Answer: 150 minutes

Answer

The clock will have accumulated a total drift of 150 minutes.
The correct answer is derived by accurately counting the exact number of days between January 14, 1900, and March 15, 1900. Since 1900 is not a leap year (it is a century year not divisible by 400), February has exactly 28 days. Adding the 17 days in January, 28 days in February, and 15 days in March gives a total of 60 days. Multiplying the 60 days by the daily drift rate of 2.5 minutes yields exactly 150 minutes.

Step-by-Step Solution

1
Calculate the days elapsed in January.
31−14=1731 - 14 = 17 days.
To find the time accumulated during the first month.
2
Determine the number of days in February 19001900.
February has 2828 days.
The year 19001900 is a century year not divisible by 400400, making it a standard non-leap year.
3
Sum the total days from January 14 to March 15.
17+28+15=6017 + 28 + 15 = 60 days.
To find the total 2424-hour periods over which the drift occurred.
4
Calculate the total drift in minutes.
60×2.5=15060 \times 2.5 = 150 minutes.
The clock loses exactly 2.52.5 minutes for every full 2424-hour period elapsed.

Key Concept

Century Leap Year Rule and Time Accumulation
Estimated Time:1m 30s
Rate this question

Topics

Clocks and Calendars

More questions from this topic