# Clocks

The problems in Clocks usually deal with finding the angle between the hour hand and the minute hand, number of times the two hands coincide, etc.

- The markings on the face of a clock are 60 spaces, one each for a minute. Every hour, the minute hand completes one round of 60 spaces and the hour hand completes one full round every 12 hours.
- In 60 minutes, the minute hand gains 55 spaces (also known as minute spaces) over the hour hand. For example, if the initial time is 12:00, then after 1 hour, the minute hand would cover 60 spaces whereas the hour hand would cover only 5 spaces. Thus, the minute hand covers 55 spaces extra than the hour hand.
- The minute hand covers 360 degrees in 60 minutes.

=> In 1 minute, the minute hand covers 360 / 60 = 6 degrees - The hour hand covers 360 degrees in 12 hours.

=> In 1 hour, the hour hand covers 360 / 12 = 30 degrees

=> In 1 minute, the hour hand covers 30 / 60 = 0.50 degrees - The angle between the minute hand and the hour hand increases by 5.50 degrees every minute. For example, after 2 minutes, angle made by the minute hand = 2 x 6 = 12 degrees and angle made by the hour hand = 2 x 0.50 = 1 degree

=> Angle between the hour hand and the minute hand after 2 minutes = 12 – 1 = 11 degrees = 2 x 5.50 degrees - In every hour, the minute hand and the hour hand coincide once.
- If the minute hand and the hour hand are in the same line, then the angle between them is either 0 degree or 180 degrees.
- The angle between the minute hand and the hour hand is 180 degrees if they are 30 spaces apart, 90 degrees if they are 15 spaces apart, and 0 degrees if they are 0-minute spaces apart.
- If the clock shows time ahead of the actual time, it is said to be running fast.

For example, if the clock is showing 12:15 PM but it is actually 12:00 PM, then the clock is said to be running 15 minutes fast. - If the clock shows time behind the actual time, it is said to be running slow.

For example, if the clock is showing 2:15 PM but it is actually 2:30 PM, then the clock is said to be running 15 minutes slow.

### Sample Problems

**Question 1 : **Find the angle between the hands of a clock at 3:20 PM.**Solution : **At 3:00 PM, angle made by the minute hand = 0 degree and angle made by the hour hand = 3 x angle made by the hour hand in one hour = 3 x 30 = 90 degrees

Now, in the next 20 minutes, angle made by the minute hand = 20 x angle made by the minute hand in 1 minute = 20 x 6 = 120 degrees and angle made by the hour hand = 20 x angle made by the hour hand in 1 minute = 20 x 0.50 = 10 degrees

=> Angle made by the minute hand at 3:20 PM = 0 + 120 = 120 degrees

=> Angle made by the hour hand at 3:20 PM = 90 + 10 = 100 degrees

Therefore, angle between the hands of the clock at 3:20 PM = 120 – 100 = 20 degrees**Another Method :**

At 3:00 PM, angle made by the minute hand = 0 degree and angle made by the hour hand = 3 x angle made by the hour hand in one hour = 3 x 30 = 90 degrees

=> Initial angle between the two hands = 90 degrees

Now, we know that the difference between the two hands of the clock increases every minute by 5.50 degrees.

=> Difference between the hands of the clock after 20 minutes = 20 x 5.50 = 110 degrees

Therefore, difference between the two hands at 3:20 PM = 110 – 90 = 20 degrees

**Question 2 : **At what time between 3 PM and 4 PM would the two hands of the clock be together ?**Solution : **At 3 PM, the hour hand would be at 15 spaces and the minute hand would be at 0 spaces. The minute hand would have to cover these extra 15 spaces in order to meet the hour hand.

Now, 55 minutes are gained by the minute hand in 60 minutes.

=> 15 minutes would be gained in (60 / 55) x 15 = 180 / 11 minutes

Thus, the two hands of the clock meet at 180 / 11 minutes past 3 PM, i.e., around 3:16:22 PM.

**Question 3 : **How many times in a day the two hands of a clock coincide?**Solution : **Between 11 to 1, the hands of the clock coincide only once, i.e., at 12. So, every 12 hours, they coincide 11 times.

Therefore, the two hands of the clock coincide 22 times in a day.

This article has been contributed by **Nishant Arora**

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