# Question #5ce04

##### 2 Answers

#### Answer:

48 times

#### Explanation:

In one hour, the clock's short and long hands form a 90° angle 2 times.There are 24 hours in a day. Hence, we have 2x24=48.

Therefore, the clock's short and long hand form 48 90° angles in one day.

#### Answer:

#### Explanation:

Take a look at how an analog clock that has

Now, the common approach here is to say that in **hour**, the hour hand and the minute hand are perpendicular to each other **times**.

This would lead to the conclusion that in **hours**, the two hands are at

#12 color(red)(cancel(color(black)("hours"))) * "2 times"/(1color(red)(cancel(color(black)("hour")))) = "24 times"#

Consequently, you would say that in **hours**, the two hands are at a

#24 color(red)(cancel(color(black)("hours"))) * "24 times"/(12color(red)(cancel(color(black)("hours")))) = "48 times"#

This is * not* the correct answer. Here's why.

As you can see, the circle is divided into

#(360^@)/"12 intervals" = 30^@"/ interval"#

An *interval* here is simply **hour**. This implies that the **hour hand** completes **per hour**. Now, you know that the minute hand completes a **full revolution** every hour.

This is equivalent to saying that the **minute hand** completes **per hour**.

This implies that the *minute hand* completes

#360^@ - 30^@ = 330^@#

in **one hour** *relative* to the hour hand. Here's where the tricky part comes in. Let's assume that you're going to keep the hour hand **locked in place** at

In this case, the minute hand will only complete **in one hour** because it's moving *relative* to a hand that is stuck in place.

In other words, the minute hand is "losing" **minutes**, for every passing hour, i.e. it doesn't complete a full revolution in

In

#12 color(red)(cancel(color(black)("hours"))) * (30^@"lost")/(1color(red)(cancel(color(black)("hour")))) = 360^@"lost"#

This means that in **full revolutions**. Since the hands are at a **twice** per hour, you will have

#overbrace(11color(red)(cancel(color(black)("hours"))))^(color(blue)("the equivalent of 12 hours when both hands are moving")) * "2 times"/(1color(red)(cancel(color(black)("hour")))) = "22 times"#

This implies that in **hours**, the two hands are at a

#24 color(red)(cancel(color(black)("hours"))) * "22 times"/(12color(red)(cancel(color(black)("hours")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("44 times")color(white)(a/a)|)))#

So, to sum this up, when **both hands** are moving, the minute hand completes **full revolutions** in **hours**.

When the hour hand is **kept in place**, the minute hand only completes **full revolutions** in **hours**. That is the case because it's moving at *relative* to the hour hand, which is being kept in place.

As a result, two perpendicular positions will be "lost" every

This is why the two hands are * not* perpendicular to each other