# Clock Problems

#### Chapter 24

5 Steps - 3 Clicks

# Clock Problems

### Introduction

Clock Problems is all about the hours hand , second hand, and minutes hand. In this chapter the problems are based mainly on the movement of the clock hands like minute spaces, minute hand, hour hand and the angles between the hands.

### Methods

Minute space: The face or dial of a watch is a circle whose circumference is divided into 60 equal parts called minute spaces.

Minutes hand: The larger hand in the clock denotes the minute hand or long hand.

Seconds hand: The hand that indicates the seconds on a clock or watch.

Hours hand: The shorter hand in the clock denotes the hours hand.

(i) In 60 minutes, the minute hand gains 55 minutes on the hour hand.

(ii) In every hour, both the hands coincide once.

(iii) The hands are same in the straight line when they are coincident or opposite to each other.

(iv) When the two hands are at right angles, they are 15 minute spaces apart.

(v) When the hands are in opposite directions, they are 30 minute spaces apart.

(vi) Angle traced by hour hand in 12 hours = 360 degrees.

(vii) Angle traced by minute hand in 60 minutes = 360 degrees.

Too fast and too slow:

(i) Too fast: If a watch or a clock indicates 2.15, when the correct time is 2, it is said to be 15 minutes too fast.

(ii) Too slow: On the other hand, if it indicates 3.45, when the correct time is 4, it is said to be 15 minutes too slow.

Example 1:
At what time between 4 and 5, will the hands of a clock coincide?

Solution:

At 4 O’clock, the hour hand has covered (4*30Â°) = 120Â°.

To catch up with the hour hand, the minute hand has to cover a relative distance of 120Â°, at a relative speed of $$5\frac{1}{2}Â°$$ per minute.

Thus, time required = $$\frac{120 \times 2}{11}$$ = $$\frac{240}{11}$$ = or $$21\frac{9}{11}$$ minutes.

Example 2:
At what time between 10 and 11 will the minute and hour hand be at right angles?

Solution:

At 10 O’clock, the hour hand has covered (10*30Â°) = 300Â°.

Considering that hour hand is at 10, to make a 90-degree angle with the hour hand, the minute hand has to be at 1 or 7.

For the first right angle, minute hand has to cover a relative distance of (1*30) = 30Â°.

For the 2nd right angle, minute hand has to cover a relative distance of (7*30)= 210Â°.

We know that the relative speed between the two hands is of $$5\frac{1}{2}Â°$$ per minute.

Hence, time required for the $$1^{st}$$ right angle = $$\frac{30 \times 2}{11}$$ = $$\frac{60}{11}$$ or $$5\frac{5}{11}$$ minutes.

Time required for the $$2^{st}$$ right angle = $$\frac{210 \times 2}{11}$$ = $$\frac{420}{11}$$ = $$38\frac{2}{11}$$ minutes.

Example 1:
The angle between the minute hand and the hour hand of a clock when the time is 4:20 is:

Solution:

At 4:00, hour hand was at 120 degrees.

Using the concept of relative distance, the minute hand will cover = $$\frac{20 \times 11}{2}$$ = 110 degrees

The angle between the hour hand and minute hand is = 120-110 = 10 degrees.

Example 2:
The angle between the minute hand and the hour hand of a clock when the time is 3:30 is:

Solution:

At 4:00, hour hand was at 90 degrees.

Using the concept of relative distance, the minute hand will cover = $$\frac{30 \times 11}{2}$$ = 165 degrees

The angle between the hour hand and minute hand is = 165-90= 75 degrees.

To learn the tricks to solve the third type of questions asked from this topic, read our article on Clocks-Gaining/Losing of Time.

### Samples

1. Find at what time between 9 and 10 o’clock will the hands of a clock be in the same straight line but not together?

Solution:

At 9 o’clock hour hand 15 minutes space apart from minute hand.

Also minute hand has gain of (30 – 15) = 15 minutes spaces apart over hour hand.

For both hand to be straight but not together they will be 30 minutes space apart.

55 minutes space gained by minute hand in 60 minutes.

So, 15 minutes spaces will be gained in = $$\frac{60}{55} * 15$$ minutes = 16$$\frac{4}{11}$$ minutes.

2. Find the angle between the hour hand and the minute hand of a clock when the time is 3.25?

Solution:

Angle traced by the hour hand in 12 hours = 360 degrees

Angle traced by it in 3 hours 25 minutes i.e. $$\frac{41}{12}$$ hours = $$(\frac{360}{12} * \frac{41}{12})^0$$ = $$({102\frac{1}{2}})^0$$

Angle traced by it in 25 minutes = $$(\frac{360}{60} * 25)^0$$ = $${150}^0$$

Therefore, required angle = $${150}^0 – ({102\frac{1}{2}})^0$$ = $$({47\frac{1}{2}})^0$$

3. A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on fourth day?

Solution:

Time from 5 a.m. on a day to 10 p.m. on fourth day = 89 hours

Now, 23 hours 44 minutes of this clock = 24 hours of correct clock.

Therefore, $$\frac{356}{15}$$ hours of this clock = 24 hours of correct clock.

89 hours of this clock = (24 x $$\frac{15}{356}$$ x 89) hours of correct clock. = 90 hours of correct clock.

So, the correct time is 11 p.m.

4. Find at what time between 4 and 5 o’clock will the hands of a clock be at right angle?

Solution:

At 4’o clock, the minute hand will be 20 min. spaces behind the hour hand.

Now, when the two hands are at right angles, both are 15 min. apart.

So, both are at right angles in two cases:

Case (i) When minute hand is 15 minutes spaces behind the hour hand:

In this case minutes hand will have to gain (20 – 15) = 5 minute spaces.

55 minutes spaces are gained by it in 60 minutes.

5 minutes spaces will be gained by it in $$\frac{60}{55} * 5$$ = 5$$\frac{5}{11}$$

Therefore, both are at right angles at 5$$\frac{5}{11}$$ minutes past 4.

Case (ii) When the minute hand is 15 minutes spaces behind the hour hand:

To be in the position, the minutes hand will have to gain (20 + 15) = 35 minute spaces.

35 minute spaces are gained in $$\frac{60}{55} * 35$$minutes = 38$$\frac{2}{11}$$ minutes.

Both are at right angles at 38$$\frac{2}{11}$$ minutes past 4.

5. At what time between 2 and 3 o’clock will the hands of a clock be together?

Solution:

At 2’o clock, the hour hand is at 2 and the minutes hand is at 12, i.e. they are 10 minutes spaces apart.

To be together, the minute hand must gain 10 minutes over the hour hand.

Now, 55 minutes are gained by it in 60 minutes.

Therefore, 10 minutes will be gained in $$\frac{60}{55} * 10$$ minutes = 10$$\frac{10}{11}$$ minutes.

Therefore, the hands will coincide at 10$$\frac{10}{11}$$ minutes past 2.