# Concept: LR - Clocks

CONTENTS

**INTRODUCTION**

Questions based on the concept of clocks are rarely asked in CAT these days. Although understanding this topic will be very helpful in other exams like SNAP, CMAT, MAT etc.

The questions can be broadly classified in two categories:

- Angle between the 2 hands of a clock
- Faulty clocks

**Angle between Hands of a Clock**

We know, that there total 12 numbers label in an analog clock. These numbers along with the center divide the whole clock in 12 sectors with each sector having a central angle of 30°.

**Example**: What is the angle between the minute and hour hands of the clock at 6 O’clock.

Solution:

We know at 6 O’clock, the hour had points exactly south while the minute hand points north.

The angle between the two hands is equivalent to 6 sectors (12-1, 1-2, 2-3, 3-4, 4-5 and 5-6).

∴ The angle between the two hands is 30° × 6 = 180°.

**Example**: What is the angle between the minute and hour hands of the clock at 9:30.

Solution:

We know at 9:30, the hour had points exactly between 9 and 10 while the minute hand points south.

The angle between the two hands is equivalent to 3.5 sectors (6-7, 7-8, 8-9 and half of 9-10).

∴ The angle between the two hands is 30° × 3.5 = 105°.

**Example**:What is the angle between the minute and hour hands of the clock at 1:20.

Solution:

We know at 1:20, the hour had points exactly between 1 and 2 while the minute hand points towards 4.

The angle between the two hands is equivalent to $3\frac{2}{3}$ = $\frac{11}{3}$ sectors (5-4, 4-3, 3-2 and a third of 2-1).

∴ The angle between the two hands is 30° × $\frac{11}{3}$ = 110°.

This was fairly easy. But what if we had to calculate the angle at say 3:48? We won’t be able to calculate the number of sectors so easily in this case.

Hence, we use a different approach for questions based on angles.

Speed of Hands of a Clock

The hour hand of a clock covers 360° in 12 hours

∴ Speed of hour hand = $\frac{360\xb0}{12\times 60\mathrm{mins}}$ = $\frac{1}{2}$ °/minute

The minute hand of a clock covers 360° in 1 hour

∴ Speed of hour hand = $\frac{360\xb0}{60\mathrm{mins}}$ = 6 °/minute

Now, if we have to calculate the angle between the two hands at h:m, we first calculate the position of both the hands after 12 O’clock.

Angle covered by hour hand till h hours and minutes after 12 O’clock

Time elapsed = (60h + m) minutes

∴ Angle covered by hour hand = (60h + m) × ½ = (30h + m/2)°

Angle covered by hour hand till h hours and minutes after 12 O’clock

Now, the minute hand comes back to its original position (north) after every complete hour. Hence, the change in it position would be due to last m minutes only.

∴ Angle covered by minute hand = m × 6 = 6m°

Now, the angle between the two hands will be the difference between the angles covered by them i.e.,

θ = |$\left|30h+\frac{m}{2}-6m\right|$|

θ = |$\left|30h-\frac{11}{2}m\right|$|

[Note: We have put a modulus sign here since we don’t whether 6m is greater or (30h + m/2).]

**Example**: What is the angle between the minute and hour hands of the clock at 3:48.

Solution:

We know, the angle between the two hands at h:m is

θ = |$\left|30h-\frac{11}{2}m\right|$|

Here, h = 3 and m = 48

⇒ θ = |$\left|30\times 3-\frac{11}{2}\times 48\right|$|

⇒ θ = |90 - 264| = 174°

**Example**: What is the angle between the minute and hour hands of the clock at 9:22.

Solution:

We know, the angle between the two hands at h:m is

θ = |$\left|30h-\frac{11}{2}m\right|$|

Here, h = 9 and m = 22

⇒ θ = |$\left|30\times 9-\frac{11}{2}\times 22\right|$|

⇒ θ = |270 - 121| = 149°

**Example**: When will the hands of a clock overlap between 4 and 5.

Solution:

We know, the angle between the two hands at h:m is

θ = |$\left|30h-\frac{11}{2}m\right|$|

Between 4 and 5, h = 4 and need to find the value of m

⇒ 0 = |$\left|30\times 4-\frac{11}{2}m\right|$|

⇒ 0 = 120 - $\frac{11m}{2}$

⇒ 120 = $\frac{11m}{2}$

⇒ m = $\frac{240}{11}$ = $21\frac{9}{11}$

∴ The hands will overlap at 4 : $21\frac{9}{11}$.

**Example**: When will the hands of a clock be opposite each other between 4 and 5.

Solution:

We know, the angle between the two hands at h:m is

θ = |$\left|30h-\frac{11}{2}m\right|$|

When the hands are opposite each other, the angle between them is 180°.

Between 4 and 5, h = 4 and need to find the value of m

⇒ 180 = |$\left|30\times 4-\frac{11}{2}m\right|$|

⇒ ± 180 = 120 - 11/2 m

**Case 1**: We take +180

⇒ 180 =120 - $\frac{11m}{2}$

⇒ -60 = $\frac{11m}{2}$

Not possible since m cannot be negative.

**Case 2**: We take -180

⇒ -180 =120 - $\frac{11m}{2}$

⇒ $\frac{11m}{2}$ = 300

⇒ m = $\frac{600}{11}$ = $54\frac{6}{11}$

∴ The hands will be opposite each other at 4 : $54\frac{6}{11}$.

**Example**: When will the hands of a clock make and angle of 60° with each other between 7 and 8.

Solution:

We know, the angle between the two hands at h:m is

θ = |$\left|30h-\frac{11}{2}m\right|$|

Between 7 and 8, h = 7 and need to find the value of m

⇒ 60 = |$\left|30\times 7-\frac{11}{2}m\right|$|

⇒ ± 60 = 210 - $\frac{11m}{2}$

**Case 1**: We take +60

⇒ 60 = 210 - $\frac{11m}{2}$

⇒ 150 = $\frac{11m}{2}$

⇒ m = $\frac{300}{11}$ = $27\frac{3}{11}$

∴ The hands make an angle of 60° at 7 : $27\frac{3}{11}$.

**Case 2**: We take -60

⇒ -60 = 210 - $\frac{11m}{2}$

⇒ 270 = $\frac{11m}{2}$

⇒ m = $\frac{540}{11}$ = $49\frac{1}{11}$

∴ The hands make an angle of 60° at 7 : $49\frac{1}{11}$.

∴ The hands make an angle of 60° at 7 : $27\frac{3}{11}$ and 7 : $49\frac{1}{11}$.