# PE 1 - Clocks | LR - Clocks

**Answer the following 4 questions based on the information given below.**

There are two faulty analog clocks, one gains 2 minutes every hour while the other loses 3 minutes every hour. Both of them show the correct time at 6 a.m. on 1^{st} January.

**PE 1 - Clocks | LR - Clocks**

When will the slower clock show correct time again?

- A.
12 noon on 10

^{th}Jan - B.
6 a.m. on 10

^{th}Jan - C.
6 a.m. on 12

^{th}Jan - D.
6 a.m. on 11

^{th}Jan

Answer: Option D

**Explanation** :

For the clock to show correct time again, the clock needs to lose 12 hours.

The slower clock loses 3 minutes every hour, hence it will lose 12 hours in $\frac{12\times 60}{3}\times 1$ = 240 hours = 10 days.

⇒ 10 days after 6 a.m. of 1^{st} January will be 6 a.m. of 11^{th} January.

Hence, option (d).

Workspace:

**PE 1 - Clocks | LR - Clocks**

When will the faster clock show correct time again?

- A.
6 a.m. of 16

^{th}January - B.
6 a.m. of 15

^{th}January - C.
10 p.m. of 16

^{th}January - D.
12 noon of 12

^{th}January - E.
None of these

Answer: Option A

**Explanation** :

For the clock to show correct time again, the clock needs to gain 12 hours.

The faster clock gains 2 minutes every hour, hence it will gain 12 hours in $\frac{12\times 60}{2}\times 1$ = 360 hours = 15 days.

⇒ 15 days after 6 a.m. of 1^{st} January will be 6 a.m. of 16^{th} January.

Hence, option (a).

Workspace:

**PE 1 - Clocks | LR - Clocks**

When will both of them show the correct time again?

- A.
12 noon on 30

^{th}January - B.
6 p.m. on 31

^{st}January - C.
6 a.m. on 31

^{st}January - D.
6 a.m. on 30

^{th}January - E.
None of these

Answer: Option C

**Explanation** :

Consider the solution to first 2 questions of this set.

The slower clock will show correct time every 10 days, while the faster clock will show correct time every 15 days.

Hence, both of them will show correct time every LCM(10, 15) = 30 days.

∴ Both of them will show correct time at 6 a.m. on 31^{st} January.

Workspace:

**PE 1 - Clocks | LR - Clocks**

When will both of them show the same time again?

- A.
12 noon on 7

^{th}Jan - B.
6 a.m. on 8

^{th}Jan - C.
6 a.m. on 7

^{th}Jan - D.
6 p.m. on 12

^{th}Jan - E.
None of these

Answer: Option C

**Explanation** :

Here it is not necessary that the same time they show should be the correct time. They may show incorrect time as long as they show the same time.

Every hour the difference in time between the 2 clocks increases by 5 minutes.

For both of them to show the same time again, the total time difference should be 12 hours.

∴ The time difference between the two clocks will becom 12 hours after = $\frac{12\times 60}{5}$ = 144 hours

⇒ 144 hours after 6 am Monday will be 6 days, i.e., 6 a.m. on 7^{th} Jan.

Hence, option (c).

Workspace:

**PE 1 - Clocks | LR - Clocks**

One evening, Aakash goes out for a drive, sometime between 3 O'clock and 4 O’clock. When he gets back, sometime between 8 O’clock and 9 O’clock, he noticed that the minute and hour hands have interchanged their positions exactly with what they were when he started out. At what time did he leave his house?

- A.
3 hours 44 minutes and 10 seconds

- B.
3 hours 42 minutes and 20 seconds

- C.
3 hours 41 minutes and 32 seconds

- D.
3 hours 43 minutes and 25 seconds

- E.
None of these

Answer: Option C

**Explanation** :

Let the angle between hour and minute hand when Aakash left the house be θ°.

Aakash leave at some time 3:x. (40 < x <4 5),

and comes back at some time 8:y. (15 < y < 20)

Now, let us calculate the angle covered by both hands when Aakash came back.

**Hour hand**:

Since the two hands exchange their position, the hour hand would cover an angle of θ°.

**Minute hand**:

Angle covered by minute hand

from 3:x till 4:x = 360°

from 4:x till 5:x = 360°

from 5:x till 6:x = 360°

from 6:x till 7:x = 360°

Now from 7:x till 8:y, the angle covered will be = (360- θ)°

∴ Total angle covered by minute hand = 360 × 5 - θ = (1800 - θ)°

Speed of hour hand is 1/2°/minute and that of minute hand is 6°/minute.

Time taken by hour hand to cover θ° is same as time taken by minute hand to cover (1440 + θ)°.

⇒ $\frac{\theta}{1/2}$ = $\frac{1800-\theta}{6}$

⇒ θ = 1800/13°

Now, we need to calculate when will the anlge between hour and minute hand be 1440/11° between 3 and 4 O'clock.

⇒ $\frac{1800}{13}$ = |$\left|30\times 3-\frac{11}{2}\mathrm{m|}\right|$

⇒ 90 - $\frac{11}{2}m$ = ± $\frac{1800}{13}$

⇒ $\frac{11}{2}m$ = 90 ± $\frac{1800}{13}$

⇒ m = 41 minutes 32 seconds

∴ Aakash left his house at 3 hours 41 minutes and 32 seconds.

Hence, option (c).

Workspace:

**PE 1 - Clocks | LR - Clocks**

How much does a watch gain or lose per day, if its hands coincide every 66 minutes?

- A.
Loses$12\frac{111}{121}$

- B.
Loses $11\frac{109}{121}$

- C.
Gains $11\frac{109}{121}$

- D.
Gains $12\frac{111}{121}$

Answer: Option B

**Explanation** :

Let us calculate time interval for hands of a normal clock to coincide.

The two hands coincide at 12 O'clock. Next they will coincide between 1 and 2 O'clock.

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

⇒ $\left|30\times 1-\frac{11}{2}\mathrm{m}\right|$ = 0

⇒ 30 - $\frac{11}{2}$m = 0

⇒ m = $\frac{60}{11}$ = 5$\frac{5}{11}$ minutes.

∴ The two hands coincide again at 1 : 5$\frac{5}{11}$.

⇒ From 12 O'clok till 1 : 5$\frac{5}{11}$ it takes 65$\frac{5}{11}$ minutes for the two hands to coincide again.

For a normal clock, hands coincide every $65\frac{5}{11}$ minutes, but for the faulty clock hands coincide every 66 minutes.

[If you look at the faulty clock you would get a sense that only $65\frac{5}{11}$ minutes have passed when actually 66 minutes have passed.]

∴ The faulty clock loses 66 - $65\frac{5}{11}$ = $\frac{6}{11}$ minutes every 66 minutes.

∴ Time lost in a day i.e., 24 hours = $\frac{\frac{6}{11}}{66}\times 24\times 60$ = $\frac{1440}{121}$ = $11\frac{109}{121}$ minutes.

Hence, option (b).

Workspace:

**PE 1 - Clocks | LR - Clocks**

If the minutes hand and seconds' hand of a clock are 45 minutes apart. What will be the angle formed between them?

[Type in your answer as the nearest possible integer in degrees.]

Answer: 270

**Explanation** :

For 60 minutes the angle covered is 360°

So for 45 minutes difference angle will be = 360/60 × 45 = 270°

Hence, 270.

Workspace:

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