# Aptitude::Clock

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Example 1 / 10

As per the formula:

Angle traced by the hour hand in 12 hours = \(360^{0}\).

Angle traced by it in 3 hrs 25 min (3 + \(\frac{25}{60}\) = \(\frac{41}{12}\) hrs) = \(\left(\frac{360}{12} \times \frac{41}{12}\right)^{0}\) = \(100\frac{1}{2}^{0}\)

Angle traced by minute hand in 60 min. = \(360^{0}\)

Angle traced by it in 25 min. = \(\left(\frac{360}{60} \times 25 \right)^{0}\)

Reqired angle = \(\left(150^{0} - 102\frac{1}{2}^{0}\right)\) = \(47\frac{1}{2}^{0}\)

Angle traced by the hour hand in 12 hours = \(360^{0}\).

Angle traced by it in 3 hrs 25 min (3 + \(\frac{25}{60}\) = \(\frac{41}{12}\) hrs) = \(\left(\frac{360}{12} \times \frac{41}{12}\right)^{0}\) = \(100\frac{1}{2}^{0}\)

Angle traced by minute hand in 60 min. = \(360^{0}\)

Angle traced by it in 25 min. = \(\left(\frac{360}{60} \times 25 \right)^{0}\)

Reqired angle = \(\left(150^{0} - 102\frac{1}{2}^{0}\right)\) = \(47\frac{1}{2}^{0}\)

**Ans.**
NA

NA

NA

Example 2 / 10

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

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

As per formula: In 60 minutes, the minute hand gains 55 minutes on the hour hand

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

so, 10 minutes will be gained in min. hand = \(\frac{60}{55} \times 10 min.\) = \(10\frac{10}{11}\) min.

The hands will coincide at \(10\frac{10}{11}\) min. past 2.

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

As per formula: In 60 minutes, the minute hand gains 55 minutes on the hour hand

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

so, 10 minutes will be gained in min. hand = \(\frac{60}{55} \times 10 min.\) = \(10\frac{10}{11}\) min.

The hands will coincide at \(10\frac{10}{11}\) min. past 2.

**Ans.**
NA

NA

NA

Example 3 / 10

At 5 o'clock, the hands are 25 min. spaces apart.

To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

As we know, 55 min. spaces are gained in 60 min

So, 40 min. spaces are gained in \(\frac{60}{55}\times 40\) min. = \(43\frac{7}{11}\) min.

=> They ar right angles at \(43\frac{7}{11}\) min. past 5.

To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

As we know, 55 min. spaces are gained in 60 min

So, 40 min. spaces are gained in \(\frac{60}{55}\times 40\) min. = \(43\frac{7}{11}\) min.

=> They ar right angles at \(43\frac{7}{11}\) min. past 5.

**Ans.**Note:

1. There are 2 right angles every hour, here only one right angle was possible because of given time i.e 5:30.

2. But in a 12 hours period, there are 22 such angles.

3. In 24 hours period, there are 44 such angles.

Please go through the next example for better understanding.

NA

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Example 4 / 10

When the two hands are at right angles, they are 15 min. space apart.

**Case 1:**

When the minute hand is 15 min. space behind the hour hand:

In this case min. hand will have to gain (25-15) = 10 min. space.

55 min. spaces are gained by it in 60 min.

10 min. spaces will be gained by it = \(\left(\frac{60}{55} \times 10\right)\) min. = \(\frac{120}{11}\) min.

So, they are at right angle at \(10\frac{10}{11}\) min. past 5.

**Case 2:**

At 5 o'clock, the hands are 25 min. spaces apart.

To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

As we know, 55 min. spaces are gained in 60 min

So, 40 min. spaces are gained in \(\frac{60}{55}\times 40\) min. = \(43\frac{7}{11}\) min.

=> They ar right angles at \(43\frac{7}{11}\) min. past 5.

**Ans.**

Note:

1. There are 2 right angles every hour, one will be before 5 and one after 5.

2. But in a 12 hours period, there are 22 such angles.

3. In 24 hours period, there are 44 such angles.

NA

NA

Example 5 / 10

**Important Point:**When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.

At 7 o'clock, they are 25 min. spaces apart.

i.e Minute hand will have to gain only 5 min. spaces more. (As we know that when min hand will move 5 min. then hour hand to move. so, here we have to find that hour hand gain when min. hand move 5 min.)

As we know,

55 min. spaces are gained in 60 min.

=> 5 min spaces are gained = \(\left(\frac{60}{55}*5\right)\) min. = \(5\frac{5}{11}\) min

so, Required time = \(5\frac{5}{11}\) min past 7

NA

NA

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Example 6 / 10

At 5 o'clock the minute hand is 25 min. spaces behind the hour hand.

**Case 1:**

When the minute hand is 3 min. space behind the hour hand:

In this case min. hand will have to gain (25-3) = 22 min. space.

55 min. spaces are gained by it in 60 min.

22 min. spaces will be gained by it = \(\left(\frac{60}{55} \times 22\right)\) min. = 24 min.

So, the hands of the clock will be 3 minutes apart at 24 min. past 5.

**Case 2:**

The minute hand is 3 min. spaces ahead of the hour hand.

In this case, the minute hand has to gain (25 + 3) = 28 min. spaces.

As we know, 55 min. spaces are gained in 60 min

So, 28 min. spaces are gained in \(\frac{60}{55}\times 28\) min. = \(31\frac{5}{11}\) min.

=> They ar right angles at \(31\frac{5}{11}\) min. past 5.

**Ans.**

NA

NA

NA

Example 7 / 10

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes over the hour

hand.

55 min. are gained in 60 min.

60 min is gained in \(\left(\frac{60}{55}\right) \times 60\) min = \(\frac{720}{11}\) min.

But, they are together after 65 min.

Gain in 65 min = \(\frac{720}{11}\)-65 = \(\frac{5}{11}\) min.

Gain in 24 hours = \(\left(\frac{5}{11} * \frac{60\times 24}{65}\right)\)min. = \(\frac{1440}{143}\) min.

The clock gains \(\frac{1440}{143}\)or \(10\frac{10}{143}\) minutes in 24 hours.

**Ans.**

NA

NA

NA

Example 8 / 10

This Sunday morning at 8:00 AM, the watch is 5 min. Slow, and the next Sunday at 8:00 PM it becomes 5 min 48 sec fast.

The watch gains \(5 + 5\frac{48}{60}\) min in a time of (7×24)+12 = 180 hours.

To show the correct time, it has to gain 5 min.

\(\frac{54}{5}\) min. → 180hours

5min ->\(\left(5/\frac{54}{2} \times 180\right)\)

\(83\frac{1}{3}\)hrs. = 72hrs + \(11\frac{1}{3}\) hrs. = 3 days + 11 hrs. + 20 min.

So the correct time will be shown on Wednesday at 7:20 PM

The watch gains \(5 + 5\frac{48}{60}\) min in a time of (7×24)+12 = 180 hours.

To show the correct time, it has to gain 5 min.

\(\frac{54}{5}\) min. → 180hours

5min ->\(\left(5/\frac{54}{2} \times 180\right)\)

\(83\frac{1}{3}\)hrs. = 72hrs + \(11\frac{1}{3}\) hrs. = 3 days + 11 hrs. + 20 min.

So the correct time will be shown on Wednesday at 7:20 PM

**Ans.**
NA

NA

NA

Example 9 / 10

As per the question, Loss of time = 16 minutes for 24 hours.

Time is taken to lost 1 hr = \(\left(\frac{24}{16} \times 60\right)\) = 90 hrs. ---(Eq.1)

Thereafter 90 hrs, the clock shows 89 hrs.

Counting hrs from 5 am to 4th day 10 pm = (24*4)-2-5 = 89 hrs.

There if the clock shows 89 hrs, then real-time is \(\left(\frac{89}{89} \times 90\right)\)= 90hrs which is 1 hr more (As per Eq. 1)

Therefore the correct time is 11 p.m.

Time is taken to lost 1 hr = \(\left(\frac{24}{16} \times 60\right)\) = 90 hrs. ---(Eq.1)

Thereafter 90 hrs, the clock shows 89 hrs.

Counting hrs from 5 am to 4th day 10 pm = (24*4)-2-5 = 89 hrs.

There if the clock shows 89 hrs, then real-time is \(\left(\frac{89}{89} \times 90\right)\)= 90hrs which is 1 hr more (As per Eq. 1)

Therefore the correct time is 11 p.m.

NA

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Example 10 / 10

Given, The clock gains 10 minutes in 24 hours.

i.e. 24 * 60= 1440 min, there is a gain of 10 mins.

So the ratio of real-time to the time of the clock is 1440:1450(gain of 10 min. added) = 144:145

Here, Time elapsed in the clock is = 29 hr i.e. 29*60 = 1740 min.

So, the real-time (a) can be found by

144:145::a:1740

a = \(\left(\frac{144 \times 1740}{145}\right)\) = 1728 min. = 28 hrs. 48 min.

So, the real-time would be found by adding 28 hrs. 48 min. at 10 a.m.

which will be 2:48 p.m.

i.e. 24 * 60= 1440 min, there is a gain of 10 mins.

So the ratio of real-time to the time of the clock is 1440:1450(gain of 10 min. added) = 144:145

Here, Time elapsed in the clock is = 29 hr i.e. 29*60 = 1740 min.

So, the real-time (a) can be found by

144:145::a:1740

a = \(\left(\frac{144 \times 1740}{145}\right)\) = 1728 min. = 28 hrs. 48 min.

So, the real-time would be found by adding 28 hrs. 48 min. at 10 a.m.

which will be 2:48 p.m.

**Ans.**
NA

NA

NA