Explore a curated selection of UPSC Prelims Topic Wise Questions focusing on Basic Numeracy, specifically delving into Clocks and Calendars. Dive into a diverse array of questions meticulously designed to enhance your understanding and preparation for the UPSC Preliminary examination in these fundamental mathematical concepts. This resource offers a structured approach to mastering key principles and techniques necessary for success in solving clock and calendar-related problems. Whether you’re a novice seeking to build foundational knowledge or an experienced candidate aiming to refine your skills, these thoughtfully crafted questions provide valuable insights and practice opportunities. With a strategic emphasis on topic-wise questions, aspirants can sharpen their problem-solving abilities and develop a deeper understanding of clock and calendar concepts. Elevate your preparation, grasp essential insights, and optimize your performance in the UPSC Prelims with proficiency in solving problems related to clocks and calendars within the realm of basic numeracy.

**Q1. The number of times in a day the hour hand and the minute hand of a clock are at right angles, is (1997)**

(a) 44

(b) 48

(c) 24

(d) 12

**Ans. (a)**

Angle formed by minute hand in an hour = 360°

Angle formed by minute hand in a minute = 6°

Angle formed by hour hand in an hour = 30°

Angle formed by hour hand in a minute = 0.5°

Now, starting from midnight, the first time difference between the two hand would be 90° is: 6x-0.5x=90°

[x is the number of minutes]

5.5x = 90°

x = 90°/5.5 = 180/11 minutes

The next time difference between the two hands would be 90° is when the minute hand would have moved 180° away from hour hand:

6x – 5x = 90° + 180 °, 5.5x = 270°

x = 270°/5.5 = 540/11 minutes

Thus, the difference in time between two consecutive moments where both hands forms a right angle is:

540/11 – 180/11 = 360/11 minutes

The two hands forms a right angle after every 360/11 minutes

Total number of minutes in a day = 24 X 60 = 1440 minutes

Number of times the two hands will form a right angle in a day = 1440+ 360/11= 44 times

Hence, the two hands will form right angle “44 times” in a day.

**Q2. An accurate clock shows the time as 3:00. After hour hand has moved 135°, the time would be (1998)**

(a) 7:30

(b) 6:30

(c) 8:00

(d) 9:30

**Ans. (a)**

According to the question,

Hour hand covers an angle of 360° in 12 h

Time taken to cover an angle of 135° = 12/360 X 135 = 4.5 h

Required time=3:00 +4.5h-7.5 h=7:30

390. An accurate clock shows 8 o’clock in the morning. Throughout how many degrees will the hour hand rotate, when the clock shown 2 o’clock in the afternoon? (2000)

(a) 150°

(b) 144°

(c) 168°

(d) 180°

Ans. (d)

As we know,

Angle covered by hour hand for 12 h = 360°

Angle covered by hour hand for 1 h = 360°/12

[As from 8’0 clock to 2’0 clock there are 6h]

Angle covered by hour hand for 6 h= 360°/12 X (6) = 180°

**Q3. When the time in the wall-clock is 3:25 pm, the acute angle between the hour hand and the minute hand is (2002)**

(a) 60°

(b) 52.5°

(c) 47.5°

(d) 42

**Ans. (c)**

According to the question,

In a clock, the angle between two successive numbers is 360°+12=30° When the time is 3:25 pm, the minute hand will be on 5 and will have moved 60° from 3 and hour hand would be between 3 and 4 and as it moves 30° in 60 min, so in 25 min.

It would move (30° X 25) / 60 = 12.5°

So, the difference between two hands will be = 60°-12.5°=47.5°

**Q4. March 1st, 2008 was Saturday. Which day was it on March 1st, 2002? (2008)**

(a) Thursday

(b) Friday

(c) Saturday

(d) Sunday

**Ans. (b)**

According to the question,

In a year, number of weeks = 52

Number of odd days = 1

From 2002 to 2008, there are 6 уг. So number of odd days = 6(1)=6

While 2004 and 2008 are leap years, having one more odd day apart from the normal odd day.

Thus, number of odd days =6+1+1=8

Out of these 8 extra days, 7 days form a week and so 1 day remains.

Hence 1st March, 2002 is 1 day less than 1st March, 2008 i.e. it is Friday.

Hence, option (b) is correct.

**Q5. How many times are an hour hand and a minute hand of a clock at right angles during their motion from 1:00 pm to 10:00 pm? (2009)**

(a) 9

(b) 16

(c) 18

(d) 20

**Ans. (b)**

Hour and minute hands are at right angle when the difference between their angles in clockwise direction are 90° or 270°.

Angle covered by minute hand in one minute = 6° Angle covered by hour hand in one minute =0.5°

90° difference happens after = 90°/(6-0.5)° = 180/11 min.

270° difference happens after = 270° / (6-0.5)° = 540/11 min.

So, hour and minute hands are at right angle 2 times in = = 180/11 + 540/11 = 720/11 min

Time duration = 10 pm-1 pm =9h=9×60 min

In 720/11 min, right angle = 2 times

In 9 x 60 min, right angle = (2 x 11/720) x 9 x 60 times = 33/2 = 16 times.

**Q6. If the 3rd day of a month is Monday, which one of the following will be the fifth day from 21st of this month? (2014)**

(a) Monday

(b) Tuesday

(c) Wednesday

(d) Friday

**Ans. (c)**

3rd day of the month = Monday

10th, 17th and 24th day will also be Monday as Monday repeats after 7 days.

24th Monday, i.e. 3rd day from 21st

5th day from 21st = Monday + 2 = Wednesday

Hence, option (c) is correct.

**Q7. Assume that 1. the hour and minute hands of a clock move without jerking. 2. the clock shows a time between 8 o’clock and 9 o’clock. 3. the two hands of the clock are one above the other. After how many minutes (nearest integer) will the two hands be again lying one above the other? (2014)**

(a) 60

(b) 62

(c) 65

(d) 67

**Ans. (c)**

In an hour, the minute hand covers a distance of 60 min spaces whereas the hour hand covers a distance of 5 min spaces. This means that in an hour, the minute hand gains 55 min spaces on the hour hand. For the two hands to be again lying one above the other, the minute hand has to gain 60 min spaces on the hour hand, which it does in (60/55 X 60) min = 65 min 27 s = 65 min

Hence, the required time = 65 min.

**Q8. Between 6 pm and 7 pm, the minute hand of a clock will be ahead of the hour hand by 3 min at (2015)**

(a) 6:15 pm

(b) 6:18 pm

(c) 6:36 pm

(d) 6:48 pm

**Ans. (c)**

We know that, exactly at 6 pm, angle between hour and minute hand is 180°.

Angle covered by minute hand in 1 min = 360°/60 = 6°

Angle covered by hour hand in 1 min = 30°/60 = 0.5°

3 min = 360°/60 X 3 = 18°

Let after x min minute hand will be 3 min ahead of hour hand.

Then,

(6°-0.5°)x=180° +18°

5.5 x = 198

x = 36 min

Required time = 6 pm + 36 min = 6:36 pm

**Q9. A class starts at 11:00 am and lasts till 2:27 pm. Four periods of equal duration are held during this interval. After every period, a rest of 5 min is given to the students. The exact duration of each period is (2016)**

(a) 48 min

(b) 50 min

(c) 51 min

(d) 53 min

**Ans. (a)**

It is given that, class starts at 11:00 am and lasts till 2:27 pm.

Total duration = 3 h and 27 min

Total number of periods = 4

Since, there is a rest of 5 min after each period, so total duration of periods =3 h 27 min-15 min

Hence, the duration of each period = 192/4 min = 48 min.

**Q11. If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of any year? (2017)**

(a) 23

(b) 22

(c) 21

(d) 20

**Ans. (b)**

February is the month in which there are minimum number of working days, i.e. 28 days,

There will be maximum four Sundays and two Saturdays off i.e. 6 days.

Minimum number of working days =28-6=22 days

**Q12. A clock strikes once at 1 o’clock, twice at 2 o’clock and thrice at 3 o’clock, and so on. If it takes 12 sec to strike at 5 o’clock, what is the time taken by it to strike at 10 o’clock? (2017)**

(a) 20 sec

(b) 24 sec

(c) 28 sec

(d) 30 sec

**Ans . (b)**

At 5 o’ clock, clock strikes 5 times in 12 sec.

Time taken for 1 strike = 12/5 sec

At 10 o’ clock, clock strikes 10 times. Therefore, total time taken by clock to strike 10 times= 1/5 x 10 sec = 24 sec.

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