**Clocks & Calendars Questions for RRB Group-D PDF**

Download Top-15 RRB Group-D Clocks & Calendars Questions PDF. RRB GROUP-D Clocks & Calendars questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

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**Question 1: **Find the mirror image of the clock when the time is 03:00

a) 08:57

b) 09:00

c) 09:03

d) 09:01

**Question 2: **If 15-06-1994 is wednesday, then what day is 19-07-1995 ?

a) Tuesday

b) Wednesday

c) Thursday

d) Friday

**Question 3: **What day is it after 178 days if it is a Monday tomorrow?

a) Tuesday

b) Saturday

c) Friday

d) Wednesday

**Question 4: **Which day is it on 1st January, 2000.

a) Monday

b) Wednesday

c) Saturday

d) Sunday

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**Question 5: **At what time the hands of the clock show a mirror image of 4:10 ?

a) 7:10

b) 7:40

c) 7:50

d) 8:10

**Question 6: **At what time between 2 O’Clock and 3 O’Clock, will the minute hand and hour hand of the clock be exactly opposite to each other?

a) $02:46$

b) $02:47\dfrac{6}{11}$

c) $02:43\dfrac{7}{11}$

d) $02:47$

**Question 7: **Find the number of days in p weeks and p days.

a) 8p

b) $7p^2$

c) $8p^2$

d) 7p

**Question 8: **Find the total number of days present in the first century.

a) 36500

b) 36512

c) 36524

d) 36525

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**Question 9: **Which of the following is not a leap year?

a) 1800

b) 1864

c) 1600

d) 1624

**Question 10: **Find the angle between the hands of the clock when the time is 10:30

a) $160^\circ$

b) $120^\circ$

c) $180^\circ$

d) $135^\circ$

**Question 11: **What day is it on 26th January, 1950?

a) Wednesday

b) Thursday

c) Friday

d) Saturday

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**Question 12: **Which of the following years has the same calendar as that of the year 1963?

a) 1971

b) 1972

c) 1973

d) 1974

**Question 13: **A clock shows 7 O’clock in the morning. By how much angle will the hours hand rotate when the clock shows 9 O’clock in the morning.

a) $40^\circ$

b) $60^\circ$

c) $45^\circ$

d) $90^\circ$

**Question 14: **Which of the following days cannot be the last day of the century?

a) Friday

b) Wednesday

c) Sunday

d) Saturday

**Question 15: **What will be the day of the week on 13th April, 2006?

a) Thursday

b) Saturday

c) Sunday

d) Tuesday

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**Answers & Solutions:**

**1) Answer (B)**

We need to subtract from 12:00 or 11:60 to get mirror image time

Mirror image of 03:00 = 12:00 – 03:00 = 09:00

**2) Answer (B)**

a non leap year has 365 days (52 weeks+1 day).

1995 is not a leap year, And 15-06-1994 is wednesday, So 15-06-1995 would be thursday.

And 19-07-1995 is 34th day after 15-06-1995 (thursday), ⇒ 34 = 7(4)+6 days, So 6th day after thursday is Wednesday.

So the answer is option B.

**3) Answer (D)**

If tomorrow is a Monday, then today is a Sunday.

178 = (7 * 25) + 3

178 days is equal to 25 weeks and 3 odd days.

3 days after a Sunday is a Wednesday.

Hence, it is a Wednesday after 178 days.

**4) Answer (C)**

The number of odd days in every 400 years = 0

=> 31st December 2000 is a Sunday.

=> 1st January 2001 is a Monday.

2000 is a leap year => It has 2 odd days.

So, the day on 1st January 2000 will be two days prior to a Monday.

Hence, 1st January 2000 is a Saturday.

**5) Answer (C)**

It should be 7:50

So the answer is option C.

**6) Answer (C)**

Angle between two hands $= \dfrac{11}{2}M – 30H$ where H is hours and M is Minutes

Here, H = 2 and Angle = $180^\circ$

$180^\circ = \dfrac{11}{2}M – 30\times2$

⇒ $\dfrac{11}{2}M = 240$

⇒ $M = \dfrac{480}{11} = 43\dfrac{7}{11}$

Hence, the required time = $02:43\dfrac{7}{11}$

**7) Answer (A)**

Each week has 7 days.

p weeks => 7p days.

So, the total number of days in p weeks and p days = 7p + p = 8p days

**8) Answer (C)**

There are 76 ordinary years and 24 leap years in the first century.

=> Number of days = (76 * 365) + (24 * 366)

= (100 * 365) + 24

= 36524

**9) Answer (A)**

We know that all the years that are a multiple of 4 but are not centuries are leap years.

Hence, 1864 and 1624 are leap years.

Also, all the years that are multiple of 400 are leap years.

Hence, 2000 is a leap year.

=> 1800 is not a leap year.

**10) Answer (D)**

Required angle $= 30H – \dfrac{11}{2}M$ where H is hours and M is Minutes

Here, H = 10 and M = 30

Hence, Required angle $= 30\times10 – \dfrac{11}{2}\times30$

$= 300 – 165 = 135^\circ$

**11) Answer (B)**

1949 = 1600 + 349

We know that after every 400 years, the number of odd days is 0.

Hence, after 1600 years, the number of odd days is 0.

350 = 300 + 50.

We know that after every 300 years, there is exactly 1 odd day.

In next 49 years, there are 12 leap years and 37 ordinary years.

=> Number of odd days until 31st December 1949 = (12 * 2) + (37 * 1) + 1 = 24 + 37 + 1 = 62

Total number of odd days until 26th January, 1950 = 26 + 62 = 88

88 = (7 * 12) + 4

=> 4 odd days

=> It is a Thursday.

**12) Answer (D)**

On counting the number of odd days from 1963 to the year that has the same calendar as that of 1963, we must get 0 odd number of days.

Let’s calculate the number of number of odd days up to 1970.

1963 has 1 odd day.

1964 has 2 odd days.

1965 has 1 odd day.

1966 has 1 odd day.

1967 has 1 odd day.

1968 has 2 odd days.

1969 has 1 odd day.

1970 has 1 odd day.

Sum of the odd days until 1970 is 10.

10 = (1 * 7) + 3 => Number of odd days is not 0 => 1971 does not have the same calendar as 1963.

1971 has 1 odd day.

Sum of the odd days until 1971 is 11.

11 = (1 * 7) + 4 => Number of odd days is not 0 => 1972 does not have the same calendar as 1963.

1972 has 2 odd days.

Sum of the odd days until 1972 is 13.

13 = (1 * 7) + 6 => Number of odd days is not 0 => 1973 does not have the same calendar as 1963.

1973 has 1 odd day.

Sum of the odd days until 1973 is 14.

14 = (2 * 7) + 0 => Number of odd days is 0 => 1974 has the same calendar as 1963.

**13) Answer (B)**

In 12 hours, the hand turns $360^\circ$.

Here, the difference between time = 2 hours

Then, Required angle $= \dfrac{360}{12}\times2 = 60^\circ$

**14) Answer (D)**

Let’s calculate the number of odd days is 100 years.

In 100 years, there are 76 ordinary years and 24 leap years.

So, in 100 years, there are (76*1) + (24*2) odd days, which is equal to 124 odd days.

124 = (7 * 17) + 5

So, 124 days is equal to 17 weeks and 5 odd days. Hence, 100 years have 5 odd days => Friday

In 200 years, there are 2*5 odd days, which is equal to 10 days.

10 = (7 * 1) + 3

So, 10 days is equal to 1 week and 3 odd days. Hence, 200 years have 3 odd days.

In 300 years, there are 3*5 odd days, which is equal to 15 days.

15 = (7 * 2) + 1

So, 15 days is equal to 2 weeks and 1 odd day. Hence, 300 years have 1 odd day.

In 400 years, there are (4 * 5 + 1) odd days. 1 is added because every 4th century is a leap year. So, there are 21 odd days in every 400 years.

21 = (7 * 3) + 0

Hence, there are 0 odd days in every 400 years => Sunday

So, only Sunday, Monday, Wednesday and Friday can be the last day of the century.

Among the given options, only Saturday cannot be the last day of the century.

**15) Answer (A)**

We know that the number of odd days in every 400 years is 0.

Hence, the number of odd days in 2000 years is 0.

In the remaining 5 years, there are 4 ordinary years and 1 leap year.

=> There are (4 * 1) + (1 * 2) = 6 odd days until December 31st, 2005.

The number of days in 2006 until 13th April is 31 + 28 + 31 + 13 = 103.

6 + 103 = 109

109 = (7 * 15) + 4.

4 odd days => It is a Thursday on 13th April 2006.

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