Quantitative Aptitude - SPLessons

Clocks and Calendars Practice Set 2

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Clocks and Calendars Practice Set 2

shape Introduction

Clock Problems is all about the hour’s hand, second hand, and minutes hand. In this chapter, the problems are based mainly on the movement of the clock hands like minute spaces, minute hand, hour hand and the angles between the hands.


Calendar deals with odd days, leap year, ordinary year, counting of odd days, and day of the week related to odd days. To find the day of the week on a given date concept of odd days is used. Clocks and Calendars Practice Set 2 Presents most important questions related to Clocks and Calendars.


shape Quiz

Q1. At what time between 9 and 10 o’clock are the hands of a clock 5 minutes spaces apart?

    A. 9.54 \(\frac {6}{11}\) , 9.43 \(\frac {7}{11}\)
    B. 9.52, 9.48 \(\frac {3}{11}\)
    C. 9.50, 9.48
    D. 9.51 \(\frac {7}{11}\) , 9.49 \(\frac {1}{11}\)

Answer: A

Explanation:
5 minute spaces = 30°


Q2. A clock is set right at 9 a.m. The clock gains 30 seconds per hour. What will be the true time when the clock indicates 3.39 p.m. on the 3rd-day afternoon?

    A. 4 p.m
    B. 3 p.m.
    C. 1 p.m.
    D. p.m.

Answer: B

Explanation:
\(\frac {30}{60}\) minutes are gained by the click in 1 hour.

24 \(\frac {1}{5}\) hour of the clock = 24 hour of the correct time

Time duration between 9 a.m. and 3 : 39 p.m. of the third day afternoon = (24 + 24 + 24 + 6 \(\frac {39}{60}\)) hour

= \(\frac {1573}{20}\) hour

Actual Time duration = \(\frac {24}{24 \frac {1}{2}} \times \frac {1573}{20}\) hour


Q3. A watch gains time uniformly. it was observed that it was 6 minutes slow at 12 o’clock in the night on Monday. On Friday at 12 o’clock in the night, it was 6 minutes 48 second fast. When was it correct?

    A. Wednesday, at 6 p.m.
    B. Wednesday, at 7 p.m.
    C. Thursday, 8 p.m.
    D. Wednesday, 9 p.m

Answer: D

Explanation:
Total time gained by the clock till 12 O’ Clock of the friday night (i. e, in 96 hours) = 6 min +6 min 48 s = 12 min 48 s

The clock was correct when itd gained 6 min

If 12 min 48 s are gained in 96 hours then 6 min would be gained in (\(\frac {96}{12 minutes} \times 6 min )\) hours


Q4. What was the day on 15th august 1947 ?

    A. Friday
    B. Saturday
    C. Sunday
    D. Thursday

Answer: A

Explanation:
Odd days in 1600 years = 0

Odd days in 300 years = 1

46 years = (35 ordinary years + 11 leap years) = (35 x 1 + 11 x 2)= 57 (8 weeks + 1 day) = 1 odd day

Jan. Feb. Mar. Apr. May. Jun. Jul. Aug

( 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 ) = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days.

Hence, as the number of odd days = 5, given day is Friday.


Q5. Today is Monday. After 61 days, it will be :

    A. Tuesday
    B. Monday
    C. Sunday
    D. Saturday

Answer: D

Explanation:
Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday.

After 61 days, it will be on Saturday.

Q1. The last day of a century cannot be

    A. Monday
    B. Wednesday
    C. Tuesday
    D. Friday

Answer: C

Explanation:
100 years contain 5 odd days

Last day of first century is Friday

200 years contain (5 \( \times 2\)) = 3 odd days

Last day of second century is Wednesday

300 years contain (5 \( \times 3\)) = 15 = 1 odd day

Last day of third century is Monday

400 years contain 0 odd days

The last day of fourth century is Sunday

This Cycle is Repeated.

Last day of acentury cannot be Tuesday, Thursday, or Saturday.


Q2. What was the day of the week on, 16th July 1776?

    A. Tuesday
    B. Wednesday
    C. Monday
    D. Saturday

Answer: A

Explanation:
16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of Odd days:

1600 years have 0 odd day.


100 years have 5 odd days.


75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. Jan Feb Mar Apr May Jun Jul

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days = (28 weeks + 2 days)

Total number of odd days = (0 + 2) = 2.

Required day was ‘Tuesday’.


Q3. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

    A. Monday
    B. Friday
    C. Sunday
    D. Tuesday

Answer: B

Explanation:
On 31st December, 2005 it was Saturday.


Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.


On 31st December 2009, it was Thursday.


Thus, on 1st Jan 2010 it is Friday


Q4. What was the day of the week on 28th May, 2006?

    A. Sunday
    B. Friday
    C. Wednesday
    D. Tuesday

Answer: A

Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)


Odd days in 1600 years = 0


Odd days in 400 years = 0


5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days


(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 dayss = (21 weeks + 1 day) = 1 odd day.


Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.


Q5. What was the day of the week on 17th June 1998 ?

    A. Monday
    B. Tuesday
    C. Wednesday
    D. Friday

Answer: C

Explanation:
17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March. April. May. June.

(31 + 28 + 31 + 30 + 31 + 17) = 168 days

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given the day is Wednesday.

Q1. Find the time between 8 and 9 o’clock when the hands of a clock will be in a straight line in opposite direction.

    A. 10 \(\frac {10}{11}\) min Past 8
    B. 12 min past 8
    C. 16 min past 8
    D. 11 min past 8

Answer: A

Explanation:
To be in a straight line the minute hand must traverse 60° over the hour hand.


Q2. At how many times pas 8 o’clock does the minute hands of a clock coincides with the hour hand between 8 p.m. and 9 p.m.?

    A. 44 \(\frac {1}{11}\)
    B. 43 \(\frac {7}{11}\)
    C. 42 \(\frac {2}{7}\)
    D. 43

Answer: B

Explanation:
In order to be coincident the minute hand must traverse 240° over the hour hand


Q3. What is the angle between the two hands of a clock when the time shown by the clock is 6.30 p.m. ?

    A. 00
    B. 50
    C. 30
    D. 150

Answer: D

Explanation:
q = \(\frac {11}{2}\) m – 30 h

= \(\frac {11}{2} \times 30 – 30 \times 6\)

= 165 – 180 = 150


Q4. How many degrees will the minute hand move, at the same time in which the second-hand move 4800 ?

    A. 60
    B. 90
    C. 40
    D. 80

Answer: D

Explanation:
28\(\times 2 \times 6 = {336}^{0}\)


Q5. By how many degrees does the minute hand move in the same time, in which the hour hand move by 280?

    A. 168
    B. 336
    C. 196
    D. 376

Answer: B

Explanation:
Minute hand covers \(\frac {480}{60}\) = 80



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