# Clock and Calendars Practice Quiz

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# Clock and Calendars Practice Quiz

### 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 Quiz Presents most important questions related to Clocks and Calendars. Clock and Calendars Practice Quiz is very useful to score maximum marks in the Clock and Calendars related sections.

### Quiz

1. An accurate clock shows 8 O’clock in the morning. Through how many degrees will the hour-hand rotate when the clock shows 2 O’clock in the afternoon

A. 30°
B. 180°
C. 90°
D. 150°

Explanation –

Angle traced by hour-hand in 6 hours = $$( 6 * \frac{360}{12})º$$ = 180º

2. At 3 : 40, the hour-hand and the minute-hand of a clock form an angle of:

A. 120°
B. 125°
C. 130°
D. 140°

Explanation –

The angle between the two hands of a clock at 3:40 is θ = $$\frac{11}{2}$$m – 30 h

Here, h = 3 and m = 40

θ = $$\frac{11}{2} * 40$$m – 30 * 3

θ = 130°

3. The angle between the minute-hand and the hour hand of a clock when the time is 4 : 20, is:

A. 10°
B. 15°
C.
D. 20°

Explanation –

The angle between the two hands of a clock at 4 : 20 is

θ = 30 h – $$\frac{11}{2}$$m

Here, m = 20 and h = 4

θ = 30 * 4 – $$\frac{11}{2} * 20$$

θ = 120 – 110

θ = 10°

4. How many times do the hands of a clock coincide in a day?

A. 24 times
B. 22 times
C. 20 times
D. 23 times

Explanation –

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e. at 12 O’clock).

i.e, The hands of a clock coincide 22 times in a day

5. How many times in a day, do the hands of a clock form a right angle?

A. 24 times
B. 48 times
C. 22 times
D. 44 times

Explanation –

In 12 hours, they are at right angles, 22 times.

i.e,In 24 hours, they are at right angles, 44 times.

1. At what time between 4O’clock and 5 O’clock will the hands of a watch point in opposite directions?

A. 4 hrs. 52 $$\frac{6}{11}$$ min.
B. 4 hrs. 32 $$\frac{6}{11}$$ min.
C. 4 hrs. 54 $$\frac{6}{11}$$ min.
D. 4 hrs. 34 $$\frac{6}{11}$$ min.

Explanation –

When t he t wo hands of t he clock ar e facing opposite directions, then the angle between them is 180°

θ = $$\frac{11}{2}$$m – 30 h

Here, θ = 180° and h = 4

180 = $$\frac{11}{2}$$m – 30 * 4

i.e, m = $$\frac{600}{11}$$ min = $$54 \frac{6}{11}$$ min

Therefor e, the hands of the clock ar e facing opposite directions at 4 hrs. $$54 \frac{6}{11}$$ min.

2. The angle between the minute-hand and the hourhand of a clock when the time is 8 : 30, is:

A. 80°
B. 75°
C. 60°
D. 105°

Explanation –

The angle between the two hands of a clock at 8:30 is

Here, $$\frac{11}{2}$$ m < 30 h

i.e, θ = 30h – $$\frac{11}{2}$$ m

θ = 30 * 8 – $$\frac{11}{2} * 30$$ m

θ = 75 º

3. At what angle are the hands of a clock inclined at 15 minutes past 5?

A. 58 $$\frac{1}{2}$$°
B. 64°
C. 7 $$\frac{1}{2}$$°
D. 72 $$\frac{1}{2}$$°

Explanation –

The angle between the hands of a clock at 15 minutes past 5 is

θ = 30h – $$\frac{11}{2}$$ m

Here, h = 5 and m = 15

θ = 30 * 5 – $$\frac{11}{2} * 15$$ m

θ = 67 $$\frac{1}{2}$$º

4. How many degrees will the minute-hand move, in the same time in which the second-hand moves 240°?

A.
B.
C.
D.

Explanation –

Second-hand moves 240° in $$(\frac{240°}{6°})$$ 40 SEC

In 60 sec the minute-hand covers $$(\frac{360°}{60°})$$ = 6°

Hence, in 40 sec. the minute-hand covers 4°

5. There are two clocks, both set to show the correct time at 10 a.m. One clock gains two minutes in one hour while the other gain one minute in one hour. If the clock which gains 2 minute shows the time as 22 minutes past 9 p. m., then what time does the other watch show?

A. 9 hrs. 33 min
B. 9 hrs. 12 min
C. 9 hrs. 11 min
D. 9 hrs. 23 min

Explanation –

Difference in minute between the two clocks in one hour = 1 minute.

Number of hours = 11 hours.

In 11 hours, one of the clock gains 22 minutes and shows the time as 9:22 p.m. The other clock which gains 1 minute per hour shows the time as 9:11 p.m.

1. Today is 1st August and the day of the week is Monday. This is a leap year. The day of the week
on this day after 3 years will be

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

Explanation –

This being a leap year none of the next 3 years is a leap year. So, the day of the week will be 3 days
beyond Monday i.e., it will be Thursday

2. Monday falls on 4th April, 1988. What was the day on $${3}^{rd}$$ November, 1987?

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

Explanation –

Counting the number of days after 3rd November, 1987 we have :

Month:Nov. Dec. Jan. Feb. Mar. Apr.

Days: 27 + 31 + 31 + 29 + 31 + 4

= 153 days, containing 6 odd days.

i.e., (7 – 6) = 1 day beyond the day on 4th April, 1988. So, the day was Tuesday

3. The year next to 1990 having the same calendar as that of 1990 is ____

A. 1998
B. 2001
C. 2002
D. 2004
E. 2000

Explanation –

We go on counting the no. of odd days from 1990 onward till the sum is exactly divisible by 7.

The number of such days are 14 upto the year 2000. So, the calendar for 1990, was repeated in the year 2001

4. Which dates of April, 2012 will be Sunday?

A. 1, 8, 15, 22, 29
B. 3, 10, 17, 24, 31
C. 2, 9, 16, 23, 30
D. 5, 12, 19, 26
E. 4, 11, 18, 25

Explanation –

$${1}^{st}$$ April, 2012:

2000 + 11 + Number of days from 1st January 2012 to 1st April, 2012.

Number of odd days in 2000 years = 0

Number of odd days in 11 years = 13

Months : Jan Feb Mar April

Odd days: 3 + 1 + 3 + 1 = 8

Total number of odd days = 8 + 13 + 0 = 21

= 0 odd days.

Hence, $${1}^{st}$$ April, 2012 is a Sunday.

$${1}^{st}$$ ,$${8}^{th}$$, $${15}^{th}$$, $${22}^{nd}$$ and $${29}^{th}$$ of April, 2012 are Sunday’s.

5. If 30 January 1989 is Monday then what was the day of week on 26 Oct 2003.

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

Explanation –

Total number of odd days from 30 January 1989 to 30 January 2003

= odd days in 3 leap year + 11 ordinary years

= 3 odd days

Total odd days from 30 January 2003 to 26 October 2003 =

Jan Feb March April
1 28 31 30
June July August Sep
30 31 31 30

= 3 odd days

So, total odd days = 6 odd days

So, day on 26 October 2003 will be Sunday.