Quantitative Aptitude - SPLessons

Clocks and Calendars Quiz 3

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Clocks and Calendars Quiz 3

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 Quiz 3 Presents most important questions related to Clocks and Calendars.


shape Quiz

1. What will be the day of the week \({15}^{th}\)July, 2001?

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


Answer: Option B


Explanation:
\({15}^{th}\) July, 2000 = (2000 years + Period 1.7.2001 to 15.7.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
1 year = (1 ordinary year) = (1 x 1) = 1 odd day.
Jan. Feb. March April May June July.
(31 + 28 + 31 + 30 + 31 + 30 + 15) = 196 days
196 days = (28 weeks) 0 odd days.
Total number of odd days = (0 + 0 + 1 + 0) = 1= 1 odd day.
Given day is Monday.


2. What was the day of the week on \({17}^{th}\) June, 1998?

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


Answer: Option C


Explanation:
\({17}^{th}\) 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


3. What was the day of the week on \({28}^{th}\) May, 2006?

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


Answer: Option D


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
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 28 ) = 148 days


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

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


Answer: Option D


Explanation:

On \({31}^{st}\) December, 2005, it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan 2010, it is Friday.


5. There are 5000 students in a school. The next year it was found that the number of boys and girls increased by 10% and 15% respectively making the total number of students in school as 5600. Find the number of girls originally in the school?

    A. 4500
    B. 2000
    C. 3000
    D. Cannot be determined


Answer: Option B


Explanation:

Let number of girls = x, then no of boys = (5000 – x). then
10% of (1000 – x) + 15% of x = (5600 – 5000)
Solve, x = 2000

1. How many times are the hands of a clock at right angle in a day?

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


Answer: Option C


Explanation:

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

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


2. In 12 hours, they are at right angles 22 times. In 24 hours, they are at right angles 44 times.

    A. 20
    B. 22
    C. 24
    D. 48


Answer: Option B


Explanation:

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clcok only).

So, in a day, the hands point in opposite directions 22 times.


3. What day will be 61 days after today, if today is Monday?

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


Answer: Option C


Explanation:
The basic concept in a problem involving days is finding the number of odd days. In 61 days:

=> 7 x 8 + 5 = 5 odd days

=> Monday + 5 days = Saturday

Hence 61 days from today will be a Saturday.


4. 15th august 2010 was which day of the week?

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


Answer: Option D


Explanation:
15th August 2010 can be written as 2009 + days from 1st January 2010 to 15th August 2010.

=> Total number of odd days in 400 years = 0

Hence, the total number of odd days in 2000 years = 0 (as 2000 is a perfect multiple of 400)

Odd in days in the period 2001-2009:
7 normal years + 2 leap yeas

=> (7*1) + (2*2) = 11

=> Odd days will be 11- (7*1) = 4

Days from January 1 to August 15 in 2010: 31+28+31+30+31+30+31+15

= 227 days.

= 32 weeks and 3 days, this gives additional 3 odd days.

=> Total odd days = 3 + 4 = 7

=> 7 odd days =1 week = 0 odd days

=> 0 odd days = Sunday

Thus, 15th August 2010 was a Sunday.


5. January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?

    A. Monday
    B. Tuesday
    C. Tuesday
    D. Thursday


Answer: Option B


Explanation:

The year 2007 is an ordinary year.
So, it has 1 odd day.1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.

1. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?

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


Answer: Option D


Explanation:

The year 2008 is a leap year.
So, it has 2 odd days.1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.


2. On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006?

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


Answer: Option A


Explanation:

The year 2006 is an ordinary year.
So, it has 1 odd day.
So, the day of 8th Dec 2007 will be 1 day beyond the day on 8th Dec 2006.
But, 8th Dec 2007 is Saturday.
8th Dec 2006 is Friday.


3. Which of the following is not a leap year?

    A. 800
    B. 700
    C. 2000
    D. 1200


Answer: Option B


Explanation:

The century divisible by 400 is a leap year.
The year 700 is not a leap year.


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

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


Answer: Option B


Explanation:

On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan 2010 it is Friday.


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

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


Answer: Option A


Explanation:

Let initially income is 100. So, expenditure = 60 and saving = 40
now income is increased by 20% i.e. 120. So, expenditure = \(\frac{70}{100}\)*120 = 84 and saving = 36
so % percent decrease in saving = \(\frac{4}{40}\)*100 = 10%

Clcoks and Calendars – Related Information
Clocks and Calendars Practice Set 1

Clocks and Calendars Practice Set 2

Book for Quantitative Aptitude