Clock and Calendar MCQ Quiz - Objective Question with Answer for Clock and Calendar - Download Free PDF

According to the question,

Statement I: Buses for Delhi leave after every 30 minutes, until 11 pm.

Here, no information regarding last bus timing from Noida is given, therefore statement I alone is not sufficient to answer the question.

Statement II: 20 minutes ago, a bus has left for Delhi.

Here, last bus timing is 6:40 pm (7:00 pm - 20 min) for Delhi but frequency of the bus cannot be determined, therefore statement I alone is not sufficient to answer the question.

Combining statement I and II: Buses for Delhi leave after every 30 minutes, until 11 pm. 20 minutes ago, a bus has left for Delhi.

Last bus timing from Noida was 6:40 pm and as buses leave after every 30 minute. Thus the next bus would be at 7:10 pm (6:40 pm + 30 min).

∴ Both statement I and statement II together are needed to answer the question.

Hence, "Option 4" is the correct answer.

According to the question,

Statement I: Buses for Ramnagar leave after every 30 minutes, till 10 p.m.

Here, no information regarding last bus timing from Ramnagar is given, therefore statement I alone is not sufficient to answer the question.

Statement II: Fifteen minutes ago, one bus has left for Ramnagar.

Here, last bus timing is 7:45 pm (8:00 pm - 15 min) for Ramnagar but frequency of the bus cannot be determined, therefore statement I alone is not sufficient to answer the question.

Combining statement I and II: Buses for Ramnagar leave after every 30 minutes, till 10 p.m. Fifteen minutes ago, one bus has left for Ramnagar.

Last bus timing from Ramnagar was 7:45 pm and as buses leave after every 30 minute. Thus the ne×t bus would be at 8:45 pm (6:40 pm + 30 min). It is mentioned that buses leave till 10:00 pm, therefore Hemant can get ne×t bus for Ramnagar from Dhanpur.

∴ Both statement I and statement II together are needed to answer the question.

Hence, "Option 4" is the correct answer.

According to the given question, we need to find the day of the 14th of the given month.

Statement 1: The last day of the month is a Wednesday.

If the last day is Wednesday we are not able to find a day on the 14th of that month because we don't know whether the month is of 28 days or 29 days or 30 days or 31 days.

Therefore, Statement-1 alone is not sufficient to answer the question.

Statement 2: The third Saturday of the month was seventeenth.

If the 17th day of the given month is Saturday then we can easily find the 14th day by subtracting 3 odd days from Saturday i.e., (6 - 3) = 3. means Wednesday.

Therefore, Statement-2 alone is sufficient to answer the question.

Hence, "Option 2" is the correct answer.

Additional InformationHere we use odd days as: Sunday = 0 , Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5, and Saturday = 6.  

From 14th September to 17th October:

Days left in September:

September has 30 days, so days remaining after the 14th = 30 - 14 = 16 days.

Days in October before the 17th:

Days = 17 days.

Total days between the two dates:

Total = 16 (September) + 17 (October) = 33 days.

Calculate the day of the week:

33 days modulo 7 gives us the remainder when divided by 7:

33 ÷ 7 = 5

32 ÷ 7 = 4 remainders 5

Starting from Tuesday (14th September):

Counting 5 days from Tuesday:

Tuesday → Wednesday (1) → Thursday (2) → Friday (3) → Saturday (4) →Sunday(5)

Thus, 17th October will be a Sunday.

Hence the correct answer is, Option(3)

To determine if a particular year is a leap year, we can follow these rules:

1. A year is a leap year if it is divisible by 4.
2. However, if the year is divisible by 100, it is not a leap year, unless:
3. The year is also divisible by 400, in which case it is a leap year.

Now, let's check each option,

1) 1600: It is a leap year because it is divisible by 400.

2) 2000: It is a leap year because it is divisible by 400.

3) 2400: It is a leap year because it is divisible by 400.

Hence, the correct answer is "All options are correct".

Statements:

I. Harish was born on an even date of the month → 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28

II. Harish’s birth date was a prime number → 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

But we know that 2 is the smallest even prime number.

According to both statements we can say that on 2nd February 2000 Harish was born.

Hence, Option (1) is correct.

Logic: If we see carefully we can find that these dates have property  

3rd April, 2005 - 03/04/05  

6th August, 2010 - 06/08/10

5th December, 2013 - 05/12/13

Similarly,

Independence day 2017 - 15 August 2017 - 15/08/17

By above logic:  

No other option satisfy the above condition. So, Indian Independence Day of 2017 has the same property.

Hence, option 1) is the correct answer.

Formula Used:

Θ = | Hours × 30^◦ -  × Minute |

Calculation:

Θ = | 2 × 30◦ -  × 13 |

Θ = | 60◦ - |

Θ = | 60◦ - 71.5◦ |

Θ = | - 11.5◦ |

Θ = 11.5◦ 

Hence, "11.5◦"  will be the acute angle between the hour-hand and the minute-hand at 2:13 p.m

Calculation:   

The angle between two digits = 30° 

⇒ At 2'o clock one hand will be on 12 and second hand will be on 2 So, 30° × 2 = 60° 

∴ At 2'o clock angles between hands will be 60°.

The correct option is 3 i.e. 60° 

Given - 

January 1, 2018 was Monday

Concept used - 

in a normal year number of days = 365

number of odd days = (365/7) = + 1

Solution - 

1 January 2018 = Monday

1 January 2019 = Monday + 1 = Tuesday

∴ January 1, 2019 = Tuesday  

Statement I: February 15 of this year was a Wednesday: We cannot get what is today’s date.

Statement II: Today is March 12: We can’t tell which day today is, it can be any day.

Taking both the statements together: We cannot tell whether it is Friday or not because we don’t know if the year is leap year or not.

Hence both the statements were not enough to give the answer.

The logic followed here is:

1600 years contain 0 odd day, 300 years contain 1 odd day. Also, 83 years contain 20 leap years and 63 ordinary years and therefore (40+0) odd days i.e., 5 odd days.

∴ 1983 years contain (0 + 1 + 5) i.e., 6 odd days.

Number of days from Jan. 1984 to 15th Oct. 1984

= (31 + 29 + 31 + 30 + 31 + 30 +31+ 31 + 30 + 15)

= 289 days = 2 odd days

∴ Total number of odd days = 6 + 2 = 1 odd days.

So, 15th Oct, 1984 was Monday.

Hence, the correct answer is "Monday".

Given: a 12-hour clock when the time is 9 ∶ 25.

Formula used:

where H represents hour and M represents minutes.

Calculation:

Thus, the approximate angle between the minute hand and the hour hand is ' '.

Given:

⇒ January 20, 2030 = Sunday

⇒ January 22, 2030 = Tuesday

As, for each month dates 1, 8, 15, 22 and 29 comes on same week day.

⇒ January 1, 2030 = Tuesday

⇒ January 4, 2030 = Friday

⇒ January 4, 2029 = Thursday

As after every leap year, dates will move two days further and 2028 is a leap year. Leap year is a year which is divisible by 4 and having 366 days.

∴ January 4, 2028 = Tuesday

Calculation 

The hands of a clock meet once every hour but

Between 11:00 and 1:00'o clock, they coincide only once.

Therefore, the hands of a clock meet 11 times in 12 hours.

∴ The correct answer is 11 points.