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# SNAP Clock and Calender Questions PDF [Most Important]

Clock and Calender is an important topic in the Clock and Calender section of the SNAP Exam. You can also download this Free Clock and Calender Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Clock and Calender questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

Question 1:Â An alarm was set at 11 AM on Monday on a clock that was set correctly at 1 AM on Sunday, but the clock started gaining 20 seconds every 24 hours. What was the actual time when the alarm went off?

a)Â 10:31:40 AM

b)Â 10:51:20 AM

c)Â 10:30:00 AM

d)Â 11:30:00 AM

Solution:

Total time between sunday at 1 AM to monday at 11 AM =24hours + 10hours = 34 hours

Now total gain in seconds in 34 hours = $\frac{20}{24}$Ã—34= $\frac{85}{3}$

The actual time at which alarm went off is

=10:59:(60-$\frac{85}{3}$)

=10:59:$\frac{95}{3}$

=10:59:32

Question 2:Â How much does a watch lose per day if hand coincides every 64 minutes?

a)Â $17\frac{5}{11}$

b)Â $32\frac{8}{11}$

c)Â $\frac{32}{11}$

d)Â $\frac{16}{11}$

Solution:

In 24 hours the hands of hour and minute coincide 22 times. 24 hours equals 24*60=1440 minutes.

1440/22 = 65 and 5/11 minutes.=> every 65 and 5/11 min, they coincide.

Now the clock in question coincide every 64 minutes; therefore 24 hours in your clock means 64*22 minutes or 1408 minutes.

So your clock should lose 1440-1408=32 minutes.

Otherwise, the clock gains 32 minutes as it is faster than usual.

Question 3:Â Find the angle between the hands at 3:30 p.m.

a)Â $120^\circ$

b)Â $75^\circ$

c)Â $90^\circ$

d)Â $105^\circ$

Solution:

At 3:30Â , the minute hand is at number 6. At 3:00 the hour handÂ was at number 3.

total 360 degree which is divided by 12 parts

hence , the angle between two number is 30 degree

so,

The angle between 3 and 6 is 90 degreeÂ but hour has moved 15 degree.(between 3 and 4 )

Hence the angle betweenthe two is ( 90 -15) = 75 degree.

Question 4:Â What is the obtuse angle formed by the hands of a clock when the time in the clock is 2:30?

a)Â $95^\circ$

b)Â $120^\circ$

c)Â $105^\circ$

d)Â $165^\circ$

Solution:

Angle between the hands of a clock is given by the formula $\dfrac{11}{2}H – 30M$ or $30M – \dfrac{11}{2}H$ where H is hours and M is minutes.
Here, Given time = 02 : 30, H = 2 and M = 30.
Angle = $\dfrac{11}{2} \times 30 – 30 \times 2 = 165 – 60 – 105^\circ$

Question 5:Â What is the measure of the smaller of the two angles formed between the hour hand and
the minute hand of a clock when it is 6:44 p.m.?

a)Â $Â 62.5^\circ$

b)Â $Â 62^\circ$

c)Â $84^\circ$

d)Â $83.5^\circ$

Solution:

A clock is a circle, and a circle always contains 360 degrees. Since there are 60 minutes on a clock, each minute mark is 6 degrees.

$\frac{360^\circ total}{60 minutes total}=6^\circ perÂ minute$

The minute hand on the clock will point at 44 minutes, allowing us to calculate it’s position on the circle.

(44 min)(6)=$264^\circ$

Since there are 12 hours on the clock, each hour mark is 30 degrees.

$\frac{360^\circ total}{12 hours total}=30^\circ perÂ hour$

We can calculate where the hour hand will be at 6:00.

$(6 hr)(30)=180^\circ$

However, the hour hand will actually be between the 6 and 7, since we are looking at 6:44 rather than an absolute hour mark. 44 minutes is equal to $\frac{44}{60}$th of an hour. Use the same equation to find the additional position of the hour hand.

$180^\circ + \frac{44}{60} \times 30 = 202^\circ$

We are looking for the smaller angle between the two hands of the clock. The will be equal to the difference between the two angle measures.

Required answer = $264^\circ – 202^\circ = 62^\circ$

So, the answer would be option b)$62^\circ$.

Question 6:Â What is the measure of the smaller of the two angles formed between the hour hand and the minute hand of a clock when it is 5:49 p.m.?

a)Â $119^\circ$

b)Â $119.5^\circ$

c)Â $120^\circ$

d)Â $120.5^\circ$

Solution:

The hour hand moves 360 degrees every 12 hours. At 5:49, its angle is $(5 + \frac{49}{60}) \times\frac{ 360}{12} = 174.5 degrees$

The minute hand moves 360 degrees each 60 minutes, so at 15 minutes past the hour it has moved $\frac{49}{60} \times 360 = 294 degrees.$

Thus, the difference between the two hands is 294 – 174.5 = 119.5 degrees.

Question 7:Â If 29 January 2003 is a Wednesday, then what day of the week will be 26 February 2005?

a)Â Thursday

b)Â Saturday

c)Â Friday

d)Â Sunday

Solution:

Given,Â 29 January 2003 is a Wednesday

Number of days between 29 January 2003 and January 29 2004 = 365 days

Number of days between 29 January 2004 and January 29 2005 = 366 days

Number of days between 29 January 2005 and 26 February 2005 = 28 days

Number of days between 29 January 2003 and 26 February 2005 = 365 + 366 + 28 = 759 days

Number of odd days between 29 January 2003 and 26 February 2005 = Remainder of $\frac{759}{7}$ = 3

$\therefore\$The day on 26 February 2005 = Wednesday + 3 = Saturday

Hence, the correct answer is Option B

Question 8:Â What day of the week was 5 February 2008?

a)Â Thursday

b)Â Monday

c)Â Tuesday

d)Â Wednesday

Solution:

Odd days in year 2007 = 6 +Â 2 = 8
($\because$ 2004 is the leap year and odd days till 2000 is 0.)
Odd days in 5 February = odd days in JanuaryÂ + 5 = 3 +Â 5 = 8
Total odd days = 8 +Â 8 = 16
Odd daysÂ = 16 = 2 weeks + 2 odd days
So dayÂ of the week was Tuesday.
$\therefore$ The correct answer is option C.

Question 9:Â What day of the week was 31 January 2007?

a)Â Monday

b)Â Tuesday

c)Â Thursday

d)Â Wednesday

Solution:

Number of odd days in 2006 = 7
($\because$ number ofÂ Leap year from 2000 to 2007 = 1(2004))
Odd days in 31 January = 31
Total odd days = (31 + 7) = 38
Odd day = 38/7 = 3 remaining
The day of the week was Wednesday.
$\therefore$ The correct answer is option D.

Question 10:Â What day of the week was 29 June 2010 ?

a)Â Sunday

b)Â Monday

c)Â Wednesday

d)Â Tuesday

Solution:

Odd day till 2009 = 7 +Â 2 + 2 = 11
($\because$ Year 2004 and 2008 are leap year and odd days till 2000 = 0)
Odd days in 29 June = Odd days in (January + February +Â March +Â April +Â May + June) =Â Â 3 + 0 +Â 3 +Â 2 +Â 3 + 1 = 12
Total odd days = 11 +Â 12 = 23
Odd days = 23/7 = 2 remaining
The daysÂ of the week was Tuesday.
$\therefore$ The correct answer is option D.

TakeÂ  SNAP mock tests here

Question 11:Â A watch reads 4:30. If the minute hand points East in which direction will the hour hand point?

a)Â North-East

b)Â North

c)Â South-West

d)Â south

Solution:

According to the question, the given information is represented as shown below

$\therefore\$The hour hand is pointing towards North-East

Hence, the correct answer is Option A

Question 12:Â If it was a Saturday on 10 November 2018, what was the day of the week on 15 August 2017?

a)Â Monday

b)Â Tuesday

c)Â Sunday

d)Â Friday

Solution:

Given,Â 10 November 2018 was Saturday

Number of days between 15 August 2017 and 15 August 2018 = 365 days

Number of days between 15 August 2018 and 10 November 2018 = 16 + 30 + 31 + 10 = 87 days

Number of days between 15 August 2017 and 10 November 2018 = 452 days

Number of odd days between 29 January 2003 and 26 February 2005 = Remainder ofÂ $\frac{452}{7}$ = 4

$\therefore\$The day on 15 August 2017 = Saturday – 4 = Tuesday

Hence, the correct answer is Option B

Question 13:Â What was the day of the week on 26 January 2012 ?

a)Â Sunday

b)Â Saturday

c)Â Thursday

d)Â Tuesday

Solution:

26 January 2012 = (2011 years + period between 1-1-2012 to 26-01-2012)

odd days in 2000 = 0

in remaining 11 years = (2 leap years + 9 normal years) = (2 * 2 + 9 * 1) = 13 – 7 = 6 odd days

Total number of days in period between 1-1-2012 to 26-01-2012 = 26Â days = 3 weeks + 5 days = 5 odd days

Total number of odd days = (6 + 5) = 11 – 7 =Â 4 days

Hence, given day is Thursday (0 odd day means sunday, 1 odd day means mon and so on).

Question 14:Â What was the day of the week on 15 August 2013?

a)Â Thursday

b)Â Monday

c)Â Wednesday

d)Â Tuesday

Solution:

15th August 2013 = (2013 years + period between 1-1-2013 to 15-8-2013)

odd days in 2000 = 0

in remaining 13 years = (3 leap years + 9 normal years) = (3 * 2 + 9 * 1) = 15 – 14 = 1 odd days

Total number of days inÂ period between 1-1-2013 to 15-8-2013 = 227 days = 32 weeks + 3 days = 3 odd days

Total number of odd days = (1 + 3) = 4 days

Hence, given day is thursday (0 odd day means sunday, 1 odd day means mon and so on )

Question 15:Â If it was a Friday on 1 January 2016, what was the day of the week on 31 December 2016?

a)Â Sunday

b)Â Saturday

c)Â Monday

d)Â Friday