MCQ: Clock and Calendar - 1 - SSC CGL MCQ

# MCQ: Clock and Calendar - 1 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Clock and Calendar - 1

MCQ: Clock and Calendar - 1 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Clock and Calendar - 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Clock and Calendar - 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Clock and Calendar - 1 below.
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MCQ: Clock and Calendar - 1 - Question 1

### Arjun asked rani that if 8th April, 2005 is Monday then what was the day of the week on 8th April, 2004?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 1

The year 2004 is a leap year. So, it has 2 odd days.
Feb 2004 is not included because we are calculating from April 2004 to April 2005.
Thus, it has 1 odd day only.
8th April, 2004 will be 1 day before the day on 8th April, 2005.
Given that, 8th March, 2005 is Monday.
So 8th March 2004 is Sunday (1 day before to 8th March, 2005).

MCQ: Clock and Calendar - 1 - Question 2

### Every Tuesday 10:00 AM the clock is set by Kevin, doing service in the clock tower which is in pune. The Clock loses 6 minutes every hour. When the faulty clock shows 3 P.M on Friday what will be the actual time?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 2

Total hours from Monday 10.00 AM to Friday 3.00 PM =101 Hours.
The clock loses 6 minutes every hour. In 101 hours it will lose 101 x 6 =606 minutes. i.e. 10 hours 06 minutes.
To find the actual time, calculate 10 hours 6 min backwards from Friday 3.00 PM which is 4 .54 AM
Hence, the actual time would be 4 .54 AM when clock shows 3.00 PM on Friday.

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MCQ: Clock and Calendar - 1 - Question 3

### Raj and Sonya were playing a game of finding same calendar years where raj asked Sonya to find out that the calendar for the year 2007 will be the same for which year?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 3

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Sum = 14 odd days 0 odd days.
Years 2018 and 2007 will the same calendar.
Option 4 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 4

Ravi in his maths class asked his teacher that how much a clock loses per day, if its hands coincide every 64 minutes.

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 4

55 min. spaces are covered in 60 min
60 min. spaces are covered
in ((60/55) x 60) min = 720/11 min.
in 64 min. the loss is = (720/11 - 64). = 16/11
In 24 hrs the loss is = (16/11 x 1/64 x 24 x 60) min = 32 x 8/11 = 256/11
Option 1 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 5

Rima while giving her final exam had faced a tough question which asked  to find out how many palindromes are there in a clock from noon to midnight (For Example 5.45 is a palindrome)?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 5

57 after 12 o’clock 12:21……= 1 from 1 to 9 it is 1.01, 1.11, 1.21, 1.31, 1.41, 1.51 = 6 similarly 2.02, 2.12… 6*9 = 54+1 = 55 after 10 o’clock 10:01, 11:11 Ans: 55 + 2 = 57
Option 3 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 6

A bet was placed between two friend ram and shyam to find out what will be the chance that a leap year selected at random contains 53 Fridays?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 6

2/7 in a leap year there are 366 days means 52 weeks and 2days. So already we have 52 Fridays. Now the rest two days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, wed), (Wednesday, Thursdays), (Thursday, Friday), (Friday, sat), (sat, sun) so, the probability of 53 Fridays= 2/7
Option 3 is the correct answer

MCQ: Clock and Calendar - 1 - Question 7

Alisha’s train left busy station X at an hour B minutes before she could catch the train. It reached station Y where Alisha had to reach at B hour C minutes on the same day where Alisha’s friend was waiting for her, after travelling C hours a minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 7

An hours + C hours = B hours … (I)
A, C and B can’t possess values greater than or equal to 24 B minutes + A minutes = C minutes … (ii)
From the above two equations, it is assumed that no value of A satisfies both the equations.
Hence, option 1 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 8

On February 20, 1999 was Saturday which was an important date in an investigation conducted by the police. December 30, 1997 was what day of the week which could help the police find the old case files regarding the investigation?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 8

This time period was 1998 which was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19th
January 1999 gives 3 odd days
19 February 1999 gives 5 odd days
1998, being ordinary year, gives 1 odd day
1997 December 30th and 31th have 2 odd days
The totalling of number of odd days = 3 + 5 + 1 + 2 = 11 days = 4 odd days
Hence December 30, 1997 will fall 4 days prior Saturday i.e. on Tuesday.
Option 2 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 9

Casio’s vintage clock in a shopping mall loses 1% time during the first week and then gains 2% time during the next one week where the mall technicians keep a track record of the clock. Clock was set at 12 noon on a Sunday. So what will be the time that the clock will show exactly 14 days from the time it was set?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 9

The clock loses 1% time during the first week.
A day comprises of 24 hours and a week has 7 days. Therefore, there are 7 * 24 = 168 hours in a week.
During the first week the clock loses 1% time, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less.
The clock gains 2% during the next week subsequently. The clock gains 2% time in the second week as it has 168 hours = 2% of 168 hours = 3.36 hours more than the actual time.
It had lost 1.68 hrs during the first week and then again gained 3.36 hrs during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.
Hence the clock will display time which is 1.68 hours more than 12 Noon two weeks from the time it was set.
1.68 hrs = 1 hr and 40.8 min = 1 hour + 40 min + 48 sec.
I.e. 1: 40: 48 P.M. Option 2 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 10

Prakash and his grandfather were solving a quiz question in a newspaper article in which it was given that on 27th February 2003 is Thursday, then what was the day on 27th February 1603?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 10

After every 400 years, the same day occurs. Before 400 years if 27th February 2003 is Thursday then on 27th February 1603 has to be Thursday.
Option 2 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 11

On 5th December 1993, Nilesh and Nitesh are twins who celebrated their birthday on Sunday. What will be the day of their birthday in 1997?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 11

Normal year jump +1 gain; Leap year jump +2 gain.

Sunday plus five days
Option 2 is the correct answer

MCQ: Clock and Calendar - 1 - Question 12

On a sports day of a school a race starts between 5pm and 6pm. The referee realizes that the hands of the clock are interchanged when the race ends between 8pm and 9pm. What was the approximate duration of the race?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 12

Let’s assume the race started at 5 hours X minutes and ended at 8 hours Y minutes. Angles made by both the hour’s hands are 150+X/2 and 240+Y/2. Angles made by the minutes hands are 6X and 6Y.

We know that the difference between the angles made by the hands of the clock is same, so we equate that: 6X – (150 + X/2) = (240 + Y/2) – 6Y => 12X – 300 – X = 480 + Y – 12 Y => 11X + 11Y = 480+300 => X+Y = (780/11) => X+Y ≈ ≈ 71

We know that the sum of the angles made by the hands of the clock is also same, so we equate that as well: 6X+150+X/2 = 240+Y/2+6Y => 13X + 300 = 480 + 13Y => 13(X-Y) = 180 => X-Y = (180/13) => X-Y = ≈ ≈ 14

Solving we two equations, we get X = 42.5 mins = 42 mins 30 secs and Y = 28.5 mins = 28 mins 30 secs

Hence, duration of the race is = 8:28-5:42 = 2 hours 46 mins

Option 2 is the correct answer

MCQ: Clock and Calendar - 1 - Question 13

Given that a person checks the time at only integral values of minutes, how many times in a day can a person not be able to find the exact time if the minutes and hours hand are identical in a clock?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 13

At any time, A hours and B minutes, the angle made by the hours hand is 30 A + B / 2. 30A+B/2. Similarly, the angle made by the minutes hand is 6B.

For a person to not be able to find the correct time, there should be another time     A1:B1 such that 30 A1 + B1 / 2 = 6B and 6B1 = 30 A + B / 2 6B1=30A+B/2

There is no integral solution existing when 0 < B < 60

Option 4 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 14

James had received a cool sport’s watch from his elder brother when he returned from Germany in which the second hand covers 3960 degrees, how many degrees does the hour hand move?

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 14

Second hand will move 360 degree in one minutes. Hence time is = 3960/360 = 11 minutes so hr hand will travel 11*30/60 = 5.5 degree.
Option 2 is the correct answer.

MCQ: Clock and Calendar - 1 - Question 15

Ramesh and Rita had formed a team for the quiz competition held at their school in which they asked each other practice questions. Rita asked Ramesh if today is Thursday, after 730 days which will be the day of the week.

Detailed Solution for MCQ: Clock and Calendar - 1 - Question 15

730/7 = 2 odd days. Hence Saturday will be the day of the week.
Option 3 is the correct answer.

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## Quantitative Aptitude for SSC CGL

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