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Three Calender Time Aptitude Questions

Question 1
History Professor Nagarajan was talking to the students about 21st century which has started with a Monday.What day India would be witnessing on the last day of the century, the Professor was posing a question. Incidently he posed a question that the last day of the century cannot be
(a)Monday (b) Tuesday (c)Wednesday (d) Friday
Can you answer the Professor ‘s question?
Answer : b) Tuesday
To solve this problem let us try to find out the number of odd days that 100 years has. When number of days in a given period of time is divided by 7 then the remainder which results represents the number of odd days in that period. For example the number of odd days in 15 days is remainder of 15/7 which is 1.
Now let us get into the solution
To start with consider a span of 100 years. Every 4th year is a leap year within a century and every 4th century year is a leap year. This means years 4,8,12 etc are leap years while 100 is not a leap year. But 400th year, 800th year etc are leap years.
By above argument 100 years contain 24 leap years and 76 non leap years. (years 4,8,12....96 are leap years and 100th year is not. Therefore number of leap years in 100 years is 100/4 - 1)
Number of days in 100 years = 24 x 366 + 76 x 365 = 36524
Dividing 36524 by 7 we will get a quotient of 5217 and remainder of 5.
Therefore 100 years i.e 36524 days has 5 odd days to end with. Since the century has started with monday, odd days in order will be Mon, Tue, Wed, Thu and Fri.
Last odd day will be the last day of the century. Hence last day of 1st century will be Friday.
By similar ways one can find that 200 years will contain 3 odd days, 300 years will contain 1 odd day and 400 years will contain 0 odd days. This means last day of 2nd century will be Wednesday, last day of 3rd century will be Monday and last day of 4th century will be Sunday.
The entire cycle will repeat for the next 400 years, thereafter next 400 years and so on.
Therefore last day of the century cannot be Tuesday, Thursday or Saturday.
Question 2
Ranjit Kumar purchased a new watch in Burma Bazaar. Sooner he found that the uniformly gaining watch starts with a lag of 2 minutes at noon on a particular Monday and it is 4 minutes 48 seconds fast at 2PM on the following Monday. Please guide Ranjit Kumar when the clock would show the true time ?
(a)2PM on Tuesday (b)2PM on Wednesday (c)3PM on Thursday (d)1PM on Friday
Answer : b)2PM on Wednesday
12 noon Monday to 2PM next Monday is 7 days and 2 hours. i.e. 7 x 24 +2 = 170 hours.
The new watch purchased by Ranjit Kumar was 2 minutes slow and by 170 hours it was fast by 4 minutes 48 seconds. (4 minutes 48 seconds can be written as 4 48/60 or 4 4/5 minutes.)
Therefore total gain in 170 hours = 2 minutes + 4 4/5 minutes = 34/5 minutes = 408 seconds
Gain per hour = 408/170 = 2.4 seconds.
It is given the watch was slow by 2 minutes (120 seconds) initially. Therefore the time at which the watch will make up for the lost 2 minutes by gaining 2 minutes or 120 seconds will be our answer.
2.4 seconds is gained in 1 hour.
120 seconds will be gained in 1/2.4 x 120 =50 hours.
So the watch will show right time in 50 hours from Monday 12.00 noon ie at Wednesday 2.00 PM.
Question 3
Dilip is a book worm and his wife Rama Dilip is also a voracious reader. Dilip reads at an average rate of 40 pages per hour, while Rama Dilip reads at an average rate of 50 pages per hour. Dilip starts reading a novel at 4.30 PM and Rama Dilip begins reading an identical copy of the same book at 5.20 PM. At 6.00 PM suddenly a family friend arrived and both husband and wife had to spend 30 minutes with the guest. Thereafter, they started reading again. At what time will both husband and wife will be reading the same page?
(a)10.00PM (b) 9.30 pm (c) 9.10 PM (d) 8.20Pm (e) 7.30 PM
Answer : c) 9.10 PM
We have to find out when Rama Dilip will catch up with Dilip. Dilip reads at the rate of 40 pages per hour and Rama Dilip reads at the rate of 50 pages per hour. Dilip starts 50 minutes ahead of his wife. Since 50 minutes is 5/6 of an hour, by the time Rama Dilip starts reading at 5.20 PM, Dilip has already read 5/6 x40=200/6 pages. Please note that Rama Dilip is faster by 10 pages per hour as compared to her husband.
Since Dilip started with 200/6 pages ahead at 5.20 PM to catch up with Dilip it should take Rama Dilip 200/6 pages / 10 pages per hour. ie 200/60 hours or 3 hours 20 minutes. And of course the common time spent by both husband and wife chatting with guest is 30 minutes. So to catch up with Dilip, Rama Dilip requires 3 hours 20 minutes + 30 minutes or 3 hours 50 minutes. Rama Dilip and Dilip will be reading the same page at 5.20 PM + 3 hours 50 minutes ie by 9.10 PM

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Aptitude Test Practice Questions - With Answers

Question 1. Which of the following is least like the others?
A.    cube
B.    sphere
C.    pyramid
D.    circle

D (because the circle is the only two-dimensional figure)

Question 2. Consider a language which uses the following set of characters:
Small set: { a b c }
Large set: { A B C }
Punctuation set: { x y }
This language must follow the following rules:
  1.    A punctuation character must end all series.
  2.    A series can have up to but no more than 4          characters,including punctuation characters.

Does the following series follow all the rules of the language defined above?
  A.    Yes
  B.    No

A (the series has only four characters and ends in a punctuation character)

Question 3. Consider the following flow chart for a customer:

The person in No.1 is:
    A.   Married, with children
    B.   Married, with at least one son
    C.   Unmarried, with at least one daughter
    D.   Unmarried, with at least one son
    E.   Unmarried, with no children