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TCS Placement Papers with Solutions - Calendars Practice Questions


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 Problems on Calendars



TCS Placement Papers - New Pattern Model Question 1
 What is the probability of a leap year selected at random contains 53 Sundays?
     a)1/7                                     b)2/7                               c)3/7                       d)4/7
Answer : B
Explanation :
   An ordinary year has 52 weeks and 1 odd day. 
 The day of the week on an extra day will be the same as the day of the week of an ordinary year.
  So an ordinary year has 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 Thursdays, 52    Fridays, 52 Saturdays and 52 Sundays + 1 extra  day
The day of the week on an extra day will be the same as the day of the week of an ordinary year.
 A leap year has 52 weeks and 2 odd days.
 A leap year has an extra day which might be a Monday or Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. 
Sample space S: { Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday- Saturday, Saturday-Sunday and  Sunday-Monday}
Number of elements in S = n(S) = 7
Let A be the set which contains the number of elements that comprises of the elements Saturday-Sunday and Sunday-Monday i.e. A : {Saturday-Sunday, Sunday-Monday}
Number of elements in set A = n(A) = 2
 Probability of a leap year selected at random contains 53 Sundays = n(A)/n(S) = 2/7

TCS Placement Papers - New Pattern Model Question 2
Valentine day 14 Feb. 2005 was celebrated by Anand and Shilpa on Monday. They were very happy. What day of the week would come on 14 February 2010?
 a) Friday                         b) Sunday                  c) Tuesday                   d) Thursday
Answer: B
Explanation :
 The given day 14th February 2005 falls on Monday.  We have to find the day of the week on 14th February 2010.
  From 14 - 2- 2005 to 14 -2- 2010, there are 5 years.
   There are  4 ordinary years and 1 leap year in these 5 years.
  Each ordinary year has 1 odd day and a leap year has 2 odd days
  So there are 4+ 2  = 6 odd days
  The day of the week on 14 th February 2010 = Monday + 6 = Sunday

TCS Placement Questions – Latest Pattern Model Question 3
In 2003 there are 28 days in February and 365 days in a year in 2004 there are 29 days in February and 366 days in the year. If the date march 11 2003 is Tuesday, then which one of the following would the date March 11 2004 would be?
a)Sunday                b)Monday                c)Thursday              d)Friday
Answer: C
Explanation :
In 2004, February month has 29 days. Then 2004 must be a leap year.
From the given question, 2004 is a leap year (since the month of February in this year has 29 days)
Now given, 11-3-2003 is Tuesday. We have to find the day of the week on 11-3-2004
The number of  days between the two dates is  366 days
The number of odd days is 2
The day of the week on 11-3-2004  => Tuesday + 2 = Thursday.

TCS Placement Papers - New Pattern Model Question 4
It was Sunday on January 1st, 2006. What was the day of the week January 1st, 2010?
a) Sunday                    b) Saturday                  c) Friday                      d) Wednesday
Answer: C
Explanation :
  From January 1st, 2006  to January 1st, 2010, there are 4 years.
   Of these 5 years, 2006, 2007, 2009 are ordinary years and 2008 is a leap year.
  Each ordinary year has 1 odd day and a leap year has 2 odd days
  So there are 3+ 2 = 5 odd days
  The day of the week 1st January 2010 = Sunday + 5 =Friday

TCS Placement Papers - New Pattern Model Question 5
In a year N, the 259th day of the year is a Saturday. In the year N+l, the 222nd day of the year is also a Saturday. What is the 119th day of the year N-l?
a) Thursday                b) Saturday                     c) Friday                      d) Tuesday
Answer: C
Explanation :
Take the Nth year.  In the Nth year, 259th is a Saturday.
                              In the N+1 the year, 222nd  day is also Saturday.
So there must be 0 odd days between these days.
But We don’t know whether Nth year is a leap year or not.
Case 1 : Let us assume Nth is an ordinary year
             Then number of days between these 2 days => (365 - 259) +222 = 328 days
             In 328 days, we have 1 odd day . So we can say that N th year is not an ordinary year.
Case 1 : Let us take Nth year is leap,
             Number of days between these 2 days => (366 - 259 ) + 222= 329 days
             In 329 days ,we have 0 odd day . So we can say that N th year is  leap year
As Nth year is leap , N-1 th year is not a leap year.
 Number of days between 259 th day of Nth year and 119 th day of N-1 th year
                    = 259 + (365-119)
                    = 505 days
The number of odd days in 505 days =>1 odd day
         Therefore , 119th day of the year N-1 => Saturday -1 = Friday
        
TCS Placement Papers - New Pattern Model Question 6 
In an year N, the 320th day of the year is Thursday. In the year N+1 the 206th day of the year is also Thursday. What is the 168th day of In the year N-1?
  a) Tuesday                 b) Thursday              c) Friday                d) Saturday
Answer :C
Explanation :
As 320th day of the N th year and 206th day of the N+1 th are same day ,i.e, Thursday, 
there must be 0 odd days between these 2 days ( The number of days between these 2 
days must be divisible by 7)
If N th is a non-leap year, the number of days between 2 days = (365 - 320 )+206 =251 days
Here the number of odd days in 251 days is 6 =>So Nth year must be an ordinary year
If Nth year is a leap , then the number of days between 2 days => (366 -320 ) + 206=  252 days
 Number of odd days in 252 days is 0 =>    So we can say that Nth year a leap year.
To find the 168 th day of the year N-1, we have to find the number of odd days between the 320th day of Nth year  and 168 th day of N-1 th year.
              => (365 -168) + 320   => 517 days
  The number of odd days in 517 => 6 odd days
  The 168th day of N-1 year is     Thursday - 6 = Friday

TCS Placement Solved Questions - New Pattern Model Question 7
302 th day of 1 year is Saturday, then what is the 15 th day of 5th  year?
 a) Friday                         b) Tuesday              c) Thursday                 d) Wednesday
Answer :B
Explanation :
To find 15 th day of 5 th year, we have to know the number of odd days 
between 302 th day of 1st year and 15 th day of 5 th year.
 302 th day of 1st year is Saturday.
In 5 consecutive years, a leap year occurs only once. 
The number of remaining days in 1st year is 365 - 302 =63
 In 63days, the number of odd days -> 0
2nd year is an ordinary year => 1 odd day
3rd year is an ordinary year => 1 odd day
4th year is a leap year => 2 odd days
In 15 days of 5 th year, there is 1 odd day
So total number of odd days = 0 + 1 + 1 + 2 +1 = 5 odd days
                                 The answer is Saturday + 5 =Thursday 

TCS Placement Questions Solved  –Model Question 8
A man has a job, which requires him to work 8 straight days and rest on a ninth day. If he started work on Monday, find the day of the week on which he gets his 12th rest day.
a) Monday                   b) Wednesday                         c)Tuesday                   d) Friday

Answer : B
Explanation :
He works for 8 consecutive days and takes rest on the 9th day.
For each rest day, 8 work days + 1 rest day= 9 days passed
So on the 12th rest day, there are 9 x 12 = 108 days passed.
We have to find the day of the week on the 12th rest day. So we have to deduct a day.
Number of odd days = (108 - 1) / 7 = 107 / 7 = 2.
He starts work on Monday.
So the 12th rest day is Monday + 2 = Wednesday.

TCS Placement Papers - New Pattern Question 9
What is the value of a+b+c , if 28a+30b+31c=365 where a,b,c are natural numbers ?
    a)10                             b)11                         c) 12                       d) 15
Answer :C
Explanation :
Given 28a + 30b+ 31c =365 .
In a calendar, there are 365 days. And 7 months have 31 days, 5 months have 30 days and 1 month ( February) has 28 days.
          (28 x  1 ) + (30 x 4) + (31 x 7) =  365 
Hence values of a,b and c are 1, 4 and 7 respectively.
           Therefore a+b+c =12

TCS Placement Papers - New Pattern Question 10
The price of a commodity (in rupees per kg) is 100 + 0.1n, on the nth day of 2007(n=1,2,3....100) and then remains constant. On the other hand, the price of another commodity( in rupees per kg) is 89 + 0.15n, on the nth day of 2007( n=1,2,3,...365). On which date in 2007 will the price of these two commodities be equal?
1) May 21             2) April 11                 3) May 20                 4)July 10
Answer: C
Explanation :
The price of a commodity (in rupees per kg) is 100 + 0.1n, on the nth day of 
2007(n=1,2,3....100) and then remains constant
Hence price of the first commodity remains constant after the 100th day (n=100).
After 100 days, its price = 100 + 0.1 x 100 = 110.
 The price of another commodity ( in rupees per kg) is 89 + 0.15n , on the nth day of 2007( n=1,2,3,...365).
  Assuming after n days, the price of both commodities will be equal
          89 + 0.15n = 110
          =>  n = 140 days
     After 140 days , the price of both commodities will be equal.
   Given the year 2007 is not a leap year.    
        January  + February + March + April + May = 31 + 28 +31 +30 +20
      Therefore, the prices of both commodities will be equal on 20th May
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