For a complete understanding of Calendars, Concept of Calendars

For more solved problems and examples, go through Problems on Calendars

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__TCS NQT 2021 Aptitude 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 Ninja Aptitude 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 NQT QUANTITATIVE APTITUDE 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
NQT APTITUDE - New Pattern Model Question 4__
It was Sunday on January 1^{st}, 2006. What was the day
of the week January 1^{st}, 2010?
a) Sunday
b) Saturday
c) Friday
d) Wednesday
**Answer: C**
**Explanation :**
From January 1st, 2006 to
January 1^{st, }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 1^{st} January 2010 =
Sunday + 5 =Friday

__TCS
NQT LATEST 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, 222^{nd} 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 NQT Latest Pattern - 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?
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 NQT latest pattern - New Pattern Model Question 7__
302 th day of 1 year is Saturday, then what is the 15 th day of
5^{th} 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 NQT 2021 APTITUDE –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 12^{th} 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
NQT 2021 - Aptitude 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
NQT APTITUDE QUESTIONS - 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

__TCS NQT 2021 Aptitude 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 Ninja Aptitude 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 NQT QUANTITATIVE APTITUDE 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 NQT APTITUDE - New Pattern Model Question 4__
It was Sunday on January 1

^{st}, 2006. What was the day of the week January 1^{st}, 2010?
a) Sunday
b) Saturday
c) Friday
d) Wednesday

**Answer: C**

**Explanation :**

From January 1st, 2006 to
January 1

^{st, }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 1

^{st}January 2010 = Sunday + 5 =Friday

__TCS NQT LATEST 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, 222

^{nd}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 NQT Latest Pattern - 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?

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 NQT latest pattern - New Pattern Model Question 7__
302 th day of 1 year is Saturday, then what is the 15 th day of
5

^{th}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 NQT 2021 APTITUDE –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

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 12

^{th}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 NQT 2021 - Aptitude 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 NQT APTITUDE QUESTIONS - 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