If 1st January, 2001 was a Monday, what was the day on 26th January, 2003?
Sunday
Calculating the Day of the Week: 1st Jan 2001 to 26th Jan 2003
This problem requires us to find the day of the week for a specific date (26th January, 2003) given the day of the week for another date (1st January, 2001). We can solve this type of problem by calculating the number of "odd days" between the two dates.
An odd day is the number of days remaining after dividing the total number of days by 7. Since there are 7 days in a week, every 7 days the cycle of days repeats. The odd days tell us how many days forward from the starting day we need to count.
Calculating Odd Days Between Years
First, let's find the number of odd days from 1st January, 2001 to 1st January, 2003. This spans two full years: 2001 and 2002.
-
Year 2001: From 1st January, 2001 to 1st January, 2002 covers the entire year 2001. The year 2001 is not a leap year (it is not divisible by 4). A normal year has 365 days.
Number of odd days in 2001 = \(365 \text{ days} \pmod{7}\)
\(365 = 52 \times 7 + 1\)
So, there is \(1\) odd day in the year 2001. -
Year 2002: From 1st January, 2002 to 1st January, 2003 covers the entire year 2002. The year 2002 is also not a leap year (it is not divisible by 4). A normal year has 365 days.
Number of odd days in 2002 = \(365 \text{ days} \pmod{7}\)
\(365 = 52 \times 7 + 1\)
So, there is \(1\) odd day in the year 2002.
Total odd days from 1st January, 2001 to 1st January, 2003 = Odd days in 2001 + Odd days in 2002
Total odd days = \(1 + 1 = 2\) odd days.
Since 1st January, 2001 was a Monday, adding 2 odd days tells us the day on 1st January, 2003.
Day on 1st January, 2003 = Monday + 2 days = Wednesday.
Calculating Odd Days in the Same Month
Now we need to find the day on 26th January, 2003, starting from 1st January, 2003, which we know was a Wednesday.
We need to count the number of days from 1st January, 2003 to 26th January, 2003.
Number of days = \(26 - 1 = 25\) days.
Next, we find the number of odd days in these 25 days.
Number of odd days = \(25 \text{ days} \pmod{7}\)
\(25 = 3 \times 7 + 4\)
So, there are \(4\) odd days in the period from 1st January, 2003 to 26th January, 2003.
Final Day Calculation
We started from 1st January, 2003, which was a Wednesday. We need to move forward by 4 odd days to find the day on 26th January, 2003.
Day on 26th January, 2003 = Day on 1st January, 2003 + 4 days
Day on 26th January, 2003 = Wednesday + 4 days
Counting forward from Wednesday:
- Day 1: Thursday
- Day 2: Friday
- Day 3: Saturday
- Day 4: Sunday
Therefore, 26th January, 2003 was a Sunday.
Revision Table: Calendar Calculations
| Concept | Explanation | Calculation Example |
|---|---|---|
| Odd Days | The remainder when the total number of days is divided by 7. Determines the shift in the day of the week. | \(30 \text{ days} \pmod{7} = 2\) odd days |
| Normal Year | 365 days. Not a leap year. | 365 days = 52 weeks and 1 day. \(1\) odd day. |
| Leap Year | 366 days. Year divisible by 4 (except for years divisible by 100 but not by 400). | 366 days = 52 weeks and 2 days. \(2\) odd days. |
| Counting Forward | Add odd days to the starting day's index (Mon=0, Tue=1, etc.) and find the resulting day. Or simply count forward. | If today is Wednesday (Day 3) and there are 4 odd days, \(3+4 = 7\). \(7 \pmod{7} = 0\), which is Monday. Or count: Thu, Fri, Sat, Sun (4 days). |
Additional Information: Calendar Concepts
Understanding calendar concepts like leap years and odd days is crucial for solving these types of reasoning problems.
- Leap Year Rule: A year is a leap year if it is divisible by 4, unless it is a century year (like 1900, 2000) in which case it must be divisible by 400. For example, 1900 was not a leap year (divisible by 100 but not 400), but 2000 was a leap year (divisible by 400). 2001 and 2002 are not divisible by 4, so they are not leap years. 2003 is not divisible by 4, so it is not a leap year.
- Cycle of Days: The days of the week repeat every 7 days. This is why we use modulo 7 to find odd days.
-
Calculating Odd Days between Dates:
- Calculate the total number of days between the two dates.
- Account for leap years in between, as they add an extra day (and thus an extra odd day).
- Find the total number of odd days by taking the total days modulo 7.
- Add the total odd days to the starting day to find the ending day.
Important Questions from Probability
A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and the second throws, and a number less than 4 in the third throw?
Two dice are thrown simultaneously. If \( X \) denotes the number of fours, then the expectation of \( X \) will be:
If the random variable \( X \) has the following distribution:
X 0 1 2 otherwise P(X) k 2k 3k 0 Match List-I with List-II:
List-I List-II (A) k (I) \(\frac{5}{6}\) (B) P(X < 2) (II) \(\frac{4}{3}\) (C) E(X) (III) \(\frac{1}{2}\) (D) P(1 ≤ X ≤ 2) (IV) \(\frac{1}{6}\) Choose the correct answer from the options given below:
Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the following form, where c is some constant:
\[ P(X = x) = \begin{cases} 0.1, & \text{if } x = 0 \\ cx, & \text{if } x = 1 \text{ or } x = 2 \\ c(5 - x), & \text{if } x = 3 \text{ or } x = 4 \\ 0, & \text{otherwise} \end{cases} \]
Match List-I with List-II:
List-I List-II (A) c (I) 0.75 (B) P(X ≤ 2) (II) 0.3 (C) P(X = 2) (III) 0.55 (D) P(X ≥ 2) (IV) 0.15 Choose the correct answer from the options given below:
For the differential equation \( (x \log_e x) dy = (\log_e x - y) dx \):
(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is \( 2y \log_e x + A = (\log_e x)^2 \), where A is an arbitrary constant.
(D) Solution is \( 2y \log_e x + A = \log_e (\log_e x) \), where A is an arbitrary constant.
Choose the correct answer from the options given below:
