Aptitude tests assess the capability of a candidate to practice fundamentals to clear up complicated situations. Every query is nothing but a scenario in which you are supposed to clear up with a fundamental grasp of maths. In the class, college students have seldom chance to experience the actual lifestyle conditions whilst on a job this is something which is required all the time.
So, this article will discuss one essential part of the aptitude test which is calendars. There are several approaches to resolve calendar-based problems. But the traditional techniques take a lot of time, which can be crucial in a time-restricted test. In this article, we will study an aptitude short trick to find the birthday of any date.
First of all, to solve this problem, you need to memorize a number sequence which is
1 4 4 0 2 5 0 3 6 1 4 6
You can remember the above sequence by breaking into pair of 3.
144, 025, 036 and 146.
122 = 144
52 = 025
62 = 036
146 = 144+2
These all numbers represent months of a year
Check the century the year belongs to and assign following values as shown below
And for days, assign the following day code values
Example 1:
Let the date be 5th April 1990 (5-4-1989)
Step 1: Write the equation in following format
Date + Month Code + Century code + Last 2 digits of year + (Quotient of (Last 2 digits of year/4))
The date is 05-04-1989
According to above equation we write it as
05 + 0 (April Month Code) +0 (19th Century code) + 89 (last 2 digits of year 1989) + 22 (89/4 =22 and reminder =1. Discard remainder)
The addition result is 116
Step 2:
Divide the addition result by 7 and check reminder.
In this case while you perform 116/7 you will get quotient =16 and remainder = 4.
Step 3: Neglect quotient and consider remainder. Compare the remainder value with day code given above.
Here remainder =4 which corresponds to Wednesday.
Therefore the day of date 5th April 1989 is Wednesday
Example 2:
Date : 21st September 1965
Date + Month Code + Century code + Last 2 digits of year + (Quotient of (Last 2 digits of year/4))
The above equation can be written as
21+ 6 (September Month Code) + 0 (19th Century code) + 65 (last 2 digits of year 1965) + 16(65/4 =22 and reminder =1. Discard remainder)
The addition result is 108
Step 2:
Divide the addition result by 7 and check reminder.
In this case while you perform 108/7 you will get quotient =15 and remainder = 3.
Step 3: Neglect quotient and take remainder. Compare the remainder value with day code given above.
Here remainder =3 which corresponds to Tuesday.
Therefore the day of date 21st September 1965 is Tuesday.
