Question
What was the day of the week on ‘2nd July
2003’?Solution
Date: 2nd July 2003
Steps to find the day of the week:
Total = (3 + 0 + 0 + 2) = 5
And 5 ÷ 7, we get 5 as remainder.
Note:
Year < 2000 – leave the remainder as it is
Year > 2000 – subtract 1 from the remainder.
So, code for the day is (5 – 1) = 4
I. p2 – 15p + 56 = 0
II. q2 + 2q – 63 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 50x + 600 = 0
Equation 2: y² - 51y + 630 = 0
I. 10x² - 11x + 3 = 0
II. 42y² - 23y – 10 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
- Determine the remainder when equation 4p³- 5p² + 2p + 1 is divided by (4p - 3).
I. x2 – 39x + 360 = 0
II. y2 – 36y + 315 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 13x² - 60x + 47 = 0
Equation 2: 17y² - 80y + 63 = 0
Equation 1: x² - 250x + 15625 = 0
Equation 2: y² - 240y + 14400 = 0
I. x2 – 10x + 21 = 0
II. y2 + 11y + 28 = 0
I. 49y2 + 35y + 6 = 0
II. 12x2 + 17 x + 6 = 0
