Calendar, clock, coding-decoding and number series MCQ - Practice Questions with Answers

The angle between the hands is |30H - 5.5M| degrees, using the hour number and the minutes after that hour. For 3:20, H is 3 and M is 20. The calculation is |30 x 3 - 5.5 x 20| = |90 - 110| = 20 degrees. Since 20 degrees is already the smaller angle, the formula statement and the numerical statement are both correct.

The clock-angle rule is angle = |30H - 5.5M|, where H is the hour on a 12-hour clock and M is the minutes after that hour. At 3:20, H = 3 and M = 20. The angle is |30 x 3 - 5.5 x 20| = |90 - 110| = 20 degrees. This already is less than 180 degrees, so no supplement is needed. The smaller angle is 20 degrees.

Calendar questions use only odd days because complete weeks do not change the weekday pattern. Here, 17 days contain 2 complete weeks and 3 odd days. Starting from Monday, moving forward 3 positions gives Tuesday, Wednesday and then Thursday. Therefore, the weekday 17 days later is Thursday. The calculation does not require counting all 17 days one by one; the important step is reducing the total by division by 7 and using the remainder as the weekday shift.

There is a two-part leap-year rule. A year divisible by 4 is usually a leap year, but a century year must also be divisible by 400. Both 1900 and 2000 are century years, so the stricter test applies. The year 2000 is divisible by 400, so it is a leap year. The year 1900 is not divisible by 400, so it is not a leap year. This distinction affects calendar and odd-day counting.

The coding notes say that alphabet positions may move forward cyclically, so after the last position the count starts again. From position 25, three forward moves are counted as 26, then 1, then 2. Therefore the assertion that position 2 is reached is true. The reason is also true because it gives the exact wrap-around rule that makes this movement possible. The reason explains the assertion, not merely a separate fact.