UPSC CSE Prelims 2026
The correct option is D - 14.
[as per provisional answerkey]We are given the jump lengths of three individuals: X = 4', Y = 6', and Z = 5'. We need to find how many marks between 195' and 1000' (inclusive) are reachable by all three individuals.
1. Identify the condition for a common mark: For a mark to be stepped on by X, Y, and Z, the mark must be a multiple of their respective jump lengths. However, the question states they all land on mark 199' at 8 AM. This implies that 199' is a common landing point for their current sequences of jumps.
2. Determine the starting point: If they all land on 199', the next common mark they will all land on depends on the Least Common Multiple (LCM) of their jump lengths.
:
5 = 5
.
3. Find the general formula for common marks: Since 199' is a common mark, all other common marks will be of the form: (where n is an integer).
4. Calculate marks between 195' and 1000':
For n = 0: ' (Valid, as 195 < 199 < 1000)
For n = -1: ' (Invalid, below 195)
To find the maximum n:
So, n can range from 0 to 13.
5. Count the values: The values of n are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
Total count = 14.
The common points in multiple periodic sequences are determined by the Least Common Multiple (LCM) of the periods, starting from a known common offset point.