X, Y and Z jump forward 4', 6' and 5', respectively. At 8 AM, they all land on mark 199'. How many times will they all land on the same mark (need not be at the same moment) between mark 195' and 1000', if all of them cross mark 1000' by 9 AM?

Why the correct option is correct: To determine how many times X, Y, and Z will land on the same mark, we must utilize the Least Common Multiple (LCM). This mathematical principle is fundamental in solving synchronized periodic events, a core component of the UPSC CSAT syllabus.

The jump lengths are 4′, 6′, and 5′. The LCM of these values (2² × 3 × 5) is 60. This establishes that all three will simultaneously land on the same mark every 60 feet.

The problem states they land together on mark 199′. We must count the total synchronized landings between the 195′ and 1000′ marks. These landing points form an Arithmetic Progression (AP) where the first valid term (a) is 199 and the common difference (d) is 60.

To find the total number of terms (n) up to the 1000′ mark, we use the AP formula: Tₙ = a + (n-1)d ≤ 1000 199 + (n-1)60 ≤ 1000 (n-1)60 ≤ 801 n-1 ≤ 13.35

Since n must be an integer, n-1 = 13, which gives n = 14. Thus, there are exactly 14 instances where they land on the same mark within the specified range. Option D is correct.

Why the incorrect options are wrong:

• Option A (11): Incorrectly underestimates the occurrences due to a miscalculated LCM or an erroneous baseline.
• Option B (12): Incorrect; usually stems from dividing the remaining distance (800/60) and dropping remainders without properly tracking the sequence.
• Option C (13): Incorrect because it fails to count the initial synchronized landing at 199′ (the 1st term), which is validly positioned between 195′ and 1000′.

Concluding Takeaway: Takeaway: For synchronized events, calculate the LCM of the intervals. To find the exact count of occurrences within a boundary, map the instances using the Arithmetic Progression formula, taking special care not to skip the first valid occurrence.