The correct option is D - 14.
[as per provisional answerkey]Solution
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.
LCM(4,6,5):
4=22
6=2×3
5 = 5
LCM=22×3×5=60.
3. Find the general formula for common marks: Since 199' is a common mark, all other common marks will be of the form: 199+60n (where n is an integer).
4. Calculate marks between 195' and 1000':
For n = 0: 199+60(0)=199' (Valid, as 195 < 199 < 1000)
For n = -1: 199−60=139' (Invalid, below 195)
To find the maximum n: 199+60n<=1000
60n<=1000−199
60n<=801
n<=801/60=13.35
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.
Why the other options are incorrect
- Option (a) - 11: This result might be reached if one incorrectly calculates the LCM or starts the sequence from a higher number, missing the initial common marks near 199.
- Option (b) - 12: This is a common error if the student calculates the number of jumps after 199 but forgets to include the mark 199 itself, or makes a minor division error (801/60).
- Option (c) - 13: This occurs if the student correctly identifies the range but fails to account for the "zero" term (n=0) in the sequence 199+60n, effectively missing one occurrence.
Key Concept
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.