$X$, $Y$ 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?
Detailed Explanation:
At 8 AM, all three are at mark 199'. Since X jumps 4', Y jumps 6', and Z jumps 5', they will all land on the same mark again when the distance from 199' is a multiple of LCM(4, 6, 5) = 60'.
Common landing marks are 199 + 60k where k = 0, 1, 2, ... Within the range 195' < m < 1000', we get: 199, 259, 319, 379, 439, 499, 559, 619, 679, 739, 799, 859, 919, 979. The next mark (1039') exceeds 1000'.
Therefore, counting from k=0 to k=13 gives 14 common marks, making Option 4 the correct answer.
Question 1 of 7 HCF and LCM
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