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UPSC Prelims 2026 CSAT, question 10

Topic: Number system

Q10

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

  1. (a) 11
  2. (b) 12
  3. (c) 13
  4. (d) 14Correct answer

Solution.

Answer: (d) 14

  1. The step lengths of X,Y,X, Y, and ZZ are 4′,6′,4', 6', and 5′5', so the common marks they all land on are separated by LCM⁡(4,6,5)=60′\operatorname{LCM}(4, 6, 5) = 60'.
  2. Since all three land on mark 199′199' at 8 AM, the common marks between 195′195' and 1000′1000' are of the form 199+60k199 + 60k for integer k≥0k \ge 0.
  3. Solving 199+60k<1000199 + 60k < 1000 gives 60k≤80060k \le 800, so k∈{0,1,2,…,13}k \in \{0, 1, 2, \dots, 13\}, which gives 1414 common marks (from 199′199' to 979′979').

Trap: Remember to include the starting common mark 199′199' (k=0k = 0), since 199′199' lies strictly between 195′195' and 1000′1000'.