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Total number of riders = 5p + 2 or 6q.

The first number that fits in is 12. Numbers following will be 12 + multiple of 30 ( LCM of 5 and 6).

Therefore, the total number of riders will be 30k+12.

42, 72, 102, 132 are the only possibilities as the number of riders should be between 29 and 150.

Among the given options (42+132 =174) is the answer.
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We are looking for a number which is divisible by 6 and leaves 2 as reminder when divided by 5. For sure, this number is even (as divisible by 6). So odd options are ruled out. Also minimum and max number will not cross 200+ so even 269 (option 1 is ruled out)
Now only 2 options are left.
First number which statisfy the condition is 12. Next number will be 12 + (LCM of 5,6) ie 12+ 30 = 42. So min number is 42. other numbers are 12+60 ie 72, 12+90 ie 102, 12+120 ie 132. So max number is 132.
42+ 132 is 174 ie option C
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If riders are let only in groups of 5, and there are 2 riders left,
Then total people waiting = 5x+2, where x can be any positive integer, and 29 <= 5x+2 <= 150.

Also, groups of 6 will completely exhaust the list.
Thus, 5x+2 is divisible by 6.

The highest value of x, such that 5x+2 <= 150 and divisible by 6 is 22.
Thus, the total number of people waiting in this case = 132.

The lowest value of x, such that 5x+2 <= 29 and divisible by 6 is 8.
Thus, the total number of people waiting, in this case, is 42.

So, the sum of least and most possible number of people = 132 + 42 = 174.

Thus, the correct option is C.
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