CEdward wrote:
robu1 wrote:
if we take 53 max as PR then three loads greater than 53 are 54,54,56
if we take 52 min as PR then three loads greater than 52 are 54,54,56 in this case too.
if we take 49 min as PR then loads greater than 49 are 50, 51, 52, 52, 54, 54, 56. ( 07 nos) which conflict the given condition: The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR.
i hope this makes clear.
I am confused...For the maximum, the problem states that 3 loads are greater than the truck's PR which would mean 56, 54, and 52. We can't count a number twice...
That would mean 51 is the max. If you take 53 as the max, then that would violate the claim that there were 3 loads greater than the truck's PR, no?
I initially made that same mistake thinking i cannot count 52 Twice or 54 Twice
But 52 CAN BE COUNTED twice.
The following suggests that
If 52 and 54 CANNOT BE counted twice as you mention above
-- Min value becomes : 51 [as you 52 (albeit 2 isntances of 52) , 54 (albeit 2 instances of 54) and 56 (one instance only)]
-- Max value becomes -- there is no max value you can come up with as you need 3 different payloads (irrespective of number of occurences)
i) lets say you try 52 -- only 54 (two instances) and 56 (one instance) -- only two payloads are seen - 54 and 56.
ii) lets say you try 53 -- only 54 (two instances) and 56 (one instance) -- only two payloads are seen - 54 and 56.
iii) lets say you try 54 -- 56 (one instance) -- only one instance of 56 is seen
But per the question stem - you should be able to get 3 payloads (irrespective of number of occurences) when choosing a max value too
But there