What is the average of all the numbers allocated to the slots?___
Sum of numbers allocated to West-end slots (all rows) = 119; Sum of numbers allocated to East-end slots (all rows) = 109; Hence sum of ‘numbers’ allocated to all slots between East and West end = 119-109 = 10.
Now, no two rows were equally long. Hence the no of slots between east and west ends for all five rows should be 0, 1, 2, 3, 4 (for a sum of 10). Hence there are 1,2,3,4 and 5 slots (inbetween) in rows 1, 2, 3, 4 and 5 respectively.
The sum of numbers allocated to Row 2 is 81. This can be sum of 1, 2, 3, 4 or 5 natural numbers.
If row 2 has only 1 or 2 slots, then it is impossible to have a sum of 109 by addition of all numbers allocated to East-ends of all rows. So the numbers allocated to Row 2 can only be 26, 27 and 28 (3 slots).
Since B has 3 slots, A and C cannot have 2 or 4 slots.
Assuming Row 1 has only 1 slot, it should be numbered as 21.
Two West-end slots, Row 2’s and Row 1’s are numbered 21 and 28. Hence two east end slots have to be numbered as 22 and 29. So the four East-end slots are numbered: 21, 22, 26, and 29. Hence the fifth West-end number is: 109-(21+22+26+29) = 11, which is not possible because in that case the sum of allocated numbers of all West-End slots would not be 119.
Hence Row 1 has 5 slots and Row 3, Row 4 and Row 5 have 1, 4 and 2 slots respectively.
Row 1 has a west-end number of 21 and Row 2 has an east-end number of 26. Thus there is a row in between Row 1 and Row2 that has 4 slots. So the other East-end numbers allocated are 17, 22 and 26. Hence the other two East-end allocated numbers are 29 and 15.
So the plot of land looks like:
RowSlot numbers:EastWestJasmine17 18 19 2021Lotus26 27 28Rose29Lavender22 23 24 25Marigold15 16
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