Bunuel wrote:
A box of chocolates has 15 rows with 10 spaces for chocolates in each row, how many chocolates are in the box?
(1) There is an average (arithmetic mean) of 6 empty spaces in the first 8 rows.
(2) There is an average (arithmetic mean) of 9 empty spaces in the last 9 rows.
So 15 rows, 10 spaces in each row. Any non-empty space is occupied by chocolate.
(1) In the first 8 rows, average empty spaces per row = 6, so total empty spaces = 6*8 = 48. So out of 80 total spaces, 48 are empty. Thus 80-48 = 32 have chocolates. But no data about other rows.
Insufficient.
(2) In the last 9 rows, average empty spaces per row = 9, so total empty spaces = 9*9 = 81. So out of 90 total spaces, 81 are empty. Thus 90-81 = 9 have chocolates. But no data about other rows.
Insufficient.
Combining the statements, we have first 8 rows and last 9 rows. 8+9 = 17, but there are total 15 rows only. So there is an overlap of 2 rows between first and second statements, of which we have no data per se. Until we know the number of chocolates in those 2 rows (7th, 8th from top Or 8th,9th from bottom) we cannot answer this question. Hence
E answer