Hi,
'r' rows and 'r+1' columns of squares = r(r+1) squares.
4th row(for that matter any row) has (r+1) squares
7th column(in fact any column) has r squares
They have 1 square in common.
So, number of squares in 4th row or 7th column = (r+1)+r-1 = 2r
So, remaining squares = r(r+1) - 2r = r^2 - r

Hence, A

i think r rows and r+1 columns will have a total of (r-1)*(r) squares
please explain why isnt it so

sumasajja wrote:i think r rows and r+1 columns will have a total of (r-1)*(r) squares
please explain why isnt it so

Hi,
That is when you are drawing r horizontal lines and (r+1) vertical lines. But, in this case, we are given that the squares are placed in r rows and (r+1) columns. So, it will be r(r+1) squares. If you are still confused draw a figure with r=3 or so. It should help you understand better.
For example, consider a chess board, it has 9 vertical lines and 9 horizontal lines, but the squares are placed in 8 rows and 8 columns. Same here.

sumasajja wrote:i think r rows and r+1 columns will have a total of (r-1)*(r) squares
please explain why isnt it so

If this were true, then a board of 1 row and 2 columns would mean 0 squares

kmittal82 wrote:

sumasajja wrote:i think r rows and r+1 columns will have a total of (r-1)*(r) squares
please explain why isnt it so

If this were true, then a board of 1 row and 2 columns would mean 0 squares

ya now i understand,thanks

IMO: A

Given, r > 10

consider r=11 and r+1 = 12
total squares = r(r+1)=132
4th row and 7th column = 11+12 = 23

total remaining squares = 109 + 1(common squarea) = 110

(r^2)-r = 110
(11)^2-11 = 110