6 equidistant vertical lines are drawn on a board. 6 equidistant horizontal lines are also drawn on the board cutting the 6 vertical lines, and the distance between any two consecutive horizontal lines is equal to that between any two consecutive vertical

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Q: 44 (IAS/2007)
6 equidistant vertical lines are drawn on a board. 6 equidistant horizontal lines are also drawn on the board cutting the 6 vertical lines, and the distance between any two consecutive horizontal lines is equal to that between any two consecutive vertical lines. What is the maximum number of squares thus formed?

question_subject: 

Logic/Reasoning

question_exam: 

IAS

stats: 

0,4,1,1,4,0,0

keywords: 

{'squares': [0, 0, 1, 0], 'equidistant horizontal lines': [0, 0, 1, 0], 'equidistant vertical lines': [0, 0, 1, 0], 'consecutive horizontal lines': [0, 0, 1, 0], 'consecutive vertical lines': [0, 0, 1, 0], 'vertical lines': [0, 0, 1, 0], 'maximum number': [1, 0, 1, 2], 'distance': [0, 3, 3, 3], 'board': [0, 1, 2, 2]}

The six vertical and horizontal lines on the board form a grid of 5 x 5 squares. Here`s why:

Each set of adjacent lines, whether vertical or horizontal, forms a square between them. Since we have six lines, there will be five spaces between them creating five squares along each line. Thus, in total we have 5*5 = 25 single squares.

Also, you can form larger squares that cover 2x2, 3x3, 4x4 until 5x5 smaller squares. For each size, the number of squares is calculated by (6-n)*(6-n), with n being the size. If we add up those values we get:

(6-2)*(6-2) for 2x2 squares = 4*4 =16

(6-3)*(6-3) for 3x3 squares = 3*3 = 9

(6-4)*(6-4) for 4x4 squares = 2*2 = 4

(6-5)*(6-5) for 5x5 squares = 1*1 = 1

Adding those values to the 25 single squares, we get 25+16+9+