Latest TCS Aptitude Question SOLUTION: The rectangle of height 100 squares and width 200 squares is drawn on a graph paper.It is colored square by square from top left corner and moving across in a spriral turning righ

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(#M40033863) TCS Numerical Ability question numerical Keep an EYE

The rectangle of height 100 squares and width 200 squares is drawn on a graph paper.It is colored square by square from top left corner and moving across in a spriral turning right whatever a side of the rectangle or a colored square is reached .Which square is colored last ( give its row and column number –the bottom right square is on row 100 and column 200)

Asked In TCS RAvi (10 years ago)

Unsolved Read Solution (7)

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jagadeesh  (10 years ago)

answer:51,50

Is this solution Helpfull? Yes (6) | No (6)  

shivam  (9 years ago)

can anyone explain this solution.............

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Lovely Munjal  (10 years ago)

row 51
column 150

Is this solution Helpfull? Yes (4) | No (3)  

Sushmita saxena  (9 years ago)

Row - 51
column - 100

Is this solution Helpfull? Yes (3) | No (1)  

kumar ashutosh anand  (10 years ago)

ans.row-50,column-100

Is this solution Helpfull? Yes (2) | No (3)  

anbu  (5 years ago)

Length(height)=100 squares
breadth(width)=200 squares
(row,col)=((length/2+1),(breadth/4))
=(100/2+1),(200/4)
=51,50

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  (4 years ago)

answer:51,150

Let's take starting index as (1,1) and after first round of moving around in a spiral we reach (2,1).....

After 1st round, we reach (2,1) , no. of rows and columns left to cover is (98,198)
After 2nd round, we reach (3,2) , rows and columns left is (96,196)
After 3rd round, we reach (4,3) , rows and columns left is (94,194)
.
.
.
So, we can observe a pattern here,after each round we reach (x+1,x) and we're left with (100-2*x,200-2*x) rows and columns respectively, so when x=50 ,(100-2*50,200-2*50)=(0,100)...So,we can infer that when we're at (51,50) we still have 100 columns left to cover,so after moving right by 100 columns we reach (51,150)

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