Life is like a sine function

Life is like a sine function,it has ups and downs.


The sine function:
    y=-|sin(x)|
                                           



                                

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How many squares are there on the chessboard?

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  Pic of question The answer is 204 squares.   This is because you have to calculate how many 1 x 1 squares, 2 x 2 square, 3 x 3 squares and so on that are on the chessboard. These numbers end up being the square numbers: 64, 49, 36, 25, 16, 9, 4, 1 respectively.  Hence, the total number of squares on the chessboard are 64+49+36+25+16+9+4+1 = 1²+2²+3²+4²+5²+6²+7²+8² = 8×9×17÷6 [∵ 1²+2²+3²+...+n² = n(n+1)(2n+1)/6 ]  i.e ., 204 Answer related pic
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Can you guess the correct answer?

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Pic of question.  We solve it by differentiate the triangles in the image into three types: small, medium, and large triangles.  Number of small triangles = 12 triangles (3 + 4 + 3 + 2) that are up facing and 12 facing down.  Sum = 24 triangles.  Number of medium triangles = 6 triangles (3 + 2 + 1) facing up and 6 facing down.  Sum = 12 triangles.  Number of large triangles = 2. One facing up and one facing down. Hence, the total number of triangles = 24 + 12 + 2 = 38 triangles .
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The rat and box puzzle.

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Pic of question.  Given conditions are (i) exactly one of the statements is true. (ii) exactly one box contains a rat. By given conditions we solve it by three cases. Case I- If first statement is true i.e., the rat is in box 1, then other two statements are false [by condition (i)] i.e., the rat is in box 2 and the rat is in box 1. Which is a contradiction by condition (ii). [Since, exactly one box can contains a rat but here box 1 as well as box 2 contains rat. ] Case II- If second statement is true i.e., the rat is not in box 2, then other two statements are false [by condition (i)] i.e., the rat is not in box 1 and the rat is in box 1. Which is again a contradiction. Case III- If third statement is true i.e., the rat is not in box 1, then other two statements are false [by condition (i)] i.e., the rat is not in box 1 and the rat is in box 2. This case satisfied the condition (ii).  Thus, the rat is in box 2. [By case III] Hence, the answer is box 2 . ...
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