Latest m4maths Aptitude Question SOLUTION: There are 50 coins. Half of them are genuine and the other half are counterfeit. Genuine coins weigh 10 grams, while counterfeit coins differ from the genuine coins by 1 gram (some

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Previous Question: Pls check the solution for puzzle Total no. of words of 5 letters formed by letters of the word BOMBAY. [You should use only 5 letters out of 6 letters of the word BOMBAY] solution given is .................. ..... (1st case all the letters are different total no. of words of this category = 5*4*3*2*1 = 120 2nd case B comes twice and 3 others total no. of words of this category = 5*4*3*2*1/2*1 = 60 so total words = 120+60 = 180) I think answer should be 360 as solution should be 1st case all the letters are different total no. of words of this category = 5*4*3*2*1 = 120 2nd case B comes twice and 3 others total no. of words of this category = 4*(5*4*3*2*1/2*1) = 60*4=240 because there will be 4 combinations of such words. so total words = 120+240 = 360 pls excuse me if I am wrong. Next Question: Place the numbers 1 through 100 on a 10x10 square table so that all consecutive numbers will be on adjacent (right, left, up, down) squares. What will be the maximum value of the sum of the 10 numbers along a diagonal of this table?

(#M40010114) M4MATHS question number of weighings Keep an EYE

There are 50 coins. Half of them are genuine and the other half are counterfeit. Genuine coins weigh 10 grams, while counterfeit coins differ from the genuine coins by 1 gram (some of them weigh 9 grams, and some of them weigh 11 grams). You have a balance scale with two pans. In each weighing the scale shows the difference of the weights placed in each pan. You will randomly choose a coin. What is the minimum number of weighings necessary to guarantee to determine whether it is genuine or not?: Asked In m4maths Brij (11 years ago); Unsolved Read Solution (1); Is this Puzzle helpful? (6) (1) Submit Your Solution

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Previous Question: Pls check the solution for puzzle Total no. of words of 5 letters formed by letters of the word BOMBAY. [You should use only 5 letters out of 6 letters of the word BOMBAY] solution given is .................. ..... (1st case all the letters are different total no. of words of this category = 5*4*3*2*1 = 120 2nd case B comes twice and 3 others total no. of words of this category = 5*4*3*2*1/2*1 = 60 so total words = 120+60 = 180) I think answer should be 360 as solution should be 1st case all the letters are different total no. of words of this category = 5*4*3*2*1 = 120 2nd case B comes twice and 3 others total no. of words of this category = 4*(5*4*3*2*1/2*1) = 60*4=240 because there will be 4 combinations of such words. so total words = 120+240 = 360 pls excuse me if I am wrong. Next Question: Place the numbers 1 through 100 on a 10x10 square table so that all consecutive numbers will be on adjacent (right, left, up, down) squares. What will be the maximum value of the sum of the 10 numbers along a diagonal of this table?

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