m4maths Maths Puzzle

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?

Read Solution (Total 1)

You must see 2 grams difference on the scale to be sure that you have two counterfreit coins. That`s why the worst scenario is :

you have 25 pieces of genuine coins each weighs 10 grams
24 pieces of counterfeit coins each weighs 9 grams
1 pieces of counterfeit coins weighs 11 grams

there are 3 posibilities

1. the first coin you take is counterfeit weighs 9 or grams, you must take 49 without seeing 2 grams difference in the scale. then the last coin in the bag and the coin tou took first is counterfreit for sure. as a result you need to take 49 coins to be sure if you took counterfreit coin in the beginning.

2. the first coin you take is genuine weighs 10 grams. you must take all coins until you will be sure that your coin is genuine because you did`t see 2 grams on the scale. (you will see 0 or 1 on the scale until the end)

The answer is 50 (you have 25 genuine coins, 25 counterfreit coins just one counterfreit coin has different weight, and you cant take it until last pick)
12 years agoHelpfull: Yes(0) No(1)

m4maths Other 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. 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?