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there are 5 boxes each of which contains coins. all the
boxes except one box contains coins whose weight is 100
grams.(each coin weight is 100gm). only one box contains
coins tat weigh 90gm each. u r given a digital balance. in
how many trys can u find out which box contains 90gm coins?
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Answer Posted / paul
Minimum is 1 try and maximum is 2 tries. That is, assuming that all boxes contain same quantity of coins each.
verify...
First, lets label the boxes a,b,c,d and e.
Then measure and balance a+b if equals to c+d.
Condition 1
If the 4 boxes balanced, then box e has the 90gm coins. Thus, answering promptly with only 1 try (Minimum).
Condition 2
If (a+b) < (c+d), then it can be observed that either box a or b has the 90 gm coins. Thus, balancing again box a vs b will give you the box with the 90 gm coins, that is, with the lower weight between the 2 boxes. Thus, obtaining the answer with 2 tries.
Condition 2
If (a+b) > (c+d), then it can be observed that either box c or d has the 90 gm coins. Thus, balancing box c against box d will give you the box with the 90 gm coins, that is, with the lower weight between the 2 boxes. Thus, obtaining the answer with 2 tries.
| Is This Answer Correct ? | 2 Yes | 6 No |
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