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 / @m@n
In 1 try..
if we take 1 coin from first box..
2 coins from 2nd box..
3 from third.. 4 4m 4th box and 5 coins 4m 5th box..
after taking the coins ... weight those coins...nw total
num of coins we take is 15...
nw if the weight will b 1490 gm thn 90gm coins in first
box.. if d weight 'll b 1480gm thn 90gm coins in 2nd box...
and similarly in case of 3,4 and 5 box...
| Is This Answer Correct ? | 69 Yes | 1 No |
Answer / paul
If the weighing tool is a digital measurement (not a balance instrument). My answer is 1 try.
My solution:
First you must collect the necessary balls from each box in an order like this:
collect 1 ball from box 1
collect 2 balls from box 2
collect 3 balls from box 3
collect 4 balls from box 4 &
collect 5 balls from box 5
Total balls you have collected is 15 balls.
To determine which box contains the 90 grm balls, you must
weigh all the 15 balls in the digital measurement. If the
result of measurement is:
5*90 + 100*10 then the answer is the 5th box
4*90 + 100*11 then the answer is the 4th box
3*90 + 100*12 then the answer is the 3rd box
2*90 + 100*13 then the answer is the 2nd box
1*90 + 100*14 then the answer is the 1st box
| Is This Answer Correct ? | 11 Yes | 3 No |
Answer / nelwin
min 1
remove any 1 box.and weigh 2 boxes each on digital bal..if
weights r same then removed box is 90gm box..
max 2
repeat same..if weights differ then remove the boxes wid
more weigs..dats is wid 100gms..again weigh each box of
separately..u wil get it..
| Is This Answer Correct ? | 9 Yes | 5 No |
Answer / bala
min 2,
weight 2 boxes on each side, if weights are same then
remaining box is 90gms. if not take the boxes from less
balanced side and weight them again.
| Is This Answer Correct ? | 7 Yes | 6 No |
Answer / 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 |
Answer / sameer k
answer :::min 2 and max 3
calc weight of 3 box, there are 2 condition
condition 1(min 2)>>if the weight of 3 boxes is (90*3)270
then, take any 1 box from other 2boxes
calc weight is weight
condition 2(max 3)>>if weight of 3 boxes is greater than 270
that means among 3 boxes there is a box whose
weight is greater than 90, then calc the weight of any 2
from them if egals 180 then third box weight is 100gm
| Is This Answer Correct ? | 6 Yes | 11 No |
what is the main reason to ask this type question in your interview sir?
A number. if it is divided by 2 reminder 1, if it is divided by 3 reminder 1, if it is divided by 4 reminder 1, if it is divided by 5 reminder 1, if it is divided by 6 reminder 1, if it is divided by 7 reminder 1, if it is divided by 8 reminder 1, if it is divided by 9 reminder 1, if it is divided by 10 reminder 1, and if it is divided by 11 reminder 0. its greater then 100. which is that number. this puzzle has been appeared in one of written test(aptitude for 20 marks).
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