There are four people who need to cross a rickety bridge in 17 minutes. Because of its dilapidated condition, the bridge can only support up to 2 people at any given time. In addition, since it is pitch black out and there is only one lantern between the four people, anybody crossing the bridge must carry the lantern (or if two people are crossing the bridge at the same time, then one of them must be carrying the lantern). Someone must always bring the lantern back to the people who haven’t crossed the bridge yet, so that when they cross, they don’t fall through the holes in the bridge. You must take into account the additional time it will take for them to cross back.
The four people can cross the bridge at the rate of 1, 2, 5, and 10 minutes, each way.
There are no tricks: you can’t carry anyone, or throw the lantern or drop it on the bridge, or swim across.
Day 1: Give worker 1 bar
Day 2: Take back 1 bar and give 2 piece bar
Day 3: Give 1 bar [Making 2 + 1 = 3]
Day 4: Take back 1 & 2 piece bar and give 4 piece bar
Day 5: Give 1 piece bar [Making 4 + 1 = 5]
Day 6: Take back 1 piece and give 2 piece
[Making 5 – 1 + 2 = 6]
Day 7: Give 1 piece bar
Mark the boxes as box1,box2,box3………box10
then from 1st box we have to take 1 ball,from 2nd box 2 ball……….from 10th box we kept 10 balls. and weight the total ball….
so if total nos ball is :n(n+1)/2 (n= total nos of box that is 10)=55
if all the ball is 10 gm.then total weight will be=550gm.
suppose total weight is 546gm.
so the difference is(550-546=4gm).
then the d defective ball is in box no:4.
similarly we can try for all possible box difference gives us the box number which contains the defective balls
case 2 : A > B