You have 9 balls...

sedm1000

sedm1000

sedm1000

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Joined
19 Oct 2002
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Why not make a collection of such puzzles with some model answers... for rainy spring days.


Q1

You have 9 balls, with one weighing more than the others, but you can't tell by just picking it up. Using a balance no more than two times, how would you find the ball that outweighs the others?
 
sedm1000 said:
Why not make a collection of such puzzles with some model answers... for rainy spring days.


Q1

You have 9 balls, with one weighing more than the others, but you can't tell by just picking it up. Using a balance no more than two times, how would you find the ball that outweighs the others?
Click to expand...

Drop them all from same height and see which one bounces differently to the rest ?
 
Split into 3 groups, weigh two of them. You've now identified the heavier of the 3. Weigh two balls from that 3 and you've identified the heaviest ball.

Win \o/

EDIT: Clarity...

Split into 3 groups of 3: If they are even, the set NOT on the scales contains the heaviest ball. If not even, then duh.
Then weighing 2 of the heaviest set: Same principle as above.
 
1 - Split the balls into three groups of three balls each.

2 - First use of the scale: Weigh any two groups against each other.

3 - If the groups weigh the same then the heavier ball is in the third group, otherwise it is in the group that weighs more.

4 - We now know that the heavy ball is one of three.

5 - Second use of the scale: Using the group of balls that we know contains the heavy ball, weight any two balls against each other.

6 - If one of the balls is heavier then we have our answer. If the balls weigh the same then the third ball is the heavy one and we again have our answer.

:p:D
 
Well, put eight balls on the balance, four on each side, and if both sides balance exactly, the one left over must be the heaviest.
If one side was heavier, select one at random from that set of four, and replace it with the ninth ball.
Now, if both sides balance exactly, you have found the heaviest ball.
If that still doesn't work, throw all the balls in turn at the windows and see which one makes the biggest hole.
 
Dj_Jestar said:
Split into 3 groups, weigh two of them. You've now identified the heavier of the 3. Weigh two balls from that 3 and you've identified the heaviest ball.

Win \o/

EDIT: Clarity...

Split into 3 groups of 3: If they are even, the set NOT on the scales contains the heaviest ball. If not even, then duh.
Then weighing 2 of the heaviest set: Same principle as above.
Click to expand...

I was going to say this but i'm slow at typing :) :p
 
Q2
You have a three gallon jug and a five gallon jug No marks on either one The goal is to fill the five gallon jug with four gallons of water How is this accomplished?
 
Does anyone remember the 'team building exercise' they sometimes do for people in new jobs (I've done this 1 for 3 jobs i've started at in the past) but can not remember it properly...

Something about 3 planes, only have 3/4 of fuel so can not go around the whole world in 1 trip, and they are carrying a parcel, the parcel has to reach 1 side of the world and do a full trip BUT the plane only has enough fuel to carry them so far before having to turn back to re-fuel...

Anyone know this 1? :s
 
_Gizka_ said:
By watching Die Hard 3?

:p:D
Click to expand...

it is probably a little easier when you haven't got the pressure of having to do it in a few minutes or getting blown up....

I don't remember the exact method from Die Hard but I can remember the principle, let's see

5 gallon full into 3 gallon empty - you now have 2 gallons in 5 gallon jug

empty the 3 gallon and fill with the 2 from the 5, so now 2 in 3

refill the 5 gallon, and use it to fill the 3 gallon, which has one gallon left

5-1=4

qed
 
DreadNiK said:
it is probably a little easier when you haven't got the pressure of having to do it in a few minutes or getting blown up....

I don't remember the exact method from Die Hard but I can remember the principle, let's see

5 gallon full into 3 gallon empty - you now have 2 gallons in 5 gallon jug

empty the 3 gallon and fill with the 2 from the 5, so now 2 in 3

refill the 5 gallon, and use it to fill the 3 gallon, which has one gallon left

5-1=4

qed
Click to expand...

Now im confused :confused:
 
You fill the 3 up and pour it in the 5. You fill the 3 again and pour it in to the 5 to leave you with 1 gallon in the 3. Empty the 5, fill it with the 1 from the 3 and then fill the 3 again. Put this into the 5 and you have 4 gallons.

I didn't google this :)
 
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