I have 8 billiards balls. They all have he same weight, except one of them is heavier than the rest. I have a balance scale that just tips to one side when that side is heavier, but doesn’t give me readings/measurements. What is the least number of times I can weigh the balls in order to find the heaviest ball? Explain
2006-11-08 06:43:04 · 5 answers · asked by sharbel_karam 2 in Science & Mathematics ➔ Mathematics
Three weighings.
Put 4 on each side and weigh
Take the 4 from the side that is heavier and
Put 2 on each side and weigh.
Take the 2 from the side that is heavier and
Put 1 on each side and weigh.
Take the one that is heaver. It is the heaviest one.
2006-11-08 06:47:24 · answer #1 · answered by DanE 7 · 0⤊ 0⤋
You can do it in 2 weighings. Put 3 balls on each side of the balance, and set two aside. If the pans balance, then your heavier ball is one of the two on the table. Put those two on the balance and you'll know which is heavier.
If your three balls don't balance, then you'll know which group of 3 has the heavier ball in it. Put any two of those on the balance, and you'll know which of the three is the heavier one.
2006-11-08 14:46:55 · answer #2 · answered by hcbiochem 7 · 2⤊ 0⤋
I'm not sure that it's the minimun. But 2^3 = 8.
So:
1. 4 vs 4, pick heaviest side
2. 2 vs 2, same
3. 1 vs 1, pick heaviest ball.
2006-11-08 14:47:41 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
3
1) put 4 balls on each side.
2) take the 4 heavier balls, split them & put 2 on each side.
3) take the 2 heqavy bals, put 1 on each side.
2006-11-08 14:54:17 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
do that hcbiochem said. But if the 6 balls are not balanced, remove one ball of each plate and when stay balanced wiil have the ball
solution 2
2006-11-08 15:06:10 · answer #5 · answered by Anonymous · 0⤊ 0⤋