Identifying the Heavier Ball

The video tutorial presents a problem involving nine identical balls, one of which is slightly heavier. The challenge is to identify the heavier ball using a scale with the minimum number of weighings. The solution is explained in detail, demonstrating that it can be achieved in two weighings by systematically ruling out 2/3 of the balls in each step. The tutorial also discusses the general principle behind the solution and explores variations of the problem, such as scenarios with different numbers of balls.

What is the main challenge in identifying the heavier ball among the nine?

• The balls are made of different materials.

• The balls are identical in appearance.

• The balls are slightly different sizes.

• The balls are different colors.

Tags

CCSS.8.EE.C.7B

How many weighings are needed to identify the heavier ball?

• Three

• Four

• Two

• One

Tags

CCSS.7.EE.B.3

What fraction of the balls can be ruled out after the first weighing?

• 3/4

• 1/2

• 1/3

• 2/3

In the first weighing, what do you do if the two groups of three balls balance?

• Conclude that the heavier ball is in the unbalanced group.

• Conclude that the heavier ball is in the balanced group.

• Weigh one ball from each group again.

• Weigh the remaining three balls against each other.

Tags

CCSS.8.EE.C.7B

What is the next step if one side is heavier in the first weighing?

• Weigh the heavier side again.

• Weigh the lighter side again.

• Weigh the remaining balls.

• Conclude the heavier ball is in the lighter side.

In the second weighing, what do you do if the two balls balance?

• The heavier ball is the one on the left.

• The heavier ball is in the balanced group.

• The heavier ball is the one not weighed.

• The heavier ball is the one on the right.

What is the outcome if one ball goes down in the second weighing?

• The ball that goes up is heavier.

• The ball that goes down is heavier.

• The ball not weighed is heavier.

• Both balls are of equal weight.

What is a general principle derived from this problem?

• You can only weigh two balls at a time.

• You can rule out 1/3 of the balls each time.

• You need more than two weighings for any number of balls.

• You can rule out 2/3 of the balls each time.

How many weighings would be needed for 27 balls?

• Four

• Five

• Three

• Two

What is a potential variation of the problem with eight balls?

• Breaking the balls into halves.

• Having two heavier balls.

• Using a different type of scale.

• Using the same method as with nine balls.