We have two red, two green and two yellow balls. For each color, one ball is heavy and the other is light. All heavy balls weigh the same. All light balls weigh the same. How many weighings on a beam balance are necessary to identify the three heavy balls?
May be three for identification of all three heavier.
three chances to identify
We have 2 Yellow(Y), 2 Reds® and 2 Greens(G)
Compare 2 Y’s.Now we will be knowing which of the balls is heavier
Compare the heavier Y to any of the Red®. If equal, it is heavier R ball, else the other one is heavier
Similar for the Blue ball
Acc to your solution, the weighing counts are going to be 4.
i am sensing answer is 2 but how?
No it is 3
1st - y and Y
2nd Y and R
3rd Y and B
We can optimise the solution to two tries
How ?? please can you share the approach !!
Take one yellow and a red on one side and the second yellow and a green one the other side.
Now watch for the results and do the rest.
Two weighings are sufficient .
weigh one red and one green ball against one yellow ball and the other green ball .if the weights are equal,then the red and yellow differ in the weight.A weighing between these two balls allows us to deduce the weights of all other balls.if the red-green combination was heavier than the yellow-green combination in the first weighing,then the green ball in the red-green combination is certainly heavy and the other gree is light .now take red from the red-green combination and yellow from the yellow-green combination .weigh these together against the remaining red and remaining yellow.the only interesting case is when this weighing is “equal”.then,the red from the red-green combinationmust be heavy and yellow from the yellow-green combination must be light.