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Interesting math problems in the ninth grade
Let the length of the left arm and the right arm of the balance be L 1 and L2, respectively.

If the mass of large clay is x, the mass of small clay is 30-X.

By the lever originally has:

L 1*X=27*L2 ……①

L 1*(30-X)=8*L2 ……②

L 1/L2=27/X comes from ①.

Substituting ② into (30-X)*27/X=8, we get X= 162/7.

So the mass of the big ball is 162/7g, and the mass of the small ball is 30- 162/7 = 48/7g.