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How many four coloured balls are there in a bag, such that all but $12$ are blue, all but $21$ are black, all but $20$ are red and all but $19$ are green

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Hint: You have 5 equations and 5 variables: $$ x+12=T \\ y+21=T\\ z+20=T\\ a+19=T\\ x+y+z+a=T $$

Solve for T by plugging in the first four to the last.

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Last step can be thought of as adding the first four equations 'column-wise' in which case we have $T + 72 = 4T$. – muzzlator Apr 2 '13 at 12:20

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