Let \(A, B, C, D\) and \(E\) weights are \(a, b, c, d\) and \(e\).
According to the ques,
\(\frac {a+b+c}{3}-\frac {a+b+c+d}{4}=x\) ……… (i)
\(\frac {a+b+c+e}{4}-\frac {a+b+c}{3}=2x\) ……… (ii)
Eq (i) + Eq (ii),
\(\frac {a+b+c}{3}-\frac {a+b+c+d}{4}+\frac {a+b+c+e}{4}-\frac {a+b+c}{3}=x+2x\)
\(\frac {e-d}{4}=3x\)
\(e-d = 12x\)
\(12x=12\)
\(x=1\)
So, the correct option is (C): \(1\)