3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?

Efficiency of 2 men = Efficiency of 1 women
1 man = $$\dfrac{1}{2}$$ woman

3 men = $$\dfrac{3}{2}$$ women

Efficiency of 2 children = Efficiency of 1 men = Efficiency of $$\dfrac{1}{2}$$ women
Efficiency of 1 child = Efficiency of $$\dfrac{1}{4}$$ women

Efficiency of 6 children = Efficiency of $$\dfrac{3}{2}$$ women

Therefore, Total efficiency of 3 men, 4 women and 6 children = $$\dfrac{3}{2} + 4 + \dfrac{3}{2} = 3+4 = 7$$ women
Therefore, 7 women are required to complete the work in 7 days.