The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is :

Let the length of each pipe be $$l$$ inches
Then, volume of water in thinner pipe :
$$\eqalign{ & = \left[ {\pi \times {{\left( {\frac{1}{2}} \right)}^2} \times 1} \right] \text{cu.inch} \cr & = \left( {\frac{{\pi l}}{4}} \right)\text{cu.inch} \cr} $$
Volume of water in thinker pipe :
$$\eqalign{ & = \left( {\pi \times {3^2} \times l} \right)\text{cu.inch} \cr & = \left( {9\pi l} \right)\text{cu.inch} \cr} $$
∴ Required number of pipes :
$$\eqalign{ & = \frac{{9\pi l}}{{\left( {\frac{{\pi l}}{4}} \right)}} \cr & = 36 \cr} $$