Volume and mass ratio questions with answers

Questions

1. Question 1Find a volume from two lengths

   Two similar cylinders have heights $5$ cm and $9$ cm. The volume of the smaller cylinder is $135$ cm³. Find the volume of the larger cylinder.

   Show the worked solution

   1. Step 1: Volumes of similar solids are in the ratio of the cubes of their lengths$\dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^3$
   2. Step 2: Put the unknown volume on top$\mathstrut \dfrac{V}{135} = \left(\dfrac{9}{5}\right)^3$$V = \left(\dfrac{9}{5}\right)^3 \times 135$$V = 787.32$ cm³
2. Question 2Find a length from two volumes

   Two similar cylinders have volumes $216$ cm³ and $512$ cm³. The height of the larger cylinder is $8$ cm. Find the height of the smaller cylinder.

   Show the worked solution

   1. Step 1: Lengths are in the ratio of the cube roots of the volumes$\dfrac{h_1}{h_2} = \sqrt[3]{\dfrac{V_1}{V_2}}$
   2. Step 2: Put the unknown height on top$\mathstrut \dfrac{h}{8} = \sqrt[3]{\dfrac{216}{512}}$$\dfrac{h}{8} = \dfrac{3}{4}$
   3. Step 3: Multiply both sides$h = \dfrac{3}{4} \times 8$$h = 6$ cm
3. Question 3Masses of similar objects

   Two similar containers are made of the same material. Their heights are $5$ cm and $6$ cm. The larger container has a mass of $54$ g. Find the mass of the smaller container.

   Show the worked solution

   1. Step 1: Same material, so the masses are in the ratio of the volumes$\dfrac{m_1}{m_2} = \left(\dfrac{l_1}{l_2}\right)^3$
   2. Step 2: Put the unknown mass on top$\mathstrut \dfrac{m}{54} = \left(\dfrac{5}{6}\right)^3$$m = \left(\dfrac{5}{6}\right)^3 \times 54$$m = 31.25$ g
4. Question 4From surface areas to volumes

   Two similar cylinders have total surface areas $450$ cm² and $1250$ cm². The volume of the smaller cylinder is $729$ cm³. Find the volume of the larger cylinder.

   Show the worked solution

   1. Step 1: From the areas, find the length ratio (square root)$\mathstrut \dfrac{l_1}{l_2} = \sqrt{\dfrac{450}{1250}}$$\dfrac{l_1}{l_2} = \dfrac{3}{5}$
   2. Step 2: Cube the length ratio for the volumes$\mathstrut \dfrac{V}{729} = \left(\dfrac{5}{3}\right)^3$$V = \left(\dfrac{5}{3}\right)^3 \times 729$$V = 3375$ cm³
5. Question 5Water in a cone

   An inverted cone of height $12$ cm has a capacity of $216$ cm³. It contains $8$ cm³ of water. Find the depth of the water.

   Show the worked solution

   1. Step 1: The water forms a cone similar to the whole cone$\dfrac{d}{H} = \sqrt[3]{\dfrac{\text{volume of water}}{\text{volume of cone}}}$
   2. Step 2: Substitute$\mathstrut \dfrac{d}{12} = \sqrt[3]{\dfrac{8}{216}}$$\dfrac{d}{12} = \dfrac{1}{3}$
   3. Step 3: Multiply both sides$d = \dfrac{1}{3} \times 12$$d = 4$ cm
6. Question 6Models: a scale of 1 : n

   A model of a statue is made to a scale of $1 : 50$. The volume of the model is $250$ cm³. Find the volume, in m³, of the real statue.

   Show the worked solution

   1. Step 1: Volumes: the ratio of lengths cubedVolume ratio $= 1 : 50^3$$= 1 : 125000$
   2. Step 2: Real volume in cm³, then in m³ (1 m³ = 1 000 000 cm³)Real volume $= 250 \times 125000$$= 31250000$ cm³$\dfrac{31250000}{1\,000\,000} = 31.25$ m³