The balls are stacked directly on top of one another, so the limiting measurement is their combined height, not their volume.

Each ball has a diameter of \(2 \times 8 = 16\) cm.

Stacked height of \(6\) balls: \(6 \times 16 = 96\) cm

The container height is \(71\) cm, which is less than \(96\) cm, so the balls do not fit inside the container.

No

The depth of cola is \(\frac{ 2 }{ 3 }\) of the full depth. The cola fills a smaller cone that is similar to the whole cup, so the volume scales by the cube of the length ratio.

\(\left(\frac{ 2 }{ 3 }\right)^3 = \frac{ 8 }{ 27 }\)

The fraction of the cola left is \(\frac{ 8 }{ 27 }\), and Ben said \(\frac{ 4 }{ 9 }\).

No

Substitute \(r = 10\) into the formula you have been given. Square the radius before multiplying by \(4\):

\(A = 4 \times \pi \times 10^2 = 4 \times \pi \times 100 = 400\pi\)

\(400\pi\) cm²

The correct method uses the radius, cubed, times the hemisphere fraction \(\frac{2}{3}\):

\(V = \frac{2}{3} \times \pi \times 12^{3} = 1152\pi\) cm³

Checking the shown working against that, line by line:

First mistake: kept the full-sphere fraction, 4/3, instead of the hemisphere fraction, 2/3 (missed the ÷2 for a hemisphere).

Second mistake: squared instead of cubing (used power 2, not power 3).

The formula uses the radius, and the sphere has been measured right across, so halve the diameter first:

\(r = 18 \div 2 = 9\) cm

Now cube the radius, then multiply:

\(r^3 = 9^3 = 729\)

\(V = \frac{4}{3} \times \pi \times 729\)

Using the \(\pi\) key on a calculator, \(V = 3053.6\) cm³ to 1 decimal place.

The whole ball first:

\(V = \frac{4}{3} \times \pi \times 6^3 = \frac{4}{3} \times \pi \times 216 = 288\pi\) cm³

A hemisphere is exactly half of that:

\(288\pi \div 2 = 144\pi\)

\(144\pi\) cm³

Half the *curved* surface of the ball is only part of the answer. Cutting the ball open makes a new flat face — a circle — and that circle counts too.

The curved part is half the sphere's surface:

\(\frac{1}{2} \times 4\pi r^2 = 2\pi \times 36 = 72\pi\) cm²

The flat face is a circle of radius \(6\) cm:

\(\pi r^2 = 36\pi\) cm²

\(72\pi + 36\pi = 108\pi\)

\(108\pi\) cm²

The radius, the vertical height and the slant height make a right-angled triangle inside the cone, with the slant height as the hypotenuse.

\(8^2 + h^2 = 17^2\)

\(64 + h^2 = 289\)

\(h^2 = 289 - 64 = 225\)

\(h = \sqrt{ 225 } = 15\) cm

The formula uses the vertical height from part (a), not the slant height. Using \(17\) here is the commonest error in this topic.

\(V = \frac{1}{3} \times \pi \times 8^2 \times 15\)

\(V = \frac{1}{3} \times \pi \times 64 \times 15 = \frac{1}{3} \times 960\pi\)

\(V = 320\pi\) cm³

Each of the cube's 6 faces has area \(x^2\), so its total surface area is \(6x^2\).

Substitute \(r = 9\) into the formula for the sphere:

\(4\pi r^2 = 4\pi \times 9^2 = 324\pi\)

The two surface areas are equal:

\(6x^2 = 324\pi\)

Divide both sides by 6:

\(x^2 = 54\pi\)

Take the square root of both sides:

\(x = \sqrt{ 54\pi }\).

First find the full volume of the ball, using the radius.

\(V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \pi \times 23^3 \approx 50965.0\) cm³

Convert to litres: \(50965.0 \div 1000 \approx 51.0\) litres

The ball is not being filled from empty — it already has \(20.9\) litres of air inside, so this must be subtracted before dividing by the rate:

Volume still needed: \(51.0 - 20.9 \approx 30.1\) litres

Extra time to fill: \(30.1 \div 2 \approx 15.1\) minutes

\(15.1\) minutes is more than \(11\) minutes, so the answer is No.

First find the volume of water poured in. It is poured at \(39.2\) cm³ per second for \(23\) seconds:

\(39.2 \times 23 \approx 901.6\) cm³

Next find the full capacity of the bowl. A hemisphere is half of a sphere of the same radius:

\(\frac{1}{2} \times \frac{4}{3} \times \pi \times 9^3 \approx 1526.8\) cm³

Now write the water poured as a percentage of that capacity:

\(901.6 \div 1526.8 \times 100 \approx 59.1\%\)

\(59.1\%\) is above \(55\%\), so the water does not fill less than \(55\%\) of the bowl.

No