Mensuration Questions and Answers

Answer: D

Work out the water needed and the rate at which it arrives.

\[ V = 200 \times 150 \times 2 = 60{,}000 \text{ m}^{3} \]

\[ \text{pipe cross-section} = 1.5 \times 1.25 = 1.875 \text{ m}^{2} \]

\[ \text{flow} = 1.875 \times 20{,}000 = 37{,}500 \text{ m}^{3}\text{/h} \]

\[ t = \frac{60{,}000}{37{,}500} = 1.6 \text{ h} = 96 \text{ minutes} \]

The speed of 20 kmph becomes 20,000 metres per hour, which is the length of water column delivered each hour.

Answer: B

The metal is what lies between the outer box and the hollow inside it. The box is open, so only the depth loses one thickness, while the length and breadth lose two.

\[ V_{\text{outer}} = 50 \times 40 \times 23 = 46{,}000 \text{ cm}^{3} \]

\[ V_{\text{inner}} = (50-6)(40-6)(23-3) = 44 \times 34 \times 20 = 29{,}920 \text{ cm}^{3} \]

\[ V_{\text{metal}} = 46{,}000 - 29{,}920 = 16{,}080 \text{ cm}^{3} \]

\[ \text{weight} = 16{,}080 \times 0.5 = 8040 \text{ g} = 8.04 \text{ kg} \]

Answer: D

The immersed cube displaces its own volume of water, which spreads over the vessel's base.

\[ V_{\text{cube}} = 15^{3} = 3375 \text{ cm}^{3} \]

\[ \text{base area} = 20 \times 15 = 300 \text{ cm}^{2} \]

\[ \text{rise} = \frac{3375}{300} = 11.25 \text{ cm} \]

The cube's own 15 cm edge is not the rise: the water spreads over a base twice the cube's footprint, so it climbs less far.

Answer: C

Pouring conserves volume, so equate the cone's with the cylinder's.

\[ V_{\text{cone}} = \frac{1}{3}\pi (12)^{2}(50) = \frac{1}{3}\pi (7200) = 2400\pi \text{ cm}^{3} \]

\[ V_{\text{cyl}} = \pi (10)^{2} h = 100\pi h \]

\[ 100\pi h = 2400\pi \quad\Longrightarrow\quad h = 24 \text{ cm} \]

The \( \pi \) cancels, so no approximation is needed. The one-third factor is what keeps the liquid from filling the wider cylinder to 50 cm.

Answer: D

Melting conserves volume, so divide the cylinder's by one bullet's. The bullet's radius is \( 1.5 \div 2 = 0.75 \) cm.

\[ V_{\text{cyl}} = \pi (6)^{2}(28) = 1008\pi \text{ cm}^{3} \]

\[ V_{\text{bullet}} = \frac{4}{3}\pi (0.75)^{3} = \frac{4}{3}\pi (0.421875) = 0.5625\pi \]

\[ n = \frac{1008}{0.5625} = 1792 \]

The diameter must be halved before cubing - using 1.5 as the radius would cut the count eightfold.

Answer: D

Surface area varies as the square of the radius.

\[ \frac{S'}{S} = \left(\frac{1.5r}{r}\right)^{2} = 2.25 \]

\[ \text{increase} = 2.25 - 1 = 1.25 = 125\% \]

The volume, which varies as the cube, would rise by \( 1.5^{3} - 1 = 2.375 \), or 237.5% - a much larger jump for the same change in radius.

Answer: C

Melting conserves volume, and both cones share the same base radius, so their volumes add through the heights.

\[ V = \frac{1}{3}\pi (2.1)^{2}(4.1 + 4.3) = \frac{1}{3}\pi (4.41)(8.4) \]

\[ = \frac{1}{3}\pi (37.044) = 12.348\pi \text{ cm}^{3} \]

Setting this equal to the sphere's volume:

\[ \frac{4}{3}\pi R^{3} = 12.348\pi \quad\Longrightarrow\quad R^{3} = 9.261 \]

\[ R = 2.1 \text{ cm} \quad\Longrightarrow\quad d = 4.2 \text{ cm} \]

The question asks for the diameter, so the radius must be doubled.

Answer: C

The liquid's volume is conserved, so divide the bowl's by one bottle's. The bottle's radius is \( 3 \div 2 = 1.5 \) cm.

\[ V_{\text{bowl}} = \frac{2}{3}\pi (9)^{3} = \frac{2}{3}\pi (729) = 486\pi \text{ cm}^{3} \]

\[ V_{\text{bottle}} = \pi (1.5)^{2}(4) = 9\pi \text{ cm}^{3} \]

\[ n = \frac{486\pi}{9\pi} = 54 \]

A hemisphere is two thirds of \( \pi r^{3} \), not one half of the sphere formula misapplied - the \( \tfrac{2}{3} \) is what makes the count come out whole.

Answer: D

An arc is the fraction of the circumference that its central angle is of a full turn.

\[ \text{arc length} = \frac{z^{\circ}}{360^{\circ}} \times \text{circumference} = \frac{z}{360} \times 2\pi r \]

The same fraction applied to the area of the circle gives the sector's area instead:

\[ \text{sector area} = \frac{z}{360} \times \pi r^{2} \]

Length comes from circumference and area from area - matching the two is what the options are testing.

Answer: C

\[ \text{arc} = \frac{\theta}{360} \times 2\pi r \]

\[ = \frac{140}{360} \times 2 \times \frac{22}{7} \times 21 \]

\[ = \frac{7}{18} \times 132 = \frac{924}{18} = 51.33 \approx 51.31 \text{ cm} \]

The whole circumference is 132 cm, and \( 140^{\circ} \) is a little under four tenths of a turn - a quick check on the size of the answer.