2022 H2 Maths Paper 2 Q2 — Worked Solution

Part (a)

1 mark

The volume of the container is base area $\text{×}$ height $\text{=}0.9h\ m^{3}$ . Since $1\ m^{3} = 1000$ litres, this is $900h$ litres. Water is poured in at a constant rate of $k$ litres per second and fills the container in $72$ seconds, so

k \times 72 = 900h

\Rightarrow k = \frac{900h}{72} = 12.5h\ \ \text{(shown)}

Part (b)

3 marks

Let $T$ be the time taken to fill the container. Water is now poured in at rate $kt$ litres per second, so the volume poured in by time $T$ is

\int_{0}^{T}kt\, dt = \left\lbrack \frac{kt^{2}}{2} \right\rbrack_{0}^{T} = \frac{kT^{2}}{2}

Setting this equal to the container’s volume, $900h$ , and substituting $k = 12.5h$ ,

\frac{kT^{2}}{2} = 900h

\Rightarrow \frac{12.5hT^{2}}{2} = 900h

\Rightarrow 6.25T^{2} = 900\ \ \left( \text{dividing by}\ h,\ \text{since}\ h \neq 0 \right)

\Rightarrow T^{2} = 144

\Rightarrow T = 12\ \ \left( \text{taking the positive root, since}\ T > 0 \right)

To check your answer: with

T = 12

,

\frac{kT^{2}}{2} = \frac{144k}{2} = 72k = 72 \times 12.5h = 900h

, matching the container’s volume ✓

$\therefore$ the container fills in $T = 12\ s$ .

Part (c)

3 marks

Water is now poured in at rate $(kt + 25)$ litres per second and fills the container in $10$ seconds, so the volume poured in is

\int_{0}^{10}(kt + 25)\, dt = \left\lbrack \frac{kt^{2}}{2} + 25t \right\rbrack_{0}^{10} = 50k + 250

Setting this equal to $900h$ and substituting $k = 12.5h$ ,

50k + 250 = 900h

\Rightarrow 50(12.5h) + 250 = 900h

\Rightarrow 625h + 250 = 900h

\Rightarrow 250 = 275h

\Rightarrow h = \frac{250}{275} = \frac{10}{11}

To check your answer: with

h = \frac{10}{11}

,

k = 12.5 \times \frac{10}{11} = \frac{125}{11}

, so

50k + 250 = \frac{6250}{11} + \frac{2750}{11} = \frac{9000}{11}

, and

900h = 900 \times \frac{10}{11} = \frac{9000}{11}

— these match ✓

$\therefore$ $h = \frac{10}{11}\ m$ .

2022 H2 Maths Paper 2, Question 2

Part (a)

Part (b)

Part (c)

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