Higher June 2017 Paper 1 Q19
19. Water in a tank is \(x\) cm deep.
The tank has a tap which is slightly open.
The rate, \(R\) cm\(^3\) per minute, at which the volume of water decreases is proportional to the square root of \(x\).
When the water is 81 cm deep, the volume of water decreases at the rate of 900 cm\(^3\) per minute.
(a) Find the rate at which the volume of water decreases when the depth of the water is 9 cm. [5]
(b) Find the depth of the water when the rate at which the volume of water decreases is 400 cm\(^3\) per minute. [2]
| Scheme | Marks | Comments |
|---|---|---|
| \(R \propto \sqrt{x}\) or \(R = k\sqrt{x}\) | M1 | |
| \(900 = k\sqrt{81}\) | M1 | Implies first M1 |
| \(k = 100\) | A1 | |
| \((R =)\ 100\sqrt9\) | M1 | FT ‘their 100’ |
| \((R =)\ 300\) (cm\(^3\) per min) | A1 | FT |
Alternative method using proportions
| Scheme | Marks | Comments |
|---|---|---|
| Sight of \(\dfrac{900}{\sqrt{81}}\) or \(\sqrt{81}\) (:) 900 Could be implied in later working | B1 | |
| \(\dfrac{900}{\sqrt{81}} = \dfrac{R}{\sqrt9}\) or compares \(\sqrt{81}\) or 9 (:) 900 with \(\sqrt9\) or 3 (:) 300 | M1 | |
| \(\dfrac{900}{\sqrt{81}} \times \sqrt9 = R\) seen or implied | M1 | |
| \(100 \times \sqrt9\ (= R)\) seen or implied | M1 | |
| \((R =)\ 300\) (cm\(^3\) per min) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(400 = 100\sqrt{x}\) or \(\dfrac{1}{100} = \dfrac{\sqrt{x}}{400}\) or equivalent or \(\left(\sqrt{x} =\right) \dfrac{400}{100}\) or \(\left(\sqrt{x} =\right) 4\) | M1 | FT ‘their 100’ |
| \((x =)\ 16\) (cm) | A1 | Not from wrong working |
| (7) |