Higher November 2020 Paper 3 Q17
17 Here are two rectangles.
Not drawn accurately
The area of the shaded rectangle is \(\dfrac{1}{4}\) the area of the large rectangle.
Work out the value of \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2(12 - x)\) or \(24 - 2x\) or \(12(x + 2)\) or \(12x + 24\) or \(12x + 2x\) or \(14x\) or \(2x + x^2 + x(12 - x)\) or \(2x + x^2 + 12x - x^2\) | M1 | oe correct area of small rectangle or large rectangle or unshaded section may be seen on diagram |
| \(\dfrac{12(x + 2)}{4} = 2(12 - x)\) or \(12x + 2x = 6(12 - x)\) | M1dep | oe equation eg \(3(x + 2) = 2(12 - x)\) \(3x + 6 = 24 - 2x\) \(12(x + 2) = 8(12 - x)\) \(12x + 24 = 96 - 8x\) |
| \(3x + 2x = 24 - 6\) or \(14x + 6x = 72\) | M1dep | oe equation with brackets expanded and terms collected eg \(5x = 18\) \(12x + 8x = 96 - 24\) \(20x = 72\) |
| \(\dfrac{18}{5}\) or \(3\dfrac{3}{5}\) or 3.6 | A1 | oe |
Additional guidance
| \(3x + 6\) | M1 |
| Trial and improvement with \(x = 3.6\) chosen | M1M1M1A1 |
| Trial and improvement without \(x = 3.6\) chosen | M0M0M0A0 |