Higher November 2022 Paper 1 Q18
18 A road has three sections, D, E and F.
The lengths of D, E and F are in the ratios
D : E = 3 : 5 E : F = 7 : 4
What fraction of the length of the road is section D? [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – combining the ratios | ||
| 21 : 35 and 35 : 20 or (3 : 5 and) \(5 : \dfrac{20}{7}\) or \(\dfrac{21}{5} : 7\) (and 7 : 4) | M1 | oe making the E term common allow as fractions with a common denominator eg \(\dfrac{21}{35}\) and \(\dfrac{20}{35}\) |
| 21 : 35 : 20 or \(3 : 5 : \dfrac{20}{7}\) or \(\dfrac{21}{5} : 7 : 4\) or \(\dfrac{21/5}{76/5}\) or \(\dfrac{3}{76/7}\) | M1dep | oe allow as integers 21 and 35 and 20 or as fractions with a common denominator eg \(\dfrac{21}{35}\) and \(\dfrac{35}{35}\) and \(\dfrac{20}{35}\) |
| \(\dfrac{21}{76}\) | A1 | |
| Alternative method 2 – based on D | ||
| \(\dfrac{5(\text{D})}{3}\) and \(\dfrac{20(\text{D})}{21}\) | M1 | oe |
| \(\dfrac{21(\text{D})}{21} + \dfrac{35(\text{D})}{21} + \dfrac{20(\text{D})}{21}\) or \(\dfrac{76(\text{D})}{21}\) | M1dep | oe with common denominator |
| \(\dfrac{21}{76}\) | A1 | |
| Alternative method 3 – based on E | ||
| \(\dfrac{3(\text{E})}{5}\) and \(\dfrac{4(\text{E})}{7}\) | M1 | oe |
| \(\dfrac{21(\text{E})}{35} + \dfrac{35(\text{E})}{35} + \dfrac{20(\text{E})}{35}\) or \(\dfrac{76(\text{E})}{35}\) | M1dep | oe with common denominator |
| \(\dfrac{21}{76}\) | A1 | |
| Alternative method 4 – based on F | ||
| \(\dfrac{21(\text{F})}{20}\) and \(\dfrac{7(\text{F})}{4}\) | M1 | oe |
| \(\dfrac{21(\text{F})}{20} + \dfrac{35(\text{F})}{20} + \dfrac{20(\text{F})}{20}\) or \(\dfrac{76(\text{F})}{20}\) | M1dep | oe with common denominator |
| \(\dfrac{21}{76}\) | A1 | |
Additional guidance
Allow unrounded decimal values throughout