Higher June 2017 Paper 1 Q25
25 In an office there are twice as many females as males.
- \(\dfrac{1}{4}\) of the females wear glasses.
- \(\dfrac{3}{8}\) of the males wear glasses.
84 people in the office wear glasses.
Work out the number of people in the office. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – based on a fraction of the number of males | ||
| \(\dfrac{1}{4} \times 2x\ (+)\ \dfrac{3}{8} \times x\) or \(\dfrac{7}{8}x\) where \(x\) is the number of males | M1 | \(\dfrac{1}{4} \times 2\ (+)\ \dfrac{3}{8}\ (\times 1)\) or \(\dfrac{7}{8}\) |
| \(\dfrac{1}{4} \times 2x + \dfrac{3}{8} \times x = 84\) or \(\dfrac{7}{8}x = 84\) or \(7x = 672\) | M1dep | oe \(\dfrac{1}{4} \times 2 + \dfrac{3}{8}\ (\times 1)\) linked to 84 or \(\dfrac{7}{8}\) linked to 84 |
| \(x = 84 \div\) their \(\dfrac{7}{8}\) or \(x = 84 \times\) their \(\dfrac{8}{7}\) or \(x = 96\) | M1dep | oe dep on M1M1 \(84 \div\) their \(\dfrac{7}{8}\) or \(84 \times\) their \(\dfrac{8}{7}\) or 96 |
| 288 | A1 | |
| Alternative method 2 - based on a fraction of the number of females | ||
| \(\dfrac{1}{4} \times y\ (+)\ \dfrac{3}{8} \times \dfrac{y}{2}\) or \(\dfrac{7}{16}y\) where \(y\) is the number of females | M1 | \(\dfrac{1}{4}\ (\times 1)\ (+)\ \dfrac{3}{8} \times \dfrac{1}{2}\) or \(\dfrac{7}{16}\) |
| \(\dfrac{1}{4} \times y + \dfrac{3}{8} \times \dfrac{y}{2} = 84\) or \(\dfrac{7}{16}y = 84\) or \(7y = 1344\) | M1dep | oe \(\dfrac{1}{4}\ (\times 1) + \dfrac{3}{8} \times \dfrac{1}{2}\) linked to 84 or \(\dfrac{7}{16}\) linked to 84 |
| \(y = 84 \div\) their \(\dfrac{7}{16}\) or \(y = 84 \times\) their \(\dfrac{16}{7}\) or \(y = 192\) | M1dep | oe dep on M1M1 \(84 \div\) their \(\dfrac{7}{16}\) or \(84 \times\) their \(\dfrac{16}{7}\) or 192 |
| 288 | A1 | |
| Alternative method 3 – based on a fraction of the total number of people | ||
| \(\dfrac{1}{4} \times \dfrac{2}{3} \times z\) or \(\dfrac{4z}{24}\) or \(\dfrac{3}{8} \times \dfrac{1}{3} \times z\) or \(\dfrac{3z}{24}\) where \(z\) is the number of people in the office | M1 | oe \(\dfrac{1}{4} \times \dfrac{2}{3}\) or \(\dfrac{4}{24}\) or \(\dfrac{3}{8} \times \dfrac{1}{3}\) or \(\dfrac{3}{24}\) |
| \(\dfrac{1}{4} \times \dfrac{2}{3} \times z + \dfrac{3}{8} \times \dfrac{1}{3} \times z = 84\) or \(\dfrac{7z}{24} = 84\) | M1dep | oe \(\;\dfrac{3}{8} \times \dfrac{1}{3} + \dfrac{1}{4} \times \dfrac{2}{3}\) linked to 84 or \(\dfrac{7}{24}\) linked to 84 |
| \(z = 84 \div\) their \(\dfrac{7}{24}\) or \(z = 84 \times\) their \(\dfrac{24}{7}\) or \(7z = 2016\) | M1dep | oe dep on M1M1 \(84 \div\) their \(\dfrac{7}{24}\) or \(84 \times\) their \(\dfrac{24}{7}\) |
| 288 | A1 | |
| Alternative method 4 – chooses numbers of females and males and factors up or down | ||
| Chooses numbers for females and males in the ratio 2 : 1 and works out the numbers of females and males wearing glasses (which should be in the ratio 4 : 3) | M1 | eg 32 females and 16 males and \(\dfrac{1}{4} \times 32\ (+)\ \dfrac{3}{8} \times 16\) or 8 and 6 or 14 |
| Works out multiplying factor by \(84 \div\) their total number of people wearing glasses | M1dep | eg \(84 \div \left(\dfrac{1}{4} \times 32 + \dfrac{3}{8} \times 16\right)\) or \(84 \div 14\ (= 6)\) |
| Multiplies their total of females and males by their multiplying factor | M1dep | eg \(32 \times\) their \(6 + 16 \times\) their 6 or \((32 + 16) \times\) their 6 |
| 288 | A1 | |
Additional guidance
If more than one method is attempted:
if an answer is given, mark the method leading to that answer
if no answer is given, mark each method and award the best mark