Foundation November 2020 Paper 1 Q6
6
(a) Write \(\dfrac{24}{40}\) as a fraction in its simplest form. (2)
(b) Write \(\dfrac{1}{5}\) as a decimal. (1)
There are only blue bricks and white bricks in a box.
The ratio of the number of blue bricks to the number of white bricks is 3 : 7
(c) What fraction of the bricks in the box are blue bricks? (1)
(d) Show that \(\;\dfrac{3}{8} + \dfrac{1}{24} = \dfrac{5}{12}\) (2)
There are 280 counters in a bag.
\(\dfrac{1}{2}\) of the counters are red.
\(\dfrac{2}{5}\) of the counters are yellow.
The rest of the counters are green.
(e) Work out the number of green counters in the bag. (3)
| Scheme | Marks |
|---|---|
| E.g. \(\dfrac{6}{10}\), \(\dfrac{9}{15}\), \(\dfrac{12}{20}\), \(\dfrac{15}{25}\), \(\dfrac{18}{30}\), \(\dfrac{21}{35}\) | M1 |
| \(\dfrac{3}{5}\) | A1 |
| (2) |
Notes
M1: for any fraction equivalent to \(\dfrac{24}{40}\) with denominator less than 40
| Scheme | Marks |
|---|---|
| 0.2 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\dfrac{3}{10}\) oe | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\dfrac{9n}{24n} + \dfrac{1n}{24n}\) or \(\dfrac{9n + 1n}{24n}\) | M1 |
eg \(\dfrac{10}{24} = \dfrac{5}{12}\) Answer: Shown | A1 |
| (2) |
Notes
M1: for correct fractions with a common denominator (multiple of 24)
A1: for a multiple of \(\dfrac{10n}{24n} = \dfrac{5}{12}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2} \times 280\) (= 140) oe or \(\dfrac{2}{5} \times 280\) (= 112) oe | M1 |
| 280 – ‘140’ – ‘112’ | M1 |
| 28 | A1 |
| (3) | |
| (9 marks) |
Notes
Alternative
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2} + \dfrac{2}{5}\) \(\left(= \dfrac{9}{10}\right)\) or 0.5 + 0.4 (= 0.9) oe | M1 |
| \(\left(1 - \text{‘}\dfrac{9}{10}\text{’}\right) \times 280\) or (1 – ‘0.9’) × 280 oe | M1 |
| 28 | A1 |