4MA0 Higher January 2017 Paper 3R Q5
5 There are 80 counters in a bag.
The counters are either red or blue.
The ratio of the number of red counters to the number of blue counters is 3 : 1
Michael takes 15% of the red counters out of the bag.
Alison takes \(\dfrac{1}{5}\) of the blue counters out of the bag.
How many counters are now in the bag?
(5)
| Scheme | Marks |
|---|---|
| 80 ÷ (3 + 1) (=20) or 20 or 60 | M1 |
| 0.15 × (3 × “20”) (=9) | M1 |
| “20” ÷ 5 (=4) | M1 |
| 80 – “9” – “4” | M1 |
| 67 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: or M1 for 0.85 × (3 × “20”) = 51
M1: or M1 for \(\dfrac{4}{5} \times\) “20” (=16)
M1: or M1 for “16” + “51”
or
| Scheme | Marks |
|---|---|
| \(\dfrac{3}{4} \times \dfrac{15}{100}\left(= \dfrac{9}{80} \text{ or } 0.1125\right)\) | M1 |
| \(\dfrac{1}{4} \times \dfrac{1}{5}\left(= \dfrac{1}{20} \text{ or } 0.05\right)\) | M1 |
| \(\text{“}\dfrac{9}{80}\text{”} + \text{“}\dfrac{1}{20}\text{”}\left(= \dfrac{13}{80}\right)\) or “0.1125” + “0.05” (=0.1625) | M1 |
| \(\left(1 - \text{“}\dfrac{13}{80}\text{”}\right) \times 80\) or (1 – “0.1625”) × 80 or \(\dfrac{67}{80}\) | M1 |
| 67 | A1 |
Notes
M1: or M1 \(\dfrac{3}{4} \times \dfrac{85}{100}\left(= \dfrac{51}{80} \text{ or } 0.6375\right)\)
M1: or M1 \(\dfrac{1}{4} \times \dfrac{4}{5}\left(= \dfrac{1}{5} \text{ or } 0.2\right)\)
M1: or M1 \(\dfrac{51}{80} + \dfrac{1}{5}\)
M1: or M1 \(\left(\dfrac{51}{80} + \dfrac{1}{5}\right) \times 80\) oe or \(\dfrac{67}{80}\)