4MA0 Foundation January 2017 Paper 1R Q14
14 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: for 0.85 × (3×”20”) = 51
M1: for \(\dfrac{4}{5} \times\) “20” (=16)
M1: for “16” + “51”
Alternative method (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 |
| “\(\dfrac{9}{80}\)” + “\(\dfrac{1}{20}\)” \(\left(= \dfrac{13}{80}\right)\) or “0.1125” + “0.05” (=0.1625) | M1 |
| (1 − “\(\dfrac{13}{80}\)”) × 80 or (1 − “0.1625”) × 80 or \(\dfrac{67}{80}\) | M1 |
| 67 | A1 |
Notes
M1: \(\dfrac{3}{4} \times \dfrac{85}{100}\ \left(= \dfrac{51}{80} \text{ or } 0.6375\right)\)
M1: \(\dfrac{1}{4} \times \dfrac{4}{5}\ \left(= \dfrac{1}{5} \text{ or } 0.2\right)\)
M1: \(\dfrac{51}{80} + \dfrac{1}{5}\)
M1: \(\left(\dfrac{51}{80} + \dfrac{1}{5}\right) \times 80\) oe or \(\dfrac{67}{80}\)