4MA0 Higher January 2014 Paper 3R Q6
6
(a) Dilip buys a painting for $ 675
Later, he sells it and makes a percentage profit of 12%.
Work out the price for which Dilip sells the painting. (3)
Later, he sells it and makes a percentage profit of 12%.
Work out the price for which Dilip sells the painting. (3)
(b) Renuka sells her car.
She makes a loss of $ 2162
Her percentage loss is 23%.
Work out the price for which Renuka sells her car. (3)
She makes a loss of $ 2162
Her percentage loss is 23%.
Work out the price for which Renuka sells her car. (3)
(c) Lin bought a computer that had a value of $ 1500
At the end of each year, the value of her computer had depreciated by 40% of its value at the start of that year.
Calculate the value of her computer at the end of 3 years. (3)
At the end of each year, the value of her computer had depreciated by 40% of its value at the start of that year.
Calculate the value of her computer at the end of 3 years. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{12}{100} \times 675\) oe or 81 | M1 |
| 675 + “81” | M1 |
| 756 | A1 |
| (3) |
Notes
M1: (dep)
A1: cao
M2 for 675 × 1.12 oe
| Scheme | Marks |
|---|---|
23% of amount = 2162 or (1% =) \(\dfrac{2162}{23}\) or 94 seen | M1 |
| “94” × 100 or 9400 or “94” × 77 | M1 |
| 7238 | A1 |
| (3) |
Notes
A1: cao
M2 for \(\dfrac{2162}{23} \times 100\) oe
| Scheme | Marks |
|---|---|
| \(\dfrac{40}{100} \times 1500\) oe or 600 | M1 |
\(\dfrac{40}{100} \times (1500 - \text{“}600\text{”})\) = 360 \(\dfrac{40}{100} \times (1500 - \text{“}600\text{”} - \text{“}360\text{”})\) = 216 1500−“600”−“360”−“216” | M1 |
| 324 | A1 |
| (3) | |
| (9 marks) |
Notes
M1: for eg \(\dfrac{40}{100} \times 1500\) or 600
M1: for completing method
A1: cao
OR \(1500 \times 0.6^3\)
M2 for \(1500 \times 0.6^3\)
(M1 for \(1500 \times 0.6\) or 900 or \(1500 \times 0.6^2\) or 540 or \(1500 \times 0.6^4\))
Accept (1− 0.4) as equivalent to 0.6