Higher June 2018 Paper 2 Q9
9 Jean invests £12 000 in an account paying compound interest for 2 years.
In the first year the rate of interest is \(x\)%
At the end of the first year the value of Jean’s investment is £12 336
In the second year the rate of interest is \(\dfrac{x}{2}\)%
What is the value of Jean’s investment at the end of 2 years? (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 12 508.7(0) | P1 | for start of process to find interest rate for year 1 eg \(12336 \div 12000\) (=1.028) or \((12336 - 12000) \div 12000\) (=0.028) OR forms a suitable equation, eg \(12000 \times \left(1 + \dfrac{x}{100}\right) = 12336\) |
| P1 | for complete process to find the interest rate for year 1 eg \((\text{``}1.028\text{''} - 1) \times 100\) (=2.8) or \(\text{``}0.028\text{''} \times 100\) (=2.8) OR correct process to solve correct equation eg \((12336 - 12000) \div 120\) (=2.8) | |
| P1 | for complete process to find the value at the end of 2 years eg \((\text{``}2.8\text{''} \div 2 + 100) \div 100 \times 12336\) | |
| A1 | accept 12508.7 to 12508.71 or 12509 |
Additional guidance
Rate of interest = 2.8, or \(x = 2.8\) implies P2
12509 must come from correct working