Higher June 2023 Paper 4 Q4

OCRHigherCurrent spec6 marksPercentages

4 Alex invests £4500 at a rate of 7.5% per year simple interest.

(a) Find the value of the investment at the end of 4 years. [3]

(b) At the end of \(t\) years, the value of the investment is over £13 500 for the first time.

Find the value of \(t\). [3]

Mark scheme (a)

Answer	Marks	Part marks and guidance
5850	3	M2 for [4500 +] 4500 × [0].075 × 4 oe or M1 for 4500 × [0].075 oe	implied by 1350 implied by 337.5[0] or 4837.5[0] If they use compound interest, put MR M2 for \(4500 \times \left(1 + \frac{7.5}{100}\right)^4\) oe implied by 6009.6[1…] or 6010 or M1 for \(4500 \times \left(1 + \frac{7.5}{100}\right)^k\) (\(k\) = 2, 3 or 5) or interest only implied by e.g. 1509.6[1…] (see appendix)

Appendix: Question 4

In trials they must have a figure that is accurate to 3 figures. Some are using compound interest so they will get

0	4500.00
1	4837.50
2	5200.31
3	5590.34
4	6009.61
5	6460.33
6	6944.86
7	7465.72
8	8025.65
9	8627.57
10	9274.64
11	9970.24
12	10718.01
13	11521.86
14	12386.00
15	13314.95
16	14313.57
17	15387.09
18	16541.12

Mark scheme (b)

Answer	Marks	Part marks and guidance
27	3	M2 for \(\frac{13\,500 - 4500}{their\ 337.50}\) oe or any correct method e.g. \(\frac{13\,500 - their\ 5850}{their\ 337.50} + 4\) or M1 for 13 500 – 4500 or 9 000 or 13 500 – their 5850 Alternative method using trials M1 for each correct trial over 4 evaluated correctly and rot to at least 3 s.f. to a maximum of M2 If 0 scored SC1 for answer of 40 or 41 from \(\frac{13\,500}{337.50}\) oe	Condone 26.66… or 26.[7] as answer for 3 marks. (table of values below) If they use compound interest, put MR M1 for each correct trial over 4 evaluated correctly and rot to at least 3 s.f. to a maximum of M2 (See appendix)

Table of values (simple interest)

5	6187.50	18	10575.00
6	6525.00	19	10912.50
7	6862.50	20	11250.00
8	7200.00	21	11587.50
9	7537.50	22	11925.00
10	7875.00	23	12262.50
11	8212.50	24	12600.00
12	8550.00	25	12937.50
13	8887.50	26	13275.00
14	9225.00	27	13612.50
15	9562.50	28	13950.00
16	9900.00	29	14287.50
17	10237.50	30	14625.00

Appendix: Question 4(b)

Note : Neat solution: log(3)÷log(1.075) = 15.19…. oe hence 16 but it is not in our specification. If you see it award M2.
OR \(1.075^t = 3\) scores M1 and a further trial will score M1

A different method is :
\(13500 = 4500\left(1 + \frac{7.5}{100} \times t\right)\)
\(3 = 1 + \frac{7.5}{100} \times t\)
\(2 = \frac{7.5}{100} \times t\)
\(200 = 7.5t\)
\(t = \frac{200}{7.5}\)
\(t = 26.6666\ldots\)