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Question 102 of 104 in Unit 4

Unit 4

5 marks · Normal

A saving scheme gives interest at 4% per year, compounded annually. The sequence of balances over five years is plotted on a graph. Simultaneously, an algebraic expression relates to the yearly deposit amount.

(a)

Amira deposits $3000. Calculate the total value of her savings after 5 years.

[2 marks]
(b)

The graph of Amira's savings, plotted against number of years n, is shown below.

Savings over time0123450500100015002000250030003500YearsSavings ($)Savings over time
Savings growth vs years (years 0–5, values not shown, curve rises)

Describe how the graph shows that the total increases by a larger amount each year.

[1 mark]
(c)

Write an algebraic expression for the value S after n years using the starting amount $3000 and the yearly multiplying factor.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) $3650 dollars

(b) The curve becomes steeper, showing yearly increases get bigger.

(c) S = 3000 × 1.04ⁿ

Common mistakes: linear instead of compound calculation; incorrect algebraic notation

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Uses S = 3000 × (1.04)^5 or step-by-step correct calculation

A1 (a) · 1 mark · Gives value to at least nearest dollar: $3650 (follow-through)

B1 (b) · 1 mark · Recognises graph is a curve, not a straight line, with increasing increment

M2 (c) · 1 mark · States correct form S = 3000 × (1.04)^n or equivalent

A2 (c) · 1 mark · Uses index notation and both correct numbers (starting amount, multiplier) (follow-through)

Worked solution

a) S = 3000 × (1.04)^5 ≈ 3000 × 1.2166529 ≈ $3649.96 ≈ $3650

b) The graph curves upward, showing the increase each year is greater than the year before.

c) S = 3000 × (1.04)^n

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