Educerie · IB Diploma · Mathematics: applications and interpretation
Topic 1 — Topic test · mark scheme
M = method, awarded for a correct approach even if the arithmetic that follows is wrong. A = accuracy, dependent on the method mark above it. R = reasoning, awarded for justification in words.
Follow through (FT) applies throughout. Accept answers to 3 s.f. or better unless stated; penalise an answer given to fewer than 3 s.f. once only, at its first occurrence.
1Accuracy — 4 marks
(a) [2]
Correct substitution into |(vA − vE)/vE| × 100, e.g. (14/254) × 100 | M1 |
|---|---|
5.51 % | A1 |
The denominator must be 254, the true value.
(14/240) × 100 = 5.83 %scores M0 A0. Accept5.5118…rounded to 3 s.f. A negative answer loses the A1 — percentage error is defined with a modulus.
(b) [2]
Correct value of a, i.e. 2.54 | A1 |
|---|---|
Correct power, giving 2.54 × 10² | A1 |
25.4 × 10¹and0.254 × 10³both score A1 A0 —amust satisfy1 ≤ a < 10.
2Geometric growth — 6 marks
(a) [2]
Recognises r = 1.032 and forms 42000 × 1.032⁵ | M1 |
|---|---|
49 200 (accept 49 164 or better) | A1 |
Using
r = 0.032scores M0. Using1.032⁶(treating the initial value asu₁atn = 1) is the other common error and also scores M0.
(b) [3]
Forms an inequality or equation, e.g. 42000 × 1.032ⁿ > 60000 or 1.032ⁿ > 1.428… | M1 |
|---|---|
Solves to obtain n > 11.3… | A1 |
12 years | A1 |
The final A1 is for the integer, correctly rounded up.
11.3left as the answer scores a maximum of 2. Award full marks for a correct answer reached by trial: candidates who shown = 11 → 59 392andn = 12 → 61 292have demonstrated the method (M1 A1) and the conclusion (A1).
(c) [1]
| Any one valid assumption, e.g. "the growth rate stays constant at 3.2 %", or "no migration, or migration is already included in the rate" | R1 |
|---|
Reject vague answers such as "the model is accurate". The assumption must be specific.
3Compound interest — 6 marks
(a) [3]
Identifies k = 4 and forms 8000(1 + 4.5/(100 × 4))^(4 × 6) or equivalent | M1 |
|---|---|
Correct exponent 24 and correct periodic rate 0.01125 | A1 |
$10 500 (accept 10 463.9… or better) | A1 |
Award the second mark only if both adjustments are made. Dividing the rate by 4 while leaving the exponent as 6, or vice versa, is the expected error and scores M1 A0 A0. A correct answer from a calculator finance solver scores full marks provided the inputs are shown.
(b) [1]
$2460 (accept 2463.9…) | A1 FT from their (a) |
|---|
(c) [2]
Recognises that inflation reduces the value, e.g. real return ≈ 4.5 − 2 = 2.5 % per year | M1 |
|---|---|
| Interprets for the investor, e.g. "the money grows in real terms, but by about 2.5 % a year rather than 4.5 %, so its purchasing power increases much more slowly than the headline figure suggests" | R1 |
The R1 is for a statement about purchasing power or real value. Restating the nominal figure, or writing only "inflation is 2 %", scores R0. A candidate who concludes the investment still gains in real terms is correct; one who concludes it loses value is not.
4Arithmetic sequence in context — 4 marks
(a) [2]
Substitutes into uₙ = u₁ + (n − 1)d, e.g. 24 + 19 × 3 | M1 |
|---|---|
81 | A1 |
24 + 20 × 3 = 84is the expected error, from usingnrather than(n − 1). M0 A0.
(b) [2]
Substitutes into either sum formula, e.g. (20/2)(24 + 81) or (20/2)(48 + 19 × 3) | M1 FT from their (a) |
|---|---|
1050 | A1 |
Mark distribution
| Split | Allocation |
|---|---|
| By topic | Accuracy and error 4 · Sequences and series 10 · Financial mathematics 6 |
| By objective | AO1 knowledge and routine technique 9 · AO2 problem solving 7 · AO3 communication and reasoning 4 |
| By command word | Find 11 · Calculate 2 · Write down 2 · Comment on 2 · State 1 · (2b integer interpretation) 2 |
Total: 20 marks
Educerie · original mark scheme for original questions, written against the published IB syllabus structure for Mathematics: applications and interpretation Topic 1, first assessment 2021. Mark-type conventions (M/A/R, FT) follow standard IB marking practice. Last reviewed 5 September 2026.