Example 1 — Percentage error

A student measures the mass of a sample as 47.2 g. The true mass is 45.8 g. Find the percentage error.

ε = | (vA − vE) / vE | × 100
  = | (47.2 − 45.8) / 45.8 | × 100
  = (1.4 / 45.8) × 100
  = 3.0567…
  = 3.06 %   (3 s.f.)

3.06 %.

The denominator is the true value, always. Dividing by 47.2 gives 2.97%, which is close enough to look right and is wrong. And the modulus means the answer is positive even when the measurement is an underestimate.

Example 2 — Geometric growth, and how many years

A colony of bacteria numbers 5000 and increases by 8% every hour. (a) Find the number after 6 hours. (b) Find the number of complete hours before it exceeds 20000.

(a) An 8% increase means r = 1.08, not 0.08. The starting value is the term at n = 0, so after 6 hours:

5000 × 1.08⁶ = 5000 × 1.586874…
             = 7934.37…
             = 7930   (3 s.f.)

(b) Solve 5000 × 1.08ⁿ > 20000:

1.08ⁿ > 4
n log 1.08 > log 4
n > log 4 / log 1.08 = 0.60206 / 0.033424 = 18.01…

So the colony first exceeds 20000 during the 19th hour, meaning 18 complete hours pass before it does.

Read the question's wording with care. "Complete hours before it exceeds" and "the hour in which it first exceeds" differ by one, and the mark is for the right one. When the boundary is this close to an integer, check by substitution: 1.08¹⁸ = 3.996, just under 4; 1.08¹⁹ = 4.316, over.

Example 3 — Compound interest with non-annual compounding

€6500 is invested at 3.8% per annum, compounded monthly. Find its value after 4 years.

Identify each quantity before substituting — this is where the marks are lost:

PV = 6500      r = 3.8      n = 4      k = 12   (monthly)

FV = 6500 × (1 + 3.8/(100 × 12))^(12 × 4)
   = 6500 × (1 + 0.00316666…)^48
   = 6500 × 1.163880…
   = 7565.22…
   = €7570   (3 s.f.)

€7570.

The rate is divided by k and the exponent is multiplied by k. Doing one and not the other is the standard error, and it produces an answer in the right region — around €7500 either way — so it does not look wrong. Note also that the 0.0031666… was not rounded before being raised to the 48th power; rounding it to 0.00317 here shifts the final answer by several euro.

Example 4 — Arithmetic sequence in context, with interpretation

A theatre has 18 seats in the front row. Each row behind has 4 more seats than the one in front. There are 25 rows. (a) Find the number of seats in the last row. (b) Find the total number of seats. (c) The manager claims that adding a 26th row would take the total past 1500. Comment.

(a) u₁ = 18, d = 4, n = 25:

u₂₅ = 18 + 24 × 4 = 18 + 96 = 114

(b) With the first and last terms both known, use the shorter sum formula:

S₂₅ = (25/2)(18 + 114) = 12.5 × 132 = 1650

(c) The total is already 1650, which is past 1500 without a 26th row. The manager's claim is therefore wrong — or rather, it is already satisfied.

114 seats; 1650 in total; the claim is mistaken because the 25 rows alone exceed 1500.

Part (c) carries a reasoning mark and it is not a calculation. The answer is a sentence about the theatre. Writing "1650 > 1500" alone will usually be condoned, but a question that says comment is asking you to say what that inequality means for the claim.

Educerie · original worked examples written against the published IB syllabus structure for Mathematics: applications and interpretation Topic 1, first assessment 2021. All scenarios and values are our own. Last reviewed 5 September 2026.