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IB® · HL/SL

IB® Math Applications & Interpretation HL/SL

Real-world mathematics: the five official IB® topics with applied focus, technology use, and HL extensions including networks, Markov chains, and matrices.

Start Unit 1 free. Topic 1: Number and Algebra is open to everyone, no account needed. Other topics are locked.

Topic 1: Number and Algebra

THE BIG PICTURE. AI Topic 1 emphasises applied algebra: scientific notation, sequences and series, financial mathematics, and basic exponents/logs. HL extends to complex numbers, matrix algebra (incl. eigenvalues), and transition matrices / Markov chains. The GDC is used heavily; AI rewards clear technology use and clear interpretation of numerical results.

NUMBER, ESTIMATION, AND PERCENTAGE ERROR

  • Scientific notation: a×10na \times 10^n with 1≤∣a∣<101 \leq |a| < 10.
  • Significant figures and rounding: be careful about precision; AI marks penalise wrong rounding.
  • Percentage error: ∣vA−vE∣vE×100%\dfrac{|v_A - v_E|}{v_E} \times 100\%, where vEv_E is the exact value and vAv_A is the approximate.

Example. A measured length is 32.432.4 cm; the true value is 3232 cm. Find the percentage error. ∣32.4−32∣32×100%=0.432×100%=1.25%.\dfrac{|32.4 - 32|}{32} \times 100\% = \dfrac{0.4}{32} \times 100\% = 1.25\%.

WORKED EXAMPLE: UPPER AND LOWER BOUNDS

A rectangular field is measured as 120 m by 85 m, each to the nearest metre. Find the bounds of its area and the largest possible percentage error in the area 120×85=10,200120 \times 85 = 10{,}200 m2^2.

  • Bounds of each measurement: 119.5≤L<120.5119.5 \le L < 120.5 and 84.5≤W<85.584.5 \le W < 85.5 (half a unit either side).
  • Lower bound of the area: 119.5×84.5=10,097.75119.5 \times 84.5 = 10{,}097.75 m2^2. Upper bound: 120.5×85.5=10,302.75120.5 \times 85.5 = 10{,}302.75 m2^2.
  • Largest percentage error: 10302.75−1020010200×100%≈1.01%\dfrac{10302.75 - 10200}{10200} \times 100\% \approx 1.01\%.
  • Method mark tip: for a product use (lower × lower) and (upper × upper); for a quotient, divide the upper bound by the lower bound to get the largest value.

SEQUENCES AND SERIES

Arithmetic: common difference dd

  • un=u1+(n−1)du_n = u_1 + (n - 1) d
  • Sn=n2(u1+un)=n2(2u1+(n−1)d)S_n = \dfrac{n}{2}(u_1 + u_n) = \dfrac{n}{2}(2 u_1 + (n - 1) d)

Example. A theatre has 25 seats in row 1, 28 in row 2, 31 in row 3. How many seats in row 20, and total seats in 20 rows? u1=25u_1 = 25, d=3d = 3. u20=25+19⋅3=82u_{20} = 25 + 19 \cdot 3 = 82. S20=202(25+82)=1,070S_{20} = \dfrac{20}{2}(25 + 82) = 1{,}070 seats.

Geometric: common ratio rr

  • un=u1⋅rn−1u_n = u_1 \cdot r^{n-1}
  • Sn=u1(1−rn)1−rS_n = \dfrac{u_1(1 - r^n)}{1 - r} for r≠1r \neq 1
  • HL: sum to infinity (∣r∣<1|r| < 1): S∞=u11−rS_\infty = \dfrac{u_1}{1 - r}

Example. A bouncing ball reaches 80%80\% of its previous peak. From a 22 m drop, total vertical distance travelled? Falls and bounces alternate. Total = drop + 2(sum of all bounce heights). Bounce heights form a geometric sequence: 1.6,1.28,1.024,…1.6, 1.28, 1.024, \ldots with u1=1.6u_1 = 1.6, r=0.8r = 0.8. S∞=1.61−0.8=8.Total distance=2+2⋅8=18 m.S_\infty = \dfrac{1.6}{1 - 0.8} = 8.\quad \text{Total distance} = 2 + 2 \cdot 8 = 18 \text{ m}.

A geometric sequence you can see Each peak is 0.8 times the one before, so the bounce heights form a geometric sequence. Every bounce is travelled up and down, which is why the total distance is 2 + 2 × S∞ = 18 m (HL: sum to infinity).

WORKED EXAMPLE: DEPRECIATION AS A GEOMETRIC SEQUENCE

A car is bought for $24,000 and loses 15%15\% of its value each year.

  • (a) Value after 5 years. Each year multiplies the value by r=0.85r = 0.85, so after 5 years it is 24000×0.855≈10,648.9324000 \times 0.85^5 \approx 10{,}648.93, about $10,648.93. (If you write the purchase price as u1u_1, the value after 5 years is u6u_6: count the multiplications, not the terms.)
  • (b) When does it first fall below $8,000? Solve 24000×0.85n<800024000 \times 0.85^n < 8000, so 0.85n<130.85^n < \tfrac{1}{3}. The GDC (table of values or graph intersection) gives n>6.76n > 6.76, so the value first drops below $8,000 after 7 whole years.
  • Marks tip: state u1u_1 and rr before calculating, and answer the question asked: a whole number of years, not 6.76.

FINANCIAL MATHEMATICS

Compound interest: FV=PV(1+r100k)kn\text{FV} = \text{PV} \left(1 + \dfrac{r}{100k}\right)^{kn}, with kk compounding periods per year for nn years at nominal rate r%r\%.

Example. $8,000 invested at 5%5\% annual interest, compounded monthly, for 7 years. FV=8000(1+51200)84≈$11,344.30.\text{FV} = 8000\left(1 + \dfrac{5}{1200}\right)^{84} \approx \$11{,}344.30. (Keep the unrounded rate in the GDC: rounding it to 1.004171.00417 before raising to the 84th power shifts the answer by about $2.)

Loans and amortisation: the same compound-interest formula, plus periodic payments. Use the GDC TVM (time-value-of-money) solver for loan and annuity problems.

Example. Borrow $15,000 at 6%6\% annual interest, compounded monthly, paid back over 4 years. The TVM solver gives a monthly payment of approximately $352.28; total paid $16,909; interest paid $1,909.

Where each loan payment goes The payment from the TVM solver never changes, but the interest part (0.5% of the current balance) shrinks, so the balance falls faster as the loan goes on.

WORKED EXAMPLE: READING AN AMORTISATION SCHEDULE

For the $15,000 loan above (monthly rate 6%÷12=0.5%6\% \div 12 = 0.5\%, payment $352.28):

  • (a) First payment: the interest is 15000×0.005=7515000 \times 0.005 = 75, so $75.00 of the first payment is interest and the other 352.28−75.00=277.28352.28 - 75.00 = 277.28, i.e. $277.28, reduces the balance.
  • (b) Balance after one year: keep the unrounded payment in the TVM solver and set N = 12: FV ≈−11,579.65\approx -11{,}579.65 (the sign only shows the direction of the cash flow), so about $11,579.65 is still owed.
  • (c) Interest paid in year 1: total paid 12×352.28≈4227.3612 \times 352.28 \approx 4227.36; principal repaid 15000−11579.65=3420.3515000 - 11579.65 = 3420.35; so interest is about $807.
  • Pattern: every payment is the same, but the interest part shrinks as the balance falls, so more of each later payment repays the loan.

PRACTICE: WHICH TOOL FOR WHICH MONEY OR SEQUENCE QUESTION?

SituationModelTool
Seats increase by 3 per rowArithmetic, d=3d = 3unu_n and SnS_n formulas
A car loses 15% of its value each yearGeometric, r=0.85r = 0.85unu_n formula or GDC table
Savings at 4% compounded quarterly, no depositsCompound interest, k=4k = 4FV formula or TVM (PMT = 0)
A loan repaid by equal monthly paymentsAmortisationTVM solver
Equal monthly deposits into a savings planAnnuityTVM solver
Total distance of a bouncing ball (HL)Infinite geometric series, −1<r<1-1 < r < 1S∞S_\infty formula

Inflation and depreciation use the same form with negative effective rate.

EXPONENTS AND LOGARITHMS (basic)

  • Exponent laws: am⋅an=am+na^m \cdot a^n = a^{m+n}, aman=am−n\dfrac{a^m}{a^n} = a^{m-n}, (am)n=amn(a^m)^n = a^{mn}, a0=1a^0 = 1, a−n=1ana^{-n} = \dfrac{1}{a^n}.
  • Definition of log: ax=b  ⇔  x=log⁡aba^x = b \;\Leftrightarrow\; x = \log_a b (with a>0a > 0, b>0b > 0, a≠1a \neq 1).
  • HL, log laws: log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy) = \log_a x + \log_a y; log⁡a(x/y)=log⁡ax−log⁡ay\log_a(x/y) = \log_a x - \log_a y; log⁡a(xm)=mlog⁡ax\log_a(x^m) = m \log_a x.
  • Natural log: ln⁡x=log⁡ex\ln x = \log_e x, with e≈2.71828e \approx 2.71828.

Example. A population P(t)=1000⋅1.04tP(t) = 1000 \cdot 1.04^t (in years). When does it reach 5,000? 1.04t=5⇒t=log⁡1.045=ln⁡5ln⁡1.04≈41.0 years.1.04^t = 5 \Rightarrow t = \log_{1.04} 5 = \dfrac{\ln 5}{\ln 1.04} \approx 41.0 \text{ years}.

SYSTEMS OF LINEAR EQUATIONS

  • Solve 2×22 \times 2 systems algebraically (substitution / elimination) or with GDC.
  • HL: 3×33 \times 3 systems via row reduction or matrix inverse.

Example. A coffee shop sells two drink sizes. 3 small + 2 large = $13.20; 2 small + 4 large = $18.40. Find each price. 3s+2L=13.20,2s+4L=18.40.3s + 2L = 13.20,\quad 2s + 4L = 18.40. Multiply first by 2: 6s+4L=26.406s + 4L = 26.40. Subtract second: 4s=8.00⇒s=24s = 8.00 \Rightarrow s = 2. So small drinks cost $2.00. Then L=(13.20−6)/2=3.60L = (13.20 - 6)/2 = 3.60, so large drinks cost $3.60.

HL EXTENSIONS

Complex numbers: z=a+biz = a + bi where i2=−1i^2 = -1. Used in AI HL mainly for roots of quadratic and polynomial equations (when Δ<0\Delta < 0) and modulus-argument form for engineering / signal-processing applications.

Example (HL). Solve x2+4x+13=0x^2 + 4x + 13 = 0. Δ=16−52=−36\Delta = 16 - 52 = -36. x=−4±−362=−2±3ix = \dfrac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i.

Matrices. Manipulate matrices for transformations and systems

  • Determinant: det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc.
  • Inverse: A−1=1det⁡A(d−b−ca)A^{-1} = \dfrac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} when det⁡A≠0\det A \neq 0.
  • Solving systems: Ax=b⇒x=A−1bA\mathbf{x} = \mathbf{b} \Rightarrow \mathbf{x} = A^{-1}\mathbf{b}.

Example (HL). Solve {2x+3y=8x−y=1\begin{cases} 2x + 3y = 8 \\ x - y = 1 \end{cases} via inverse matrix. A=(231−1)A = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}, det⁡A=−2−3=−5\det A = -2 - 3 = -5. A−1=1−5(−1−3−12)=(0.20.60.2−0.4)A^{-1} = \dfrac{1}{-5}\begin{pmatrix} -1 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 0.2 & 0.6 \\ 0.2 & -0.4 \end{pmatrix}. x=A−1(81)=(2.21.2)\mathbf{x} = A^{-1}\begin{pmatrix} 8 \\ 1 \end{pmatrix} = \begin{pmatrix} 2.2 \\ 1.2 \end{pmatrix}.

Eigenvalues / Eigenvectors (HL): Av=λvA\mathbf{v} = \lambda \mathbf{v}. Find eigenvalues from det⁡(A−λI)=0\det(A - \lambda I) = 0.

Markov chains (HL). A transition matrix TT describes probabilities of moving between states. State sn\mathbf{s}_n after nn steps: sn=Tns0\mathbf{s}_n = T^n \mathbf{s}_0. Steady-state vector v\mathbf{v}: solves Tv=vT\mathbf{v} = \mathbf{v} (eigenvalue 1).

Example (HL Markov). Two coffee shops A and B. Each week, 80%80\% of A's customers stay and 20%20\% switch to B. 30%30\% of B's customers switch to A and 70%70\% stay. Initially A has 60%60\%, B has 40%40\%. Find market share next week. T=(0.80.30.20.7)T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}, s0=(0.60.4)\mathbf{s}_0 = \begin{pmatrix} 0.6 \\ 0.4 \end{pmatrix}. s1=Ts0=(0.60.4)\mathbf{s}_1 = T\mathbf{s}_0 = \begin{pmatrix} 0.6 \\ 0.4 \end{pmatrix}. The market share is unchanged: this initial distribution is already the steady-state vector (the eigenvector of TT with eigenvalue 1).

Markov chains forget where they started (HL) Columns of T give the probabilities of moving from each state. Whatever the starting split, T to the power n times the start tends to the steady state (0.6, 0.4), the eigenvector for eigenvalue 1.

WORKED EXAMPLE: LONG-TERM BEHAVIOUR FROM EIGENVALUES (HL)

Use the same TT, but now shop A opens first and has every customer: s0=(10)\mathbf{s}_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.

  • Eigenvalues: det⁡(T−λI)=(0.8−λ)(0.7−λ)−0.06=λ2−1.5λ+0.5=0\det(T - \lambda I) = (0.8 - \lambda)(0.7 - \lambda) - 0.06 = \lambda^2 - 1.5\lambda + 0.5 = 0, so λ=1\lambda = 1 or λ=0.5\lambda = 0.5.
  • Eigenvectors: for λ=1\lambda = 1, 0.3b=0.2a0.3b = 0.2a gives (32)\begin{pmatrix} 3 \\ 2 \end{pmatrix}; for λ=0.5\lambda = 0.5, (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix}.
  • Diagonalise: T=PDP−1T = PDP^{-1} with P=(312−1)P = \begin{pmatrix} 3 & 1 \\ 2 & -1 \end{pmatrix} and D=(1000.5)D = \begin{pmatrix} 1 & 0 \\ 0 & 0.5 \end{pmatrix}, so Tn=PDnP−1T^n = PD^nP^{-1} and

sn=Tns0=(0.6+0.4(0.5)n0.4−0.4(0.5)n).\mathbf{s}_n = T^n\mathbf{s}_0 = \begin{pmatrix} 0.6 + 0.4(0.5)^n \\ 0.4 - 0.4(0.5)^n \end{pmatrix}.

  • Check: n=3n = 3 gives 0.6+0.05=0.650.6 + 0.05 = 0.65, the same as T3s0T^3\mathbf{s}_0 on the GDC.
  • Interpret: as n→∞n \to \infty, (0.5)n→0(0.5)^n \to 0, so A's share tends to 0.60.6 whatever the starting split. The eigenvalue 1 gives the steady state; the other eigenvalue (0.50.5) controls how quickly the chain gets there.

EXAM CONNECTIONS.

  • Paper 1 (with GDC): financial calculations, sequence problems, percentage error in real contexts.
  • Paper 2 (with GDC): larger applied problems with multiple sub-questions; expect modelling.
  • Paper 3 (HL only): Paper 3 can draw on any AI HL syllabus area; Markov chains/transition matrices, matrix work, sequences, and financial scenarios are plausible Paper 3-style contexts, but no topic is guaranteed in any given session.

Key Terms

Scientific Notation

a×10na \times 10^n where 1≤∣a∣<101 \leq |a| < 10 and n∈Zn \in \mathbb{Z}. Used to compactly express very large or very small numbers.

Percentage Error

∣vA−vE∣vE×100%\dfrac{|v_A - v_E|}{v_E} \times 100\%, where vEv_E is the exact value and vAv_A is the approximate. Always positive.

Arithmetic Sequence

Sequence with constant common difference dd. un=u1+(n−1)du_n = u_1 + (n-1) d; Sn=n2(u1+un)S_n = \dfrac{n}{2}(u_1 + u_n).

Geometric Sequence

Sequence with constant common ratio rr. un=u1rn−1u_n = u_1 r^{n-1}. Sum to infinity exists only when ∣r∣<1|r| < 1 (HL).

Compound Interest

FV=PV(1+r100k)kn\text{FV} = \text{PV}\left(1 + \dfrac{r}{100k}\right)^{kn}. Future value with kk compounding periods/year for nn years at rate r%r\%.

Amortisation

Gradual repayment of a loan via regular payments covering interest + principal. Solve with the GDC TVM (time-value-of-money) solver.

Annuity

Series of equal payments at regular intervals. Used for retirement income, loan repayments. Calculated via TVM.

Logarithm

log⁡ab=x  ⇔  ax=b\log_a b = x \;\Leftrightarrow\; a^x = b. Used to solve exponential equations. Natural log: ln⁡x=log⁡ex\ln x = \log_e x.

Determinant of 2×22 \times 2 Matrix (HL)

det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc. Zero determinant means matrix is singular (no inverse).

Eigenvalue (HL)

Scalar λ\lambda such that Av=λvA\mathbf{v} = \lambda \mathbf{v} for some non-zero v\mathbf{v}. Found via det⁡(A−λI)=0\det(A - \lambda I) = 0.

Markov Chain (HL)

Sequence of states with probabilistic transitions described by a transition matrix TT. State after nn steps: sn=Tns0\mathbf{s}_n = T^n \mathbf{s}_0.

Steady-State Vector (HL)

Vector v\mathbf{v} satisfying Tv=vT\mathbf{v} = \mathbf{v}. Eigenvector with eigenvalue 1. Long-run distribution of a Markov chain.

Exam Tips

  • GDC mastery: AI is technology-heavy. Learn the TVM solver and statistics functions cold: most marks come from these.
  • Show inputs, not just answers: when using TVM, list PV, FV, I%, N, P/Y, C/Y so the examiner can verify.
  • Significant figures: AI penalises wrong precision more than other math papers. Read the question for required sig figs.
  • Real-world interpretation: every numerical answer should be followed by a sentence explaining what it means in context.
  • Currency/units: include them in every monetary or measured answer.
  • HL Markov chains: state the transition matrix clearly (with row/column labels), then compute. Steady state via eigenvector method or by solving the linear system Tv=vT\mathbf{v} = \mathbf{v}.

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