Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A property is expected to earn net rent of €10 000 next year. The rent is expected to grow by 1.5% per year forever. The required return is 5.5% per year. Find the value PV of the property today.
You save €250 at the end of every month for 6 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €12 000, €7 000, €15 000, €13 000. The cost of capital is 12%. Find the net present value NPV.
An investment of €10 000 returns €2 000 at the end of each year for 9 years. Find its internal rate of return IRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 8% and matures in 7 years. Its yield to maturity changes from 5% to 7%. By how much ΔP does the bond price change?
A bond with a face value of €1 000 and an annual coupon of 8% matures in 3 years and sells for €940. Find its yield to maturity YTM in percent.
A company has equity worth €12 000 000 and debt worth €5 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 5% and the tax rate is 20%. Find the WACC.
A company has 4 000 000 shares at €12.50 each and 40 000 bonds trading at €980 each. The pre-tax cost of debt is 3.5% and the tax rate is 20%. The share’s beta is 0.6, the risk-free rate 4% and the market return 8.5%. Find the WACC.
Find the sample standard deviation s of the data: 17, 48, 38, 17, 47, 60.
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 72?
For the distribution shown, find the variance Var(X) of X.
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 4 correct answers?
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 245.25 and 255.25 g? Give P as a decimal.
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
The battery life of two phone models is compared. Model 1: 60 phones, mean 44 h, known standard deviation 7 h. Model 2: 100 phones, mean 55.5 h, known standard deviation 20 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
Two fertilisers are tested. Plot group 1: 8 plots, mean yield 68 kg, standard deviation 4 kg. Plot group 2: 9 plots, mean 76 kg, standard deviation 6.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 1.52 and the standard deviation of the differences is 5.7. Find the paired test statistic t for H₀: μd = 0.
A sample of 81 measurements has mean 99.2. The population standard deviation is 5. Find the upper limit U of the 95% confidence interval for the mean.
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 61) is made. Find its residual e = y − ŷ.
Answer: PV = 250000 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Answer: FV = 20431 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
You paid in €18 000 yourself; the rest is interest.
Answer: NPV = 5233 €
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
NPV is positive, so the project earns more than the required return.
Answer: IRR = 13.7 %
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
The annuity factor falls as the rate rises; bracket the rate and narrow down.
Answer: ΔP = −119.7 €
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Answer: YTM = 10.43 %
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Answer: WACC = 11.06 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Answer: WACC = 4.986 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Answer: s = 17.59
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Answer: P = 0.1587
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Answer: Var(X) = 1.24
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Answer: P = 0.00744
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Answer: P = 0.6821
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
Answer: P = 0.7881
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Answer: upper limit = −7.89 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
The interval does not contain 0, so the means differ significantly.
Answer: upper limit = −3.334 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Answer: t = 0.8
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
With df = 8, the two-sided 5% critical value is 2.306; |t| is smaller, so H₀ is not rejected.
Answer: U = 100.3
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Answer: r = 0.9498
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Answer: e = 5.495
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
The point lies above the line.
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