You invest €8 000 at a nominal rate of 6% per year, compounded quarterly. Find the value FV of the investment after 5 years.
PV = €8 0006% / 4n = 5
FV = €
2.Time value of moneyHard
A mortgage of €150 000 is repaid with equal monthly payments over 25 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 15? Find the outstanding balance B.
P = €150 0002.4% / 12N = 300k = 180
B = €
3.NPV & IRRMedium
A machine costs €40 000 today, lasts 6 years and costs €1 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €40 000M = €1 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
A project costs €20 000 and returns €10 500, €13 000, €11 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 10 500CF2 = 13 000CF3 = 11 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 8% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 8%n = 9
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 3 years and sells for €975. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 3P = €975
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.75. Its cost of equity is 12%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.
D/E = 0.75re = 12%rd = 4.5%T = 21%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares trading at €20.00; the book value of its equity is €24 000 000. It just paid a dividend of €3.00 per share, and dividends grow by 2% per year. Its debt has a market value of €24 000 000 and a pre-tax cost of 4.5%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 9 of them is below 801 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 9< 801 kg
P =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
A sample of 9 items has mean 111 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 9x̄ = 111s = 1490%t(8) = 1.86
E =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 12 plots, mean yield 29.5 kg, standard deviation 8.5 kg. Plot group 2: 9 plots, mean 31.5 kg, standard deviation 5.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 140 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 80 parts has mean 143.6 mm, known standard deviation 15 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 81 measurements has mean 50. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 50σ = 4n = 8195%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 13?
n = 5ŷ = 13
x =
3Solutions
1.Time value of money
Answer: FV = 10775 €
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.
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.
Monthly payment
i = 0.002, N = 300
A = 150 000·0.002/(1 − (1 + 0.002)−300) = €665.40
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 300 − 180 = 120
B = 665.4·(1 − (1 + 0.002)−120)/0.002 = €70 925.74
Answer
B = €70 925.74
3.NPV & IRR
Answer: EAC = 10729 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.12−6)/0.12 = 4.111
Annualised purchase price
40 000/4.111 = 9 729.03
EAC
EAC = 9 729.03 + 1 000 = €10 729.03
Answer
EAC = €10 729.03
4.NPV & IRR
Answer: IRR = 32.71 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(30%) = 776.06
NPV(35%) = −618.30
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 30 + 5·776.06/(776.06 − (−618.30)) = 32.78%
exact: NPV(32.71%) = 0 → IRR = 32.71%
Answer
IRR = 32.71%
5.Bond pricing & yield
Answer: P = 778.5 €
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.
Per period
C = €22.50, i = 0.04, N = 18
Present value of the coupons
€22.50·(1 − 0.4936)/0.04 = €284.83
Present value of the face value
€1 000·0.4936 = €493.63
Price
P = €778.46
Answer
P = €778.46
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 4.917 %
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.
Equation for the yield
€975 = €40·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(4%) = €1 000.00, P(5%) = €972.77
YTM = 4.917%
Answer
YTM = 4.917%
7.CAPM & WACC
Answer: WACC = 8.381 %
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).
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).
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.
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.
Standardise
z = (190 − 170)/20 = 1
Answer
z = 1
11.Probability
Answer: P = 0.4667
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.
First and second draw
P = 7/10 · 6/9 = 42/90 = 0.4667
Answer
P = 0.4667
12.Probability
Answer: σY = 58.9
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.
Spread of X
μ = 22.5, Var(X) = 138.8, σX = 11.78
Linear change
σY = |5|·11.78 = 58.9
Answer
σY = 58.9
The mean does move: E(Y) = 5·22.5 + 4 = 116.5.
13.Normal distribution
Answer: P = 0.8413
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).
Distribution of the total
μT = 765 kg, σT = 12·√9 = 36 kg
Standardise
z = (801 − 765)/36 = 1
Probability
P = Φ(1) = 0.8413
Answer
P = 0.8413
14.Normal distribution
Answer: P = 0.7389
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).
Distribution of the difference
μ = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: E = 8.68
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 8, t = 1.86
Margin of error
E = 1.86·14/√9 = 8.68
Answer
E = 8.68
16.Confidence intervals
Answer: lower limit = −7.632 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.
Pooled standard deviation
sp² = (11·8.5² + 8·5.5²)/19 = 54.57
sp = 7.387
Margin of error
E = 1.729·7.387·√(1/12 + 1/9) = 5.632
Interval
−2 ± 5.632 → [−7.632, 3.632]
Answer
lower limit = −7.632 kg
17.Hypothesis tests & confidence intervals
Answer: z = −1.006
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Standard error of the difference
SE = √(20²/40 + 15²/80) = √12.81 = 3.579
Test statistic
z = (140 − 143.6)/3.579 = −1.006
Answer
z = −1.006
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 49.13
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).
Margin of error
E = 1.96·4/√81 = 0.8711
Interval limit
L = 50 − 0.8711 = 49.13
Answer
L = 49.13
19.Linear regression & correlation
Answer: r = 0.9744
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.
Means
x̄ = 21/6 = 3.5
ȳ = 349/6 = 58.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 87.5
Syy = Σ(y − ȳ)² = 460.8
Correlation coefficient
r = 87.5/√(17.5·460.8) = 0.9744
Answer
r = 0.9744
20.Linear regression & correlation
Answer: x = 3.564
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.
Means
x̄ = 15/5 = 3
ȳ = 76/5 = 15.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −39
Regression line
b = −39/10 = −3.9
a = 15.2 − (−3.9)·3 = 26.9
ŷ = 26.9 − 3.9·x
Solve for x
x = (13 − 26.9)/(−3.9) = 3.564
Answer
x = 3.564
This is independent practice material. ExamProblems is not affiliated with or endorsed by the organisation that runs this exam, and the exam name is used only to describe the content. CFA® and Chartered Financial Analyst® are trademarks of CFA Institute, which is not affiliated with ExamProblems and does not endorse it or vouch for its accuracy.