You invest €12 000 at a nominal rate of 6% per year, compounded quarterly. Find the value FV of the investment after 4 years.
PV = €12 0006% / 4n = 4
FV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 20 years at 2.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0002.4% / 12N = 240
I = €
3.NPV & IRRMedium
A project costs €500 000 today. It generates €60 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 12%. Find the net present value NPV.
C0 = €500 000CF1 = €60 000g = 2%r = 12%
NPV = €
4.NPV & IRRHard
A project costs €10 000 and returns €6 500, €6 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 6 500CF2 = 6 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 10 years and pays annual coupons. It sells for €765 when the yield to maturity is 8%. Find the coupon rate c in percent.
F = €1 000n = 10P = €765YTM = 8%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 6 years and sells for €875. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 6P = €875
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 000 000 and debt worth €3 000 000. The cost of equity is 10%, the cost of preferred stock 6% and the pre-tax cost of debt 5%. The tax rate is 30%. Find the WACC.
A company's business has an unlevered (asset) beta of 1. It plans a debt-to-equity ratio of 0.5. The tax rate is 30%, the risk-free rate is 2.5% and the market risk premium is 7%. Find the cost of equity re at the new capital structure.
βU = 1D/E = 0.5T = 30%rf = 2.5%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 54, 44, 27, 55, 35.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 90.
μ = 100σ = 20x = 90
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.35, 0.15, 0.35, 0.15
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = 5X + 1. Find the standard deviation σY of Y.
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 29.5 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 29.5 min
P =
14.Normal distributionHard
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.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.1 and standard deviation 3. Find the lower limit of the 95% confidence interval for the mean difference.
n = 13d̄ = −5.1sd = 395%t(12) = 2.179
lower limit = mmHg
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 22 plots, mean yield 60 kg, standard deviation 10 kg. Plot group 2: 9 plots, mean 54 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
A sample of 49 students has a mean score of 72.79 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 72.79s = 5n = 49
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 90 of 200 patients on treatment A recover, and 150 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 90n₁ = 200x₂ = 150n₂ = 300
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.
n = 5x = 3
ŷ =
3Solutions
1.Time value of money
Answer: FV = 15228 €
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.
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.
Value of the cash flows
PV = 60 000/(0.12 − 0.02) = 600 000.00
NPV
NPV = 600 000.00 − 500 000 = €100 000.00
Answer
NPV = €100 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 16.5 %
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(15%) = 189.04
NPV(20%) = −416.67
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 15 + 5·189.04/(189.04 − (−416.67)) = 16.56%
exact: NPV(16.5%) = 0 → IRR = 16.5%
Answer
IRR = 16.5%
5.Bond pricing & yield
Answer: c = 4.498 %
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.
Present value of the face value
1000·(1 + 0.08)−10 = €463.19
Coupon
C = (765 − 463.19)/6.71 = €44.98
Coupon rate
c = 44.98/1000 = 4.498%
Answer
c = 4.498%
6.Bond pricing & yield
Answer: YTM = 5.502 %
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
€875 = €30·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(5%) = €898.49, P(6%) = €852.48
YTM = 5.502%
Answer
YTM = 5.502%
7.CAPM & WACC
Answer: WACC = 7.65 %
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).
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Levered beta
βL = 1·(1 + 0.7·0.5) = 1.35
Cost of equity (CAPM)
re = 2.5% + 1.35·7% = 11.95%
Answer
re = 11.95%
9.Descriptive statistics
Answer: s = 12.86
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 = (90 − 100)/20 = −0.5
Answer
z = −0.5
11.Probability
Answer: Var(X) = 4.84
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.
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 = 2, Var(X) = 2.1, σX = 1.449
Linear change
σY = |5|·1.449 = 7.246
Answer
σY = 7.246
The mean does move: E(Y) = 5·2 + 1 = 11.
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 sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (29.5 − 32)/2.5 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
14.Normal distribution
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).
Distribution of the difference
μ = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: lower limit = −6.913 mmHg
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.
Critical value
df = 12, t = 2.179
Margin of error
E = 2.179·3/√13 = 1.813
Interval
−5.1 ± 1.813 → [−6.913, −3.287]
Answer
lower limit = −6.913 mmHg
16.Confidence intervals
Answer: lower limit = −0.3039 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 = (21·10² + 8·7.5²)/29 = 87.93
sp = 9.377
Margin of error
E = 1.699·9.377·√(1/22 + 1/9) = 6.304
Interval
6 ± 6.304 → [−0.3039, 12.3]
Answer
lower limit = −0.3039 kg
17.Hypothesis tests & confidence intervals
Answer: t = −3.094
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).
Standard error
SE = 5/√49 = 5/7 = 0.7143
Test statistic
t = (72.79 − 75)/0.7143 = −3.094
Answer
t = −3.094
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −1.096
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.
Sample proportions
p̂₁ = 90/200 = 0.45
p̂₂ = 150/300 = 0.5
p̂ = 240/500 = 0.48
Standard error under H₀
SE = √(0.48·0.52·(1/200 + 1/300)) = 0.04561
Test statistic
z = (0.45 − 0.5)/0.04561 = −1.096
Answer
z = −1.096
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.9574
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 = 80/5 = 16
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30
Syy = Σ(y − ȳ)² = 94
Coefficient of determination
r² = −30²/(10·94) = 0.9574
Answer
r² = 0.9574
About 95.74% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 54.7
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 = 30/5 = 6
ȳ = 332/5 = 66.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156
Regression line
b = 156/40 = 3.9
a = 66.4 − 3.9·6 = 43
ŷ = 43 + 3.9·x
Prediction
ŷ = 43 + 3.9·3 = 54.7
Answer
ŷ = 54.7
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.