A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 2% per year forever. The required return is 4% per year. Find the value PV of the property today.
C1 = €5 000g = 2%r = 4%
PV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 6 years. The interest rate is 7.2% per year, compounded monthly. Find the monthly payment A.
P = €25 0007.2% / 12N = 72
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €15 000, €23 000, €19 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.
A project costs €20 000 and returns €11 500, €13 500, €13 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 11 500CF2 = 13 500CF3 = 13 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €1 349 when the yield to maturity is 2.5%. Find the coupon rate c in percent.
F = €1 000n = 7P = €1 349YTM = 2.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 12 years. It sells for €1 258. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 5% (2×)n = 12P = €1 258
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 500 000 and debt worth €6 000 000. The cost of equity is 10%, the cost of preferred stock 7.5% and the pre-tax cost of debt 3%. The tax rate is 25%. Find the WACC.
A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25%, the risk-free rate 3.5% and the market risk premium 5.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 43, 34, 25, 22, 54.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 20?
μ = 50σ = 15x = 20
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.15, 0.35
σ =
12.ProbabilityHard
7 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 7p = 0.25
P =
13.Normal distributionMedium
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 30 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h25%
x = h
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 7.5. What is the margin of error E₂ of a 90% interval based on 450 observations?
95%n₁ = 50E₁ = 7.590%n₂ = 450
E₂ =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0199%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 50.96 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 50.96s = 4n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 27. Find the minimum sample size n.
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 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 35?
n = 6ŷ = 35
x =
3Solutions
1.Time value of money
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.
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 rate and number of payments
i = 0.07200000000000001/12 = 0.006
N = 12·6 = 72
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.65
A = 25 000·0.006/(1 − 0.65) = €428.63
Answer
A = €428.63
3.NPV & IRR
Answer: NPV = −10383 €
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.
Present value of each cash flow
CF0 = −60 000
CF1/(1 + 0.07)1 = 15 000/1.07 = 14 018.69
CF2/(1 + 0.07)2 = 23 000/1.145 = 20 089.09
CF3/(1 + 0.07)3 = 19 000/1.225 = 15 509.66
Add them up
NPV = €−10 382.56
Answer
NPV = €−10 382.56
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 39.38 %
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(35%) = 1 209.67
NPV(40%) = −160.35
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 + 0.025)−7 = €841.27
Coupon
C = (1349 − 841.27)/6.349 = €79.97
Coupon rate
c = 79.97/1000 = 7.997%
Answer
c = 7.997%
6.Bond pricing & yield
Answer: YTM = 2.498 %
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.
Per period
C = €25.00, N = 24
Solve numerically
i = 1.249% per half-year
YTM
YTM = 2·1.249% = 2.498%
Answer
YTM = 2.498%
7.CAPM & WACC
Answer: WACC = 6.278 %
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.
Unlever the beta
βU = 1.5/(1 + 0.75·0.5) = 1.091
Relever at the new D/E
βL = 1.091·(1 + 0.75·1) = 1.909
Cost of equity (CAPM)
re = 3.5% + 1.909·5.5% = 14%
Answer
re = 14%
9.Descriptive statistics
Answer: s = 13.16
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
11.Probability
Answer: σ = 1.244
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.
Mean
μ = 1·0.3 + 2·0.2 + 3·0.15 + 4·0.35 = 2.55
Mean of the squares
E(X²) = 1·0.3 + 4·0.2 + 9·0.15 + 16·0.35 = 8.05
Variance
Var(X) = 8.05 − 2.55² = 1.548
σ = √1.548 = 1.244
Answer
σ = 1.244
12.Probability
Answer: P = 0.4449
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.
Zero and one failure
P(X = 0) = 0.757 = 0.1335
P(X = 1) = 7·0.25·0.756 = 0.3115
Add them
P = 0.1335 + 0.3115 = 0.4449
Answer
P = 0.4449
13.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 sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (30 − 32)/2.5 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
14.Normal distribution
Answer: x = 473 h
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 500 − 0.67·40 = 473 h
Answer
x = 473 h
15.Confidence intervals
Answer: E₂ = 2.098
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.
Ratio of margins
E₂/E₁ = (1.645/1.96)·√(50/450) = 0.2798
New margin
E₂ = 7.5·0.2798 = 2.098
Answer
E₂ = 2.098
16.Confidence intervals
Answer: n = 16590
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.5·0.5/0.01² = 16589
Round up
n = 16590
Answer
n = 16590
17.Hypothesis tests & confidence intervals
Answer: t = 2.4
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 = 4/√100 = 4/10 = 0.4
Test statistic
t = (50.96 − 50)/0.4 = 2.4
Answer
t = 2.4
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 250
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 1.282)·27/5)² = 249.8
Round up
n = 250
Answer
n = 250
19.Linear regression & correlation
Answer: r = 0.9257
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 = 346/6 = 57.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 67
Syy = Σ(y − ȳ)² = 299.3
Correlation coefficient
r = 67/√(17.5·299.3) = 0.9257
Answer
r = 0.9257
20.Linear regression & correlation
Answer: x = 11.06
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 = 42/6 = 7
ȳ = 131/6 = 21.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 227
Regression line
b = 227/70 = 3.243
a = 21.83 − 3.243·7 = −0.8667
ŷ = −0.8667 + 3.243·x
Solve for x
x = (35 − (−0.8667))/3.243 = 11.06
Answer
x = 11.06
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.