CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
years012€1 000FV = ?

You invest €1 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 2 years.

PV = €1 0005% / 4n = 2
FV = €
2.Time value of moneyHard
months0123…192A?AA€75 000

You want €75 000 in 16 years. You save the same amount at the end of every month in an account paying 7.2% per year, compounded monthly. Find the monthly deposit A.

FV = €75 0007.2% / 12N = 192
A = €
3.NPV & IRRMedium
years0123…8−30 0005 5005 5005 50011 500

A machine costs €30 000 today. It generates €5 500 at the end of each year for 8 years and can then be sold for €6 000. The cost of capital is 10%. Find the net present value NPV.

C0 = €30 000CF = €5 500S = €6 000n = 8r = 10%
NPV = €
4.NPV & IRRHard
years01234−80 00028 00028 00028 00028 000

An investment of €80 000 returns €28 000 at the end of each year for 4 years. Find its internal rate of return IRR in percent.

C0 = €80 000CF = €28 000n = 4
IRR = %
5.Bond pricing & yieldMedium
half-years0123456€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 3 years. The yield to maturity is 4% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 4%n = 3
P = €
6.Bond pricing & yieldHard
years0123456€1 075€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 6 years and sells for €1 075. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 6P = €1 075
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€6 000 000re​ = 8%debt€2 000 000rd​ = 3.5%T = 25.8%

A company has equity worth €6 000 000 and debt worth €2 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.

E = €6 000 000D = €2 000 000re = 8%rd = 3.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
rf​ = 2%, rm​ − rf​ = 6.5%no debtequityβU​ = 1D/E = 0.6equityre​ = ?debtT = 21%

A company's business has an unlevered (asset) beta of 1. It plans a debt-to-equity ratio of 0.6. The tax rate is 21%, the risk-free rate is 2% and the market risk premium is 6.5%. Find the cost of equity re at the new capital structure.

βU = 1D/E = 0.6T = 21%rf = 2%rm − rf = 6.5%
re = %
9.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 58, 37, 37, 10, 32.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 66?

μ = 50σ = 8x = 66
P =
11.ProbabilityMedium
red: 7blue: 7

A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 7
P =
12.ProbabilityHard
xp(x)00.3510.3520.3μ

X has the distribution shown, and Y = 10X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 10X − 3
σY =
13.Normal distributionMedium
120 kg118.7 kgT = X₁ + … + X₁₀?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 118.7 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 10< 118.7 kg
P =
14.Normal distributionHard
10 cm0X − Y?

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
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 60 phones, mean 95 h, known standard deviation 16 h. Model 2: 100 phones, mean 96.5 h, known standard deviation 17 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 95 hσ₁ = 16 hn₂ = 100x̄₂ = 96.5 hσ₂ = 17 h95%
E = h
16.Confidence intervalsHard
n = 10σ = ?Us = 7.895%

The diameters of 10 machined shafts have a sample standard deviation of 7.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 7.8 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 123.9 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 123.9s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 195 of 300 patients on treatment A recover, and 92 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 195n₁ = 300x₂ = 92n₂ = 120
z =
19.Linear regression & correlationMedium
xy165918632068227324852683xy2025607080

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy1533625817959100xy56080100x = 6

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6, 90) is made. Find its residual e = y − ŷ.

n = 5(6, 90)
e =

3Solutions

1.Time value of money

Answer: FV = 1104 €

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.

Rate and number of periods

i = r/m = 0.05/4 = 0.0125
N = m·n = 4·2 = 8

Future value

FV = 1 000·(1 + 0.0125)8 = 1 000·1.104 = €1 104.49

Answer

FV = €1 104.49
2.Time value of money

Answer: A = 208.9 €

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 deposits

i = 0.07200000000000001/12 = 0.006
N = 12·16 = 192

Deposit

A = FV·i/((1 + i)N − 1) = 75 000·0.006/(3.154 − 1) = €208.95

Answer

A = €208.95
3.NPV & IRR

Answer: NPV = 2141 €

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 the yearly cash flows

5 500·5.335 = 29 342.09

Present value of the salvage value

6 000/2.144 = 2 799.04

NPV

NPV = −30 000 + 29 342.09 + 2 799.04 = €2 141.14

Answer

NPV = €2 141.14
4.NPV & IRR

Answer: IRR = 14.96 %

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.

Required annuity factor

(1 − (1 + i)−4)/i = 80 000/28 000 = 2.857

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 14%: 2.914
i = 15%: 2.855
IRR = 14.96%

Answer

IRR = 14.96%
5.Bond pricing & yield

Answer: P = 1084 €

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 = €35.00, i = 0.02, N = 6

Present value of the coupons

€35.00·(1 − 0.888)/0.02 = €196.05

Present value of the face value

€1 000·0.888 = €887.97

Price

P = €1 084.02

Answer

P = €1 084.02

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTM = 4.544 %

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

€1 075 = €60·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(4%) = €1 104.84, P(5%) = €1 050.76
YTM = 4.544%

Answer

YTM = 4.544%
7.CAPM & WACC

Answer: WACC = 6.649 %

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).

Weights

V = €8 000 000, E/V = 0.75, D/V = 0.25

WACC

WACC = 0.75·8% + 0.25·3.5%·(1 − 0.258) = 6.649%

Answer

WACC = 6.649%
8.CAPM & WACC

Answer: re = 11.58 %

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 + 0.79·0.6) = 1.474

Cost of equity (CAPM)

re = 2% + 1.474·6.5% = 11.58%

Answer

re = 11.58%
9.Descriptive statistics

Answer: s = 17.11

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.

Mean

x̄ = 174/5 = 34.8

Squared deviations

Σ(x − x̄)² = 23.2² + 2.2² + 2.2² + (−24.8)² + (−2.8)² = 1171

Standard deviation

s = √(1171/4) = 17.11

Answer

s = 17.11
10.Descriptive statistics

Answer: P = 0.9772

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) = 0.9772
P = 0.9772

Answer

P = 0.9772
11.Probability

Answer: P = 0.2308

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 = 7/14 · 6/13 = 42/182 = 0.2308

Answer

P = 0.2308
12.Probability

Answer: σY = 8.047

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 = 0.95, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |10|·0.8047 = 8.047

Answer

σY = 8.047

The mean does move: E(Y) = 10·0.95 − 3 = 6.5.

13.Normal distribution

Answer: P = 0.305

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 = 120 kg, σT = 0.8·√10 = 2.53 kg

Standardise

z = (118.7 − 120)/2.53 = −0.51

Probability

P = Φ(−0.51) = 0.305

Answer

P = 0.305
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 = 5.243 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.

Point estimate

x̄₁ − x̄₂ = 95 − 96.5 = −1.5

Standard error

SE = √(16²/60 + 17²/100) = 2.675

Margin of error

E = 1.96·2.675 = 5.243

Answer

E = 5.243 h

The interval contains 0, so the data do not show a significant difference.

16.Confidence intervals

Answer: σ = 5.365 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 9·7.8² = 547.6

Limits for σ²

547.6/19.023 = 28.78
547.6/2.7 = 202.8

Limit for σ

σ = √28.78 = 5.365 µm

Answer

σ = 5.365 µm

95% interval for σ: [5.365, 14.24] µm.

17.Hypothesis tests & confidence intervals

Answer: t = 2.925

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 = 12/√81 = 12/9 = 1.333

Test statistic

t = (123.9 − 120)/1.333 = 2.925

Answer

t = 2.925
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −2.322

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 = 195/300 = 0.65
p̂₂ = 92/120 = 0.7667
p̂ = 287/420 = 0.6833

Standard error under H₀

SE = √(0.6833·0.3167·(1/300 + 1/120)) = 0.05024

Test statistic

z = (0.65 − 0.7667)/0.05024 = −2.322

Answer

z = −2.322

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.9674

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 = 126/6 = 21
ȳ = 431/6 = 71.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 191
Syy = Σ(y − ȳ)² = 556.8

Correlation coefficient

r = 191/√(70·556.8) = 0.9674

Answer

r = 0.9674
20.Linear regression & correlation

Answer: e = 5.45

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 = 25/5 = 5
ȳ = 391/5 = 78.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 254

Regression line

b = 254/40 = 6.35
a = 78.2 − 6.35·5 = 46.45
ŷ = 46.45 + 6.35·x

Predicted value

ŷ = 46.45 + 6.35·6 = 84.55

Residual

e = 90 − 84.55 = 5.45

Answer

e = 5.45

The point lies above the line.

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