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
years0123…15€10 000FV = ?

You invest €10 000 at 7% per year with continuous compounding. Find the value FV after 15 years.

PV = €10 000r = 7%t = 15
FV = €
2.Time value of moneyHard
months0123…108A?AA€30 000

You want €30 000 in 9 years. You save the same amount at the end of every month in an account paying 2.4% per year, compounded monthly. Find the monthly deposit A.

FV = €30 0002.4% / 12N = 108
A = €
3.NPV & IRRMedium
years01234−60 00022 00027 00021 00015 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €22 000, €27 000, €21 000, €15 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −60 000CF1 = 22 000CF2 = 27 000CF3 = 21 000CF4 = 15 000
NPV = €
4.NPV & IRRHard
years0123−10 0004 5004 5005 000

A project costs €10 000 and returns €4 500, €4 500, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 4 500CF2 = 4 500CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 10 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 8%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 7.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 2%n = 10y0 = 8%y1 = 7.5%
HPR = %
6.Bond pricing & yieldHard

A bond trades at €1 025 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 9.3 years. Use the modified duration to estimate the new price P if the yield rises by 150 basis points.

P0 = €1 025y = 5%D = 9.3Δy = +150 bp
P = €
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 10%debt€1 000 000rd​ = 4.5%T = 28%

A company has equity worth €8 000 000 and debt worth €1 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the WACC.

E = €8 000 000D = €1 000 000re = 10%rd = 4.5%T = 28%
WACC = %
8.CAPM & WACCHard
rf​ = 3.5%, rm​ − rf​ = 4.5%D/E = 0.2equityβL​ = 1debtD/E = 0.75equityre​ = ?debtT = 21%

A company's shares have a beta of 1 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 21%, the risk-free rate 3.5% and the market risk premium 4.5%. Find the cost of equity re after the change.

βL = 1D/E1 = 0.2D/E2 = 0.75T = 21%rf = 3.5%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 59, 58, 21, 16, 35.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 580?

μ = 500σ = 40x = 580
P =
11.ProbabilityMedium
red: 7blue: 8

A bag holds 7 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 8
P =
12.ProbabilityHard
n = 9, p = 0.5: P(X = 3) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?

n = 9p = 0.5k = 3
P =
13.Normal distributionMedium
340 kg388 kgT = X₁ + … + X₄?

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 4 of them exceeds 388 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 4> 388 kg
P =
14.Normal distributionHard
235240245250255260265gbottom 75%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g75%
x = g
15.Confidence intervalsMedium
n = 21, s = 17LU = ?x̄ = 106E90%

A sample of 21 items has mean 106 and standard deviation s = 17. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 21x̄ = 106s = 1790%t(20) = 1.725
upper limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 64.5 kg, standard deviation 8 kg. Plot group 2: 21 plots, mean 76 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 64.5s₁ = 8n₂ = 21x̄₂ = 76s₂ = 1095%t(29) = 2.045
upper limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 103 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 103s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.01βσ = 7, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 7. Find the minimum sample size n.

α = 0.01power = 80%δ = 2σ = 7zα = 2.326zβ = 0.842
n =
19.Linear regression & correlationMedium
xy251355462566677779xy24650607080

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
xy125223317417514xy24152025x = 2.5

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. A new observation (2.5, 16) is made. Find its residual e = y − ŷ.

n = 5(2.5, 16)
e =

3Solutions

1.Time value of money

Answer: FV = 28577 €

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.

Exponent

r·t = 0.07·15 = 1.05
e1.05 = 2.858

Future value

FV = 10 000·2.858 = €28 576.51

Answer

FV = €28 576.51
2.Time value of money

Answer: A = 249.1 €

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 = 0.024/12 = 0.002
N = 12·9 = 108

Deposit

A = FV·i/((1 + i)N − 1) = 30 000·0.002/(1.241 − 1) = €249.13

Answer

A = €249.13
3.NPV & IRR

Answer: NPV = 1930 €

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 + 0.15)1 = 22 000/1.15 = 19 130.43
CF2/(1 + 0.15)2 = 27 000/1.322 = 20 415.88
CF3/(1 + 0.15)3 = 21 000/1.521 = 13 807.84
CF4/(1 + 0.15)4 = 15 000/1.749 = 8 576.30

Add them up

NPV = €1 930.45

Answer

NPV = €1 930.45

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 18.54 %

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%) = 603.27
NPV(20%) = −231.48

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 15 + 5·603.27/(603.27 − (−231.48)) = 18.61%
exact: NPV(18.54%) = 0 → IRR = 18.54%

Answer

IRR = 18.54%
5.Bond pricing & yield

Answer: HPR = 12.01 %

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.

Purchase price

C = €20.00, i = 0.08, N = 10
€20.00·(1 − 0.4632)/0.08 + €1 000·0.4632 = €134.20 + €463.19 = €597.40

Selling price

C = €20.00, i = 0.075, N = 9
€20.00·(1 − 0.5216)/0.075 + €1 000·0.5216 = €127.58 + €521.58 = €649.16

Holding period return

HPR = (20 + 649.16 − 597.40)/597.40 = 12.01%

Answer

HPR = 12.01%
6.Bond pricing & yield

Answer: P = 888.8 €

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Modified duration

Dmod = 9.3/1.05 = 8.857

Estimated price change

%ΔP ≈ −8.857·(0.015) = −13.29%
P ≈ 1025·(1 − 0.1329) = €888.82

Answer

P = €888.82

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 9.249 %

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 = €9 000 000, E/V = 0.8889, D/V = 0.1111

WACC

WACC = 0.8889·10% + 0.1111·4.5%·(1 − 0.28) = 9.249%

Answer

WACC = 9.249%
8.CAPM & WACC

Answer: re = 9.688 %

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/(1 + 0.79·0.2) = 0.8636

Relever at the new D/E

βL = 0.8636·(1 + 0.79·0.75) = 1.375

Cost of equity (CAPM)

re = 3.5% + 1.375·4.5% = 9.688%

Answer

re = 9.688%
9.Descriptive statistics

Answer: s = 20.14

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 = 189/5 = 37.8

Squared deviations

Σ(x − x̄)² = 21.2² + 20.2² + (−16.8)² + (−21.8)² + (−2.8)² = 1623

Standard deviation

s = √(1623/4) = 20.14

Answer

s = 20.14
10.Descriptive statistics

Answer: P = 0.02275

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 = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
11.Probability

Answer: P = 0.2

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 = 7/15 · 6/14 = 42/210 = 0.2

Answer

P = 0.2
12.Probability

Answer: P = 0.1641

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.

Binomial formula

P(X = 3) = 84·0.53·0.56 = 84·0.125·0.01563 = 0.1641

Answer

P = 0.1641
13.Normal distribution

Answer: P = 0.02275

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 = 340 kg, σT = 12·√4 = 24 kg

Standardise

z = (388 − 340)/24 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
14.Normal distribution

Answer: x = 253.4 g

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 250 + 0.67·5 = 253.4 g

Answer

x = 253.4 g
15.Confidence intervals

Answer: upper limit = 112.4

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 = 20, t = 1.725

Margin of error

E = 1.725·17/√21 = 6.399

Interval

106 ± 6.399 → [99.6, 112.4]

Answer

upper limit = 112.4
16.Confidence intervals

Answer: upper limit = −4.095 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 = (9·8² + 20·10²)/29 = 88.83
sp = 9.425

Margin of error

E = 2.045·9.425·√(1/10 + 1/21) = 7.405

Interval

−11.5 ± 7.405 → [−18.91, −4.095]

Answer

upper limit = −4.095 kg
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 = 5/√16 = 5/4 = 1.25

Test statistic

t = (103 − 100)/1.25 = 2.4

Answer

t = 2.4
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 123

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 = ((2.326 + 0.842)·7/2)² = 122.9

Round up

n = 123

Answer

n = 123
19.Linear regression & correlation

Answer: r = 0.9875

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 = 27/6 = 4.5
ȳ = 390/6 = 65

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 105
Syy = Σ(y − ȳ)² = 646

Correlation coefficient

r = 105/√(17.5·646) = 0.9875

Answer

r = 0.9875
20.Linear regression & correlation

Answer: e = −4.6

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 = 96/5 = 19.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −28

Regression line

b = −28/10 = −2.8
a = 19.2 − (−2.8)·3 = 27.6
ŷ = 27.6 − 2.8·x

Predicted value

ŷ = 27.6 + (−2.8)·2.5 = 20.6

Residual

e = 16 − 20.6 = −4.6

Answer

e = −4.6

The point lies below the line.

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