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
years0123PV = ?€25 000

You need €25 000 in 3 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €25 0004% / 12n = 3
PV = €
2.Time value of moneyHard
months0123…72€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 6 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.

P = €8 0004.8% / 12N = 72
A = €
3.NPV & IRRMedium
years01234€40 000€1 000€1 000€1 000€1 000r = 12%, EAC = ?

A machine costs €40 000 today, lasts 4 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 = 4r = 12%
EAC = €
4.NPV & IRRHard
years012345−25 0007 0007 0007 0007 0007 000

An investment of €25 000 returns €7 000 at the end of each year for 5 years. Find its internal rate of return IRR in percent.

C0 = €25 000CF = €7 000n = 5
IRR = %
5.Bond pricing & yieldMedium
years0123…12€957C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €957 when the yield to maturity is 5.5%. Find the coupon rate c in percent.

F = €1 000n = 12P = €957YTM = 5.5%
c = %
6.Bond pricing & yieldHard

A bond trades at €1 070 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 3.6 years. Use the modified duration to estimate the new price P if the yield falls by 150 basis points.

P0 = €1 070y = 7%D = 3.6Δy = −150 bp
P = €
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 13%debt€2 000 000rd​ = 7%T = 30%

A company has equity worth €2 000 000 and debt worth €2 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 7% and the tax rate is 30%. Find the WACC.

E = €2 000 000D = €2 000 000re = 13%rd = 7%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €31.00β = 1.5rf​ = 1.5%, rm​ = 7.5%debt40 000 × €1 000rd​ = 7%T = 30%

A company has 1 000 000 shares at €31.00 each and 40 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 1.5% and the market return 7.5%. Find the WACC.

β = 1.5rf = 1.5%rm = 7.5%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 50, 15, 54, 48, 31, 29.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 65.

μ = 70σ = 10x = 65
z =
11.ProbabilityMedium
red: 7blue: 5

A bag holds 7 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 5
P =
12.ProbabilityHard
n = 12, p = 0.1: P(X = 1) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 1 correct answers?

n = 12p = 0.1k = 1
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8324

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8324. Find z to two decimals.

A = 0.8324
z =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 25, s = 15L = ?Ux̄ = 100.5E95%

A sample of 25 items has mean 100.5 and standard deviation s = 15. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 25x̄ = 100.5s = 1595%t(24) = 2.064
lower limit =
16.Confidence intervalsHard
n = 20σ = ?Us = 395%

The diameters of 20 machined shafts have a sample standard deviation of 3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 3 µmχ² = 8.907, 32.852
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 8 students, mean score 33.5, standard deviation 11.5. Group 2: 13 students, mean score 24, standard deviation 4.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 8x̄₁ = 33.5s₁ = 11.5n₂ = 13x̄₂ = 24s₂ = 4.5
t =
18.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 390%
n =
19.Linear regression & correlationMedium
xy15821890211082413127139xy15202580100120140

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy2346688121017xy51051015x = 5

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (5, 3) is made. Find its residual e = y − ŷ.

n = 5(5, 3)
e =

3Solutions

1.Time value of money

Answer: PV = 22177 €

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 = 0.04/12 = 0.003333
N = 12·3 = 36

Present value

PV = FV/(1 + i)N = 25 000/1.127 = €22 177.44

Answer

PV = €22 177.44
2.Time value of money

Answer: A = 128.1 €

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 payments

i = 0.048/12 = 0.004
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 = 0.7502
A = 8 000·0.004/(1 − 0.7502) = €128.10

Answer

A = €128.10
3.NPV & IRR

Answer: EAC = 14169 €

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 − 1.12−4)/0.12 = 3.037

Annualised purchase price

40 000/3.037 = 13 169.38

EAC

EAC = 13 169.38 + 1 000 = €14 169.38

Answer

EAC = €14 169.38
4.NPV & IRR

Answer: IRR = 12.38 %

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.

Required annuity factor

(1 − (1 + i)−5)/i = 25 000/7 000 = 3.571

Solve for the rate

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

i = 12%: 3.605
i = 13%: 3.517
IRR = 12.38%

Answer

IRR = 12.38%
5.Bond pricing & yield

Answer: c = 5.001 %

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 + 0.055)−12 = €525.98

Coupon

C = (957 − 525.98)/8.619 = €50.01

Coupon rate

c = 50.01/1000 = 5.001%

Answer

c = 5.001%
6.Bond pricing & yield

Answer: P = 1124 €

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 = 3.6/1.07 = 3.364

Estimated price change

%ΔP ≈ −3.364·(−0.015) = 5.047%
P ≈ 1070·(1 + 0.05047) = €1 124.00

Answer

P = €1 124.00

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

7.CAPM & WACC

Answer: WACC = 8.95 %

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 = €4 000 000, E/V = 0.5, D/V = 0.5

WACC

WACC = 0.5·13% + 0.5·7%·(1 − 0.3) = 8.95%

Answer

WACC = 8.95%
8.CAPM & WACC

Answer: WACC = 7.345 %

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).

Market values

E = 1 000 000·€31.00 = €31 000 000
D = 40 000·€1 000 = €40 000 000

Cost of equity (CAPM)

re = 1.5% + 1.5·6% = 10.5%

Weights

V = €71 000 000, E/V = 0.4366, D/V = 0.5634

WACC

WACC = 0.4366·10.5% + 0.5634·7%·(1 − 0.3) = 7.345%

Answer

WACC = 7.345%
9.Descriptive statistics

Answer: s = 15.22

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 = 227/6 = 37.83

Squared deviations

Σ(x − x̄)² = 12.17² + (−22.83)² + 16.17² + 10.17² + (−6.833)² + (−8.833)² = 1159

Standard deviation

s = √(1159/5) = 15.22

Answer

s = 15.22
10.Descriptive statistics

Answer: z = −0.5

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 = (65 − 70)/10 = −0.5

Answer

z = −0.5
11.Probability

Answer: P = 0.3182

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 = 7/12 · 6/11 = 42/132 = 0.3182

Answer

P = 0.3182
12.Probability

Answer: P = 0.3766

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 = 1) = 12·0.11·0.911 = 12·0.1·0.3138 = 0.3766

Answer

P = 0.3766
13.Normal distribution

Answer: z = 1.38

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 of z

Φ(z) = (1 + 0.8324)/2 = 0.9162

Read the z-table backwards

Find 0.9162 inside the table and read off z.

z = 1.38

Answer

z = 1.38
14.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
15.Confidence intervals

Answer: lower limit = 94.31

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 = 24, t = 2.064

Margin of error

E = 2.064·15/√25 = 6.192

Interval

100.5 ± 6.192 → [94.31, 106.7]

Answer

lower limit = 94.31
16.Confidence intervals

Answer: σ = 2.281 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 = 19·3² = 171

Limits for σ²

171/32.852 = 5.205
171/8.907 = 19.2

Limit for σ

σ = √5.205 = 2.281 µm

Answer

σ = 2.281 µm

95% interval for σ: [2.281, 4.382] µm.

17.Hypothesis tests & confidence intervals

Answer: t = 2.696

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.

Pooled variance

sp² = (7·11.5² + 12·4.5²)/19 = 61.51
sp = 7.843

Standard error

SE = 7.843·√(1/8 + 1/13) = 3.524

Test statistic

t = (33.5 − 24)/3.524 = 2.696

Answer

t = 2.696

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 5

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).

Solve for n

n = (1.645·4/3)² = 4.811

Round up

n = 5

Answer

n = 5
19.Linear regression & correlation

Answer: r = 0.9862

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 = 105/5 = 21
ȳ = 550/5 = 110

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 465
Syy = Σ(y − ȳ)² = 2470

Correlation coefficient

r = 465/√(90·2470) = 0.9862

Answer

r = 0.9862
20.Linear regression & correlation

Answer: e = −4.5

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 = 46/5 = 9.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 68

Regression line

b = 68/40 = 1.7
a = 9.2 − 1.7·6 = −1
ŷ = −1 + 1.7·x

Predicted value

ŷ = −1 + 1.7·5 = 7.5

Residual

e = 3 − 7.5 = −4.5

Answer

e = −4.5

The point lies below the line.

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