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
years0123456PV = ?€2 000€2 000€2 000€2 000

An annuity pays €2 000 at the end of each year for 6 years. The interest rate is 3% per year. Find the present value PV of the annuity.

A = €2 000i = 3%n = 6
PV = €
2.Time value of moneyHard
months0123…60€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 5 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.

P = €8 0004.8% / 12N = 60
A = €
3.NPV & IRRMedium
years0123…9−250 00014 000+3%+3%…r = 10%, NPV = ?

A project costs €250 000 today. It generates €14 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 10%. Find the net present value NPV.

C0 = €250 000CF1 = €14 000g = 3%r = 10%
NPV = €
4.NPV & IRRHard
years012−25 00010 50018 500

A project costs €25 000 and returns €10 500, €18 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 10 500CF2 = 18 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€10€10€10 + €1 000P = ?

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 8% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 8%n = 9
P = €
6.Bond pricing & yieldHard
years0123€1 055€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 3 years and sells for €1 055. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 3P = €1 055
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 10%debtD = 0.75·Erd​ = 4.5%T = 25.8%

A company has a debt-to-equity ratio (at market values) of 0.75. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 25.8%. Find the WACC.

D/E = 0.75re = 10%rd = 4.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €20.00β = 0.9rf​ = 4%, rm​ = 9%debt20 000 × €1 050rd​ = 3%T = 30%

A company has 2 500 000 shares at €20.00 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 30%. The share’s beta is 0.9, the risk-free rate 4% and the market return 9%. Find the WACC.

β = 0.9rf = 4%rm = 9%rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
38404244464850525456

Find the sample standard deviation s of the data: 52, 37, 56, 51, 52, 49.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 60.4?

μ = 70σ = 8x = 60.4
P =
11.ProbabilityMedium
red: 8blue: 4

A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 4
P =
12.ProbabilityHard
xp(x)10.1520.2530.240.2550.15μ

X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.15, 0.25, 0.2, 0.25, 0.15Y = 5X + 10
σY =
13.Normal distributionMedium
32 min29.5 minX + 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
235240245250255260265gbottom 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
15.Confidence intervalsMedium
n = 9, s = 11L = ?Ux̄ = 96E90%

A sample of 9 items has mean 96 and standard deviation s = 11. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 9x̄ = 96s = 1190%t(8) = 1.86
lower limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 89 kg, standard deviation 8 kg. Plot group 2: 16 plots, mean 88 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 89s₁ = 8n₂ = 16x̄₂ = 88s₂ = 4.595%t(19) = 2.093
E = kg
17.Hypothesis tests & confidence intervalsMedium
2.4p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 9. A random sample of 36 observations has a mean of 503.6. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 9n = 36x̄ = 503.6H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 100 measurements has mean 47.5. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 47.5σ = 15n = 10090%
L =
19.Linear regression & correlationMedium
xy125223321418513611xy24610152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy237337448548658xy246405060ŷ = 43

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 43?

n = 5ŷ = 43
x =

3Solutions

1.Time value of money

Answer: PV = 10834 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.8375)/0.03 = 5.417

Present value

PV = 2 000·5.417 = €10 834.38

Answer

PV = €10 834.38
2.Time value of money

Answer: A = 150.2 €

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 = 12·5 = 60

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 = 0.787
A = 8 000·0.004/(1 − 0.787) = €150.24

Answer

A = €150.24
3.NPV & IRR

Answer: NPV = −50000 €

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 = 14 000/(0.1 − 0.03) = 200 000.00

NPV

NPV = 200 000.00 − 250 000 = −€50 000.00

Answer

NPV = −€50 000.00

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 9.549 %

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(5%) = 1 780.05
NPV(10%) = −165.29

Narrow down

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

IRR ≈ 5 + 5·1 780.05/(1 780.05 − (−165.29)) = 9.575%
exact: NPV(9.549%) = 0 → IRR = 9.549%

Answer

IRR = 9.549%
5.Bond pricing & yield

Answer: P = 620.2 €

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 = €10.00, i = 0.04, N = 18

Present value of the coupons

€10.00·(1 − 0.4936)/0.04 = €126.59

Present value of the face value

€1 000·0.4936 = €493.63

Price

P = €620.22

Answer

P = €620.22

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: YTM = 5.945 %

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

€1 055 = €80·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(5%) = €1 081.70, P(6%) = €1 053.46
YTM = 5.945%

Answer

YTM = 5.945%
7.CAPM & WACC

Answer: WACC = 7.145 %

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 from D/E

E/V = 1/(1 + 0.75) = 0.5714
D/V = 0.75/(1 + 0.75) = 0.4286

WACC

WACC = 0.5714·10% + 0.4286·4.5%·(1 − 0.258) = 7.145%

Answer

WACC = 7.145%
8.CAPM & WACC

Answer: WACC = 6.607 %

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 = 2 500 000·€20.00 = €50 000 000
D = 20 000·€1 050 = €21 000 000

Cost of equity (CAPM)

re = 4% + 0.9·5% = 8.5%

Weights

V = €71 000 000, E/V = 0.7042, D/V = 0.2958

WACC

WACC = 0.7042·8.5% + 0.2958·3%·(1 − 0.3) = 6.607%

Answer

WACC = 6.607%
9.Descriptive statistics

Answer: s = 6.535

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 = 297/6 = 49.5

Squared deviations

Σ(x − x̄)² = 2.5² + (−12.5)² + 6.5² + 1.5² + 2.5² + (−0.5)² = 213.5

Standard deviation

s = √(213.5/5) = 6.535

Answer

s = 6.535
10.Descriptive statistics

Answer: P = 0.1151

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 0.1151

Answer

P = 0.1151
11.Probability

Answer: P = 0.4242

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 = 8/12 · 7/11 = 56/132 = 0.4242

Answer

P = 0.4242
12.Probability

Answer: σY = 6.519

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 = 3, Var(X) = 1.7, σX = 1.304

Linear change

σY = |5|·1.304 = 6.519

Answer

σY = 6.519

The mean does move: E(Y) = 5·3 + 10 = 25.

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: x = 248.8 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 250 − 0.25·5 = 248.8 g

Answer

x = 248.8 g
15.Confidence intervals

Answer: lower limit = 89.18

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 = 8, t = 1.86

Margin of error

E = 1.86·11/√9 = 6.82

Interval

96 ± 6.82 → [89.18, 102.8]

Answer

lower limit = 89.18
16.Confidence intervals

Answer: E = 5.82 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 = (4·8² + 15·4.5²)/19 = 29.46
sp = 5.428

Margin of error

E = 2.093·5.428·√(1/5 + 1/16) = 5.82

Answer

E = 5.82 kg
17.Hypothesis tests & confidence intervals

Answer: p = 0.008198

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.

Test statistic

SE = σ/√n = 9/√36 = 1.5
z = (503.6 − 500)/1.5 = 2.4

Tail area

Φ(2.4) = 0.9918
1 − Φ(2.4) = 0.0082

p-value

p = 0.008198

Answer

p = 0.008198

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: L = 45.03

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).

Margin of error

E = 1.645·15/√100 = 2.468

Interval limit

L = 47.5 − 2.468 = 45.03

Answer

L = 45.03
19.Linear regression & correlation

Answer: r = −0.9872

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 = 111/6 = 18.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −51.5
Syy = Σ(y − ȳ)² = 155.5

Correlation coefficient

r = −51.5/√(17.5·155.5) = −0.9872

Answer

r = −0.9872
20.Linear regression & correlation

Answer: x = 3.509

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 = 20/5 = 4
ȳ = 228/5 = 45.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 53

Regression line

b = 53/10 = 5.3
a = 45.6 − 5.3·4 = 24.4
ŷ = 24.4 + 5.3·x

Solve for x

x = (43 − 24.4)/5.3 = 3.509

Answer

x = 3.509

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