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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
periods01211 + EAR = ?25% / 2

A credit card charges a nominal interest rate of 25% per year, compounded semi-annually. Find the effective annual rate EAR in percent.

r = 25%m = 2
EAR = %
2.Time value of moneyHard
months0123…216A?AA€20 000

You want €20 000 in 18 years. You save the same amount at the end of every month in an account paying 5.4% per year, compounded monthly. Find the monthly deposit A.

FV = €20 0005.4% / 12N = 216
A = €
3.NPV & IRRMedium
years01234−20 00010 0007 00010 0005 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €10 000, €7 000, €10 000, €5 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −20 000CF1 = 10 000CF2 = 7 000CF3 = 10 000CF4 = 5 000
NPV = €
4.NPV & IRRHard
years0123−20 00014 5009 0009 500

A project costs €20 000 and returns €14 500, €9 000, €9 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 14 500CF2 = 9 000CF3 = 9 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…14€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 9%n = 7
P = €
6.Bond pricing & yieldHard
years01234P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 4 years. The yield to maturity is 3.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 5%n = 4YTM = 3.5%
D = years
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 10%debt€4 000 000rd​ = 5%T = 25%

A company has equity worth €12 000 000 and debt worth €4 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.

E = €12 000 000D = €4 000 000re = 10%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €21.50β = 1.1rf​ = 2.5%, rm​ = 8%debt10 000 × €960rd​ = 4%T = 28%

A company has 5 000 000 shares at €21.50 each and 10 000 bonds trading at €960 each. The pre-tax cost of debt is 4% and the tax rate is 28%. The share’s beta is 1.1, the risk-free rate 2.5% and the market return 8%. Find the WACC.

β = 1.1rf = 2.5%rm = 8%rd = 4%T = 28%
WACC = %
9.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 10, 30, 28, 24, 55, 26.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 88.

μ = 70σ = 12x = 88
z =
11.ProbabilityMedium
red: 6blue: 3

A bag holds 6 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 3
P =
12.ProbabilityHard
€p(x)−30.7990.21470.01

A game costs €3 to play. You win €12 with probability 0.2 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.2€150: p = 0.01
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7967

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7967. Find z to two decimals.

A = 0.7967
z =
14.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 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 = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 30, s = 11.5LU = ?x̄ = 113E90%

A sample of 30 items has mean 113 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 30x̄ = 113s = 11.590%t(29) = 1.699
upper limit =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0390%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.03?

p = 0.5E = 0.0390%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −0.53 and the standard deviation of the differences is 2.3. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = −0.53sd = 2.3
t =
18.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 199%
n =
19.Linear regression & correlationMedium
xy224861181510201226xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy2941162082510261232xy510102030x = 3

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. A new observation (3, 15) is made. Find its residual e = y − ŷ.

n = 6(3, 15)
e =

3Solutions

1.Time value of money

Answer: EAR = 26.56 %

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 per period

r/m = 0.25/2 = 0.125

Compound over one year

EAR = (1 + 0.125)2 − 1 = 1.266 − 1 = 26.56%

Answer

EAR = 26.56%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: A = 54.96 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·18 = 216

Deposit

A = FV·i/((1 + i)N − 1) = 20 000·0.0045/(2.637 − 1) = €54.96

Answer

A = €54.96
3.NPV & IRR

Answer: NPV = 8625 €

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 = −20 000
CF1/(1 + 0.05)1 = 10 000/1.05 = 9 523.81
CF2/(1 + 0.05)2 = 7 000/1.103 = 6 349.21
CF3/(1 + 0.05)3 = 10 000/1.158 = 8 638.38
CF4/(1 + 0.05)4 = 5 000/1.216 = 4 113.51

Add them up

NPV = €8 624.90

Answer

NPV = €8 624.90

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

4.NPV & IRR

Answer: IRR = 33.11 %

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(30%) = 803.37
NPV(35%) = −459.79

Narrow down

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

IRR ≈ 30 + 5·803.37/(803.37 − (−459.79)) = 33.18%
exact: NPV(33.11%) = 0 → IRR = 33.11%

Answer

IRR = 33.11%
5.Bond pricing & yield

Answer: P = 846.7 €

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 = €30.00, i = 0.045, N = 14

Present value of the coupons

€30.00·(1 − 0.54)/0.045 = €306.68

Present value of the face value

€1 000·0.54 = €539.97

Price

P = €846.66

Answer

P = €846.66

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

6.Bond pricing & yield

Answer: D = 3.731 years

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.

Present value of each cash flow

t = 1: 50/1.035 = 48.31, t·PV = 48.31
t = 2: 50/1.071 = 46.68, t·PV = 93.35
t = 3: 50/1.109 = 45.10, t·PV = 135.29
t = 4: 1050/1.148 = 915.01, t·PV = 3 660.06

Duration

P = 1 055.10
D = 3 937.01/1 055.10 = 3.731

Answer

D = 3.731 years
7.CAPM & WACC

Answer: WACC = 8.438 %

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 = €16 000 000, E/V = 0.75, D/V = 0.25

WACC

WACC = 0.75·10% + 0.25·5%·(1 − 0.25) = 8.438%

Answer

WACC = 8.438%
8.CAPM & WACC

Answer: WACC = 8.085 %

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 = 5 000 000·€21.50 = €107 500 000
D = 10 000·€960 = €9 600 000

Cost of equity (CAPM)

re = 2.5% + 1.1·5.5% = 8.55%

Weights

V = €117 100 000, E/V = 0.918, D/V = 0.08198

WACC

WACC = 0.918·8.55% + 0.08198·4%·(1 − 0.28) = 8.085%

Answer

WACC = 8.085%
9.Descriptive statistics

Answer: s = 14.65

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 = 173/6 = 28.83

Squared deviations

Σ(x − x̄)² = (−18.83)² + 1.167² + (−0.8333)² + (−4.833)² + 26.17² + (−2.833)² = 1073

Standard deviation

s = √(1073/5) = 14.65

Answer

s = 14.65
10.Descriptive statistics

Answer: z = 1.5

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 = (88 − 70)/12 = 1.5

Answer

z = 1.5
11.Probability

Answer: P = 0.4167

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 = 6/9 · 5/8 = 30/72 = 0.4167

Answer

P = 0.4167
12.Probability

Answer: E(G) = 0.9 €

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.

Net outcomes

G: 9 (p = 0.2), 147 (p = 0.01), −3 (p = 0.79)

Expected value

E(G) = 9·0.2 + 147·0.01 − 3·0.79 = 0.9 €

Answer

E(G) = 0.9 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = −0.83

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 − 0.7967 = 0.2033

Read the z-table backwards

Find 0.2033 inside the table and read off z.

z = −0.83

Answer

z = −0.83
14.Normal distribution

Answer: P = 0.9207

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 = 25.03 − 25 = 0.03 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.03)/0.02121 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
15.Confidence intervals

Answer: upper limit = 116.6

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 = 29, t = 1.699

Margin of error

E = 1.699·11.5/√30 = 3.567

Interval

113 ± 3.567 → [109.4, 116.6]

Answer

upper limit = 116.6
16.Confidence intervals

Answer: n = 752

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 = 1.645²·0.5·0.5/0.03² = 751.7

Round up

n = 752

Answer

n = 752
17.Hypothesis tests & confidence intervals

Answer: t = −0.7982

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 of the mean difference

SE = 2.3/√12 = 0.664

Test statistic

t = −0.53/0.664 = −0.7982

Answer

t = −0.7982

With df = 11, the two-sided 5% critical value is 2.201; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 166

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 = (2.576·5/1)² = 165.9

Round up

n = 166

Answer

n = 166
19.Linear regression & correlation

Answer: r² = 0.9902

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 = 82/6 = 13.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 160
Syy = Σ(y − ȳ)² = 369.3

Coefficient of determination

r² = 160²/(70·369.3) = 0.9902

Answer

r² = 0.9902

About 99.02% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 3.929

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 = 123/6 = 20.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 165

Regression line

b = 165/70 = 2.357
a = 20.5 − 2.357·7 = 4
ŷ = 4 + 2.357·x

Predicted value

ŷ = 4 + 2.357·3 = 11.07

Residual

e = 15 − 11.07 = 3.929

Answer

e = 3.929

The point lies above the line.

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.

ExamProblems · examproblems.com/courses/cfa-level-1-quant-and-fixed-income