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
years0123…9PV = ?€2 000+2%+2%…r = 5%

A property is expected to earn net rent of €2 000 next year. The rent is expected to grow by 2% per year forever. The required return is 5% per year. Find the value PV of the property today.

C1 = €2 000g = 2%r = 5%
PV = €
2.Time value of moneyHard
months0123…N = ?€20 000€275€275€275€2754.8% / 12

You borrow €20 000 at 4.8% per year, compounded monthly, and repay €275 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €20 000A = €2754.8% / 12
N = payments
3.NPV & IRRMedium
years012345−50 00012 00012 00012 00012 00019 500

A machine costs €50 000 today. It generates €12 000 at the end of each year for 5 years and can then be sold for €7 500. The cost of capital is 8%. Find the net present value NPV.

C0 = €50 000CF = €12 000S = €7 500n = 5r = 8%
NPV = €
4.NPV & IRRHard
years0123…8−100 00012 500+4%+4%+4%r = 12%, NPV = ?

A project costs €100 000 today. Its cash flow is €12 500 at the end of year 1 and grows by 4% per year until the last cash flow at the end of year 8. The cost of capital is 12%. Find the net present value NPV.

C0 = €100 000CF1 = €12 500g = 4%n = 8r = 12%
NPV = €
5.Bond pricing & yieldMedium
half-years0123456€22.50€22.50€22.50 + €1 000P = ?

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 3 years. The yield to maturity is 6% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 6%n = 3
P = €
6.Bond pricing & yieldHard
years012345P€70€70€70 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 5 years. The yield to maturity is 5.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 7%n = 5YTM = 5.5%
D = years
7.CAPM & WACCMedium
WACC = 9.31%equity€5 000 000re​ = ?debt€3 000 000rd​ = 4%T = 21%

A company has equity worth €5 000 000 and debt worth €3 000 000. Its WACC is 9.31%, its pre-tax cost of debt is 4% and the tax rate is 21%. Find the cost of equity re.

E = €5 000 000D = €3 000 000WACC = 9.31%rd = 4%T = 21%
re = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €40.00β = 1.2rf​ = 2%, rm​ = 8.5%debt40 000 × €980rd​ = 7%T = 20%

A company has 1 000 000 shares at €40.00 each and 40 000 bonds trading at €980 each. The pre-tax cost of debt is 7% and the tax rate is 20%. The share’s beta is 1.2, the risk-free rate 2% and the market return 8.5%. Find the WACC.

β = 1.2rf = 2%rm = 8.5%rd = 7%T = 20%
WACC = %
9.Descriptive statisticsMedium
3540455055

Find the sample standard deviation s of the data: 48, 57, 58, 45, 31, 37, 49.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 166.

μ = 170σ = 8x = 166
z =
11.ProbabilityMedium
xp(x)10.220.130.240.250.3μ = 3.3

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.2, 0.2, 0.3
σ =
12.ProbabilityHard
n = 9, p = 0.4: P(X = 5) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?

n = 9p = 0.4k = 5
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9500

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9500. Find z to two decimals.

A = 0.9500
z =
14.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
n = 11, s = 19.5LUx̄ = 37E = ?99%

A sample of 11 items has mean 37 and standard deviation s = 19.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 11x̄ = 37s = 19.599%t(10) = 3.169
E =
16.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 27. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 4?

σ = 27E = 490%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 95 mm, and the line's known standard deviation is 6 mm. Line 2: a sample of 100 parts has mean 97.1 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 95 mmσ₁ = 6 mmn₂ = 100x̄₂ = 97.1 mmσ₂ = 9 mm
z =
18.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 2.595%
n =
19.Linear regression & correlationMedium
xy36957778299611104xy51080100

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy127221319416514612xy246152025x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 66667 €

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.

Growing perpetuity

PV = C1/(r − g) = 2 000/(0.05 − 0.02) = 2 000/0.03 = €66 666.67

Answer

PV = €66 666.67
2.Time value of money

Answer: N = 87 payments

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.

Rearrange the annuity formula

i = 0.004
(1 + i)−N = 1 − 20 000·0.004/275 = 0.7091

Solve with logarithms

N = −ln(0.7091)/ln(1.004) = 86.11

Round up to whole payments

86 full payments leave a small balance, which the last, smaller payment clears.

N = 87

Answer

N = 87 payments
3.NPV & IRR

Answer: NPV = 3017 €

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

12 000·3.993 = 47 912.52

Present value of the salvage value

7 500/1.469 = 5 104.37

NPV

NPV = −50 000 + 47 912.52 + 5 104.37 = €3 016.89

Answer

NPV = €3 016.89
4.NPV & IRR

Answer: NPV = −30116 €

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 growing annuity

((1.04)/(1.12))8 = 0.5527
PV = 12 500/(0.12 − 0.04)·(1 − 0.5527) = 69 884.05

NPV

NPV = 69 884.05 − 100 000 = −€30 115.95

Answer

NPV = −€30 115.95

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

5.Bond pricing & yield

Answer: P = 959.4 €

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 = €22.50, i = 0.03, N = 6

Present value of the coupons

€22.50·(1 − 0.8375)/0.03 = €121.89

Present value of the face value

€1 000·0.8375 = €837.48

Price

P = €959.37

Answer

P = €959.37

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

6.Bond pricing & yield

Answer: D = 4.408 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: 70/1.055 = 66.35, t·PV = 66.35
t = 2: 70/1.113 = 62.89, t·PV = 125.78
t = 3: 70/1.174 = 59.61, t·PV = 178.84
t = 4: 70/1.239 = 56.51, t·PV = 226.02
t = 5: 1070/1.307 = 818.69, t·PV = 4 093.47

Duration

P = 1 064.05
D = 4 690.46/1 064.05 = 4.408

Answer

D = 4.408 years
7.CAPM & WACC

Answer: re = 13 %

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 = 0.625, D/V = 0.375

Debt part of the WACC

0.375·4%·(1 − 0.21) = 1.185%

Cost of equity

re = (9.31% − 1.185%)/0.625 = 13%

Answer

re = 13%
8.CAPM & WACC

Answer: WACC = 7.721 %

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·€40.00 = €40 000 000
D = 40 000·€980 = €39 200 000

Cost of equity (CAPM)

re = 2% + 1.2·6.5% = 9.8%

Weights

V = €79 200 000, E/V = 0.5051, D/V = 0.4949

WACC

WACC = 0.5051·9.8% + 0.4949·7%·(1 − 0.2) = 7.721%

Answer

WACC = 7.721%
9.Descriptive statistics

Answer: s = 9.863

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 = 325/7 = 46.43

Squared deviations

Σ(x − x̄)² = 1.571² + 10.57² + 11.57² + (−1.429)² + (−15.43)² + (−9.429)² + 2.571² = 583.7

Standard deviation

s = √(583.7/6) = 9.863

Answer

s = 9.863
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 = (166 − 170)/8 = −0.5

Answer

z = −0.5
11.Probability

Answer: σ = 1.487

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.

Mean

μ = 1·0.2 + 2·0.1 + 3·0.2 + 4·0.2 + 5·0.3 = 3.3

Mean of the squares

E(X²) = 1·0.2 + 4·0.1 + 9·0.2 + 16·0.2 + 25·0.3 = 13.1

Variance

Var(X) = 13.1 − 3.3² = 2.21
σ = √2.21 = 1.487

Answer

σ = 1.487
12.Probability

Answer: P = 0.1672

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 = 5) = 126·0.45·0.64 = 126·0.01024·0.1296 = 0.1672

Answer

P = 0.1672
13.Normal distribution

Answer: z = 1.96

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 + 0.9500)/2 = 0.9750

Read the z-table backwards

Find 0.9750 inside the table and read off z.

z = 1.96

Answer

z = 1.96
14.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
15.Confidence intervals

Answer: E = 18.63

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 = 10, t = 3.169

Margin of error

E = 3.169·19.5/√11 = 18.63

Answer

E = 18.63
16.Confidence intervals

Answer: n = 124

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.

Solve for n

n = (1.645·27/4)² = 123.3

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 124

Answer

n = 124
17.Hypothesis tests & confidence intervals

Answer: z = −1.606

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.

Standard error of the difference

SE = √(6²/40 + 9²/100) = √1.71 = 1.308

Test statistic

z = (95 − 97.1)/1.308 = −1.606

Answer

z = −1.606

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 10

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 = (1.96·4/2.5)² = 9.834

Round up

n = 10

Answer

n = 10
19.Linear regression & correlation

Answer: r = 0.9894

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 = 35/5 = 7
ȳ = 428/5 = 85.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 178
Syy = Σ(y − ȳ)² = 809.2

Correlation coefficient

r = 178/√(40·809.2) = 0.9894

Answer

r = 0.9894
20.Linear regression & correlation

Answer: ŷ = 21

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 = 109/6 = 18.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49.5

Regression line

b = −49.5/17.5 = −2.829
a = 18.17 − (−2.829)·3.5 = 28.07
ŷ = 28.07 − 2.829·x

Prediction

ŷ = 28.07 + (−2.829)·2.5 = 21

Answer

ŷ = 21

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