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
periods0123…1211 + EAR = ?24% / 12

A credit card charges a nominal interest rate of 24% per year, compounded monthly. Find the effective annual rate EAR in percent.

r = 24%m = 12
EAR = %
2.Time value of moneyHard
months0123…156A?AA€10 000

You want €10 000 in 13 years. You save the same amount at the end of every month in an account paying 6% per year, compounded monthly. Find the monthly deposit A.

FV = €10 0006% / 12N = 156
A = €
3.NPV & IRRMedium
years0123…10€60 000€2 000€2 000€2 000€2 000r = 10%, EAC = ?

A machine costs €60 000 today, lasts 10 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.

C0 = €60 000M = €2 000n = 10r = 10%
EAC = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €12 000 and returns €3 000, €5 500, €6 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −10 000, 5 500, 4 000, 2 500B: −12 000, 3 000, 5 500, 6 000
r* = %
5.Bond pricing & yieldMedium

You buy a bond with 8 years to maturity (face value €1 000, annual coupon 6%) when its yield to maturity is 3%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 3.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 6%n = 8y0 = 3%y1 = 3.5%
HPR = %
6.Bond pricing & yieldHard
years0123…9€1 080€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 9 years and sells for €1 080. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 9P = €1 080
YTM = %
7.CAPM & WACCMedium
WACC = 7.9%equity€6 000 000re​ = ?debt€3 000 000rd​ = 5%T = 25.8%

A company has equity worth €6 000 000 and debt worth €3 000 000. Its WACC is 7.9%, its pre-tax cost of debt is 5% and the tax rate is 25.8%. Find the cost of equity re.

E = €6 000 000D = €3 000 000WACC = 7.9%rd = 5%T = 25.8%
re = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €38.50β = 0.9rf​ = 3%, rm​ = 10%debt5 000 × €980rd​ = 4.5%T = 25%

A company has 2 000 000 shares at €38.50 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.9, the risk-free rate 3% and the market return 10%. Find the WACC.

β = 0.9rf = 3%rm = 10%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 37, 14, 58, 27, 13, 28, 58.

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

Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 600.

μ = 500σ = 100x = 600
z =
11.ProbabilityMedium
red: 5blue: 8

A bag holds 5 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 8
P =
12.ProbabilityHard
xp(x)100.35200.1300.3400.25μ

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

x: 10, 20, 30, 40p(x): 0.35, 0.1, 0.3, 0.25Y = 10X + 10
σY =
13.Normal distributionMedium
462548P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 548 h? Give P as a decimal.

μ = 500σ = 40x1 = 462x2 = 548
P =
14.Normal distributionHard
35455565758595pointstop 80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points80%
x = points
15.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 59 h, known standard deviation 19 h. Model 2: 30 phones, mean 47.5 h, known standard deviation 8 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 59 hσ₁ = 19 hn₂ = 30x̄₂ = 47.5 hσ₂ = 8 h90%
upper limit = h
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 31. 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 0.5?

σ = 31E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 497.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 497.4s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 135 of 300 patients on treatment A recover, and 153 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 135n₁ = 300x₂ = 153n₂ = 300
z =
19.Linear regression & correlationMedium
xy3615637749861187xy51060708090

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
xy21041362283110341243xy5102040x = 9

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. Predict ŷ for x = 9.

n = 6x = 9
ŷ =

3Solutions

1.Time value of money

Answer: EAR = 26.82 %

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 = 0.24/12 = 0.02

Compound over one year

EAR = (1 + 0.02)12 − 1 = 1.268 − 1 = 26.82%

Answer

EAR = 26.82%

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

2.Time value of money

Answer: A = 42.47 €

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 = 0.06/12 = 0.005
N = 12·13 = 156

Deposit

A = FV·i/((1 + i)N − 1) = 10 000·0.005/(2.177 − 1) = €42.47

Answer

A = €42.47
3.NPV & IRR

Answer: EAC = 11765 €

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 − 1.1−10)/0.1 = 6.145

Annualised purchase price

60 000/6.145 = 9 764.72

EAC

EAC = 9 764.72 + 2 000 = €11 764.72

Answer

EAC = €11 764.72
4.NPV & IRR

Answer: r* = 5.034 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −12 000 − (−10 000) = −2 000
ΔCF1 = 3 000 − 5 500 = −2 500
ΔCF2 = 5 500 − 4 000 = 1 500
ΔCF3 = 6 000 − 2 500 = 3 500

IRR of the difference

NPVΔ(5%) = 3.02
NPVΔ(6%) = −84.83
r* = 5.034%

Answer

r* = 5.034%

Check: at 5.034% both NPVs are €1 019.71.

5.Bond pricing & yield

Answer: HPR = 0.1877 %

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 = €60.00, i = 0.03, N = 8
€60.00·(1 − 0.7894)/0.03 + €1 000·0.7894 = €421.18 + €789.41 = €1 210.59

Selling price

C = €60.00, i = 0.035, N = 7
€60.00·(1 − 0.786)/0.035 + €1 000·0.786 = €366.87 + €785.99 = €1 152.86

Holding period return

HPR = (60 + 1 152.86 − 1 210.59)/1 210.59 = 0.1877%

Answer

HPR = 0.1877%
6.Bond pricing & yield

Answer: YTM = 2.974 %

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

€1 080 = €40·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(2%) = €1 163.24, P(3%) = €1 077.86
YTM = 2.974%

Answer

YTM = 2.974%
7.CAPM & WACC

Answer: re = 9.995 %

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 = 0.6667, D/V = 0.3333

Debt part of the WACC

0.3333·5%·(1 − 0.258) = 1.237%

Cost of equity

re = (7.9% − 1.237%)/0.6667 = 9.995%

Answer

re = 9.995%
8.CAPM & WACC

Answer: WACC = 8.946 %

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 = 2 000 000·€38.50 = €77 000 000
D = 5 000·€980 = €4 900 000

Cost of equity (CAPM)

re = 3% + 0.9·7% = 9.3%

Weights

V = €81 900 000, E/V = 0.9402, D/V = 0.05983

WACC

WACC = 0.9402·9.3% + 0.05983·4.5%·(1 − 0.25) = 8.946%

Answer

WACC = 8.946%
9.Descriptive statistics

Answer: s = 18.64

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 = 235/7 = 33.57

Squared deviations

Σ(x − x̄)² = 3.429² + (−19.57)² + 24.43² + (−6.571)² + (−20.57)² + (−5.571)² + 24.43² = 2086

Standard deviation

s = √(2086/6) = 18.64

Answer

s = 18.64
10.Descriptive statistics

Answer: z = 1

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 = (600 − 500)/100 = 1

Answer

z = 1
11.Probability

Answer: P = 0.1282

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 = 5/13 · 4/12 = 20/156 = 0.1282

Answer

P = 0.1282
12.Probability

Answer: σY = 120.3

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 = 24.5, Var(X) = 144.8, σX = 12.03

Linear change

σY = |10|·12.03 = 120.3

Answer

σY = 120.3

The mean does move: E(Y) = 10·24.5 + 10 = 255.

13.Normal distribution

Answer: P = 0.7139

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (462 − 500)/40 = −0.95
z₂ = (548 − 500)/40 = 1.2

Area between

P = Φ(1.2) − Φ(−0.95) = 0.8849 − 0.1711 = 0.7139

Answer

P = 0.7139
14.Normal distribution

Answer: x = 56.6 points

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 65 − 0.84·10 = 56.6 points

Answer

x = 56.6 points
15.Confidence intervals

Answer: upper limit = 17.69 h

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.

Point estimate

x̄₁ − x̄₂ = 59 − 47.5 = 11.5

Standard error

SE = √(19²/30 + 8²/30) = 3.764

Margin of error

E = 1.645·3.764 = 6.192

Interval

11.5 ± 6.192 → [5.308, 17.69]

Answer

upper limit = 17.69 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 10402

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 = (1.645·31/0.5)² = 10402

Round up

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

n = 10402

Answer

n = 10402
17.Hypothesis tests & confidence intervals

Answer: t = −2.167

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 = 12/√100 = 12/10 = 1.2

Test statistic

t = (497.4 − 500)/1.2 = −2.167

Answer

t = −2.167
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −1.471

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.

Sample proportions

p̂₁ = 135/300 = 0.45
p̂₂ = 153/300 = 0.51
p̂ = 288/600 = 0.48

Standard error under H₀

SE = √(0.48·0.52·(1/300 + 1/300)) = 0.04079

Test statistic

z = (0.45 − 0.51)/0.04079 = −1.471

Answer

z = −1.471

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

19.Linear regression & correlation

Answer: r = 0.966

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 = 371/5 = 74.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 150
Syy = Σ(y − ȳ)² = 602.8

Correlation coefficient

r = 150/√(40·602.8) = 0.966

Answer

r = 0.966
20.Linear regression & correlation

Answer: ŷ = 32.27

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 = 153/6 = 25.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 237

Regression line

b = 237/70 = 3.386
a = 25.5 − 3.386·7 = 1.8
ŷ = 1.8 + 3.386·x

Prediction

ŷ = 1.8 + 3.386·9 = 32.27

Answer

ŷ = 32.27

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