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
years0123456PV = ?€25 000

You need €25 000 in 6 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €25 0004% / 12n = 6
PV = €
2.Time value of moneyHard
months0123…300€500€500€500FV = ?

You save €500 at the end of every month for 25 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €5004.2% / 12N = 300
FV = €
3.NPV & IRRMedium
years01234−20 0008 0005 0009 0008 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €8 000, €5 000, €9 000, €8 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −20 000CF1 = 8 000CF2 = 5 000CF3 = 9 000CF4 = 8 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €20 000 and returns €11 000, €8 000, €4 500 at the end of years 1 to 3. Project B costs €25 000 and returns €6 500, €9 500, €14 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −20 000, 11 000, 8 000, 4 500B: −25 000, 6 500, 9 500, 14 000
r* = %
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 4% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 4%n = 7
P = €
6.Bond pricing & yieldHard
years01234€1 228€70€70€70 + €1 070

A bond with 12 years to maturity (face value €1 000, annual coupon 7%) sells for €1 228. The issuer can call it in 4 years at a call price of €1 070. Find the yield to call YTC in percent.

F = €1 000coupon = 7%P = €1 228call price = €1 070call in 4 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€6 000 000re​ = 13%debt€3 000 000rd​ = 5%T = 21%

A company has equity worth €6 000 000 and debt worth €3 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.

E = €6 000 000D = €3 000 000re = 13%rd = 5%T = 21%
WACC = %
8.CAPM & WACCHard
rf​ = 1.5%, rm​ − rf​ = 7%no debtequityβU​ = 0.8D/E = 1.5equityre​ = ?debtT = 28%

A company's business has an unlevered (asset) beta of 0.8. It plans a debt-to-equity ratio of 1.5. The tax rate is 28%, the risk-free rate is 1.5% and the market risk premium is 7%. Find the cost of equity re at the new capital structure.

βU = 0.8D/E = 1.5T = 28%rf = 1.5%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 30, 43, 50, 18, 25, 50.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 78.75?

μ = 60σ = 15x = 78.75
P =
11.ProbabilityMedium
xp(x)100.35200.3300.25400.1μ = 21

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

x: 10, 20, 30, 40p(x): 0.35, 0.3, 0.25, 0.1
σ =
12.ProbabilityHard
€p(x)−30.74120.25270.01

A game costs €3 to play. You win €15 with probability 0.25 and €30 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.25€30: p = 0.01
E(G) = €
13.Normal distributionMedium
105 s108.2 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 108.2 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 108.2 s
P =
14.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h
15.Confidence intervalsMedium
x̄E₁ = 1.990%, n = 50x̄E₂ = ?90%, n = 450

A 90% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.9. What is the margin of error E₂ if the same study uses 450 observations instead?

90%n₁ = 50E₁ = 1.9n₂ = 450
E₂ =
16.Confidence intervalsHard
p ≈ 0.7, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.7E = 0.0190%
n =
17.Hypothesis tests & confidence intervalsMedium
2p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 83. Test H₀: μ = 80 against H₁: μ > 80. Find the p-value p.

μ₀ = 80σ = 15n = 100x̄ = 83H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 101.5. The population standard deviation is 5. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 101.5σ = 5n = 8199%
U =
19.Linear regression & correlationMedium
xy3605727779871110013112xy5106080100

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy120218319415512xy241520x = 2.5

The table gives 5 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 = 5x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 19674 €

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 = 12·6 = 72

Present value

PV = FV/(1 + i)N = 25 000/1.271 = €19 673.55

Answer

PV = €19 673.55
2.Time value of money

Answer: FV = 264631 €

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 = 0.042/12 = 0.0035
N = 12·25 = 300

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 500·(2.852 − 1)/0.0035 = €264 631.03

Answer

FV = €264 631.03

You paid in €150 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 5890 €

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 + 0.06)1 = 8 000/1.06 = 7 547.17
CF2/(1 + 0.06)2 = 5 000/1.124 = 4 449.98
CF3/(1 + 0.06)3 = 9 000/1.191 = 7 556.57
CF4/(1 + 0.06)4 = 8 000/1.262 = 6 336.75

Add them up

NPV = €5 890.47

Answer

NPV = €5 890.47

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

4.NPV & IRR

Answer: r* = 6.315 %

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 = −25 000 − (−20 000) = −5 000
ΔCF1 = 6 500 − 11 000 = −4 500
ΔCF2 = 9 500 − 8 000 = 1 500
ΔCF3 = 14 000 − 4 500 = 9 500

IRR of the difference

NPVΔ(6%) = 66.09
NPVΔ(7%) = −140.62
r* = 6.315%

Answer

r* = 6.315%

Check: at 6.315% both NPVs are €1 169.12.

5.Bond pricing & yield

Answer: P = 1121 €

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 = 0.02, N = 14

Present value of the coupons

€30.00·(1 − 0.7579)/0.02 = €363.19

Present value of the face value

€1 000·0.7579 = €757.88

Price

P = €1 121.06

Answer

P = €1 121.06

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTC = 2.607 %

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

1228 = 70·(1 − (1 + y)−4)/y + 1070/(1 + y)4

Solve numerically

y = 2%: €1 255.06
y = 3%: €1 210.88
YTC = 2.607%

Answer

YTC = 2.607%

The yield to maturity is 4.5%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: WACC = 9.983 %

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

WACC

WACC = 0.6667·13% + 0.3333·5%·(1 − 0.21) = 9.983%

Answer

WACC = 9.983%
8.CAPM & WACC

Answer: re = 13.15 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Levered beta

βL = 0.8·(1 + 0.72·1.5) = 1.664

Cost of equity (CAPM)

re = 1.5% + 1.664·7% = 13.15%

Answer

re = 13.15%
9.Descriptive statistics

Answer: s = 13.58

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 = 216/6 = 36

Squared deviations

Σ(x − x̄)² = (−6)² + 7² + 14² + (−18)² + (−11)² + 14² = 922

Standard deviation

s = √(922/5) = 13.58

Answer

s = 13.58
10.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
11.Probability

Answer: σ = 9.95

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 = 10·0.35 + 20·0.3 + 30·0.25 + 40·0.1 = 21

Mean of the squares

E(X²) = 100·0.35 + 400·0.3 + 900·0.25 + 1600·0.1 = 540

Variance

Var(X) = 540 − 21² = 99
σ = √99 = 9.95

Answer

σ = 9.95
12.Probability

Answer: E(G) = 1.05 €

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: 12 (p = 0.25), 27 (p = 0.01), −3 (p = 0.74)

Expected value

E(G) = 12·0.25 + 27·0.01 − 3·0.74 = 1.05 €

Answer

E(G) = 1.05 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.3085

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (108.2 − 105)/6.403 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
14.Normal distribution

Answer: x = 578.4 h

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 for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h
15.Confidence intervals

Answer: E₂ = 0.6333

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.

Ratio of margins

E₂/E₁ = (1.645/1.645)·√(50/450) = 0.3333

New margin

E₂ = 1.9·0.3333 = 0.6333

Answer

E₂ = 0.6333
16.Confidence intervals

Answer: n = 5683

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 = 1.645²·0.7·0.3/0.01² = 5683

Round up

n = 5683

Answer

n = 5683
17.Hypothesis tests & confidence intervals

Answer: p = 0.02275

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 = 15/√100 = 1.5
z = (83 − 80)/1.5 = 2

Tail area

Φ(2) = 0.9772
1 − Φ(2) = 0.0228

p-value

p = 0.02275

Answer

p = 0.02275

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

18.Hypothesis tests & confidence intervals

Answer: U = 102.9

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 = 2.576·5/√81 = 1.431

Interval limit

U = 101.5 + 1.431 = 102.9

Answer

U = 102.9
19.Linear regression & correlation

Answer: r = 0.9931

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 = 48/6 = 8
ȳ = 508/6 = 84.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 354
Syy = Σ(y − ȳ)² = 1815

Correlation coefficient

r = 354/√(70·1815) = 0.9931

Answer

r = 0.9931
20.Linear regression & correlation

Answer: ŷ = 17.75

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 = 15/5 = 3
ȳ = 84/5 = 16.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19

Regression line

b = −19/10 = −1.9
a = 16.8 − (−1.9)·3 = 22.5
ŷ = 22.5 − 1.9·x

Prediction

ŷ = 22.5 + (−1.9)·2.5 = 17.75

Answer

ŷ = 17.75

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