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 = ?22% / 12

A credit card charges a nominal interest rate of 22% per year, compounded monthly. Find the effective annual rate EAR in percent.

r = 22%m = 12
EAR = %
2.Time value of moneyHard
months0123…204€50€50€50FV = ?

You save €50 at the end of every month for 17 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €507.2% / 12N = 204
FV = €
3.NPV & IRRMedium
years012345−50 00011 00011 00011 00011 00016 000

A machine costs €50 000 today. It generates €11 000 at the end of each year for 5 years and can then be sold for €5 000. The cost of capital is 10%. Find the net present value NPV.

C0 = €50 000CF = €11 000S = €5 000n = 5r = 10%
NPV = €
4.NPV & IRRHard
years0123−25 00013 00012 00010 000

A project costs €25 000 and returns €13 000, €12 000, €10 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.

r = 8%CF0 = −25 000CF1 = 13 000CF2 = 12 000CF3 = 10 000
MIRR = %
5.Bond pricing & yieldMedium
years0123…12P€60€60€60 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 12 years. Its yield to maturity changes from 9% to 8.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 6%n = 12y1 = 9%y2 = 8.5%
ΔP = €
6.Bond pricing & yieldHard
years0123456P€30€30€30 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 6 years. The yield to maturity is 7%. Find the Macaulay duration D in years.

F = €1 000coupon = 3%n = 6YTM = 7%
D = years
7.CAPM & WACCMedium
βr (%)SMLrf​ = 4%β = 1rm​ = ?β = 2re​ = 13%

A share with a beta of 2 has a required return of 13%. The risk-free rate is 4%. What expected market return rm does CAPM imply?

β = 2re = 13%rf = 4%
rm = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €18.00β = 0.8rf​ = 3.5%, rm​ = 8%debt40 000 × €1 020rd​ = 4.5%T = 20%

A company has 1 000 000 shares at €18.00 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 4.5% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 3.5% and the market return 8%. Find the WACC.

β = 0.8rf = 3.5%rm = 8%rd = 4.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 26, 47, 18, 35, 49.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 620.

μ = 500σ = 60x = 620
z =
11.ProbabilityMedium
xp(x)00.1520.2540.360.3μ = 3.5

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

x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.3, 0.3
σ =
12.ProbabilityHard
n = 9, p = 0.1: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 9p = 0.1
P =
13.Normal distributionMedium
750 g759.8 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 759.8 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 759.8 g
P =
14.Normal distributionHard
μ = 500x = ?20%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 20% of them lie (the 20th percentile).

μ = 500σ = 40P = 20%
x = h
15.Confidence intervalsMedium
n = 16, s = 17.5LUx̄ = 17.5E = ?95%

A sample of 16 items has mean 17.5 and standard deviation s = 17.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 16x̄ = 17.5s = 17.595%t(15) = 2.131
E =
16.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 15 plots, mean yield 88 kg, standard deviation 4.5 kg. Plot group 2: 7 plots, mean 84.5 kg, standard deviation 3 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 = 15x̄₁ = 88s₁ = 4.5n₂ = 7x̄₂ = 84.5s₂ = 395%t(20) = 2.086
E = kg
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 202.5 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 202.5s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 18 of 120 patients on treatment A recover, and 50 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 18n₁ = 120x₂ = 50n₂ = 250
z =
19.Linear regression & correlationMedium
xy125223324420515xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 107Σx² = 55Σy² = 2355Σxy = 298
r =
20.Linear regression & correlationHard
xy2474586678711080xy5106080ŷ = 69

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 = 69?

n = 5ŷ = 69
x =

3Solutions

1.Time value of money

Answer: EAR = 24.36 %

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 = 0.22/12 = 0.01833

Compound over one year

EAR = (1 + 0.01833)12 − 1 = 1.244 − 1 = 24.36%

Answer

EAR = 24.36%

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

2.Time value of money

Answer: FV = 19903 €

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 = 0.07200000000000001/12 = 0.006
N = 12·17 = 204

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 50·(3.388 − 1)/0.006 = €19 902.90

Answer

FV = €19 902.90

You paid in €10 200 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = −5197 €

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

11 000·3.791 = 41 698.65

Present value of the salvage value

5 000/1.611 = 3 104.61

NPV

NPV = −50 000 + 41 698.65 + 3 104.61 = −€5 196.74

Answer

NPV = −€5 196.74
4.NPV & IRR

Answer: MIRR = 15.1 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

13 000·1.082 = 15 163.20
12 000·1.081 = 12 960.00
10 000·1.080 = 10 000.00
TV = 38 123.20

MIRR

MIRR = (38 123.20/25 000)1/3 − 1 = 15.1%

Answer

MIRR = 15.1%
5.Bond pricing & yield

Answer: ΔP = 31.2 €

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.

Price at the old yield

C = €60.00, i = 0.09, N = 12
€60.00·(1 − 0.3555)/0.09 + €1 000·0.3555 = €429.64 + €355.53 = €785.18

Price at the new yield

C = €60.00, i = 0.085, N = 12
€60.00·(1 − 0.3757)/0.085 + €1 000·0.3757 = €440.68 + €375.70 = €816.38

Change

ΔP = 816.38 − 785.18 = €31.20

Answer

ΔP = €31.20
6.Bond pricing & yield

Answer: D = 5.524 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: 30/1.07 = 28.04, t·PV = 28.04
t = 2: 30/1.145 = 26.20, t·PV = 52.41
t = 3: 30/1.225 = 24.49, t·PV = 73.47
t = 4: 30/1.311 = 22.89, t·PV = 91.55
t = 5: 30/1.403 = 21.39, t·PV = 106.95
t = 6: 1030/1.501 = 686.33, t·PV = 4 117.99

Duration

P = 809.34
D = 4 470.40/809.34 = 5.524

Answer

D = 5.524 years
7.CAPM & WACC

Answer: rm = 8.5 %

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 risk premium

rm − rf = (13% − 4%)/2 = 4.5%

Market return

rm = 4% + 4.5% = 8.5%

Answer

rm = 8.5%
8.CAPM & WACC

Answer: WACC = 4.671 %

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·€18.00 = €18 000 000
D = 40 000·€1 020 = €40 800 000

Cost of equity (CAPM)

re = 3.5% + 0.8·4.5% = 7.1%

Weights

V = €58 800 000, E/V = 0.3061, D/V = 0.6939

WACC

WACC = 0.3061·7.1% + 0.6939·4.5%·(1 − 0.2) = 4.671%

Answer

WACC = 4.671%
9.Descriptive statistics

Answer: s = 13.32

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 = 175/5 = 35

Squared deviations

Σ(x − x̄)² = (−9)² + 12² + (−17)² + 0² + 14² = 710

Standard deviation

s = √(710/4) = 13.32

Answer

s = 13.32
10.Descriptive statistics

Answer: z = 2

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 = (620 − 500)/60 = 2

Answer

z = 2
11.Probability

Answer: σ = 2.086

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 = 0·0.15 + 2·0.25 + 4·0.3 + 6·0.3 = 3.5

Mean of the squares

E(X²) = 0·0.15 + 4·0.25 + 16·0.3 + 36·0.3 = 16.6

Variance

Var(X) = 16.6 − 3.5² = 4.35
σ = √4.35 = 2.086

Answer

σ = 2.086
12.Probability

Answer: P = 0.7748

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.

Zero and one failure

P(X = 0) = 0.99 = 0.3874
P(X = 1) = 9·0.1·0.98 = 0.3874

Add them

P = 0.3874 + 0.3874 = 0.7748

Answer

P = 0.7748
13.Normal distribution

Answer: P = 0.1587

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (759.8 − 750)/9.849 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
14.Normal distribution

Answer: x = 466.3 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) = 0.2 → z = −0.8416

Back to x

x = 500 + −0.8416·40 = 466.34 h

Answer

x = 466.34 h
15.Confidence intervals

Answer: E = 9.323

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 = 15, t = 2.131

Margin of error

E = 2.131·17.5/√16 = 9.323

Answer

E = 9.323
16.Confidence intervals

Answer: E = 3.922 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 = (14·4.5² + 6·3²)/20 = 16.88
sp = 4.108

Margin of error

E = 2.086·4.108·√(1/15 + 1/7) = 3.922

Answer

E = 3.922 kg
17.Hypothesis tests & confidence intervals

Answer: t = 2.083

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 = (202.5 − 200)/1.2 = 2.083

Answer

t = 2.083
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 = −1.162

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 = 18/120 = 0.15
p̂₂ = 50/250 = 0.2
p̂ = 68/370 = 0.1838

Standard error under H₀

SE = √(0.1838·0.8162·(1/120 + 1/250)) = 0.04301

Test statistic

z = (0.15 − 0.2)/0.04301 = −1.162

Answer

z = −1.162

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

19.Linear regression & correlation

Answer: r = −0.9007

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.

Centred sums

Sxy = 298 − 15·107/5 = −23
Sxx = 55 − 15²/5 = 10
Syy = 2355 − 107²/5 = 65.2

Correlation coefficient

r = −23/√(10·65.2) = −0.9007

Answer

r = −0.9007
20.Linear regression & correlation

Answer: x = 7.114

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 = 30/5 = 6
ȳ = 323/5 = 64.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 158

Regression line

b = 158/40 = 3.95
a = 64.6 − 3.95·6 = 40.9
ŷ = 40.9 + 3.95·x

Solve for x

x = (69 − 40.9)/3.95 = 7.114

Answer

x = 7.114

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