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
years01234567€20 000FV = ?

You invest €20 000 at 8% per year with continuous compounding. Find the value FV after 7 years.

PV = €20 000r = 8%t = 7
FV = €
2.Time value of moneyHard
months0123…N = ?€5 000€100€100€100€1003% / 12

You borrow €5 000 at 3% per year, compounded monthly, and repay €100 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 = €5 000A = €1003% / 12
N = payments
3.NPV & IRRMedium
years01234−30 00012 00013 00012 00011 000

An investment of €30 000 gives these cash flows at the end of years 1 to 4: €12 000, €13 000, €12 000, €11 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −30 000CF1 = 12 000CF2 = 13 000CF3 = 12 000CF4 = 11 000
NPV = €
4.NPV & IRRHard
years0123−20 00012 50014 50012 000

A project costs €20 000 and returns €12 500, €14 500, €12 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 12 500CF2 = 14 500CF3 = 12 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…12€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 7%n = 6
P = €
6.Bond pricing & yieldHard
years01234€1 098€90€90€90 + €1 090

A bond with 20 years to maturity (face value €1 000, annual coupon 9%) sells for €1 098. The issuer can call it in 4 years at a call price of €1 090. Find the yield to call YTC in percent.

F = €1 000coupon = 9%P = €1 098call price = €1 090call in 4 years
YTC = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 2.5%β = 1rm​ = ?β = 0.6re​ = 6.4%

A share with a beta of 0.6 has a required return of 6.4%. The risk-free rate is 2.5%. What expected market return rm does CAPM imply?

β = 0.6re = 6.4%rf = 2.5%
rm = %
8.CAPM & WACCHard
rf​ = 2%, rm​ − rf​ = 6%no debtequityβU​ = 0.6D/E = 0.75equityre​ = ?debtT = 30%

A company's business has an unlevered (asset) beta of 0.6. It plans a debt-to-equity ratio of 0.75. The tax rate is 30%, the risk-free rate is 2% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.

βU = 0.6D/E = 0.75T = 30%rf = 2%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 29, 35, 29, 27, 52.

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

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 545?

μ = 500σ = 75x = 545
P =
11.ProbabilityMedium
n = 7k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.

n = 7k = 2
N =
12.ProbabilityHard
n = 8, p = 0.3: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 8p = 0.3
P =
13.Normal distributionMedium
750 g764.8 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 764.8 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 764.8 g
P =
14.Normal distributionHard
557085100115130145pointstop 80%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points80%
x = points
15.Confidence intervalsMedium
x̄E₁ = 3.795%, n = 36x̄E₂ = ?99%, n = 144

A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 3.7. What is the margin of error E₂ of a 99% interval based on 144 observations?

95%n₁ = 36E₁ = 3.799%n₂ = 144
E₂ =
16.Confidence intervalsHard
n = 30σ = ?Us = 10.695%

The diameters of 30 machined shafts have a sample standard deviation of 10.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 10.6 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

16 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 −4.4 and the standard deviation of the differences is 8. Find the paired test statistic t for H₀: μd = 0.

n = 16d̄ = −4.4sd = 8
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 88 of 250 patients on treatment A recover, and 36 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 88n₁ = 250x₂ = 36n₂ = 120
z =
19.Linear regression & correlationMedium
xy294196258361045xy5102040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 134Σx² = 220Σy² = 4388Σxy = 982
r =
20.Linear regression & correlationHard
xy2104176258301037xy51010203040ŷ = 20

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 20?

n = 5ŷ = 20
x =

3Solutions

1.Time value of money

Answer: FV = 35013 €

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.

Exponent

r·t = 0.08·7 = 0.56
e0.56 = 1.751

Future value

FV = 20 000·1.751 = €35 013.45

Answer

FV = €35 013.45
2.Time value of money

Answer: N = 54 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 = 0.0025
(1 + i)−N = 1 − 5 000·0.0025/100 = 0.875

Solve with logarithms

N = −ln(0.875)/ln(1.002) = 53.48

Round up to whole payments

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

N = 54

Answer

N = 54 payments
3.NPV & IRR

Answer: NPV = 12636 €

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 = −30 000
CF1/(1 + 0.05)1 = 12 000/1.05 = 11 428.57
CF2/(1 + 0.05)2 = 13 000/1.103 = 11 791.38
CF3/(1 + 0.05)3 = 12 000/1.158 = 10 366.05
CF4/(1 + 0.05)4 = 11 000/1.216 = 9 049.73

Add them up

NPV = €12 635.73

Answer

NPV = €12 635.73

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

4.NPV & IRR

Answer: IRR = 42.74 %

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(40%) = 699.71
NPV(45%) = −546.56

Narrow down

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

IRR ≈ 40 + 5·699.71/(699.71 − (−546.56)) = 42.81%
exact: NPV(42.74%) = 0 → IRR = 42.74%

Answer

IRR = 42.74%
5.Bond pricing & yield

Answer: P = 1048 €

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 = €40.00, i = 0.035, N = 12

Present value of the coupons

€40.00·(1 − 0.6618)/0.035 = €386.53

Present value of the face value

€1 000·0.6618 = €661.78

Price

P = €1 048.32

Answer

P = €1 048.32

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

6.Bond pricing & yield

Answer: YTC = 8.035 %

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

1098 = 90·(1 − (1 + y)−4)/y + 1090/(1 + y)4

Solve numerically

y = 8%: €1 099.27
y = 9%: €1 063.76
YTC = 8.035%

Answer

YTC = 8.035%

The yield to maturity is 8.002%. 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: rm = 9 %

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 = (6.4% − 2.5%)/0.6 = 6.5%

Market return

rm = 2.5% + 6.5% = 9%

Answer

rm = 9%
8.CAPM & WACC

Answer: re = 7.49 %

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 = 0.6·(1 + 0.7·0.75) = 0.915

Cost of equity (CAPM)

re = 2% + 0.915·6% = 7.49%

Answer

re = 7.49%
9.Descriptive statistics

Answer: s = 10.29

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 = 172/5 = 34.4

Squared deviations

Σ(x − x̄)² = (−5.4)² + 0.6² + (−5.4)² + (−7.4)² + 17.6² = 423.2

Standard deviation

s = √(423.2/4) = 10.29

Answer

s = 10.29
10.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
11.Probability

Answer: N = 21

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.

Combinations

C(7, 2) = (7·6)/(2·1) = 21

Answer

N = 21
12.Probability

Answer: P = 0.2553

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) = 0.78 = 0.05765
P(X = 1) = 8·0.3·0.77 = 0.1977

Add them

P = 0.05765 + 0.1977 = 0.2553

Answer

P = 0.2553
13.Normal distribution

Answer: P = 0.06681

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 = (764.8 − 750)/9.849 = 1.5

Upper tail

P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
14.Normal distribution

Answer: x = 87.4 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 = 100 − 0.84·15 = 87.4 points

Answer

x = 87.4 points
15.Confidence intervals

Answer: E₂ = 2.431

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 = (2.576/1.96)·√(36/144) = 0.6571

New margin

E₂ = 3.7·0.6571 = 2.431

Answer

E₂ = 2.431
16.Confidence intervals

Answer: σ = 8.442 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 29·10.6² = 3258

Limits for σ²

3258/45.722 = 71.27
3258/16.047 = 203.1

Limit for σ

σ = √71.27 = 8.442 µm

Answer

σ = 8.442 µm

95% interval for σ: [8.442, 14.25] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 8/√16 = 2

Test statistic

t = −4.4/2 = −2.2

Answer

t = −2.2

With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: z = 0.9919

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 = 88/250 = 0.352
p̂₂ = 36/120 = 0.3
p̂ = 124/370 = 0.3351

Standard error under H₀

SE = √(0.3351·0.6649·(1/250 + 1/120)) = 0.05242

Test statistic

z = (0.352 − 0.3)/0.05242 = 0.9919

Answer

z = 0.9919

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

19.Linear regression & correlation

Answer: r = 0.997

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 = 982 − 30·134/5 = 178
Sxx = 220 − 30²/5 = 40
Syy = 4388 − 134²/5 = 796.8

Correlation coefficient

r = 178/√(40·796.8) = 0.997

Answer

r = 0.997
20.Linear regression & correlation

Answer: x = 4.866

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 = 119/5 = 23.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134

Regression line

b = 134/40 = 3.35
a = 23.8 − 3.35·6 = 3.7
ŷ = 3.7 + 3.35·x

Solve for x

x = (20 − 3.7)/3.35 = 4.866

Answer

x = 4.866

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