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.

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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
years01234€12 000FV = ?

You invest €12 000 at a nominal rate of 6% per year, compounded quarterly. Find the value FV of the investment after 4 years.

PV = €12 0006% / 4n = 4
FV = €
2.Time value of moneyHard
months0123…240€200 000AAAA2.4% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 20 years at 2.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0002.4% / 12N = 240
I = €
3.NPV & IRRMedium
years0123…9−500 00060 000+2%+2%…r = 12%, NPV = ?

A project costs €500 000 today. It generates €60 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 12%. Find the net present value NPV.

C0 = €500 000CF1 = €60 000g = 2%r = 12%
NPV = €
4.NPV & IRRHard
years012−10 0006 5006 000

A project costs €10 000 and returns €6 500, €6 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 6 500CF2 = 6 000
IRR = %
5.Bond pricing & yieldMedium
years0123…10€765C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 10 years and pays annual coupons. It sells for €765 when the yield to maturity is 8%. Find the coupon rate c in percent.

F = €1 000n = 10P = €765YTM = 8%
c = %
6.Bond pricing & yieldHard
years0123456€875€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 6 years and sells for €875. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 6P = €875
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 10%preferred shares€1 000 000rp​ = 6%debt€3 000 000rd​ = 5%T = 30%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 000 000 and debt worth €3 000 000. The cost of equity is 10%, the cost of preferred stock 6% and the pre-tax cost of debt 5%. The tax rate is 30%. Find the WACC.

E = €6 000 000Ps = €1 000 000D = €3 000 000re = 10%rp = 6%rd = 5%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 2.5%, rm​ − rf​ = 7%no debtequityβU​ = 1D/E = 0.5equityre​ = ?debtT = 30%

A company's business has an unlevered (asset) beta of 1. It plans a debt-to-equity ratio of 0.5. The tax rate is 30%, the risk-free rate is 2.5% and the market risk premium is 7%. Find the cost of equity re at the new capital structure.

βU = 1D/E = 0.5T = 30%rf = 2.5%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 26, 54, 44, 27, 55, 35.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 90.

μ = 100σ = 20x = 90
z =
11.ProbabilityMedium
xp(x)00.3520.1540.3560.15μ = 2.6

For the distribution shown, find the variance Var(X) of X.

x: 0, 2, 4, 6p(x): 0.35, 0.15, 0.35, 0.15
Var(X) =
12.ProbabilityHard
xp(x)00.2510.1520.130.3540.15μ

X has the distribution shown, and Y = 5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.15, 0.1, 0.35, 0.15Y = 5X + 1
σY =
13.Normal distributionMedium
32 min29.5 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 29.5 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 29.5 min
P =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
n = 13, sd​ = 30L = ?Ud̄ = −5.1E95%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.1 and standard deviation 3. Find the lower limit of the 95% confidence interval for the mean difference.

n = 13d̄ = −5.1sd = 395%t(12) = 2.179
lower limit = mmHg
16.Confidence intervalsHard
μ₁ − μ₂, t(29) = 1.6990L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 22 plots, mean yield 60 kg, standard deviation 10 kg. Plot group 2: 9 plots, mean 54 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 22x̄₁ = 60s₁ = 10n₂ = 9x̄₂ = 54s₂ = 7.590%t(29) = 1.699
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 72.79 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 72.79s = 5n = 49
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 90 of 200 patients on treatment A recover, and 150 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 90n₁ = 200x₂ = 150n₂ = 300
z =
19.Linear regression & correlationMedium
xy12221831741459xy24101520

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy2524546708761080xy51050607080x = 3

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.

n = 5x = 3
ŷ =

3Solutions

1.Time value of money

Answer: FV = 15228 €

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 = r/m = 0.06/4 = 0.015
N = m·n = 4·4 = 16

Future value

FV = 12 000·(1 + 0.015)16 = 12 000·1.269 = €15 227.83

Answer

FV = €15 227.83
2.Time value of money

Answer: I = 52021 €

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 payment

i = 0.002, N = 240
A = 200 000·0.002/(1 − (1 + 0.002)−240) = €1 050.09

Total interest

I = N·A − P = 240·1050 − 200 000 = €52 021.48

Answer

I = €52 021.48
3.NPV & IRR

Answer: NPV = 100000 €

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.

Value of the cash flows

PV = 60 000/(0.12 − 0.02) = 600 000.00

NPV

NPV = 600 000.00 − 500 000 = €100 000.00

Answer

NPV = €100 000.00

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

4.NPV & IRR

Answer: IRR = 16.5 %

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(15%) = 189.04
NPV(20%) = −416.67

Narrow down

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

IRR ≈ 15 + 5·189.04/(189.04 − (−416.67)) = 16.56%
exact: NPV(16.5%) = 0 → IRR = 16.5%

Answer

IRR = 16.5%
5.Bond pricing & yield

Answer: c = 4.498 %

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.

Present value of the face value

1000·(1 + 0.08)−10 = €463.19

Coupon

C = (765 − 463.19)/6.71 = €44.98

Coupon rate

c = 44.98/1000 = 4.498%

Answer

c = 4.498%
6.Bond pricing & yield

Answer: YTM = 5.502 %

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

€875 = €30·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(5%) = €898.49, P(6%) = €852.48
YTM = 5.502%

Answer

YTM = 5.502%
7.CAPM & WACC

Answer: WACC = 7.65 %

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 = 6 000 000 + 1 000 000 + 3 000 000 = €10 000 000
E/V = 0.6, Ps/V = 0.1, D/V = 0.3

WACC

WACC = 0.6·10% + 0.1·6% + 0.3·5%·(1 − 0.3) = 7.65%

Answer

WACC = 7.65%
8.CAPM & WACC

Answer: re = 11.95 %

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 = 1·(1 + 0.7·0.5) = 1.35

Cost of equity (CAPM)

re = 2.5% + 1.35·7% = 11.95%

Answer

re = 11.95%
9.Descriptive statistics

Answer: s = 12.86

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 = 241/6 = 40.17

Squared deviations

Σ(x − x̄)² = (−14.17)² + 13.83² + 3.833² + (−13.17)² + 14.83² + (−5.167)² = 826.8

Standard deviation

s = √(826.8/5) = 12.86

Answer

s = 12.86
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 = (90 − 100)/20 = −0.5

Answer

z = −0.5
11.Probability

Answer: Var(X) = 4.84

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 = 0·0.35 + 2·0.15 + 4·0.35 + 6·0.15 = 2.6

Mean of the squares

E(X²) = 0·0.35 + 4·0.15 + 16·0.35 + 36·0.15 = 11.6

Variance

Var(X) = 11.6 − 2.6² = 4.84

Answer

Var(X) = 4.84
12.Probability

Answer: σY = 7.246

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 = 2, Var(X) = 2.1, σX = 1.449

Linear change

σY = |5|·1.449 = 7.246

Answer

σY = 7.246

The mean does move: E(Y) = 5·2 + 1 = 11.

13.Normal distribution

Answer: P = 0.8413

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (29.5 − 32)/2.5 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
14.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: lower limit = −6.913 mmHg

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.

Critical value

df = 12, t = 2.179

Margin of error

E = 2.179·3/√13 = 1.813

Interval

−5.1 ± 1.813 → [−6.913, −3.287]

Answer

lower limit = −6.913 mmHg
16.Confidence intervals

Answer: lower limit = −0.3039 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 = (21·10² + 8·7.5²)/29 = 87.93
sp = 9.377

Margin of error

E = 1.699·9.377·√(1/22 + 1/9) = 6.304

Interval

6 ± 6.304 → [−0.3039, 12.3]

Answer

lower limit = −0.3039 kg
17.Hypothesis tests & confidence intervals

Answer: t = −3.094

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 = 5/√49 = 5/7 = 0.7143

Test statistic

t = (72.79 − 75)/0.7143 = −3.094

Answer

t = −3.094
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −1.096

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 = 90/200 = 0.45
p̂₂ = 150/300 = 0.5
p̂ = 240/500 = 0.48

Standard error under H₀

SE = √(0.48·0.52·(1/200 + 1/300)) = 0.04561

Test statistic

z = (0.45 − 0.5)/0.04561 = −1.096

Answer

z = −1.096

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

19.Linear regression & correlation

Answer: r² = 0.9574

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 = 80/5 = 16

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30
Syy = Σ(y − ȳ)² = 94

Coefficient of determination

r² = −30²/(10·94) = 0.9574

Answer

r² = 0.9574

About 95.74% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 54.7

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 = 332/5 = 66.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156

Regression line

b = 156/40 = 3.9
a = 66.4 − 3.9·6 = 43
ŷ = 43 + 3.9·x

Prediction

ŷ = 43 + 3.9·3 = 54.7

Answer

ŷ = 54.7

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