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
years0123456€1 000FV = ?

You invest €1 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 6 years.

PV = €1 0004.8% / 12n = 6
FV = €
2.Time value of moneyHard
months0123…N = ?€12 000€175€175€175€1757.2% / 12

You borrow €12 000 at 7.2% per year, compounded monthly, and repay €175 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 = €12 000A = €1757.2% / 12
N = payments
3.NPV & IRRMedium
years0123−20 0008 0007 0008 000

An investment of €20 000 gives these cash flows at the end of years 1 to 3: €8 000, €7 000, €8 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −20 000CF1 = 8 000CF2 = 7 000CF3 = 8 000
NPV = €
4.NPV & IRRHard
years0123…8−50 0005 500+3%+3%+3%r = 12%, NPV = ?

A project costs €50 000 today. Its cash flow is €5 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 8. The cost of capital is 12%. Find the net present value NPV.

C0 = €50 000CF1 = €5 500g = 3%n = 8r = 12%
NPV = €
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 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 3.5%n = 7
P = €
6.Bond pricing & yieldHard
years0123P€45€45€45 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 3 years. The yield to maturity is 4%. Find the Macaulay duration D in years.

F = €1 000coupon = 4.5%n = 3YTM = 4%
D = years
7.CAPM & WACCMedium
βr (%)SMLrf​ = 2%β = 1rm​ = ?β = 1.2re​ = 9.8%

A share with a beta of 1.2 has a required return of 9.8%. The risk-free rate is 2%. What expected market return rm does CAPM imply?

β = 1.2re = 9.8%rf = 2%
rm = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €49.50D0​ = €2.10g = 4%debt€39 600 000rd​ = 4.5%T = 21%

A company has 1 000 000 shares trading at €49.50; the book value of its equity is €24 800 000. It just paid a dividend of €2.10 per share, and dividends grow by 4% per year. Its debt has a market value of €39 600 000 and a pre-tax cost of 4.5%. The tax rate is 21%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €49.50book equity = €24 800 000D0 = €2.10g = 4%D = €39 600 000rd = 4.5%T = 21%
WACC = %
9.Descriptive statisticsMedium
20253035

Find the sample standard deviation s of the data: 22, 34, 24, 37, 18, 16.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 40?

μ = 50σ = 5x = 40
P =
11.ProbabilityMedium
xp(x)10.2520.2530.340.1550.05μ = 2.5

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

x: 1, 2, 3, 4, 5p(x): 0.25, 0.25, 0.3, 0.15, 0.05
Var(X) =
12.ProbabilityHard
n = 10, p = 0.3: P(X = 3) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 3 correct answers?

n = 10p = 0.3k = 3
P =
13.Normal distributionMedium
178183.6P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 178 and 183.6 cm? Give P as a decimal.

μ = 178σ = 7x1 = 178x2 = 183.6
P =
14.Normal distributionHard
μ = 250x = ?99%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).

μ = 250σ = 5P = 99%
x = g
15.Confidence intervalsMedium
x̄E₁ = 4.199%, n = 50x̄E₂ = ?90%, n = 50

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.1. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 50E₁ = 4.190%
E₂ =
16.Confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 5. 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 = 5E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.0862.086t = ?df = 20

Two teaching methods are compared. Group 1: 15 students, mean score 21, standard deviation 7.5. Group 2: 7 students, mean score 12.6, standard deviation 3.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 15x̄₁ = 21s₁ = 7.5n₂ = 7x̄₂ = 12.6s₂ = 3.5
t =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 1.595%
n =
19.Linear regression & correlationMedium
xy124220316412512xy24152025

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

n = 5
r =
20.Linear regression & correlationHard
xy186421792486279430102xy20253080100x = 19.5

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19.5, 74) is made. Find its residual e = y − ŷ.

n = 5(19.5, 74)
e =

3Solutions

1.Time value of money

Answer: FV = 1333 €

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 = 0.048/12 = 0.004
N = m·n = 12·6 = 72

Future value

FV = 1 000·(1 + 0.004)72 = 1 000·1.333 = €1 332.99

Answer

FV = €1 332.99
2.Time value of money

Answer: N = 89 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 = 0.006
(1 + i)−N = 1 − 12 000·0.006/175 = 0.5886

Solve with logarithms

N = −ln(0.5886)/ln(1.006) = 88.61

Round up to whole payments

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

N = 89

Answer

N = 89 payments
3.NPV & IRR

Answer: NPV = 494.1 €

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 = 7 000/1.124 = 6 229.98
CF3/(1 + 0.06)3 = 8 000/1.191 = 6 716.95

Add them up

NPV = €494.10

Answer

NPV = €494.10

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

4.NPV & IRR

Answer: NPV = −20155 €

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 growing annuity

((1.03)/(1.12))8 = 0.5116
PV = 5 500/(0.12 − 0.03)·(1 − 0.5116) = 29 845.01

NPV

NPV = 29 845.01 − 50 000 = −€20 154.99

Answer

NPV = −€20 154.99

NPV is negative, so the project earns less than the required return.

5.Bond pricing & yield

Answer: P = 1154 €

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 = 0.0175, N = 14

Present value of the coupons

€30.00·(1 − 0.7844)/0.0175 = €369.66

Present value of the face value

€1 000·0.7844 = €784.36

Price

P = €1 154.03

Answer

P = €1 154.03

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

6.Bond pricing & yield

Answer: D = 2.874 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: 45/1.04 = 43.27, t·PV = 43.27
t = 2: 45/1.082 = 41.61, t·PV = 83.21
t = 3: 1045/1.125 = 929.00, t·PV = 2 787.00

Duration

P = 1 013.88
D = 2 913.48/1 013.88 = 2.874

Answer

D = 2.874 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 = (9.8% − 2%)/1.2 = 6.5%

Market return

rm = 2% + 6.5% = 8.5%

Answer

rm = 8.5%
8.CAPM & WACC

Answer: WACC = 6.253 %

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).

Cost of equity

D1 = 2.1·1.04 = €2.18
re = 2.184/49.5 + 0.04 = 8.412%

Market values

E = 1 000 000·€49.50 = €49 500 000
V = €89 100 000, E/V = 0.5556, D/V = 0.4444

WACC

WACC = 0.5556·8.412% + 0.4444·4.5%·(1 − 0.21) = 6.253%

Answer

WACC = 6.253%
9.Descriptive statistics

Answer: s = 8.542

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 = 151/6 = 25.17

Squared deviations

Σ(x − x̄)² = (−3.167)² + 8.833² + (−1.167)² + 11.83² + (−7.167)² + (−9.167)² = 364.8

Standard deviation

s = √(364.8/5) = 8.542

Answer

s = 8.542
10.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
11.Probability

Answer: Var(X) = 1.35

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 = 1·0.25 + 2·0.25 + 3·0.3 + 4·0.15 + 5·0.05 = 2.5

Mean of the squares

E(X²) = 1·0.25 + 4·0.25 + 9·0.3 + 16·0.15 + 25·0.05 = 7.6

Variance

Var(X) = 7.6 − 2.5² = 1.35

Answer

Var(X) = 1.35
12.Probability

Answer: P = 0.2668

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.

Binomial formula

P(X = 3) = 120·0.33·0.77 = 120·0.027·0.08235 = 0.2668

Answer

P = 0.2668
13.Normal distribution

Answer: P = 0.2881

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 = (178 − 178)/7 = 0
z₂ = (183.6 − 178)/7 = 0.8

Area between

P = Φ(0.8) − Φ(0) = 0.7881 − 0.5 = 0.2881

Answer

P = 0.2881
14.Normal distribution

Answer: x = 261.6 g

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) = 0.99 → z = 2.326

Back to x

x = 250 + 2.326·5 = 261.63 g

Answer

x = 261.63 g
15.Confidence intervals

Answer: E₂ = 2.618

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 = (1.645/2.576)·√(50/50) = 0.6386

New margin

E₂ = 4.1·0.6386 = 2.618

Answer

E₂ = 2.618
16.Confidence intervals

Answer: n = 271

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 = (1.645·5/0.5)² = 270.6

Round up

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

n = 271

Answer

n = 271
17.Hypothesis tests & confidence intervals

Answer: t = 2.797

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.

Pooled variance

sp² = (14·7.5² + 6·3.5²)/20 = 43.05
sp = 6.561

Standard error

SE = 6.561·√(1/15 + 1/7) = 3.003

Test statistic

t = (21 − 12.6)/3.003 = 2.797

Answer

t = 2.797

With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 246

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).

Solve for n

n = (1.96·12/1.5)² = 245.9

Round up

n = 246

Answer

n = 246
19.Linear regression & correlation

Answer: r = −0.9701

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) = −32
Syy = Σ(y − ȳ)² = 108.8

Correlation coefficient

r = −32/√(10·108.8) = −0.9701

Answer

r = −0.9701
20.Linear regression & correlation

Answer: e = 2.65

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 = 120/5 = 24
ȳ = 425/5 = 85

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 273

Regression line

b = 273/90 = 3.033
a = 85 − 3.033·24 = 12.2
ŷ = 12.2 + 3.033·x

Predicted value

ŷ = 12.2 + 3.033·19.5 = 71.35

Residual

e = 74 − 71.35 = 2.65

Answer

e = 2.65

The point lies above the line.

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