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
years0123PV = ?€50 000

You need €50 000 in 3 years. An account pays 4% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €50 0004% / 4n = 3
PV = €
2.Time value of moneyHard
months0123…324€150€150€150FV = ?

You save €150 at the end of every month for 27 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1505.4% / 12N = 324
FV = €
3.NPV & IRRMedium
years01234−75 00030 00030 00023 00037 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €30 000, €30 000, €23 000, €37 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −75 000CF1 = 30 000CF2 = 30 000CF3 = 23 000CF4 = 37 000
NPV = €
4.NPV & IRRHard
years012−25 00015 00011 500

A project costs €25 000 and returns €15 000, €11 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 15 000CF2 = 11 500
IRR = %
5.Bond pricing & yieldMedium
years01234€935C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 4 years and pays annual coupons. It sells for €935 when the yield to maturity is 9%. Find the coupon rate c in percent.

F = €1 000n = 4P = €935YTM = 9%
c = %
6.Bond pricing & yieldHard

A bond trades at €1 005 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 6.4 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.

P0 = €1 005y = 5%D = 6.4Δy = +25 bp
P = €
7.CAPM & WACCMedium
WACC = ?equityEre​ = 13%debtD = 1·Erd​ = 6%T = 30%

A company has a debt-to-equity ratio (at market values) of 1. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the WACC.

D/E = 1re = 13%rd = 6%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 5%D/E = 0.5equityβL​ = 1.3debtD/E = 0.75equityre​ = ?debtT = 25%

A company's shares have a beta of 1.3 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 25%, the risk-free rate 4% and the market risk premium 5%. Find the cost of equity re after the change.

βL = 1.3D/E1 = 0.5D/E2 = 0.75T = 25%rf = 4%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 52, 40, 11, 60, 16, 54.

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

Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 167.5.

μ = 170σ = 5x = 167.5
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
12.ProbabilityHard
xp(x)00.110.1520.3530.3540.05μ

X has the distribution shown, and Y = 2X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.1, 0.15, 0.35, 0.35, 0.05Y = 2X + 5
σY =
13.Normal distributionMedium
248.75256P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.75 and 256 g? Give P as a decimal.

μ = 250σ = 5x1 = 248.75x2 = 256
P =
14.Normal distributionHard
μ = 178x = ?75%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).

μ = 178σ = 7P = 75%
x = cm
15.Confidence intervalsMedium
n = 8, sd​ = 3.90L = ?Ud̄ = 4E90%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 4 and standard deviation 3.9. Find the lower limit of the 90% confidence interval for the mean difference.

n = 8d̄ = 4sd = 3.990%t(7) = 1.895
lower limit = mmHg
16.Confidence intervalsHard
n = 10Lσ = ?s = 5.595%

The diameters of 10 machined shafts have a sample standard deviation of 5.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 5.5 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 499 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 499s = 5n = 81
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 97.3. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 97.3σ = 10n = 1690%
U =
19.Linear regression & correlationMedium
xy154717551969217423782579xy15202550607080

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 120Σy = 402Σx² = 2470Σy² = 27796Σxy = 8274
r =
20.Linear regression & correlationHard
xy1477178520108231202614029151xy15202530100150x = 21.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21.5, 112) is made. Find its residual e = y − ŷ.

n = 6(21.5, 112)
e =

3Solutions

1.Time value of money

Answer: PV = 44372 €

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 = 0.04/4 = 0.01
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 50 000/1.127 = €44 372.46

Answer

PV = €44 372.46
2.Time value of money

Answer: FV = 109444 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·27 = 324

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 150·(4.283 − 1)/0.0045 = €109 444.13

Answer

FV = €109 444.13

You paid in €48 600 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 19618 €

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 = −75 000
CF1/(1 + 0.1)1 = 30 000/1.1 = 27 272.73
CF2/(1 + 0.1)2 = 30 000/1.21 = 24 793.39
CF3/(1 + 0.1)3 = 23 000/1.331 = 17 280.24
CF4/(1 + 0.1)4 = 37 000/1.464 = 25 271.50

Add them up

NPV = €19 617.85

Answer

NPV = €19 617.85

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

4.NPV & IRR

Answer: IRR = 4.162 %

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(0%) = 1 500.00
NPV(5%) = −283.45

Narrow down

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

IRR ≈ 0 + 5·1 500.00/(1 500.00 − (−283.45)) = 4.205%
exact: NPV(4.162%) = 0 → IRR = 4.162%

Answer

IRR = 4.162%
5.Bond pricing & yield

Answer: c = 6.994 %

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 + 0.09)−4 = €708.43

Coupon

C = (935 − 708.43)/3.24 = €69.94

Coupon rate

c = 69.94/1000 = 6.994%

Answer

c = 6.994%
6.Bond pricing & yield

Answer: P = 989.7 €

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.

Modified duration

Dmod = 6.4/1.05 = 6.095

Estimated price change

%ΔP ≈ −6.095·(0.0025) = −1.524%
P ≈ 1005·(1 − 0.01524) = €989.69

Answer

P = €989.69

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 8.6 %

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 from D/E

E/V = 1/(1 + 1) = 0.5
D/V = 1/(1 + 1) = 0.5

WACC

WACC = 0.5·13% + 0.5·6%·(1 − 0.3) = 8.6%

Answer

WACC = 8.6%
8.CAPM & WACC

Answer: re = 11.39 %

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.

Unlever the beta

βU = 1.3/(1 + 0.75·0.5) = 0.9455

Relever at the new D/E

βL = 0.9455·(1 + 0.75·0.75) = 1.477

Cost of equity (CAPM)

re = 4% + 1.477·5% = 11.39%

Answer

re = 11.39%
9.Descriptive statistics

Answer: s = 20.73

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 = 233/6 = 38.83

Squared deviations

Σ(x − x̄)² = 13.17² + 1.167² + (−27.83)² + 21.17² + (−22.83)² + 15.17² = 2149

Standard deviation

s = √(2149/5) = 20.73

Answer

s = 20.73
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 = (167.5 − 170)/5 = −0.5

Answer

z = −0.5
11.Probability

Answer: P = 0.5981

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.

Complement

P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
12.Probability

Answer: σY = 2.088

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 = 2.1, Var(X) = 1.09, σX = 1.044

Linear change

σY = |2|·1.044 = 2.088

Answer

σY = 2.088

The mean does move: E(Y) = 2·2.1 + 5 = 9.2.

13.Normal distribution

Answer: P = 0.4836

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 = (248.75 − 250)/5 = −0.25
z₂ = (256 − 250)/5 = 1.2

Area between

P = Φ(1.2) − Φ(−0.25) = 0.8849 − 0.4013 = 0.4836

Answer

P = 0.4836
14.Normal distribution

Answer: x = 182.7 cm

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) = 0.75 → z = 0.6745

Back to x

x = 178 + 0.6745·7 = 182.72 cm

Answer

x = 182.72 cm
15.Confidence intervals

Answer: lower limit = 1.387 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 = 7, t = 1.895

Margin of error

E = 1.895·3.9/√8 = 2.613

Interval

4 ± 2.613 → [1.387, 6.613]

Answer

lower limit = 1.387 mmHg
16.Confidence intervals

Answer: σ = 10.04 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 = 9·5.5² = 272.3

Limits for σ²

272.3/19.023 = 14.31
272.3/2.7 = 100.8

Limit for σ

σ = √100.8 = 10.04 µm

Answer

σ = 10.04 µm

95% interval for σ: [3.783, 10.04] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −1.8

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/√81 = 5/9 = 0.5556

Test statistic

t = (499 − 500)/0.5556 = −1.8

Answer

t = −1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 101.4

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 = 1.645·10/√16 = 4.112

Interval limit

U = 97.3 + 4.112 = 101.4

Answer

U = 101.4
19.Linear regression & correlation

Answer: r = 0.9526

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 = 8274 − 120·402/6 = 234
Sxx = 2470 − 120²/6 = 70
Syy = 27796 − 402²/6 = 862

Correlation coefficient

r = 234/√(70·862) = 0.9526

Answer

r = 0.9526
20.Linear regression & correlation

Answer: e = −1.5

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 = 129/6 = 21.5
ȳ = 681/6 = 113.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 820.5

Regression line

b = 820.5/157.5 = 5.21
a = 113.5 − 5.21·21.5 = 1.495
ŷ = 1.495 + 5.21·x

Predicted value

ŷ = 1.495 + 5.21·21.5 = 113.5

Residual

e = 112 − 113.5 = −1.5

Answer

e = −1.5

The point lies below 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